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Hamilton Jacobi Equations and Viscosity Solutions

1 · Prerequisites

2 · Summary

This page develops the viscosity theory of first-order Hamilton--Jacobi equations from the test-function definition. It fixes the Cauchy problem, its classical solutions, the upper and lower semicontinuous envelopes and their least-majorant properties, and the equivalence between test-function and first-order-jet formulations, with the quartic strictification device used to localise contacts. Finite maxima of subsolutions and minima of supersolutions, stability under locally uniform convergence and the half-relaxed-limit stability theorem with vanishing perturbations are proved, followed by the doubling-of-variables localisation lemma, the time-penalisation lemma and comparison for the Cauchy problem in the two settings of the design, including the autonomous convex superlinear case. Uniqueness and sup-norm contraction are derived, and Perron's method is proved between the two explicit barriers with the upper-envelope and local-bump lemmas. The convex-duality half of the page defines the Legendre transform, proves biconjugacy for finite convex Hamiltonians without choice, and develops the Hopf--Lax operator: localisation of its infima, sup-norm contraction, preservation of a modulus of continuity, the dynamic-programming semigroup, the characteristic Euler relation, the Hopf--Lax solution of the Cauchy problem and vanishing viscosity. Finite speed of dependence and the value-function orientation close the page. No item of the page consumes the Axiom of Choice; Countable Choice is declared exactly where the half-relaxed-limit stability theorem extracts sequences under its sequential hypothesis.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The Hamilton--Jacobi Cauchy problem and its classical solutions

Definition

Let $n \geq 1$ , let $O \subseteq R^{n}$ be nonempty and open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), let $T > 0$ , and let $H : O \times [0, T] \times R^{n} \to R$ be continuous (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form). Write $Z : = O \times (0, T), \overline{Z} : = \overline{O} \times [0, T], Γ : = (O \times {0}) \cup (\partial O \times [0, T])$ for the open space--time cylinder, its closure in $R^{n + 1}$ , and the parabolic boundary. Given a continuous $u_{0} : O \to R$ , the Hamilton--Jacobi Cauchy problem is $u_{t} + H (x, t, D u) = 0 in Z, u = u_{0} on O \times {0},$ where $D u$ denotes the spatial gradient and $u_{t}$ the time derivative of the unknown function (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder).

A classical solution of this problem is a function $u \in C^{0} (\overline{Z}) \cap C^{1} (Z)$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space) with $u (x, 0) = u_{0} (x)$ for every $x \in O$ and $u_{t} (x, t) + H (x, t, D u (x, t)) = 0 for every (x, t) \in Z .$ The continuity requirement $u \in C^{0} (\overline{Z})$ ensures continuous attainment of the initial datum and also continuity on the lateral and terminal faces; no condition on the lateral face $\partial O \times [0, T]$ is imposed unless it is stated explicitly.

Stationary specialization. Suppose that $H$ does not depend on $t$ and that $v \in C^{1} (O)$ satisfies $H (x, D v (x)) = 0$ for every $x \in O$ ; such a $v$ is called a classical solution of the stationary Hamilton--Jacobi equation $H (x, D v) = 0$ . If in addition $v$ extends continuously to $\overline{O}$ , then $u (x, t) : = v (x)$ is a classical solution of the Cauchy problem with datum $u_{0} : = v ∣_{O}$ , because $u \in C^{0} (\overline{Z}) \cap C^{1} (Z)$ and $u_{t} \equiv 0$ on $Z$ .

Remarks

What the definition fixes. The open cylinder $Z = O \times (0, T)$ , its closure, the parabolic boundary $Γ$ , the initial face $O \times {0}$ on which the datum is read, the Hamiltonian domain $O \times [0, T] \times R^{n}$ with its continuity, and the regularity class $C^{0} (\overline{Z}) \cap C^{1} (Z)$ of a classical solution are the data used by every viscosity notion on this page. In particular, "the initial datum is attained" means continuous attainment on the initial face, not a merely pointwise boundary value on a larger set.
No lateral condition, no choice. The definition imposes no condition on $\partial O \times [0, T]$ ; statements about bounded domains add whatever boundary comparison they need explicitly. Nothing is selected anywhere in the definition, so no choice principle is used.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Upper and lower semicontinuous envelopes by local limsup and liminf

Definition

Let $A \subseteq R^{m}$ be nonempty (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) and let $u : A \to R$ . For $x \in A$ and $r > 0$ put $M_{r} (x) : = \sup {u (y) : y \in A, ∣ y - x ∣ \leq r}, m_{r} (x) : = \inf {u (y) : y \in A, ∣ y - x ∣ \leq r},$ the supremum and infimum being taken in $\overline{R} = R \cup {\pm \infty}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Greatest lower bound (infimum), Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

The upper semicontinuous envelope of $u$ is $u^{*} (x) : = \inf_{r > 0} M_{r} (x) = \lim_{r ↓ 0} \sup_{\begin{array}{c} y \in A \\ ∣ y - x ∣ \leq r \end{array}} u (y),$ and the lower semicontinuous envelope of $u$ is $u_{*} (x) : = \sup_{r > 0} m_{r} (x) = \lim_{r ↓ 0} \inf_{\begin{array}{c} y \in A \\ ∣ y - x ∣ \leq r \end{array}} u (y),$ both with values in $\overline{R}$ .

Remarks

The limits exist. For fixed $x \in A$ the map $r \mapsto M_{r} (x)$ is nondecreasing in $r$ (the set it is taken over grows with $r$ ) and the map $r \mapsto m_{r} (x)$ is nonincreasing in $r$ , so the one-sided limits as $r ↓ 0$ exist in $\overline{R}$ , with $\inf_{r > 0} M_{r} (x) = \lim_{r ↓ 0} M_{r} (x)$ and $\sup_{r > 0} m_{r} (x) = \lim_{r ↓ 0} m_{r} (x)$ . Since $A$ is nonempty, the sets over which the suprema and infima are taken are nonempty; infinite values are kept rather than discarded.
Comparison with $u$ . For every $x \in A$ one has $m_{r} (x) \leq u (x) \leq M_{r} (x)$ for all $r > 0$ , hence $u_{*} (x) \leq u (x) \leq u^{*} (x)$ pointwise. If $u$ is bounded on $A$ , then all values $M_{r} (x), m_{r} (x)$ lie between $\inf_{A} u$ and $\sup_{A} u$ , so $u^{*}$ and $u_{*}$ are real-valued and bounded on $A$ ; this is a sufficient hypothesis for real-valued envelopes. Local boundedness near each point also suffices, because only arbitrarily small radii affect the defining infimum and supremum.
Scope. These are the envelopes of $u$ on the Euclidean set $A$ . The space--time cylinder of The Hamilton--Jacobi Cauchy problem and its classical solutions is used with $m = n + 1$ and $A = Z$ , and the envelope notation is the one appearing in the stability, Perron and comparison statements of this page (Upper and lower semicontinuity on subsets of $R^{n}$ is the underlying mode of semicontinuity). No choice is used: each envelope is the value of a monotone limit indexed by $r$ , hence by $r = 1 / k$ , and no sequence or point is selected.

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The envelopes are the least upper and greatest lower semicontinuous functions

Statement

Let $A \subseteq R^{m}$ be nonempty and let $u : A \to R$ be bounded, with envelopes $u^{*}, u_{*}$ as in Upper and lower semicontinuous envelopes by local limsup and liminf. Then $u^{*}$ is upper semicontinuous on $A$ , $u_{*}$ is lower semicontinuous on $A$ , and $u_{*} \leq u \leq u^{*}$ pointwise. Moreover: (1) $u^{*}$ is the least upper semicontinuous function $w : A \to \overline{R}$ with $w \geq u$ pointwise, and $u^{*} = u$ if and only if $u$ is upper semicontinuous; (2) $u_{*}$ is the greatest lower semicontinuous function $w : A \to \overline{R}$ with $w \leq u$ pointwise, and $u_{*} = u$ if and only if $u$ is lower semicontinuous; (3) $(u^{*})^{*} = u^{*}$ and $(u_{*})_{*} = u_{*}$ . No choice principle is used.

Facts & Assumptions

Given: A nonempty set $A \subseteq R^{m}$ , a bounded function $u : A \to R$ , the local suprema $M_{r} (x) = \sup {u (y) : y \in A, ∣ y - x ∣ \leq r}$ and local infima $m_{r} (x) = \inf {u (y) : y \in A, ∣ y - x ∣ \leq r}$ for $x \in A$ , $r > 0$ , all computed in $\overline{R}$ , and the envelopes $u^{*} (x) = \inf_{r > 0} M_{r} (x)$ , $u_{*} (x) = \sup_{r > 0} m_{r} (x)$ .

[F1]

For every $x \in A$ the sets ${M_{r} (x) : r > 0}$ and ${m_{r} (x) : r > 0}$ are nonempty and bounded (by the bounds of $u$ ); $u^{*} (x) = \inf_{r > 0} M_{r} (x)$ is the greatest lower bound of the first set, $u_{*} (x) = \sup_{r > 0} m_{r} (x)$ is the least upper bound of the second, and $m_{r} (x) \leq u (x) \leq M_{r} (x)$ for every $r > 0$ . In particular $u^{*} (x) \leq M_{r} (x)$ and $m_{r} (x) \leq u_{*} (x)$ for every $r > 0$ (Upper and lower semicontinuous envelopes by local limsup and liminf).

[F2]

For real-valued functions, upper and lower semicontinuity have the local $ε$ characterizations: near $a$ , respectively $f (x) < f (a) + ε$ and $f (x) > f (a) - ε$ (Upper and lower semicontinuity on subsets of $R^{n}$ ). For an extended-real upper semicontinuous $w$ , we use the standard strict-sublevel convention ${w < c}$ open for every real $c$ ; if $w (a)$ is finite, applying it with $c = w (a) + ε$ gives the same local upper bound. The dual strict-superlevel convention for lower semicontinuity gives the local lower bound when $w (a)$ is finite. These are exactly the finite-value cases used in steps 1.3 and 1.4; the infinite endpoint cases are disposed of there directly.

[F3]

If $S \subseteq R$ is nonempty and $ℓ = \inf S$ , then $ℓ \leq s$ for every $s \in S$ , and $ℓ^{'} \leq ℓ$ for every lower bound $ℓ^{'}$ of $S$ (Greatest lower bound (infimum)).

[F4]

If $S \subseteq R$ is nonempty, bounded above and $v \in R$ is an upper bound of $S$ , then $v = \sup S$ if and only if for every $ε > 0$ there is $s \in S$ with $v - ε < s$ ; in particular a supremum of $S$ in $R$ is an upper bound of $S$ (Epsilon characterisation of the supremum).

Proof

technique · monotone localisation; the envelopes are compared with a competing semicontinuous function by testing the defining infimum and supremum

1.1F1F2F3F4algebra

$u^{*}$ is upper semicontinuous on $A$ and $u_{*} \leq u \leq u^{*}$ . Fix $x \in A$ and $ε > 0$ . Since $u^{*} (x) + 2^{- 1} ε > u^{*} (x)$ is not a lower bound of the nonempty set ${M_{r} (x) : r > 0}$ by the leastness clause of [F3], there is $r > 0$ with $M_{r} (x) < u^{*} (x) + 2^{- 1} ε$ . For $z \in A$ with $∣ z - x ∣ < r$ put $δ : = r - ∣ z - x ∣ > 0$ ; every $w \in A$ with $∣ w - z ∣ \leq δ$ satisfies $∣ w - x ∣ \leq ∣ w - z ∣ + ∣ z - x ∣ \leq r$ , so the set defining $M_{δ} (z)$ is contained in the set defining $M_{r} (x)$ and therefore $M_{δ} (z) \leq M_{r} (x) < u^{*} (x) + 2^{- 1} ε$ ; since $u^{*} (z) \leq M_{δ} (z)$ by [F1], we get $u^{*} (z) < u^{*} (x) + ε$ . Thus $u^{*}$ is upper semicontinuous at $x$ by [F2], and $x$ was arbitrary. Next, $u (x)$ is a lower bound of ${M_{r} (x) : r > 0}$ by [F1], so $u (x) \leq u^{*} (x)$ by the greatest-lower-bound clause of [F3]. Finally $u (x)$ is an upper bound of ${m_{r} (x) : r > 0}$ by [F1]; if $u_{*} (x) > u (x)$ held, then $η : = u_{*} (x) - u (x) > 0$ and [F4] would give $r > 0$ with $u_{*} (x) - η < m_{r} (x)$ , that is $u (x) < m_{r} (x)$ , contradicting $m_{r} (x) \leq u (x)$ ; hence $u_{*} (x) \leq u (x)$ .

1.2F1F2F4algebra

$u_{*}$ is lower semicontinuous on $A$ . Fix $x \in A$ and $ε > 0$ . By [F4] applied to the nonempty bounded-above set ${m_{r} (x) : r > 0}$ with supremum $u_{*} (x)$ , there is $r > 0$ with $u_{*} (x) - ε < m_{r} (x)$ . For $z \in A$ with $∣ z - x ∣ < r$ put $δ : = r - ∣ z - x ∣ > 0$ ; every $w \in A$ with $∣ w - z ∣ \leq δ$ satisfies $∣ w - x ∣ \leq r$ , so $m_{δ} (z) \geq m_{r} (x)$ , and $u_{*} (z) \geq m_{δ} (z)$ by [F1]; hence $u_{*} (z) > u_{*} (x) - ε$ , which is lower semicontinuity at $x$ by [F2].

1.3F1F2F3algebra

Least upper semicontinuous majorant. Let $w : A \to \overline{R}$ be upper semicontinuous with $w \geq u$ , fix $x \in A$ and $ε > 0$ . Since $w \geq u$ and $u$ is real-valued, $w$ takes no value $- \infty$ ; if $w (x) = + \infty$ then $u^{*} (x) \leq w (x)$ holds because $u^{*}$ is real-valued by [F1] and boundedness of $u$ , so assume $w (x) \in R$ . By [F2] there is $r_{0} > 0$ with $w (y) < w (x) + ε$ for all $y \in A$ with $∣ y - x ∣ < r_{0}$ . For $y \in A$ with $∣ y - x ∣ \leq r_{0} / 2$ we then have $u (y) \leq w (y) < w (x) + ε$ , so $w (x) + ε$ is an upper bound of the set defining $M_{r_{0} / 2} (x)$ , whence $M_{r_{0} / 2} (x) \leq w (x) + ε$ . Since $u^{*} (x) = \inf_{r > 0} M_{r} (x) \leq M_{r_{0} / 2} (x)$ by [F3], we get $u^{*} (x) \leq w (x) + ε$ , and letting $ε ↓ 0$ gives $u^{*} (x) \leq w (x)$ . Hence $u^{*} \leq w$ for every upper semicontinuous majorant $w$ of $u$ .

1.4F1F2algebra

Greatest lower semicontinuous minorant. Let $w : A \to \overline{R}$ be lower semicontinuous with $w \leq u$ , fix $x \in A$ and $ε > 0$ . Since $w \leq u$ , $w (x) \in R \cup {- \infty}$ ; if $w (x) = - \infty$ then $u_{*} (x) \geq w (x)$ is automatic, so assume $w (x) \in R$ . By [F2] there is $r_{0} > 0$ with $w (y) > w (x) - ε$ for all $y \in A$ with $∣ y - x ∣ < r_{0}$ . For $y \in A$ with $∣ y - x ∣ \leq r_{0} / 2$ we have $w (y) > w (x) - ε$ and $w (y) \leq u (y)$ , so $w (x) - ε$ is a lower bound of the set defining $m_{r_{0} / 2} (x)$ , whence $m_{r_{0} / 2} (x) \geq w (x) - ε$ . Since $u_{*} (x)$ is an upper bound of ${m_{r} (x) : r > 0}$ by [F1], we get $u_{*} (x) \geq m_{r_{0} / 2} (x) \geq w (x) - ε$ , and letting $ε ↓ 0$ gives $u_{*} (x) \geq w (x)$ .

2.1step 1.1step 1.2step 1.3step 1.4

The two equivalences. If $u$ is upper semicontinuous, then $u$ is an upper semicontinuous majorant of itself, so $u^{*} \leq u$ by step 1.3; with $u \leq u^{*}$ from step 1.1 this gives $u^{*} = u$ . Conversely, if $u^{*} = u$ , then $u$ is upper semicontinuous because $u^{*}$ is, by step 1.1. The same two lines with step 1.4 and step 1.2 show that $u_{*} = u$ if and only if $u$ is lower semicontinuous.

3.1step 1.1step 1.2step 2.1∎

Idempotence. The function $u^{*}$ is bounded and upper semicontinuous on $A$ by step 1.1, and it is its own upper semicontinuous majorant; applying the equivalence of step 2.1 to $u^{*}$ in place of $u$ gives $(u^{*})^{*} = u^{*}$ . Likewise $u_{*}$ is bounded and lower semicontinuous by steps 1.1 and 1.2, so $(u_{*})_{*} = u_{*}$ .

Remarks

Where boundedness is used. Boundedness of $u$ keeps every $M_{r} (x)$ and $m_{r} (x)$ in $R$ , so the infimum and supremum over $r$ are taken in the ordered field and the elementary leastness arguments of steps 1.1--1.4 apply directly. For unbounded $u$ the envelopes can be infinite; idempotence in that setting must use the same local formulas extended to extended-valued inputs, whereas the present statement and Upper and lower semicontinuous envelopes by local limsup and liminf take real-valued input.
Strictness is not needed. The proof nowhere requires the contact or the majorant to be strict: the least-majorant property is proved by a direct pointwise comparison against an arbitrary upper semicontinuous majorant.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem

Definition

Let $n \geq 1$ , let $U \subseteq R^{n + 1}$ be nonempty open with coordinates $(x, t) \in R^{n} \times R$ , and let $H : U \times R^{n} \to R$ be continuous. A function $u : U \to R$ that is upper semicontinuous on $U$ (Upper and lower semicontinuity on subsets of $R^{n}$ ) is a viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $U$ if for every $ϕ \in C^{1} (U)$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space) and every point $z_{0} \in U$ at which $u - ϕ$ has a local maximum, $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0.$ A function $v : U \to R$ that is lower semicontinuous on $U$ is a viscosity supersolution if for every $ϕ \in C^{1} (U)$ and every point $z_{0} \in U$ at which $v - ϕ$ has a local minimum, $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \geq 0.$ A viscosity solution of the equation in $U$ is a function that is both a viscosity subsolution and a viscosity supersolution.

Cauchy problem. Let $Z = O \times (0, T)$ and $H : O \times [0, T] \times R^{n} \to R$ be as in The Hamilton--Jacobi Cauchy problem and its classical solutions, and let $u_{0} : O \to R$ . A viscosity subsolution of the Cauchy problem is an upper semicontinuous $u : Z \to R$ satisfying the subsolution inequality at every $z_{0} \in Z$ together with the relaxed initial condition ${lim sup}_{\begin{array}{c} (y, s) \to (x, 0) \\ s > 0, y \in O \end{array}} u (y, s) \leq u_{0} (x) for every x \in O;$ a viscosity supersolution $v$ satisfies the supersolution inequality at every $z_{0} \in Z$ together with ${lim inf}_{\begin{array}{c} (y, s) \to (x, 0) \\ s > 0, y \in O \end{array}} v (y, s) \geq u_{0} (x) for every x \in O .$ A viscosity solution of the Cauchy problem is a function that is both a subsolution and a supersolution of the Cauchy problem; a continuous solution of the Cauchy problem satisfies the initial data pointwise, $u (x, 0) = u_{0} (x)$ for every $x \in O$ .

Remarks

Direction of the contact. The subsolution test is taken at a local maximum of $u - ϕ$ , equivalently where $ϕ$ touches $u$ from above, and the supersolution test at a local minimum; reversing the extremum reverses the direction of the required inequality. This is exactly the sign asymmetry that A strict subsolution can fail the supersolution lower-test condition ↗ records.
Semicontinuity is part of the definition. The subsolution must be upper semicontinuous and the supersolution lower semicontinuous: those are the classes in which the tested extrema behave well under localisation and limits. No continuity of $u$ on $U$ and no almost-everywhere class is assumed; a merely locally bounded function is handled through the two envelopes in Discontinuous viscosity solutions through the two envelopes.
The initial condition is relaxed. The limsup/liminf condition is imposed at points of the initial face through sequences with $s > 0$ ; it does not require $u$ to extend to $t = 0$ . For a continuous $u$ the two relaxed conditions force $u (x, s) \to u_{0} (x)$ as $(y, s) \to (x, 0)$ , hence the pointwise equality. No choice principle is used.

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Strictification of a viscosity test function by a quartic perturbation

Statement

Let $U \subseteq R^{m}$ be open, let $v : U \to R$ , let $ϕ \in C^{1} (U)$ , suppose $v - ϕ$ has a local maximum at $z_{0} \in U$ , and fix $r > 0$ with $\overline{B} (z_{0}, r) \subseteq U$ and $v (z) - ϕ (z) \leq v (z_{0}) - ϕ (z_{0})$ for every $z \in \overline{B} (z_{0}, r)$ . For $ε > 0$ define $ϕ_{ε} (z) : = ϕ (z) + ε ∣ z - z_{0} ∣^{4} .$ Then $ϕ_{ε} \in C^{1} (U)$ agrees with $ϕ$ to first order at the contact, $ϕ_{ε} (z_{0}) = ϕ (z_{0}), D ϕ_{ε} (z_{0}) = D ϕ (z_{0}),$ and $v - ϕ_{ε}$ has a strict maximum over $\overline{B} (z_{0}, r)$ at $z_{0}$ : $v (z) - ϕ_{ε} (z) < v (z_{0}) - ϕ_{ε} (z_{0}) for every z \in \overline{B} (z_{0}, r) ∖ {z_{0}} .$ The same statement with $ϕ_{ε} : = ϕ - ε ∣ z - z_{0} ∣^{4}$ strictifies a local minimum contact of a test function, and the first jet at the contact is again unchanged. No choice principle is used.

Facts & Assumptions

Given: An open $U \subseteq R^{m}$ , functions $v : U \to R$ and $ϕ \in C^{1} (U)$ , a point $z_{0} \in U$ , a radius $r > 0$ with $\overline{B} (z_{0}, r) \subseteq U$ and $v (z) - ϕ (z) \leq v (z_{0}) - ϕ (z_{0})$ for all $z \in \overline{B} (z_{0}, r)$ , and for $ε > 0$ the function $ϕ_{ε} (z) = ϕ (z) + ε ∣ z - z_{0} ∣^{4}$ .

[F1]

The map $q (z) : = ∣ z - z_{0} ∣^{4}$ is a polynomial in the coordinates of $z$ , hence of class $C^{1}$ on $R^{m}$ , with $q (z_{0}) = 0$ and $D q (z_{0}) = 0$ ; more precisely $D q (z) = 4 ∣ z - z_{0} ∣^{2} (z - z_{0})$ for every $z$ , the derivative being the total derivative in the sense of The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder and its components the partial derivatives of Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ .

[F2]

If $f, g$ are $C^{1}$ on an open set, then so is $f + g$ , with $D (f + g) = D f + D g$ and $(f + g) (z_{0}) = f (z_{0}) + g (z_{0})$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

Proof

technique · an explicit perturbation whose first jet vanishes at the contact

1.1F1F2algebra

Regularity and first jet. The function $q (z) = ∣ z - z_{0} ∣^{4}$ is a polynomial with $q (z_{0}) = 0$ and $D q (z_{0}) = 0$ by [F1]; adding it to the $C^{1}$ function $ϕ$ with coefficient $ε > 0$ gives $ϕ_{ε} \in C^{1} (U)$ with $ϕ_{ε} (z_{0}) = ϕ (z_{0}) + ε q (z_{0}) = ϕ (z_{0})$ and $D ϕ_{ε} (z_{0}) = D ϕ (z_{0}) + ε D q (z_{0}) = D ϕ (z_{0})$ by [F2].

2.1step 1.1algebra

Strict maximum. Let $z \in \overline{B} (z_{0}, r)$ with $z \neq z_{0}$ . Then $∣ z - z_{0} ∣^{4} > 0$ , and the hypothesised maximum inequality gives $v (z) - ϕ (z) \leq v (z_{0}) - ϕ (z_{0})$ ; subtracting the positive quantity $ε ∣ z - z_{0} ∣^{4}$ from the left-hand side and using $ϕ_{ε} = ϕ + ε ∣ \cdot - z_{0} ∣^{4}$ and $ϕ_{ε} (z_{0}) = ϕ (z_{0})$ , we get $v (z) - ϕ_{ε} (z) \leq v (z_{0}) - ϕ (z_{0}) - ε ∣ z - z_{0} ∣^{4} < v (z_{0}) - ϕ (z_{0}) = v (z_{0}) - ϕ_{ε} (z_{0})$ . Hence $z_{0}$ is the strict maximum of $v - ϕ_{ε}$ over $\overline{B} (z_{0}, r)$ .

3.1step 2.1F1F2algebra

Minimum case. If $v - ϕ$ has a local minimum at $z_{0}$ with $v (z) - ϕ (z) \geq v (z_{0}) - ϕ (z_{0})$ on $\overline{B} (z_{0}, r)$ and $ϕ_{ε}^{-} : = ϕ - ε ∣ z - z_{0} ∣^{4}$ , the same two computations with signs reversed give $ϕ_{ε}^{-} (z_{0}) = ϕ (z_{0})$ , $D ϕ_{ε}^{-} (z_{0}) = D ϕ (z_{0})$ and $v (z) - ϕ_{ε}^{-} (z) > v (z_{0}) - ϕ_{ε}^{-} (z_{0})$ for every $z \in \overline{B} (z_{0}, r) ∖ {z_{0}}$ .

4.1step 1.1step 2.1step 3.1∎

Conclusion. Step 1.1 and step 2.1 give the upper-contact statement, and step 3.1 gives the lower-contact statement; the proof used only the polynomial computation [F1] and additivity [F2], so it selects nothing and uses no choice principle.

Remarks

Why the quartic. The perturbation has value $0$ and gradient $0$ at the contact, so it changes neither the value nor the first jet tested in the viscosity inequalities, while it is strictly positive away from the contact and therefore turns a nonstrict contact into a strict one. This is the device that lets the stability and supremum-envelope arguments localise a maximum on a closed ball without losing the tested jet.
Scope. The statement is pointwise in the ball and does not require $v$ to be semicontinuous, bounded or measurable; the compactness and extreme-value suppliers For $n \geq 1$ , every Euclidean closed ball and every Euclidean sphere of positive radius is compact and Semicontinuous extreme value theorem on compact Euclidean sets are available for applications that patch such a ball maximum into a global one, and they are not needed for the computation above.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Discontinuous viscosity solutions through the two envelopes

Definition

Let $n \geq 1$ , let $O \subseteq R^{n}$ be open, $T > 0$ , let $H : O \times [0, T] \times R^{n} \to R$ be continuous, $Z = O \times (0, T)$ , and let $u_{0} : O \to R$ . A locally bounded function $u : Z \to R$ is a viscosity solution of the Cauchy problem if its upper semicontinuous envelope $u^{*}$ is a viscosity subsolution and its lower semicontinuous envelope $u_{*}$ is a viscosity supersolution in the sense of Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem, each carrying the initial datum $u_{0}$ in the relaxed limsup/liminf sense. Both envelopes are taken over $Z$ as a subset of $R^{n + 1}$ (Upper and lower semicontinuous envelopes by local limsup and liminf); the subsolution inequalities are imposed at every point of $Z$ and the relaxed initial conditions at every point of $O$ .

A continuous viscosity solution is a viscosity solution $u$ that is continuous on $Z$ . For such a $u$ one has $u^{*} = u_{*} = u$ (apply The envelopes are the least upper and greatest lower semicontinuous functions on a small closed ball about each interior point, where continuity makes $u$ bounded). The relaxed initial conditions give a continuous extension to the initial face with value $u_{0}$ , so the definition specializes to the one for continuous test functions. In the opposite direction, the envelope formulation is forced whenever $u$ is not continuous: it keeps the subsolution inequality attached to $u^{*}$ and the supersolution inequality to $u_{*}$ , and never asks a single discontinuous function to satisfy both.

Remarks

Why the two envelopes. A merely locally bounded $u$ need not be upper or lower semicontinuous, so the test-function definition of Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem cannot be applied to $u$ directly. The envelopes are respectively the least upper semicontinuous majorant and the greatest lower semicontinuous minorant of $u$ , so requiring $u^{*}$ to be a subsolution and $u_{*}$ a supersolution is the weakest formulation in which the two one-sided inequalities can be tested; a function is a viscosity solution exactly when both envelopes solve their respective one-sided problems.
Where it is used. This is the notion under which the half-relaxed limits of a locally bounded family are sub- and supersolutions (Half-relaxed limits of a locally bounded family, Half-relaxed limits of sub- and supersolutions with vanishing perturbations) and under which the perron-type and comparison statements of the page are formulated when continuity of the produced object is not known in advance (The upper envelope of a locally bounded supremum of subsolutions is a subsolution). The domain conventions $Z = O \times (0, T)$ and the initial face $O \times {0}$ are those of The Hamilton--Jacobi Cauchy problem and its classical solutions. No choice principle is used.

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Viscosity testing by first-order jets, and closure of the jet inequality

Statement

Let $U \subseteq R^{n + 1}$ be open and let $H : U \times R^{n} \to R$ be continuous. For $w : U \to R$ and $z_{0} \in U$ define the first-order superjet and subjet $D^{+} w (z_{0}) : = {p = (p_{x}, p_{t}) \in R^{n} \times R : w (z) \leq w (z_{0}) + ⟨ p, z - z_{0} ⟩ + o (∣ z - z_{0} ∣) (z \to z_{0})},$ $D^{-} w (z_{0}) : = - D^{+} (- w) (z_{0}) = {p : w (z) \geq w (z_{0}) + ⟨ p, z - z_{0} ⟩ + o (∣ z - z_{0} ∣) (z \to z_{0})} .$ Then: (1) an upper semicontinuous $u : U \to R$ is a viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $U$ if and only if $p_{t} + H (z_{0}, p_{x}) \leq 0 for all z_{0} \in U and all p \in D^{+} u (z_{0});$ (2) a lower semicontinuous $v$ is a viscosity supersolution if and only if $p_{t} + H (z_{0}, p_{x}) \geq 0$ for all $z_{0} \in U$ and all $p \in D^{-} v (z_{0})$ ; (3) if $u$ is such a subsolution, $z_{k} \to z_{0}$ in $U$ with $u (z_{k}) \to u (z_{0})$ , and $p_{k} \in D^{+} u (z_{k})$ with $p_{k} \to p$ , then $p_{t} + H (z_{0}, p_{x}) \leq 0$ ; the analogous closure statement holds for supersolutions and subjets. The closure statement does not assert that $p$ itself belongs to $D^{+} u (z_{0})$ ; that membership may fail, and it is not needed. No choice principle is used.

Facts & Assumptions

Given: An open $U \subseteq R^{n + 1}$ , continuous $H : U \times R^{n} \to R$ , an upper semicontinuous $u : U \to R$ , a lower semicontinuous $v : U \to R$ , and the superjet and subjet of the statement.

[F1]

$u$ is a viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $U$ when $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ holds for every $ϕ \in C^{1} (U)$ and every $z_{0} \in U$ at which $u - ϕ$ has a local maximum; $v$ is a viscosity supersolution when the reverse inequality holds at every local minimum of $v - ϕ$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

A map $f$ defined near $a$ is totally differentiable at $a$ with derivative $D f (a)$ exactly when $f (a + h) = f (a) + D f (a) h + r (h)$ with $∥ r (h) ∥ / ∥ h ∥ \to 0$ as $h \to 0$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

[F3]

If $g : [0, ρ] \to R$ is continuous and $G (r) : = \int_{0}^{r} g$ , then $G$ is differentiable at every $r \in [0, ρ]$ with $G^{'} (r) = g (r)$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive), the integral existing because a continuous function on a closed bounded interval is Riemann integrable (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[F4]

Closed bounded Euclidean sets are compact, and a finite-valued upper semicontinuous function on a nonempty compact set is bounded above (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent, Semicontinuous extreme value theorem on compact Euclidean sets).

Proof

technique · a test function gives its own jet directly; conversely a jet is converted into a $C^1$ majorant whose radial part is built by integrating a continuous dyadic modulus

1.1F1F2algebra

Test functions produce jets. Suppose $ϕ \in C^{1} (U)$ and $u - ϕ$ has a local maximum at $z_{0}$ ; then for $z$ near $z_{0}$ we have $u (z) - u (z_{0}) \leq ϕ (z) - ϕ (z_{0}) = ⟨ D ϕ (z_{0}), z - z_{0} ⟩ + o (∣ z - z_{0} ∣)$ by [F2], so $D ϕ (z_{0}) \in D^{+} u (z_{0})$ . If instead $v - ϕ$ has a local minimum at $z_{0}$ , the same computation with the inequality reversed gives $D ϕ (z_{0}) \in D^{-} v (z_{0})$ . Hence the jet inequality for all jets implies the test-function inequality of [F1], in both the sub- and the supersolution case.

1.2F1F2F3F4algebra

Jets produce test functions. Assume $u$ is an upper semicontinuous subsolution in the test-function sense and let $p \in D^{+} u (z_{0})$ . Choose $ρ > 0$ with $\overline{B} (z_{0}, ρ) \subseteq U$ and put $R (h) : = u (z_{0} + h) - u (z_{0}) - ⟨ p, h ⟩$ for $∣ h ∣ \leq ρ$ . By the definition of $D^{+} u (z_{0})$ , ${lim sup}_{h \to 0} R (h) / ∣ h ∣ \leq 0$ , so $r^{+} (h) : = \max {R (h), 0} = o (∣ h ∣)$ . Define $μ (s) : = \sup {r^{+} (h) / ∣ h ∣ : 0 < ∣ h ∣ \leq s}$ for $0 < s \leq ρ$ . This supremum is finite: the jet condition bounds the quotient near $h = 0$ , and on every remaining closed annulus $u (z_{0} + h)$ is bounded above by upper semicontinuity and compactness [F4]. Thus $μ$ is finite and nondecreasing, $μ (s) \to 0$ as $s ↓ 0$ , and $r^{+} (h) \leq μ (∣ h ∣) ∣ h ∣$ . The function $μ$ need not be continuous, so we first construct a continuous majorant. Put $r_{k} : = 2^{- k} ρ$ and $a_{k} : = μ (r_{k})$ for $k \geq 0$ . Define $ν (0) : = 0$ , set $ν (r_{k}) : = a_{k - 1}$ for $k \geq 1$ , interpolate linearly on each $[r_{k + 1}, r_{k}]$ , and set $ν (s) : = a_{0}$ on $[ρ / 2, ρ]$ . The definitions agree at $r_{1} = ρ / 2$ , $ν$ is continuous with $ν (s) \to 0$ at $0$ , and $ν (s) \geq μ (s)$ : on $[r_{k + 1}, r_{k}]$ , $ν (s) \geq a_{k} = μ (r_{k}) \geq μ (s)$ , while on $[ρ / 2, ρ]$ it equals $μ (ρ)$ . Let $σ (s) : = \max {0, \min {1, 2 (ρ - s) / ρ}}$ , continuous with $σ = 1$ on $[0, ρ / 2]$ and $σ (ρ) = 0$ , and put $g (s) : = 2 ν (\min {2 s, ρ}) σ (s)$ ; this is continuous on $[0, ρ]$ , with $g (0) = g (ρ) = 0$ . Define $λ (r) : = \int_{0}^{r} g (s) d s$ for $0 \leq r \leq ρ$ and $λ (r) : = λ (ρ)$ for $r > ρ$ . By [F3], $λ^{'} (r) = g (r)$ on $[0, ρ]$ , and $g (ρ) = 0$ makes the constant extension $C^{1}$ . For $0 < r \leq ρ / 2$ , $λ (r) \geq \int_{r / 2}^{r} 2 ν (2 s) d s \geq \int_{r / 2}^{r} 2 μ (r) d s = r μ (r),$ because $2 s \geq r$ , $ν (2 s) \geq μ (2 s) \geq μ (r)$ and $σ = 1$ there. Hence $λ (∣ h ∣) \geq r^{+} (h) \geq R (h)$ for $0 < ∣ h ∣ \leq ρ / 2$ . Define $ϕ (z) : = u (z_{0}) + ⟨ p, z - z_{0} ⟩ + λ (∣ z - z_{0} ∣)$ on $U$ . Then $u - ϕ$ has a local maximum at $z_{0}$ , and $D ϕ (z_{0}) = p$ because $λ (r) / r \leq \max_{0 \leq s \leq r} g (s) \to 0$ . The radial term has gradient $g (∣ z - z_{0} ∣) (z - z_{0}) / ∣ z - z_{0} ∣$ off $z_{0}$ , which tends to $0$ there; thus $ϕ \in C^{1} (U)$ . The subsolution inequality [F1] gives $p_{t} + H (z_{0}, p_{x}) \leq 0$ . The subjet case applies this construction to $- v$ and $- p$ , then negates the resulting test function, giving $p_{t} + H (z_{0}, p_{x}) \geq 0$ for every $p \in D^{-} v (z_{0})$ .

2.1step 1.1step 1.2F1

Parts (1) and (2). Step 1.1 shows that the jet inequalities imply the test-function inequalities. Step 1.2 proves the reverse implication by constructing a $C^{1}$ test for every prescribed jet, with signs reversed for subjets. Hence both equivalences (1) and (2) hold.

3.1step 2.1algebra∎

Closure. Let $u$ be a subsolution in the test-function sense, let $z_{k} \to z_{0}$ in $U$ , and let $p_{k} \in D^{+} u (z_{k})$ with $p_{k} \to p$ . Fix $k$ . By step 2.1, part (1), applied at $z_{k}$ , we have $p_{k}^{t} + H (z_{k}, p_{k}^{x}) \leq 0$ . Since $(z_{k}, p_{k}) \to (z_{0}, p)$ and $H$ is continuous, passing to the limit in the inequality gives $p_{t} + H (z_{0}, p_{x}) \leq 0$ , which is the closure statement for subsolutions; the same argument with the inequalities reversed and subjets in place of superjets gives the supersolution statement. The limit $p$ need not lie in $D^{+} u (z_{0})$ , and this is not used.

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Classical solutions are viscosity solutions and differentiable viscosity solutions solve the equation pointwise

Statement

Let $O \subseteq R^{n}$ be open, $T > 0$ , let $H : O \times [0, T] \times R^{n} \to R$ be continuous, let $u_{0} : O \to R$ be continuous, and let $Z = O \times (0, T)$ . (1) If $u \in C^{1} (Z) \cap C^{0} (\overline{Z})$ satisfies $u_{t} + H (x, t, D u) = 0$ pointwise on $Z$ and $u (x, 0) = u_{0} (x)$ on $O$ (a classical solution in the sense of The Hamilton--Jacobi Cauchy problem and its classical solutions), then $u$ is a viscosity solution of the Cauchy problem. (2) Conversely, if $u : Z \to R$ is a continuous viscosity solution and $u$ is differentiable at a point $z_{0} \in Z$ , then $u_{t} (z_{0}) + H (z_{0}, D u (z_{0})) = 0.$ In particular, a viscosity solution of class $C^{1} (Z)$ is a classical solution of the equation on $Z$ (its initial trace being part of the Cauchy-problem notion). No choice principle is used.

Facts & Assumptions

Given: Open $O \subseteq R^{n}$ , $T > 0$ , continuous $H : O \times [0, T] \times R^{n} \to R$ , continuous $u_{0} : O \to R$ , $Z = O \times (0, T)$ .

[F1]

$w$ is a viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $Z$ when $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ at every local maximum of $w - ϕ$ , $ϕ \in C^{1} (Z)$ ; a supersolution satisfies the reverse inequality at every local minimum; a subsolution of the Cauchy problem also satisfies ${lim sup}_{(y, s) \to (x, 0), s > 0, y \in O} w (y, s) \leq u_{0} (x)$ and a supersolution ${lim inf}_{(y, s) \to (x, 0), s > 0, y \in O} w (y, s) \geq u_{0} (x)$ at every $x \in O$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

If a real-valued function on an open set is differentiable at a point where it has a local maximum or a local minimum, then its total derivative vanishes there (Fermat's theorem: an interior differentiable local extremum has zero gradient).

[F3]

For an upper semicontinuous $u$ : $u$ is a viscosity subsolution of the equation in $Z$ if and only if $p_{t} + H (z_{0}, p_{x}) \leq 0$ for every $z_{0} \in Z$ and every $p \in D^{+} u (z_{0})$ ; for a lower semicontinuous $v$ : $v$ is a supersolution if and only if $p_{t} + H (z_{0}, p_{x}) \geq 0$ for every $p \in D^{-} v (z_{0})$ (Viscosity testing by first-order jets, and closure of the jet inequality).

[F4]

Total differentiability of $u$ at $z_{0}$ with derivative $D u (z_{0})$ means $u (z_{0} + h) = u (z_{0}) + D u (z_{0}) h + o (∣ h ∣)$ ; consequently $D u (z_{0}) \in D^{+} u (z_{0}) \cap D^{-} u (z_{0})$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

Proof

technique · Fermat's theorem at a $C^1$ contact for the classical direction, and the jet characterisation for the converse

1.1F1F2algebra

Classical solutions are viscosity solutions. Let $u \in C^{1} (Z) \cap C^{0} (\overline{Z})$ solve the equation pointwise, and let $ϕ \in C^{1} (Z)$ with $u - ϕ$ having a local maximum at $z_{0} \in Z$ . Then $u - ϕ$ is differentiable at $z_{0}$ and has a local extremum there, so $D ϕ (z_{0}) = D u (z_{0})$ by [F2]; hence $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) = u_{t} (z_{0}) + H (z_{0}, D u (z_{0})) = 0 \leq 0$ . At a local minimum the same computation gives $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) = 0 \geq 0$ . Since $u$ extends continuously to $\overline{Z}$ with $u (x, 0) = u_{0} (x)$ , the relaxed initial conditions of [F1] hold; so $u$ is both a subsolution and a supersolution of the Cauchy problem.

1.2F1F3F4

Differentiable viscosity solutions solve the equation pointwise. Let $u : Z \to R$ be continuous and differentiable at $z_{0} \in Z$ . By [F4], $D u (z_{0}) \in D^{+} u (z_{0}) \cap D^{-} u (z_{0})$ ; since $u$ is a continuous viscosity solution, it equals its envelopes (Discontinuous viscosity solutions through the two envelopes), so $u$ is both an upper semicontinuous subsolution and a lower semicontinuous supersolution in the sense of [F1]. Applying [F3] at $z_{0}$ with $p = D u (z_{0})$ gives both $u_{t} (z_{0}) + H (z_{0}, D u (z_{0})) \leq 0$ and $u_{t} (z_{0}) + H (z_{0}, D u (z_{0})) \geq 0$ , hence equality.

2.1step 1.1step 1.2F1∎

Conclusion. If in addition $u \in C^{1} (Z)$ , then the pointwise equation holds at every $z_{0} \in Z$ by step 1.2, and by step 1.1 the classical solution is a viscosity solution; the initial trace of a Cauchy-problem viscosity solution is the datum by [F1], so a $C^{1}$ viscosity solution solves the equation classically on $Z$ and extends continuously to the initial face. To be a classical solution of the Cauchy problem in the stronger sense of The Hamilton--Jacobi Cauchy problem and its classical solutions, it must additionally extend continuously to all of $\overline{Z}$ .

Remarks

What is not claimed. Part (2) presupposes that the viscosity solution is differentiable at the point; viscosity solutions of Hamilton--Jacobi equations are typically not differentiable everywhere, and the proposition says nothing about the nondifferentiable set.
Choice. Both directions are pointwise computations with the definitions; no selection principle occurs.

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Finite maxima of subsolutions and finite minima of supersolutions

Statement

Let $n \geq 1$ and $k \geq 1$ , let $U \subseteq R^{n + 1}$ be open, and let $H : U \times R^{n} \to R$ be continuous. (1) If $u_{1}, \dots, u_{k}$ are viscosity subsolutions of $u_{t} + H (x, t, D u) = 0$ in $U$ , then $u : = \max (u_{1}, \dots, u_{k})$ is a viscosity subsolution in $U$ . (2) If $v_{1}, \dots, v_{k}$ are viscosity supersolutions, then $v : = \min (v_{1}, \dots, v_{k})$ is a viscosity supersolution in $U$ . (3) For the Cauchy problem on $Z = O \times (0, T)$ the same statements hold when all the functions carry the same continuous initial datum $u_{0}$ in the relaxed sense; the maximum of subsolutions then also satisfies the relaxed initial condition for $u_{0}$ . The mixed operations are not asserted: a finite minimum of subsolutions and a finite maximum of supersolutions need not preserve the corresponding inequality when the equation has a zero-order term. No choice principle is used.

Facts & Assumptions

Given: An open $U \subseteq R^{n + 1}$ , continuous $H : U \times R^{n} \to R$ , viscosity subsolutions $u_{1}, \dots, u_{k}$ and supersolutions $v_{1}, \dots, v_{k}$ of $u_{t} + H (x, t, D u) = 0$ in $U$ , and $u = \max (u_{1}, \dots, u_{k})$ , $v = \min (v_{1}, \dots, v_{k})$ .

[F1]

Each $u_{i}$ is upper semicontinuous and satisfies $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ at every local maximum $z_{0}$ of $u_{i} - ϕ$ , $ϕ \in C^{1} (U)$ ; each $v_{j}$ is lower semicontinuous and satisfies the reverse inequality at every local minimum of $v_{j} - ϕ$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

For all reals $a_{1}, \dots, a_{k}$ the set ${a_{1}, \dots, a_{k}}$ has a maximum and a minimum, and its maximum equals one of the $a_{i}$ (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

[F3]

A real-valued function has a local maximum at $z_{0}$ when it is defined on a neighbourhood of $z_{0}$ and its value there is at most its value at $z_{0}$ (Local and strict local extrema for scalar fields on Euclidean open sets).

Proof

technique · select the active index at the contact

1.1F1F2F3algebra

Finite maxima of subsolutions. The function $u = \max (u_{1}, \dots, u_{k})$ is the pointwise maximum of finitely many upper semicontinuous functions and is therefore upper semicontinuous. Let $ϕ \in C^{1} (U)$ and let $u - ϕ$ have a local maximum at $z_{0} \in U$ ; by [F2] there is an index $i$ with $u_{i} (z_{0}) = u (z_{0})$ . Since $u_{i} \leq u$ pointwise, for every $z$ in a neighbourhood of $z_{0}$ we have $u_{i} (z) - ϕ (z) \leq u (z) - ϕ (z) \leq u (z_{0}) - ϕ (z_{0}) = u_{i} (z_{0}) - ϕ (z_{0})$ ; hence $u_{i} - ϕ$ has a local maximum at $z_{0}$ by [F3], and the subsolution inequality for $u_{i}$ gives $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ . Therefore $u$ is a viscosity subsolution.

1.2F1F2F3algebra

Finite minima of supersolutions. If $v = \min (v_{1}, \dots, v_{k})$ and $ϕ \in C^{1} (U)$ with $v - ϕ$ having a local minimum at $z_{0}$ , [F2] gives an index $j$ with $v_{j} (z_{0}) = v (z_{0})$ ; since $v \leq v_{j}$ and $v (z_{0}) = v_{j} (z_{0})$ , for $z$ near $z_{0}$ we have $v_{j} (z) - ϕ (z) \geq v (z) - ϕ (z) \geq v (z_{0}) - ϕ (z_{0}) = v_{j} (z_{0}) - ϕ (z_{0})$ , so $v_{j} - ϕ$ has a local minimum at $z_{0}$ and $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \geq 0$ by [F1]. Hence $v$ is a viscosity supersolution, and it is lower semicontinuous as a finite minimum of lower semicontinuous functions.

2.1step 1.1step 1.2F1algebra

The Cauchy problem. Suppose all $u_{i}$ and $v_{j}$ satisfy the relaxed initial conditions with the same continuous datum $u_{0}$ . Fix $x \in O$ and write $L_{i} : = {lim sup}_{(y, s) \to (x, 0), s > 0} u_{i} (y, s) \leq u_{0} (x)$ . For any real $c > \max_{i} L_{i}$ , the definition of each limsup gives a neighbourhood $N_{i}$ of $(x, 0)$ on which $u_{i} < c$ in $Z$ ; the finite intersection of these neighbourhoods then has $\max_{i} u_{i} < c$ , so $lim sup \max_{i} u_{i} \leq \max_{i} L_{i} \leq u_{0} (x)$ . The reverse inequality $lim sup \max_{i} u_{i} \geq \max_{i} L_{i}$ follows from $\max_{i} u_{i} \geq u_{i}$ for each $i$ . Dually, put $M_{j} : = {lim inf}_{(y, s) \to (x, 0), s > 0} v_{j} (y, s) \geq u_{0} (x)$ . For any real $c < \min_{j} M_{j}$ , each liminf gives a neighbourhood $N_{j}$ on which $v_{j} > c$ in $Z$ ; on their finite intersection $\min_{j} v_{j} > c$ , so $lim inf \min_{j} v_{j} \geq \min_{j} M_{j} \geq u_{0} (x)$ . The reverse inequality follows from $\min_{j} v_{j} \leq v_{j}$ for every $j$ . These finite-neighbourhood arguments also cover infinite relaxed limits and use no sequence extraction. With steps 1.1 and 1.2, the maximum of the subsolutions and minimum of the supersolutions satisfy the relaxed Cauchy conditions.

3.1step 1.1step 1.2step 2.1∎

Conclusion. Steps 1.1 and 1.2 prove the interior statements (1) and (2), and step 2.1 proves (3). Nothing is selected beyond a finite index, supplied by [F2], and no envelope or infinite supremum is used; the mixed operations are not claimed.

Remarks

The passage from finite families to arbitrary suprema requires upper regularisation and local boundedness, which is treated in The upper envelope of a locally bounded supremum of subsolutions is a subsolution.

Why the mixed operations fail. The active-index argument requires the function that touches $ϕ$ to be the same function that satisfies the one-sided inequality; for a maximum of subsolutions the active function is a subsolution, for a minimum of supersolutions it is a supersolution, and no argument of this shape covers a minimum of subsolutions. The companion counterexample Minima of viscosity subsolutions need not be subsolutions ↗ shows the failure is genuine for an equation with a zero-order term.
Choice. Only finitely many indices are involved and the selection is made inside a finite set, so no choice principle is used.

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Stability of viscosity sub-, super- and solutions under locally uniform convergence

Statement

Let $O \subseteq R^{n}$ be open, $T > 0$ , $Z = O \times (0, T)$ , and $Z_{0} = O \times [0, T)$ . Let $H_{k}, H : O \times [0, T] \times R^{n} \to R$ be continuous with $H_{k} \to H$ uniformly on every compact subset as $k \to \infty$ . For each $k$ , let $u_{k} : Z_{0} \to R$ be continuous, with $u_{k}$ restricted to $Z$ a viscosity subsolution of $(u_{k})_{t} + H_{k} (x, t, D u_{k}) = 0$ and with $u_{k} (x, 0) = u_{0}^{(k)} (x)$ . Suppose $u_{0}^{(k)} \to u_{0}$ locally uniformly on $O$ and $u_{k} \to u$ locally uniformly on $Z_{0}$ . Then $u$ is a viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $Z$ and its continuous initial trace is $u_{0}$ . The same statement holds for supersolutions, and combining the two, locally uniform limits of viscosity solutions are viscosity solutions. The conclusion is insensitive to the sign of the approximation: no differentiability and no monotonicity of convergence is used, only local uniformity up to the initial face. No choice principle is used.

Facts & Assumptions

Given: Open $O \subseteq R^{n}$ , $T > 0$ , $Z = O \times (0, T)$ , $Z_{0} = O \times [0, T)$ , continuous $H_{k}, H$ , continuous $u_{k} : Z_{0} \to R$ with $u_{k} \to u$ locally uniformly on $Z_{0}$ , continuous data $u_{0}^{(k)} \to u_{0}$ locally uniformly on $O$ , and $u_{k} ∣_{Z}$ a viscosity subsolution of $(u_{k})_{t} + H_{k} (x, t, D u_{k}) = 0$ .

[F1]

$u_{k} ∣_{Z}$ is upper semicontinuous and satisfies $ϕ_{t} (z_{0}) + H_{k} (z_{0}, D ϕ (z_{0})) \leq 0$ at every $z_{0} \in Z$ at which $u_{k} - ϕ$ has a local maximum, $ϕ \in C^{1} (Z)$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

If $v - ϕ$ has a local maximum at $z_{0}$ for a $C^{1}$ test $ϕ$ , and $r > 0$ is such that $v - ϕ \leq v (z_{0}) - ϕ (z_{0})$ on $\overline{B} (z_{0}, r) \subseteq U$ , then for every $ε > 0$ the function $ϕ_{ε} : = ϕ + ε ∣ z - z_{0} ∣^{4}$ is $C^{1}$ with $ϕ_{ε} (z_{0}) = ϕ (z_{0})$ , $D ϕ_{ε} (z_{0}) = D ϕ (z_{0})$ and $v - ϕ_{ε}$ strictly maximised over $\overline{B} (z_{0}, r)$ at $z_{0}$ (Strictification of a viscosity test function by a quartic perturbation).

[F3]

A nonempty subset of $R^{m}$ is compact if and only if it is closed and bounded, and every continuous real-valued function on a nonempty compact subset attains a maximum and a minimum there (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[F4]

In this statement, local uniform convergence on $Z_{0}$ means that for every compact $K \subseteq Z_{0}$ and every $η > 0$ there is $N$ such that $∣ u_{k} (z) - u (z) ∣ < η$ for all $k \geq N$ and $z \in K$ ; the Hamiltonians and data have the analogous meaning on their stated domains. Compactness here is intrinsic (Open cover, subcover, compact metric space, and compact subset of a metric space). In particular $u$ is continuous: on a small compact relative neighbourhood of any point, approximate $u$ within $η / 3$ by one continuous $u_{k}$ , then use continuity of that $u_{k}$ and the triangle inequality to bound the variation of $u$ by $η$ . This convergence condition is an explicit convention here.

Proof

technique · compact maximum localisation and passage to the limit in the test inequality

1.1F1F3F4algebra

The strict-contact case. Let $ϕ \in C^{1} (Z)$ and suppose $u - ϕ$ has a strict local maximum at $z_{0} \in Z$ . Choose $r > 0$ with $K : = \overline{B} (z_{0}, r) \subseteq Z$ and strict inequality at every point of $K ∖ {z_{0}}$ . For each $k$ , let $M_{k}$ be the nonempty compact set of maximisers of $u_{k} - ϕ$ on $K$ ; it is compact because $u_{k} - ϕ$ is continuous. For every $r_{0} \in (0, r)$ , the upper semicontinuous function $u - ϕ$ has a strict gap below its value at $z_{0}$ on the compact annulus $K ∖ B (z_{0}, r_{0})$ . Uniform convergence on $K$ therefore puts every point of $M_{k}$ inside $B (z_{0}, r_{0})$ for all sufficiently large $k$ . Since $r_{0}$ is arbitrary, $\sup_{z \in M_{k}} ∣ z - z_{0} ∣ \to 0$ , without choosing a maximiser for each $k$ . Every point $z \in M_{k}$ is then an interior local maximum of $u_{k} - ϕ$ and satisfies [F1]. If the desired residual $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0}))$ were positive, continuity would make $ϕ_{t} (z) + H (z, D ϕ (z))$ positive on a neighbourhood of $z_{0}$ ; uniform convergence of $H_{k}$ to $H$ on the compact set $K \times D ϕ (K)$ would make $ϕ_{t} (z) + H_{k} (z, D ϕ (z)) > 0$ there for all large $k$ . This contradicts [F1] at every point of the nonempty set $M_{k}$ . Hence the subsolution inequality holds at $z_{0}$ .

2.1F1F4algebra

The initial trace and the supersolution case. The local uniform convergence on $Z_{0}$ makes $u$ continuous on $Z_{0}$ with $u (x, 0) = \lim_{k} u_{k} (x, 0) = \lim_{k} u_{0}^{(k)} (x) = u_{0} (x)$ for every $x \in O$ , the convergence on compact subsets of $O$ being uniform; hence $u$ has the continuous initial trace $u_{0}$ and satisfies the relaxed initial condition ${lim sup}_{(y, s) \to (x, 0), s > 0} u (y, s) = u_{0} (x)$ . The same argument as in step 1.1 with local minima in place of local maxima, and the supersolution inequality of [F1] in place of the subsolution inequality, shows that a locally uniform limit of supersolutions is a supersolution with the same initial trace; no sign of the convergence is used, only that the test function is fixed.

3.1step 1.1step 2.1F2∎

Removal of strictness and conclusion. If $u - ϕ$ merely has a (nonstrict) local maximum at $z_{0}$ , fix $r > 0$ with $\overline{B} (z_{0}, r) \subseteq Z$ on which the maximum inequality holds and strictify: by [F2], $ϕ_{ε} = ϕ + ε ∣ z - z_{0} ∣^{4}$ satisfies $D ϕ_{ε} (z_{0}) = D ϕ (z_{0})$ , $\partial_{t} ϕ_{ε} (z_{0}) = ϕ_{t} (z_{0})$ and makes $u - ϕ_{ε}$ strictly maximised at $z_{0}$ over $\overline{B} (z_{0}, r)$ ; step 1.1 applied to $ϕ_{ε}$ gives $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ . Hence $u$ is a viscosity subsolution of the limit equation in $Z$ , and by step 2.1 it carries the datum $u_{0}$ ; the supersolution statement and the solution statement follow by step 2.1 and by combining the two one-sided conclusions.

Remarks

Where local uniformity up to $t = 0$ is needed. The interior equation only uses convergence on compact subsets of $Z$ ; the initial trace uses the convergence on compact subsets of $Z_{0}$ , which includes the initial face. Interior convergence alone would not imply the boundary conclusion, and the theorem does not claim it.
Choice. The proof uses the sets of maximisers on compact balls and the uniform strict gap away from the limiting contact; it selects no sequence of points and uses no choice principle.

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Half-relaxed limits of a locally bounded family

Definition

Let $U \subseteq R^{m}$ be open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement) and let $(u_{ε})_{ε \in (0, 1)}$ be a family of real-valued functions on $U$ that is locally bounded: for every compact $K \subseteq U$ there is $M_{K} < \infty$ with $∣ u_{ε} (y) ∣ \leq M_{K}$ for all $ε \in (0, 1)$ and $y \in K$ . The upper half-relaxed limit and the lower half-relaxed limit of the family, computed in $\overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Greatest lower bound (infimum), Epsilon characterisation of the supremum), are $\overline{u} (z) : = {lim sup}_{\begin{array}{c} ε ↓ 0 \\ U ∋ y \to z \end{array}} u_{ε} (y) : = \inf_{δ > 0} \sup {u_{ε} (y) : 0 < ε < δ, y \in U, ∣ y - z ∣ < δ},$ $\underline{u} (z) : = {lim inf}_{\begin{array}{c} ε ↓ 0 \\ U ∋ y \to z \end{array}} u_{ε} (y) : = \sup_{δ > 0} \inf {u_{ε} (y) : 0 < ε < δ, y \in U, ∣ y - z ∣ < δ} .$ The quantifiers range over the index $ε$ and the point $y$ simultaneously, so the limit records the behaviour of the whole family near $z$ , not the limit of the single family of values $(u_{ε} (z))_{ε}$ ; both envelopes are local and depend only on the germ of the family at $z$ .

Remarks

Basic properties. If the family converges locally uniformly on $U$ to a continuous function $u$ , then $\overline{u} = \underline{u} = u$ . If the family is only locally bounded, then $\underline{u} \leq \overline{u}$ pointwise, and $\overline{u}$ is upper semicontinuous while $\underline{u}$ is lower semicontinuous on $U$ : the expressions are again monotone limits of local suprema and infima over families, and the proof of The envelopes are the least upper and greatest lower semicontinuous functions applies verbatim with the family indexed by $δ$ . Every value is kept in $\overline{R}$ ; local boundedness makes both envelopes real-valued on each compact subset of $U$ .
Why the joint limit. Under Countable Choice, there are pairs $(ε_{j}, y_{j}) \to (0, z)$ with $u_{ε_{j}} (y_{j}) \to \underline{u} (z)$ , and likewise for $\overline{u} (z)$ . For finite lower limit, take the infima over $0 < ε < 1 / j$ , $∣ y - z ∣ < 1 / j$ and choose a point within $1 / j$ of each infimum; these infima increase to $\underline{u} (z)$ . The upper case is dual, with suprema decreasing to $\overline{u} (z)$ . Infinite values use diverging finite thresholds. The definition itself is set-based and selects no subsequence or point; only this sequential characterization uses Countable Choice. This is the limit notion consumed by Half-relaxed limits of sub- and supersolutions with vanishing perturbations.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Half-relaxed limits of sub- and supersolutions with vanishing perturbations

Statement

Let $O \subseteq R^{n}$ be open, $T > 0$ , $Z = O \times (0, T)$ , let $H_{ε}, H : O \times [0, T] \times R^{n} \to R$ be continuous for $ε \in (0, 1)$ , and let $(u_{ε})$ be a locally bounded family of real functions on $Z$ that are upper semicontinuous in the subsolution case. Assume one of the two forms of the Hamiltonian condition: (a) $H_{ε} \to H$ uniformly on compact subsets of $O \times [0, T] \times R^{n}$ ; or (b) the exact limit-inferior condition: for all sequences $ε_{j} ↓ 0$ , $z_{j} \to z \in Z$ and $p_{j} \to p$ one has ${lim inf}_{j} H_{ε_{j}} (z_{j}, p_{j}) \geq H (z, p)$ . Suppose moreover that for every $ϕ \in C^{1} (Z)$ and every local maximum point $z_{ε} \in Z$ of $u_{ε} - ϕ$ there holds $ϕ_{t} (z_{ε}) + H_{ε} (z_{ε}, D ϕ (z_{ε})) \leq c_{ε} (z_{ε}),$ where $c_{ε} : Z \to [0, \infty)$ is locally bounded with $c_{ε} \to 0$ locally uniformly. Then the upper half-relaxed limit $\overline{u}$ (Half-relaxed limits of a locally bounded family) is a viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $Z$ . If in addition $u_{ε} (x, 0) \leq u_{0}^{(ε)} (x)$ in the relaxed sense with $u_{0}^{(ε)} \to u_{0}$ locally uniformly on $O$ and the family is locally equicontinuous up to the initial face, then $\overline{u}$ carries the initial datum $u_{0}$ in the relaxed sense. The dual statement with ${lim sup}_{j} H_{ε_{j}} (z_{j}, p_{j}) \leq H (z, p)$ , $\geq - c_{ε}$ and the lower half-relaxed limit holds for supersolutions. Choice. Under hypothesis (a) the proof is choice-free. Under hypothesis (b) it extracts a sequence of near-maximisers at the relaxed limit and therefore uses Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), which is declared as a dependency; the extraction is the only place where the principle is consumed.

Facts & Assumptions

Given: The open sets $O \subseteq R^{n}$ , $Z = O \times (0, T)$ , continuous Hamiltonians $H_{ε}, H$ , a locally bounded family $(u_{ε})$ of real functions on $Z$ , locally bounded perturbations $c_{ε} \geq 0$ with $c_{ε} \to 0$ locally uniformly, and the half-relaxed limits $\overline{u}, \underline{u}$ of Half-relaxed limits of a locally bounded family.

[F1]

$\overline{u} (z) = \inf_{δ > 0} \sup {u_{ε} (y) : 0 < ε < \min {1, δ}, y \in Z, ∣ y - z ∣ < δ}$ and $\underline{u} (z) = \sup_{δ > 0} \inf {u_{ε} (y) : 0 < ε < \min {1, δ}, y \in Z, ∣ y - z ∣ < δ}$ ; local boundedness makes both real-valued on compact subsets of $Z$ ; $\overline{u}$ is upper semicontinuous and $\underline{u}$ lower semicontinuous (Half-relaxed limits of a locally bounded family).

[F2]

At every local maximum of $u_{ε} - ϕ$ with $ϕ \in C^{1} (Z)$ the assumed inequality $ϕ_{t} (z_{ε}) + H_{ε} (z_{ε}, D ϕ (z_{ε})) \leq c_{ε} (z_{ε})$ holds; a viscosity subsolution of the limit equation is a function that satisfies $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ at every local maximum of the function and test function (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F3]

If $v - ϕ$ has a local maximum at $z_{0}$ and $\overline{B} (z_{0}, r) \subseteq U$ is a ball on which $v - ϕ \leq v (z_{0}) - ϕ (z_{0})$ , then for every $ε > 0$ the perturbed test $ϕ_{ε} = ϕ + ε ∣ z - z_{0} ∣^{4}$ has the same value and first jet as $ϕ$ at $z_{0}$ and makes $v - ϕ_{ε}$ strictly maximised over $\overline{B} (z_{0}, r)$ at $z_{0}$ (Strictification of a viscosity test function by a quartic perturbation).

[F4]

Every upper semicontinuous real-valued function on a nonempty compact subset of $R^{m}$ attains its maximum there (Semicontinuous extreme value theorem on compact Euclidean sets).

[F5]

Countable Choice is the principle that every sequence $(S_{j})_{j \geq 1}$ of nonempty sets has a sequence of choices $(s_{j})$ with $s_{j} \in S_{j}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F6]

A compact metric space has a finite subcover from every intrinsic open cover (Open cover, subcover, compact metric space, and compact subset of a metric space); for a compact Euclidean subset, every family of ambient open balls covering it has finitely many members covering it, also with their indices retained (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, clauses 2--3).

[F7]

Local equicontinuity up to the initial face gives each $u_{ε}$ a continuous trace $u_{ε}^{0} (x) : = \lim_{(y, s) \to (x, 0), s > 0} u_{ε} (y, s)$ and a common local modulus there; the relaxed initial inequality implies $u_{ε}^{0} (x) \leq u_{0}^{(ε)} (x)$ .

Proof

technique · separate the uniform case (a) from the sequential case (b); in case (a) the argument uses a single near-maximal pair and one compact maximiser, and in case (b) a sequence of near-maximisers is extracted at the relaxed limit

1.1F1F2F4F6algebra

Case (a): the strict-contact case, choice-free. Let $ϕ \in C^{1} (Z)$ and let $\overline{u} - ϕ$ have a strict local maximum at $z_{0} = (x_{0}, t_{0}) \in Z$ . Choose $ρ > 0$ with $\overline{B} (z_{0}, ρ) \subseteq Z$ and strict inequality away from $z_{0}$ . Assume for contradiction that $θ : = ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) > 0$ . By continuity there is $r \in (0, ρ / 2)$ such that $ϕ_{t} (z) + 2 (t - t_{0}) + H (z, D ϕ (z) + 2 (x - x_{0})) > 3 θ / 4$ for $∣ z - z_{0} ∣ \leq r$ . The compact annulus $A_{r} : = {z : r / 2 \leq ∣ z - z_{0} ∣ \leq r}$ has a strict gap $g : = \overline{u} (z_{0}) - ϕ (z_{0}) - \max_{A_{r}} (\overline{u} - ϕ) > 0$ by [F1, F4]. Write $z = (x, t)$ and $z_{0} = (x_{0}, t_{0})$ . For each $z \in A_{r}$ , the defining infimum for $\overline{u} (z)$ gives $δ_{z} > 0$ such that $u_{ε} (y) < \overline{u} (z) + g / 8$ whenever $0 < ε < δ_{z}$ and $∣ y - z ∣ < δ_{z}$ ; shrink $δ_{z}$ so also $∣ ϕ (y) + ∣ y - z_{0} ∣^{2} - ϕ (z) - ∣ z - z_{0} ∣^{2} ∣ < g / 8$ there. Use the collection of all pairs $(z, δ)$ satisfying these bounds; their balls $B (z, δ / 2)$ cover $A_{r}$ without choosing one radius at each point. By compactness and [F6], finitely many such balls $B (z_{i}, δ_{i} / 2)$ cover $A_{r}$ ; let $ε_{A} : = \min_{i} δ_{i}$ . For every $0 < ε < ε_{A}$ and $y \in A_{r}$ , these bounds give $u_{ε} (y) - ϕ (y) - ∣ y - z_{0} ∣^{2} < \overline{u} (z_{i}) - ϕ (z_{i}) - ∣ z_{i} - z_{0} ∣^{2} + g / 4 \leq \overline{u} (z_{0}) - ϕ (z_{0}) - 3 g / 4$ for some $i$ . Uniform convergence $H_{ε} \to H$ on compact subsets and local uniform convergence $c_{ε} \to 0$ provide $ε_{H} > 0$ such that for $ε < ε_{H}$ and $∣ z - z_{0} ∣ \leq r$ , the corresponding upper-test residual with gradient $D ϕ (z) + 2 (x - x_{0})$ is $> θ / 2$ and $c_{ε} (z) < θ / 2$ . Choose $0 < η < g / 12$ and then $δ < \min {r / 4, ε_{A}, ε_{H}}$ so that $∣ ϕ (y) - ϕ (z_{0}) ∣ < η$ and $∣ y - z_{0} ∣^{2} < η$ when $∣ y - z_{0} ∣ < δ$ . By [F1] there is one pair $(ε, y)$ with $0 < ε < δ$ , $∣ y - z_{0} ∣ < δ$ and $u_{ε} (y) > \overline{u} (z_{0}) - η$ . Let $z_{ε}$ maximise the upper semicontinuous function $u_{ε} (z) - ϕ (z) - ∣ z - z_{0} ∣^{2}$ on the compact ball $\overline{B} (z_{0}, r)$ , possible by [F4]. Its value is $> \overline{u} (z_{0}) - ϕ (z_{0}) - 3 η$ , so the uniform annulus bound forces $∣ z_{ε} - z_{0} ∣ < r / 2$ . Thus $u_{ε} - (ϕ + ∣ z - z_{0} ∣^{2})$ has a local maximum at $z_{ε}$ . Writing $z_{0} = (x_{0}, t_{0})$ and $z_{ε} = (x_{ε}, t_{ε})$ , the test has time derivative $ϕ_{t} (z_{ε}) + 2 (t_{ε} - t_{0})$ and spatial gradient $D ϕ (z_{ε}) + 2 (x_{ε} - x_{0})$ ; its residual is $> θ / 2$ while $c_{ε} (z_{ε}) < θ / 2$ , contradicting the assumed subsolution inequality. Hence $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ at every strict local maximum.

1.2F1F7algebra

The initial trace. Assume the relaxed initial inequality and data convergence of the statement, and use the traces of [F7]. Fix $x \in O$ and $η > 0$ . By local equicontinuity and continuity of $u_{0}^{(ε)} \to u_{0}$ there is a neighbourhood $V$ of $(x, 0)$ such that, for all sufficiently small $ε$ and $(y, s) \in V \cap Z$ , $u_{ε} (y, s) \leq u_{ε}^{0} (x) + η \leq u_{0}^{(ε)} (x) + η \leq u_{0} (x) + 2 η .$ Shrink to a neighbourhood $V^{'}$ whose closure lies in $V$ . For each $z^{'} \in V^{'} \cap Z$ sufficiently close to $(x, 0)$ , the neighborhoods in the definition of $\overline{u} (z^{'})$ can be taken inside $V$ , so the same bound gives $\overline{u} (z^{'}) \leq u_{0} (x) + 2 η$ . Therefore ${lim sup}_{(y, s) \to (x, 0), s > 0} \overline{u} (y, s) \leq u_{0} (x) + 2 η$ ; letting $η ↓ 0$ proves the relaxed subsolution initial condition. The lower-limit argument is the dual one.

2.1F1F2F4F5F6algebra

Case (b): the strict-contact case with near-maximiser extraction. Assume the exact limit-inferior condition and let $ϕ$ and $z_{0}$ be as in step 1.1. For each $j \geq 1$ , consider triples $(ε, y, z)$ with $0 < ε < \min {1, 1 / j}$ , $y \in Z$ , $∣ y - z_{0} ∣ < \min {ρ / 2, 1 / j}$ , $u_{ε} (y) > \overline{u} (z_{0}) - 1 / j$ , and $z$ a maximiser of $u_{ε} (\cdot) - ϕ (\cdot) - ∣ \cdot - z_{0} ∣^{2}$ on $\overline{B} (z_{0}, ρ)$ . This set is nonempty by [F1] and [F4]; Countable Choice [F5] selects triples $(ε_{j}, y_{j}, z_{j})$ . Then $ε_{j} \to 0$ , $y_{j} \to z_{0}$ and the maximal values satisfy $u_{ε_{j}} (z_{j}) - ϕ (z_{j}) - ∣ z_{j} - z_{0} ∣^{2} \geq \overline{u} (z_{0}) - ϕ (z_{0}) - o (1)$ . Fix any $r \in (0, ρ)$ . The strict maximum of $\overline{u} - ϕ$ gives a positive gap on the compact annulus $r / 2 \leq ∣ z - z_{0} ∣ \leq ρ$ ; the finite-cover argument of step 1.1 then bounds $u_{ε} (z) - ϕ (z) - ∣ z - z_{0} ∣^{2}$ strictly below $\overline{u} (z_{0}) - ϕ (z_{0})$ on this annulus for all sufficiently small $ε$ . Since $ε_{j} \to 0$ and the maximizing values are at least that limit minus $o (1)$ , eventually $∣ z_{j} - z_{0} ∣ < r / 2$ . As $r > 0$ was arbitrary, $z_{j} \to z_{0}$ . By taking the canonical strictly decreasing subsequence of $(ε_{j})$ (at each stage use the least later index with smaller $ε$ , which exists because $ε_{j} \to 0$ ) and relabelling, we may assume $ε_{j} ↓ 0$ ; then $z_{j} \to z_{0}$ still. Eventually $z_{j}$ is interior to $\overline{B} (z_{0}, ρ)$ , so $ψ_{j} : = ϕ + ∣ z - z_{0} ∣^{2}$ is a local upper test for $u_{ε_{j}}$ there. Thus $ϕ_{t} (z_{j}) + 2 (t_{j} - t_{0}) + H_{ε_{j}} (z_{j}, D ϕ (z_{j}) + 2 (x_{j} - x_{0})) \leq c_{ε_{j}} (z_{j}) .$ Writing $z_{j} = (x_{j}, t_{j})$ , the extra time derivative $2 (t_{j} - t_{0})$ tends to zero. Since the gradients converge and $c_{ε_{j}} (z_{j}) \to 0$ , taking the limit inferior of the displayed inequality and using (b) gives $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ . Condition (a) implies (b) by uniform convergence on compact sets, so this proves the strict-contact case.

3.1step 1.1step 1.2step 2.1F3∎

General contacts, the dual statement and conclusion. If $\overline{u} - ϕ$ merely has a local maximum at $z_{0}$ , strictify with [F3] and apply the strict-contact conclusion of steps 1.1 or 2.1 to the strictified test; the perturbed test has the same value and first jet at $z_{0}$ , so the resulting inequality is exactly $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ . Hence $\overline{u}$ is a viscosity subsolution of the limit equation, and by step 1.2 it carries the initial datum when the additional hypotheses hold. The dual argument, replacing $u_{ε}$ by $- u_{ε}$ and local maxima by local minima, shows that $\underline{u}$ is a viscosity supersolution with the dual initial condition. The half-relaxed limits themselves are computed as infima and suprema over sets, and only step 2.1 involves a countable selection, so under hypothesis (a) no choice principle is used and under hypothesis (b) Countable Choice is used exactly as declared.

Remarks

The role of $c_{ε}$ . The vanishing perturbation $c_{ε}$ is the fixed-test mechanism used for the viscous equation $u_{t} + H (x, t, D u) = ε Δ u$ , where the extra term $ε Δ ϕ$ is locally bounded and tends to $0$ uniformly on compact sets for a fixed $C^{1, 2}$ test. This is not directly the theorem's hypothesis for all $C^{1}$ tests with one common error function; Vanishing viscosity selects the viscosity solution supplies the fixed-test argument and smooth approximation needed there.
Choice ledger. Case (a), which is the case used by the vanishing-viscosity argument of this page, is choice-free: a single near-maximal pair and a single compact maximiser suffice. Case (b) needs Countable Choice to turn the defining infimum-of-suprema at the relaxed limit into a sequence of near-maximisers; this is the use of choice declared in the statement.

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Doubling variables: existence, relative contacts at the maximiser and localisation

Statement

Let $n \geq 1$ , $T > 0$ , let $u : R^{n} \times [0, T] \to R$ be bounded above and upper semicontinuous, and let $v : R^{n} \times [0, T] \to R$ be bounded below and lower semicontinuous. Fix $α, ρ > 0$ and define $Φ (x, y, t, s) : = u (x, t) - v (y, s) - \frac{α}{2} ∣ x - y ∣^{2} - \frac{α}{2} ∣ t - s ∣^{2} - ρ (∣ x ∣^{2} + ∣ y ∣^{2})$ on $R^{n} \times R^{n} \times [0, T]^{2}$ , with $M_{α, ρ} : = \sup Φ$ . Then: (1) $M_{α, ρ}$ is finite and attained, and at every maximiser $(x_{α}, y_{α}, t_{α}, s_{α})$ the $C^{1}$ test functions $ψ_{1} (x, t) : = \frac{α}{2} ∣ x - y_{α} ∣^{2} + \frac{α}{2} ∣ t - s_{α} ∣^{2} + ρ ∣ x ∣^{2}, ψ_{2} (y, s) : = - \frac{α}{2} ∣ x_{α} - y ∣^{2} - \frac{α}{2} ∣ t_{α} - s ∣^{2} - ρ ∣ y ∣^{2}$ satisfy: $u - ψ_{1}$ has a local maximum at $(x_{α}, t_{α})$ and $v - ψ_{2}$ has a local minimum at $(y_{α}, s_{α})$ , with $p_{α} : = D_{x} ψ_{1} (x_{α}) = α (x_{α} - y_{α}) + 2 ρ x_{α}, q_{α} : = D_{y} ψ_{2} (y_{α}) = α (x_{α} - y_{α}) - 2 ρ y_{α}, p_{α}^{0} : = \partial_{t} ψ_{1} (t_{α}) = \partial_{s} ψ_{2} (s_{α}) = α (t_{α} - s_{α});$ (2) for each fixed $ρ > 0$ , along any sequence of maximisers as $α \to \infty$ , $α (∣ x_{α} - y_{α} ∣^{2} + ∣ t_{α} - s_{α} ∣^{2}) ⟶ 0, (1 + ∣ p_{α} ∣) (∣ x_{α} - y_{α} ∣ + ∣ t_{α} - s_{α} ∣) ⟶ 0, ∣ x_{α} - y_{α} ∣ + ∣ t_{α} - s_{α} ∣ ⟶ 0,$ while $∣ q_{α} - p_{α} ∣ = 2 ρ ∣ x_{α} + y_{α} ∣$ is bounded by $2 ρ (∣ x_{α} ∣ + ∣ y_{α} ∣)$ and is not claimed to vanish for fixed $ρ$ ; (3) at every maximiser $\frac{α}{2} (∣ x_{α} - y_{α} ∣^{2} + ∣ t_{α} - s_{α} ∣^{2}) + ρ (∣ x_{α} ∣^{2} + ∣ y_{α} ∣^{2}) \leq \sup u - \inf v - M_{α, ρ},$ so whenever the maximiser values $M_{α, ρ}$ are bounded below by $m$ on a set of parameters, the corresponding maximisers satisfy $ρ (∣ x_{α} ∣^{2} + ∣ y_{α} ∣^{2}) \leq \sup u - \inf v - m$ and $∣ q_{α} - p_{α} ∣ \leq 2 (2 ρ (\sup u - \inf v - m))^{1 / 2}$ . No choice principle is used.

Facts & Assumptions

Given: Bounded-above upper semicontinuous $u$ and bounded-below lower semicontinuous $v$ on $R^{n} \times [0, T]$ , parameters $α, ρ > 0$ , and the function $Φ$ of the statement.

[F1]

Upper semicontinuity of $u$ and lower semicontinuity of $v$ mean that every superlevel set of $u$ and every sublevel set of $v$ is relatively closed; equivalently, $- v$ is upper semicontinuous (Upper and lower semicontinuity on subsets of $R^{n}$ ).

[F2]

Every upper semicontinuous real-valued function on a nonempty compact subset of $R^{n}$ is bounded above and attains its maximum (Semicontinuous extreme value theorem on compact Euclidean sets).

[F3]

A subset of $R^{n}$ is compact if and only if it is closed and bounded (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F4]

If $S \subseteq R$ is nonempty, bounded above and $w$ is an upper bound of $S$ with the property that for every $ε > 0$ there is $s \in S$ with $w - ε < s$ , then $w = \sup S$ ; in particular $\sup S - ε < s$ for some $s \in S$ and every $ε > 0$ (Epsilon characterisation of the supremum).

Proof

technique · coercive weight for existence, monotonicity in $\alpha$ for the localisation, and explicit $C^1$ contacts for the two tests

1.1F1F2F3algebra

Existence and finiteness of $M_{α, ρ}$ . On $R^{n} \times R^{n} \times [0, T]^{2}$ the function $Φ$ is upper semicontinuous, being $u$ plus the upper semicontinuous $- v$ plus continuous terms by [F1]; it is bounded above by $\sup u - \inf v < \infty$ because the quadratic and weight terms are nonpositive. Pick any point $p_{0} : = (0, 0, 0, 0)$ and put $m_{0} : = Φ (p_{0}) \in R$ ; the superlevel set $S_{0} : = {Φ \geq m_{0}}$ is nonempty, closed by upper semicontinuity, and bounded because $ρ (∣ x ∣^{2} + ∣ y ∣^{2}) \leq \sup u - \inf v - m_{0}$ on it; hence $S_{0}$ is compact by [F3]. On $S_{0}$ the restriction of $Φ$ is real-valued and upper semicontinuous, so it attains a maximum by [F2]; that maximum is a global maximum of $Φ$ because every point outside $S_{0}$ has value $< m_{0} \leq \max_{S_{0}} Φ$ . Hence $M_{α, ρ}$ is finite and attained.

2.1step 1.1algebra

The relative contacts and their derivatives. Let $(x_{α}, y_{α}, t_{α}, s_{α})$ be a maximiser. Fixing $(y, s) = (y_{α}, s_{α})$ , the inequality $Φ (x, y_{α}, t, s_{α}) \leq Φ (x_{α}, y_{α}, t_{α}, s_{α})$ for all $(x, t)$ reads $u (x, t) - ψ_{1} (x, t) \leq u (x_{α}, t_{α}) - ψ_{1} (x_{α}, t_{α})$ , so $u - ψ_{1}$ has a local maximum at $(x_{α}, t_{α})$ ; fixing $(x, t) = (x_{α}, t_{α})$ similarly gives $v (y, s) - ψ_{2} (y, s) \geq v (y_{α}, s_{α}) - ψ_{2} (y_{α}, s_{α})$ , a local minimum of $v - ψ_{2}$ at $(y_{α}, s_{α})$ . The displayed gradients and time derivatives are the derivatives of the two quadratic test functions: $D_{x} ψ_{1} = α (x - y_{α}) + 2 ρ x$ evaluated at $x_{α}$ gives $p_{α}$ , $D_{y} ψ_{2} = α (x_{α} - y) - 2 ρ y$ evaluated at $y_{α}$ gives $q_{α}$ , and $\partial_{t} ψ_{1} = α (t - s_{α})$ , $\partial_{s} ψ_{2} = α (t_{α} - s)$ both equal $p_{α}^{0}$ at the maximiser.

3.1step 2.1algebra

The weight bound (3). At a maximiser, $Φ (x_{α}, y_{α}, t_{α}, s_{α}) = M_{α, ρ}$ , that is $u (x_{α}, t_{α}) - v (y_{α}, s_{α}) - \frac{α}{2} d_{α}^{2} - ρ (∣ x_{α} ∣^{2} + ∣ y_{α} ∣^{2}) = M_{α, ρ}$ ; since $u (x_{α}, t_{α}) \leq \sup u$ and $- v (y_{α}, s_{α}) \leq - \inf v$ , the sum of the two penalty terms is at most $\sup u - \inf v - M_{α, ρ}$ , which is (3). If in addition $M_{α, ρ} \geq m$ then $ρ (∣ x_{α} ∣^{2} + ∣ y_{α} ∣^{2}) \leq \sup u - \inf v - m$ , and $∣ q_{α} - p_{α} ∣ \leq 2 ρ (∣ x_{α} ∣ + ∣ y_{α} ∣) \leq 2 ρ \sqrt{2 (∣ x_{α} ∣^{2} + ∣ y_{α} ∣^{2})} \leq 2 (2 ρ (\sup u - \inf v - m))^{1 / 2}$ .

4.1step 2.1step 3.1F4algebra

Localisation as $α \to \infty$ . Fix any sequence $α_{j} \to \infty$ and any corresponding sequence of maximisers $(x_{j}, y_{j}, t_{j}, s_{j})$ ; these are exactly the sequences quantified in part (2). For fixed $ρ$ , $α \mapsto M_{α, ρ}$ is nonincreasing and bounded below by $Φ (0, 0, 0, 0)$ , so it converges. Put $d_{j}^{2} : = ∣ x_{j} - y_{j} ∣^{2} + ∣ t_{j} - s_{j} ∣^{2}$ . Evaluating the $α_{j} / 2$ -function at this same maximiser gives $M_{α_{j} / 2, ρ} \geq M_{α_{j}, ρ} + \frac{α_{j}}{4} d_{j}^{2}$ , hence $α_{j} d_{j}^{2} \leq 4 (M_{α_{j} / 2, ρ} - M_{α_{j}, ρ}) \to 0$ . In particular $∣ x_{j} - y_{j} ∣ + ∣ t_{j} - s_{j} ∣ \to 0$ . By step 3.1, $∣ x_{j} ∣$ is uniformly bounded for fixed $ρ$ . With $p_{j} = α_{j} (x_{j} - y_{j}) + 2 ρ x_{j}$ , the bounds $α_{j} ∣ x_{j} - y_{j} ∣ (∣ x_{j} - y_{j} ∣ + ∣ t_{j} - s_{j} ∣) \leq 2 α_{j} d_{j}^{2} \to 0$ and $2 ρ ∣ x_{j} ∣ (∣ x_{j} - y_{j} ∣ + ∣ t_{j} - s_{j} ∣) \to 0$ give $(1 + ∣ p_{j} ∣) (∣ x_{j} - y_{j} ∣ + ∣ t_{j} - s_{j} ∣) \to 0$ . Finally $q_{j} - p_{j} = - 2 ρ (x_{j} + y_{j})$ and step 3.1 gives the asserted bound on $∣ q_{j} - p_{j} ∣$ , which need not vanish for fixed $ρ$ . The argument applies to every given sequence of maximisers and selects none.

5.1step 1.1step 2.1step 3.1step 4.1∎

Conclusion. Part (1) is steps 1.1 and 2.1, part (2) is step 4.1, and part (3) is step 3.1; the maximiser is obtained from the compactness of a closed bounded superlevel set and the extreme-value property for upper semicontinuous functions, and no sequence, point or index is selected in the construction.

Remarks

The contacts in part (1) are relative to $R^{n} \times [0, T]$ . A maximiser may have $t_{α} = 0$ or $s_{α} = 0$ (for example, $u (x, t) = - t$ , $v (y, s) = s$ ), or lie on a terminal face. The corresponding derivative belongs to the first-order superjet or subjet of Viscosity testing by first-order jets, and closure of the jet inequality only when the contact time is in $(0, T)$ , where the restricted domain is open. Viscosity inequalities in the interior therefore require a separate exclusion of time-boundary contacts.

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Time penalisation moves a doubling-variables maximum away from the terminal boundary

Statement

Let $T > 0$ , $Z = R^{n} \times (0, T)$ , and let $H : R^{n} \times [0, T] \times R^{n} \to R$ be continuous. Suppose $u, v : R^{n} \times [0, T) \to R$ , where $u$ is upper semicontinuous and bounded above, $v$ is lower semicontinuous and bounded below, and their restrictions to $Z$ are respectively a viscosity subsolution and a viscosity supersolution of $u_{t} + H (x, t, D u) = 0$ . For $η, η^{'} > 0$ put ${\tilde{u}}_{η} (x, t) = u (x, t) - η / (T - t)$ and ${\tilde{v}}_{η^{'}} (x, t) = v (x, t) + η^{'} / (T - t)$ for $t < T$ . Then: (1) every $C^{1}$ upper contact $ϕ$ for ${\tilde{u}}_{η}$ at $z_{0} = (x_{0}, t_{0}) \in Z$ satisfies $ϕ_{t} (z_{0}) + H (x_{0}, t_{0}, D ϕ (z_{0})) \leq - \frac{η}{(T - t_{0})^{2}} < 0;$ (2) every $C^{1}$ lower contact $ϕ$ for ${\tilde{v}}_{η^{'}}$ at $z_{0} \in Z$ satisfies $ϕ_{t} (z_{0}) + H (x_{0}, t_{0}, D ϕ (z_{0})) \geq \frac{η^{'}}{(T - t_{0})^{2}} > 0;$ moreover ${\tilde{u}}_{η} \to - \infty$ and ${\tilde{v}}_{η^{'}} \to + \infty$ uniformly in $x$ as $t ↑ T$ ; (3) for every $α, ρ > 0$ , define on $R^{n} \times R^{n} \times [0, T]^{2}$ $Φ_{α, ρ} (x, y, t, s) = {\tilde{u}}_{η} (x, t) - {\tilde{v}}_{η^{'}} (y, s) - \frac{α}{2} ∣ x - y ∣^{2} - \frac{α}{2} ∣ t - s ∣^{2} - ρ (∣ x ∣^{2} + ∣ y ∣^{2})$ when $t, s < T$ , and set $Φ_{α, ρ} = - \infty$ when $t = T$ or $s = T$ . Then $Φ_{α, ρ}$ attains a finite maximum, and every maximiser has $t, s < T$ . A maximum may occur on an initial face (that is, with $t = 0$ or $s = 0$ ). No choice principle is used.

Facts & Assumptions

Given: $T > 0$ , continuous $H : R^{n} \times [0, T] \times R^{n} \to R$ , an upper semicontinuous function $u : R^{n} \times [0, T) \to R$ bounded above, a lower semicontinuous $v$ bounded below, whose restrictions to $Z = R^{n} \times (0, T)$ are a viscosity subsolution and supersolution of $u_{t} + H (x, t, D u) = 0$ , the functions ${\tilde{u}}_{η} = u - η / (T - t)$ , ${\tilde{v}}_{η^{'}} = v + η^{'} / (T - t)$ for $η, η^{'} > 0$ , and the functions $Φ_{α, ρ}$ of the statement, read in $\overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[F1]

A viscosity subsolution $w$ of $u_{t} + H (x, t, D u) = 0$ in $Z$ satisfies $ψ_{t} (z_{0}) + H (z_{0}, D ψ (z_{0})) \leq 0$ at every local maximum $z_{0} \in Z$ of $w - ψ$ with $ψ \in C^{1} (Z)$ ; a viscosity supersolution satisfies the reverse inequality $\geq 0$ at every local minimum of $w - ψ$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

The function $t \mapsto η / (T - t)$ is $C^{1}$ on $(- \infty, T)$ with derivative $η / (T - t)^{2}$ ; sums of $C^{1}$ functions are $C^{1}$ with the sum of the total derivatives (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder), and a local maximum of ${\tilde{u}}_{η} - ϕ$ is a local maximum of $u - (ϕ + η / (T - t))$ because the two differences are the same function.

[F3]

Upper semicontinuity of $u$ and lower semicontinuity of $v$ are the relative notions on the Euclidean set $R^{n} \times [0, T)$ (Upper and lower semicontinuity on subsets of $R^{n}$ ); $w$ is upper semicontinuous exactly when every superlevel set ${w \geq c}$ is closed.

[F4]

[F5]

A metric space is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection; no choice principle is used (A metric space is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection).

Proof

technique · add the two explicit time-boundary penalties, then use spatial coercivity and compact superlevel sets

1.1F1F2algebra

The subsolution penalty. Let $ϕ \in C^{1} (Z)$ and suppose ${\tilde{u}}_{η} - ϕ$ has a local maximum at $z_{0} = (x_{0}, t_{0}) \in Z$ . Then $u - ψ$ has a local maximum at $z_{0}$ for $ψ (z) : = ϕ (z) + η / (T - t)$ , which is $C^{1}$ on $Z$ with $ψ_{t} = ϕ_{t} + η / (T - t)^{2}$ and $D ψ = D ϕ$ by [F2]; the subsolution inequality [F1] gives $ϕ_{t} (z_{0}) + η / (T - t_{0})^{2} + H (x_{0}, t_{0}, D ϕ (z_{0})) \leq 0$ , that is $ϕ_{t} + H (x_{0}, t_{0}, D ϕ (z_{0})) \leq - η / (T - t_{0})^{2} < 0$ .

1.2F1F2algebra

The supersolution penalty and the uniform terminal limits. If ${\tilde{v}}_{η^{'}} - ϕ$ has a local minimum at $z_{0} \in Z$ , then $v - ψ$ has a local minimum at $z_{0}$ for $ψ : = ϕ - η^{'} / (T - t)$ , $C^{1}$ with $ψ_{t} = ϕ_{t} - η^{'} / (T - t)^{2}$ ; the supersolution inequality [F1] gives $ϕ_{t} + H (x_{0}, t_{0}, D ϕ (z_{0})) \geq η^{'} / (T - t_{0})^{2} > 0$ . For the terminal limits, ${\tilde{u}}_{η} (x, t) \leq \sup u - η / (T - t)$ and ${\tilde{v}}_{η^{'}} (x, t) \geq \inf v + η^{'} / (T - t)$ for every $x$ , and the right-hand sides are independent of $x$ and tend to $- \infty$ , respectively $+ \infty$ , as $t ↑ T$ .

1.3F3F4F5algebra

Existence and finiteness of the maximum of $Φ_{α, ρ}$ . On the product space $R^{n} \times R^{n} \times [0, T]^{2}$ the function $Φ_{α, ρ}$ is upper semicontinuous: it is built from the upper semicontinuous ${\tilde{u}}_{η}$ , the function $- {\tilde{v}}_{η^{'}}$ , which is upper semicontinuous because $v$ is lower semicontinuous, and continuous terms, and at a sequence with $t \to T$ or $s \to T$ it tends to $- \infty$ uniformly, since ${\tilde{u}}_{η} (x, t) \leq \sup u - η / (T - t)$ and $- {\tilde{v}}_{η^{'}} (y, s) \leq - \inf v - η^{'} / (T - s)$ ; hence it takes the value $- \infty$ on the terminal faces in the upper-semicontinuous sense fixed in [F3]. It is bounded above by $\sup u - \inf v < \infty$ , and its value at any diagonal point $(0, 0, τ, τ)$ with $0 < τ < T$ is finite, so $M : = \sup Φ_{α, ρ} \in R$ . For each $k \geq 1$ the set $A_{k} : = {(x,y,t,s) : Φ_{α, ρ} \geq M - 1 / k}$ is nonempty by the definition of $M$ , closed by upper semicontinuity [F3], bounded because $ρ (∣ x ∣^{2} + ∣ y ∣^{2}) \leq \sup u - \inf v - M + 1 / k \leq \sup u - \inf v - M + 1$ on $A_{k}$ , and disjoint from the terminal faces because $M - 1 / k$ is finite there; hence each $A_{k}$ is a compact subset of $A_{1}$ by [F4]. The family ${A_{k}}_{k \geq 1}$ is nested, so it has the finite intersection property, and [F5] applied in the compact set $A_{1}$ gives a point of $⋂_{k} A_{k}$ , at which $Φ_{α, ρ} \geq M - 1 / k$ for every $k$ , hence $Φ_{α, ρ} \geq M$ ; since $M$ is an upper bound, $Φ_{α, ρ} = M$ there. Thus the maximum is attained and equals the finite number $M$ , and every maximiser has $t, s < T$ because the terminal faces carry the value $- \infty$ .

2.1step 1.1step 1.2step 1.3∎

Conclusion. Parts (1) and (2) of the statement are steps 1.1 and 1.2, and part (3) is step 1.3, whose construction nowhere selects a sequence or a point: the maximiser is obtained from the finite intersection property, which the cited lemma proves choice-free. Nothing in the argument rules out a maximiser with $t = 0$ or $s = 0$ , since only the terminal faces $t = T$ , $s = T$ carry the value $- \infty$ .

Remarks

Why the penalties are the right shape. Each penalty is continuous on $[0, T)$ with derivative diverging at $T$ , so it produces the exact interior residual shift $η / (T - t)^{2}$ , whose magnitude is at least $η / T^{2}$ (and similarly for $η^{'}$ ), and pushes every doubling maximum off the terminal face. The initial faces carry finite values and are deliberately allowed: the comparison theorem treats them separately with the pointwise initial inequality.
Choice. The only compactness input is the finite-intersection characterisation [F5], which is choice-free.

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Comparison for first-order Hamilton--Jacobi equations

Statement

Comparison for first-order Hamilton--Jacobi equations, in the two settings of the design. (a) The case $O = R^{n}$ . Let $T > 0$ and let $H : R^{n} \times [0, T] \times R^{n} \to R$ be continuous for which there is $C > 0$ with $∣ H (x, t, p) - H (y, s, p) ∣ \leq C (1 + ∣ p ∣) (∣ x - y ∣ + ∣ t - s ∣), ∣ H (x, t, p) - H (x, t, q) ∣ \leq C ∣ p - q ∣$ for all $x, y, p, q \in R^{n}$ and $t, s \in [0, T]$ . Let $u$ be a bounded upper semicontinuous viscosity subsolution and $v$ a bounded lower semicontinuous viscosity supersolution of the Cauchy problem in $Z = R^{n} \times (0, T)$ , each defined on the closed slab $R^{n} \times [0, T)$ and satisfying the pointwise initial inequality $u (x, 0) \leq v (x, 0)$ for every $x \in R^{n}$ . Then $u \leq v$ on $Z$ . (b) The compact-cylinder case. Let $O \subseteq R^{n}$ be bounded and open, $T > 0$ , and let $H : \overline{O} \times [0, T] \times R^{n} \to R$ be continuous and uniformly continuous in $(x, t)$ uniformly on bounded $p$ -sets: there is a nondecreasing modulus $ω : [0, \infty) \to [0, \infty)$ with $ω (0 +) = 0$ such that $∣ H (x, t, p) - H (y, s, p) ∣ \leq ω ((1 + ∣ p ∣) (∣ x - y ∣ + ∣ t - s ∣))$ for all $(x, t), (y, s) \in \overline{O} \times [0, T]$ and all $p \in R^{n}$ . Let $u, v$ be continuous on $\overline{Z} = \overline{O} \times [0, T]$ , $u$ a viscosity subsolution and $v$ a viscosity supersolution of $u_{t} + H (x, t, D u) = 0$ in $Z$ , with $u \leq v$ on the parabolic boundary $Γ = (O \times {0}) \cup (\partial O \times [0, T])$ . Then $u \leq v$ on $\overline{Z}$ . No growth hypothesis on $H$ in the momentum variable is imposed in this case; the modulus condition replaces it. For an $x$ -independent autonomous Hamiltonian $H (p)$ , it holds with the zero modulus. No choice principle is used.

Facts & Assumptions

Given: The two settings of the statement; parameters $η, η^{'}, ρ, α > 0$ ; the time penalties ${\tilde{u}}_{η} = u - η / (T - t)$ and ${\tilde{v}}_{η^{'}} = v + η^{'} / (T - t)$ (case (a)) or ${\tilde{v}}_{η^{'}} = v + η^{'} / (T - s)$ read at the respective time variable; the doubling functions $Φ_{α} (x, y, t, s) : = {\tilde{u}}_{η} (x, t) - {\tilde{v}}_{η^{'}} (y, s) - \frac{α}{2} ∣ x - y ∣^{2} - \frac{α}{2} ∣ t - s ∣^{2}$ in case (b) and the same with the additional weight $- ρ (∣ x ∣^{2} + ∣ y ∣^{2})$ in case (a); their suprema $M_{α}$ .

[F1]

Every $C^{1}$ upper contact $ϕ$ of ${\tilde{u}}_{η}$ at an interior point satisfies $ϕ_{t} + H (x, t, D ϕ) \leq - η / (T - t)^{2} \leq - η / T^{2}$ , and every $C^{1}$ lower contact of ${\tilde{v}}_{η^{'}}$ satisfies the reverse with $η^{'}$ ; moreover ${\tilde{u}}_{η} (x, t) \leq \sup u - η / (T - t) \to - \infty$ and ${\tilde{v}}_{η^{'}} (x, t) \geq \inf v + η^{'} / (T - t) \to + \infty$ uniformly at the terminal time (Time penalisation moves a doubling-variables maximum away from the terminal boundary).

[F2]

At every maximiser of the doubling function, the two test functions displayed in Doubling variables: existence, relative contacts at the maximiser and localisation are $C^{1}$ contacts for ${\tilde{u}}_{η}$ and ${\tilde{v}}_{η^{'}}$ with the jets $p = α (x - y) + 2 ρ x$ (case (a)) or $p = α (x - y)$ (case (b)), $q = α (x - y) - 2 ρ y$ or $q = α (x - y)$ , and common time derivative $p_{0} = α (t - s)$ ; and the weight bound $\frac{α}{2} (∣ x - y ∣^{2} + ∣ t - s ∣^{2}) + ρ (∣ x ∣^{2} + ∣ y ∣^{2}) \leq \sup {\tilde{u}}_{η} - \inf {\tilde{v}}_{η^{'}} - M_{α}$ holds at every maximiser, a bound that in case (a) restricts every maximiser by $ρ (∣ x ∣^{2} + ∣ y ∣^{2}) \leq \sup u - \inf v - M_{α}$ (Doubling variables: existence, relative contacts at the maximiser and localisation).

[F3]

In case (a), $u$ is upper semicontinuous and $v$ lower semicontinuous on the closed slab, so $u (x, 0) - v (y, s)$ and $u (x, t) - v (y, 0)$ are upper semicontinuous in their variables. In case (b), $u, v$ are continuous on the compact set $\overline{Z}$ , hence uniformly continuous by Heine--Cantor; choose a common space-time modulus $ϖ$ with $∣ u (x, t) - u (y, s) ∣, ∣ v (x, t) - v (y, s) ∣ \leq ϖ (∣ x - y ∣ + ∣ t - s ∣)$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem, Upper and lower semicontinuity on subsets of $R^{n}$ , Uniform continuity of a map of metric spaces: one $δ$ serving every point, Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[F4]

Closed bounded subsets of finite-dimensional Euclidean space are compact; compactness implies the finite-intersection property for nested nonempty closed subsets (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent, Open cover, subcover, compact metric space, and compact subset of a metric space, A metric space is compact if and only if every family of closed subsets with the finite intersection property has nonempty intersection).

[F5]

A finite-valued upper semicontinuous function has closed superlevel sets; if $f$ is upper semicontinuous and $g$ lower semicontinuous, then $f - g$ is upper semicontinuous, by applying their local one-sided bounds with half the tolerance (Upper and lower semicontinuity on subsets of $R^{n}$ ).

[F6]

An upper semicontinuous real-valued function on a nonempty compact Euclidean set is bounded above and attains a maximum (Semicontinuous extreme value theorem on compact Euclidean sets).

[F7]

A continuous function on a compact metric space is uniformly continuous (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

Proof

technique · doubling with terminal-time penalisation; all localisation estimates are proved for every maximiser, so that no sequence of maximisers and no choice principle is used

1.1F1F2F4algebra

Case (a): setup and uniform localisation estimates. Assume $σ : = \sup_{Z} (u - v) > 0$ and fix $(x_{1}, t_{1}) \in Z$ with $u (x_{1}, t_{1}) - v (x_{1}, t_{1}) > 3 σ / 4$ . Choose $η, η^{'} > 0$ with $(η + η^{'}) / (T - t_{1}) < σ / 4$ and put $B : = η + η^{'}$ ; then ${\tilde{u}}_{η} (x_{1}, t_{1}) - {\tilde{v}}_{η^{'}} (x_{1}, t_{1}) > σ / 2$ . Fix $ρ > 0$ with $2 ρ ∣ x_{1} ∣^{2} < σ / 8$ ; then for the doubling function of case (a) with these penalties, $M_{α} \geq Φ_{α} (x_{1}, x_{1}, t_{1}, t_{1}) > σ / 2 - σ / 8 = 3 σ / 8$ for every $α$ . Every maximiser $(x, y, t, s)$ of $Φ_{α}$ has $t, s < T$ , and by [F2] the weight bound gives $ρ (∣ x ∣^{2} + ∣ y ∣^{2}) \leq \sup u - \inf v - 3 σ / 8 = : K_{ρ}$ , so $∣ x ∣, ∣ y ∣ \leq C_{ρ} : = \sqrt{K_{ρ} / ρ}$ . Writing $d^{2} : = ∣ x - y ∣^{2} + ∣ t - s ∣^{2}$ and evaluating $Φ_{α / 2}$ at any maximiser of $Φ_{α}$ (the penalty difference is $\frac{α}{4} d^{2}$ ) gives $M_{α / 2} \geq M_{α} + \frac{α}{4} d^{2}$ , hence $d \leq δ_{α} : = (4 (M_{α / 2} - M_{α}) / α)^{1 / 2}$ for every maximiser; since $M_{α} ↓ M_{\infty}$ boundedly as $α \to \infty$ , $δ_{α} \to 0$ . With $p = α (x - y) + 2 ρ x$ as in [F2] we get $∣ p ∣ \leq α d + 2 ρ C_{ρ}$ and therefore $(1 + ∣ p ∣) d \leq d + α d^{2} + 2 ρ C_{ρ} d \leq δ_{α} + 4 (M_{α / 2} - M_{α}) + 2 ρ C_{ρ} δ_{α} \to 0$ and $∣ q - p ∣ = 2 ρ ∣ x + y ∣ \leq 4 ρ C_{ρ}$ uniformly over all maximisers, both limits being as $α \to \infty$ with $ρ$ fixed.

1.2F1F2F4F6algebra

Case (b): setup and uniform localisation estimates. Let $σ : = \max_{\overline{Z}} (u - v) > 0$ ; the maximum is attained by compactness and continuity, and if it were attained on $Γ$ it would be $\leq 0$ , then continuity supplies a point $(x_{1}, t_{1}) \in Z$ with $u (x_{1}, t_{1}) - v (x_{1}, t_{1}) > 3 σ / 4$ , even if the maximum occurs at $t = T$ . Choose $η, η^{'} > 0$ with $B : = η + η^{'} < σ (T - t_{1}) / 4$ ; then ${\tilde{u}}_{η} (x_{1}, t_{1}) - {\tilde{v}}_{η^{'}} (x_{1}, t_{1}) > σ / 2$ , and the doubling function of case (b) satisfies $M_{α} \geq σ / 2$ for every $α$ . It is upper semicontinuous on the compact box $\overline{O} \times \overline{O} \times [0, T]^{2}$ (the penalties tend to $- \infty$ at the terminal faces, where the value is declared $- \infty$ ); a nonempty compact superlevel set and [F6] give a finite attained maximum, and every maximiser has $t, s < T$ . Evaluating $Φ_{α / 2}$ at a maximiser of $Φ_{α}$ again gives $\frac{α}{4} d^{2} \leq M_{α / 2} - M_{α}$ , hence $d \leq δ_{α} \to 0$ uniformly over maximisers, and with $p = q = α (x - y)$ in case (b) one has $(1 + ∣ p ∣) d \leq d + α d^{2} \leq δ_{α} + 4 (M_{α / 2} - M_{α}) \to 0$ . Since $∣ x - y ∣ + ∣ t - s ∣ \leq 2 d$ , it follows that $(1 + ∣ p ∣) (∣ x - y ∣ + ∣ t - s ∣) \leq 2 δ_{α} + 8 (M_{α / 2} - M_{α}) \to 0$ .

2.1step 1.2F1F2F3F6F7algebra

Case (b): exclusion of the parabolic boundary and the contradiction. Let $α$ be large enough that $ϖ (2 δ_{α}) < B / T$ . If a maximiser had $t = 0$ , then $(x, 0) \in Γ$ gives $u (x, 0) \leq v (x, 0)$ and $∣ x - y ∣ + s \leq 2 d \leq 2 δ_{α}$ , so $Φ_{α} \leq u (x, 0) - v (y, s) - \frac{B}{T} \leq ϖ (2 δ_{α}) - \frac{B}{T} < 0,$ contradicting $M_{α} \geq σ / 2$ . If $x \in \partial O$ with $t, s > 0$ , then $(x, t) \in Γ$ gives $u (x, t) \leq v (x, t)$ and $∣ x - y ∣ + ∣ t - s ∣ \leq 2 d \leq 2 δ_{α}$ , so $Φ_{α} \leq ϖ (2 δ_{α}) - B / T < 0$ ; the case $y \in \partial O$ is symmetric, as is $s = 0$ using the initial inequality at $(y, 0)$ and the modulus of $u$ . Thus all maximisers for large $α$ have $0 < t, s < T$ and $x, y \in O$ . At such a maximiser the penalty inequalities of [F1] hold at the jets $p = q$ , $p_{0}$ of case (b), and subtracting them gives $\frac{B}{T^{2}} \leq H (y, s, p) - H (x, t, p) \leq ω ((1 + ∣ p ∣) (∣ x - y ∣ + ∣ t - s ∣)) \leq ω (2 δ_{α} + 8 (M_{α / 2} - M_{α})),$ which tends to $0$ as $α \to \infty$ by step 1.2 and $ω (0 +) = 0$ ; this contradicts $B / T^{2} > 0$ . Hence $σ \leq 0$ , that is $u \leq v$ on $\overline{Z}$ .

2.2step 1.1F1F2F3F4F5F6algebra

Case (a): exclusion of the initial faces and the contradiction. Fix the $ρ$ of step 1.1 and let $K$ be the closed ball containing every spatial coordinate of every maximiser. On the compact set $K \times K \times [0, T / 2]$ , the functions $f_{0} (x, y, s) : = u (x, 0) - v (y, s)$ and $f_{1} (x, y, t) : = u (x, t) - v (y, 0)$ are upper semicontinuous by [F3, F5], and both are nonpositive on the diagonal sets $(z, z, 0)$ by the initial inequality. There is $δ_{0} > 0$ such that $f_{0} (x, y, s) < 3 σ / 8$ whenever $∣ x - y ∣ + s < δ_{0}$ , and likewise $δ_{1} > 0$ for $f_{1}$ whenever $∣ x - y ∣ + t < δ_{1}$ : otherwise the closed superlevel sets ${f_{i} \geq 3 σ / 8}$ intersected with the nested closed sets where the corresponding distance is at most $1 / m$ would be nonempty compact sets with the finite-intersection property, so [F4] would give a point $(z, z, 0)$ in the superlevel set, a contradiction. For $α$ large enough that $2 δ_{α} < \min {δ_{0}, δ_{1}, T / 2}$ , if a maximiser had $t = 0$ , then $Φ_{α} \leq f_{0} (x, y, s)$ because all remaining penalties are nonpositive, while $∣ x - y ∣ + s \leq 2 d \leq 2 δ_{α}$ ; this contradicts $Φ_{α} = M_{α} \geq 3 σ / 8$ . If $s = 0$ , similarly $Φ_{α} \leq f_{1} (x, y, t)$ and $∣ x - y ∣ + t \leq 2 δ_{α}$ , again a contradiction. Hence for all sufficiently large $α$ every maximiser has $0 < t, s < T$ . At such a maximiser the contact inequalities of [F1] apply at the jets $p, q, p_{0}$ of [F2]: $p_{0} + H (x, t, p) \leq - η / T^{2}$ and $p_{0} + H (y, s, q) \geq η^{'} / T^{2}$ . Subtracting and using the two Lipschitz conditions of case (a) gives $B / T^{2} \leq H (y, s, q) - H (x, t, p) \leq C ∣ q - p ∣ + C (1 + ∣ p ∣) (∣ x - y ∣ + ∣ t - s ∣)$ , and by step 1.1 the right-hand side is at most $4 ρ C C_{ρ} + 2 C (δ_{α} + 4 (M_{α / 2} - M_{α}) + 2 ρ C_{ρ} δ_{α})$ , using $∣ x - y ∣ + ∣ t - s ∣ \leq 2 d$ for every maximiser with $α$ large. Letting $α \to \infty$ gives $B / T^{2} \leq 4 ρ C C_{ρ} = 4 C (ρ (\sup u - \inf v - 3 σ / 8))^{1 / 2}$ , and then letting $ρ ↓ 0$ gives $B / T^{2} \leq 0$ , a contradiction. Thus $σ \leq 0$ and $u \leq v$ on $Z$ .

3.1step 2.1step 2.2∎

Conclusion. Case (a) is step 2.2 and case (b) is step 2.1; in both cases the contradiction is obtained by uniform estimates over the maximiser sets, so no maximiser, subsequence or index is selected and no choice principle is used.

Remarks

Autonomy. In case (b) an $x$ -independent autonomous Hamiltonian $H (p)$ satisfies the modulus condition with $ω \equiv 0$ . A general autonomous $H (x, p)$ still needs the stated spatial modulus condition; in case (a) the two Lipschitz conditions are exactly what the subtracted inequality consumes.
What each hypothesis is for. The terminal-time penalties give the strict margin $B / T^{2}$ ; the localisation $α d^{2} \to 0$ makes the momentum gap $q - p = 2 ρ (x + y)$ and the space-time displacement disappear after $α \to \infty$ and $ρ ↓ 0$ ; the pointwise initial inequality (case (a)) or the boundary inequality (case (b)) excludes the initial and lateral faces.

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Uniqueness and sup-norm contraction for the Cauchy problem

Statement

Let $n \geq 1$ , $T > 0$ , and let $H : R^{n} \times [0, T] \times R^{n} \to R$ satisfy the Lipschitz conditions of part (a) of Comparison for first-order Hamilton--Jacobi equations. (1) If $u, v$ are bounded viscosity solutions of the Cauchy problem in $Z = R^{n} \times (0, T)$ with the same bounded continuous initial datum $u_{0}$ , then $u = v$ on $Z$ ; in particular the classical and the Hopf--Lax solutions of later sections are the unique ones in the bounded class whenever $H$ satisfies those conditions. (2) More generally, if $u, v$ are bounded viscosity solutions with initial data $u_{0}, v_{0}$ , then for every $T > 0$ $\sup_{z \in Z} (u - v) \leq \max (0, \sup_{x \in R^{n}} (u_{0} (x) - v_{0} (x))),$ and applying the same bound to $v - u$ gives $\sup_{Z} ∣ u - v ∣ \leq \sup_{R^{n}} ∣ u_{0} - v_{0} ∣$ when the initial difference is bounded. In particular the solution operator is a contraction in the supremum norm on the bounded initial data. No choice principle is used.

Facts & Assumptions

Given: A Hamiltonian $H$ satisfying the Lipschitz conditions of comparison case (a), $T > 0$ , bounded viscosity solutions $u, v$ of the Cauchy problem in $Z = R^{n} \times (0, T)$ with bounded continuous data $u_{0}, v_{0}$ .

[F1]

Comparison, case (a): if $U$ is a bounded upper semicontinuous subsolution and $V$ a bounded lower semicontinuous supersolution on the closed slab with $U (x, 0) \leq V (x, 0)$ pointwise, then $U \leq V$ on $Z$ (Comparison for first-order Hamilton--Jacobi equations).

[F2]

For a bounded viscosity solution $u$ , its upper envelope $u^{*}$ is a bounded upper semicontinuous subsolution and its lower envelope $u_{*}$ is a bounded lower semicontinuous supersolution, each satisfying the corresponding relaxed initial inequality (Discontinuous viscosity solutions through the two envelopes). Adding a constant $K$ shifts both envelopes by $K$ and preserves their one-sided viscosity inequalities because $H$ is independent of the unknown (Discontinuous viscosity solutions through the two envelopes, Comparison for first-order Hamilton--Jacobi equations for the equation class).

[F3]

Boundedness of $u, v$ and of their continuous initial data is assumed in the statement and Given, so the displayed suprema are finite. The relaxed joint initial limsup/liminf conditions are part of Discontinuous viscosity solutions through the two envelopes, as recorded in [F2]. No uniform-continuity hypothesis is needed for this comparison consequence.

Proof

technique · comparison applied twice, plus the constant-shift invariance of the equation

1.1F1F2F3

Uniqueness. Let $u, v$ be bounded viscosity solutions with the same datum $u_{0}$ . By [F2], $u^{*}$ is a bounded upper semicontinuous subsolution and $v_{*}$ is a bounded lower semicontinuous supersolution. Extend them to the initial face by $U (x, 0) = u_{0} (x)$ and $V (x, 0) = u_{0} (x)$ . Their relaxed initial inequalities and continuity of $u_{0}$ make $U$ upper semicontinuous and $V$ lower semicontinuous on the closed slab, with ordered pointwise initial values. Comparison [F1] gives $u^{*} \leq v_{*}$ on $Z$ . Since $u \leq u^{*}$ and $v_{*} \leq v$ , this yields $u \leq v$ . Applying the same argument to $(v, u)$ gives $v \leq u$ , hence $u = v$ on $Z$ ; in fact all four envelopes and functions coincide. In particular, whenever a classical or Hopf--Lax solution is known to be a bounded viscosity solution of the same Cauchy problem, it is the unique bounded solution.

1.2F1F2F3algebra

The one-sided bound for general data. Put $K : = \max {0, \sup_{R^{n}} (u_{0} - v_{0})} < \infty$ . By [F2], $u^{*}$ is a bounded upper semicontinuous subsolution and $(v + K)_{*} = v_{*} + K$ is a bounded lower semicontinuous supersolution. Extend these envelopes to $t = 0$ by $u_{0}$ and $v_{0} + K$ , respectively; their relaxed initial inequalities and continuity of the data make the extensions semicontinuous on the closed slab with ordered pointwise initial values. Comparison [F1] gives $u^{*} \leq v_{*} + K$ on $Z$ . Since $u \leq u^{*}$ and $v_{*} \leq v$ , this implies $u - v \leq K$ ; taking the supremum over $Z$ gives $\sup_{Z} (u - v) \leq K$ .

2.1step 1.1step 1.2∎

Conclusion. Applying step 1.2 to the pair $(u, v)$ and to the exchanged pair $(v, u)$ gives $\sup_{Z} (u - v) \leq \max (0, \sup (u_{0} - v_{0}))$ and $\sup_{Z} (v - u) \leq \max (0, \sup (v_{0} - u_{0}))$ ; when the initial difference is bounded, both right-hand sides are at most $\sup_{R^{n}} ∣ u_{0} - v_{0} ∣$ , hence $\sup_{Z} ∣ u - v ∣ \leq \sup_{R^{n}} ∣ u_{0} - v_{0} ∣$ and the solution operator is a contraction in the supremum norm.

Remarks

Domain. The corollary is stated on $O = R^{n}$ because comparison case (a) is; on a bounded domain without lateral data uniqueness fails, as the companion counterexample shows.
Choice. Only comparison and the constant shift are used, both choice-free.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The upper envelope of a locally bounded supremum of subsolutions is a subsolution

Statement

Let $U \subseteq R^{n + 1}$ be open, let $H : U \times R^{n} \to R$ be continuous, and let $F$ be a nonempty family of real-valued upper semicontinuous viscosity subsolutions of $u_{t} + H (x, t, D u) = 0$ in $U$ . Put $w (z) : = \sup_{w^{'} \in F} w^{'} (z)$ for $z \in U$ and assume that $w$ is locally bounded above: $w (z) < \infty$ for every $z \in U$ and $w$ is bounded above on every compact subset of $U$ . Then the upper semicontinuous envelope $w^{*}$ (Upper and lower semicontinuous envelopes by local limsup and liminf) is a viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $U$ . No choice principle is used.

Facts & Assumptions

Given: An open $U \subseteq R^{n + 1}$ , continuous $H : U \times R^{n} \to R$ , a nonempty family $F$ of upper semicontinuous viscosity subsolutions, $w = \sup_{w^{'} \in F} w^{'}$ , locally bounded above, and its upper envelope $w^{*}$ .

[F1]

$w^{*} (z) = \inf_{r > 0} M_{r} (z)$ with $M_{r} (z) : = \sup {w (y) : ∣ y - z ∣ \leq r, y \in U}$ , and $w^{*} (z) = \lim_{r ↓ 0} M_{r} (z)$ ; if $w$ is locally bounded above then $w^{*}$ is real-valued on $U$ . The envelope is upper semicontinuous: for $z$ and $η > 0$ choose $r > 0$ with $M_{r} (z) \leq w^{*} (z) + η$ ; then for $∣ z^{'} - z ∣ < r$ one has $M_{r - ∣ z^{'} - z ∣} (z^{'}) \leq M_{r} (z)$ , hence $w^{*} (z^{'}) \leq w^{*} (z) + η$ (Upper and lower semicontinuous envelopes by local limsup and liminf).

[F2]

Each $w^{'} \in F$ satisfies $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) \leq 0$ at every $z_{0} \in U$ at which $w^{'} - ϕ$ has a local maximum, $ϕ \in C^{1} (U)$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem, Viscosity testing by first-order jets, and closure of the jet inequality).

[F3]

Every upper semicontinuous real-valued function on a nonempty compact subset of $R^{m}$ attains its maximum there (Semicontinuous extreme value theorem on compact Euclidean sets).

[F4]

If $v - ϕ$ has a local maximum at $z_{0}$ and $\overline{B} (z_{0}, r) \subseteq U$ is a ball on which $v - ϕ \leq v (z_{0}) - ϕ (z_{0})$ , then for every $ε > 0$ the test $ϕ_{ε} = ϕ + ε ∣ z - z_{0} ∣^{4}$ has the same value and first jet as $ϕ$ at $z_{0}$ and makes $v - ϕ_{ε}$ strictly maximised over $\overline{B} (z_{0}, r)$ at $z_{0}$ (Strictification of a viscosity test function by a quartic perturbation).

Proof

technique · near-maximal selection from the supremum plus strict test perturbation

1.1F1F2F3algebra

Strict-contact case. Let $ϕ \in C^{1} (U)$ touch $w^{*}$ from above at $z_{0} = (x_{0}, t_{0})$ with a strict local maximum of $w^{*} - ϕ$ , and suppose $θ : = ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) > 0$ . Choose $r > 0$ so that $\overline{B} (z_{0}, r) \subseteq U$ , the contact is strict on this ball, and $ϕ_{t} (z) + 2 (t - t_{0}) + H (z, D ϕ (z) + 2 (x - x_{0})) > θ / 2$ throughout it, by continuity. On the compact annulus $A = {r / 2 \leq ∣ z - z_{0} ∣ \leq r}$ , [F1, F3] give the positive gap $g = (w^{*} - ϕ) (z_{0}) - \max_{A} (w^{*} - ϕ)$ . Choose $0 < η < g / 4$ and $0 < δ < r / 2$ such that $∣ ϕ (y) - ϕ (z_{0}) ∣ < η$ and $∣ y - z_{0} ∣^{2} < η$ for $∣ y - z_{0} ∣ < δ$ . The two supremum definitions in [F1] supply one pair $(w^{'}, y)$ with $w^{'} \in F$ , $∣ y - z_{0} ∣ < δ$ , and $w^{'} (y) > w^{*} (z_{0}) - η$ (use a closed radius smaller than $δ$ ). By [F3], $w^{'} - ϕ - ∣ z - z_{0} ∣^{2}$ attains a maximum on $\overline{B} (z_{0}, r)$ , of value greater than $(w^{*} - ϕ) (z_{0}) - 3 η$ . Its value on $A$ is at most $(w^{*} - ϕ) (z_{0}) - g$ , since $w^{'} \leq w \leq w^{*}$ . Thus any maximiser $z_{*} = (x_{*}, t_{*})$ lies in $B (z_{0}, r / 2)$ and is an interior upper contact for $ψ = ϕ + ∣ z - z_{0} ∣^{2}$ . Its derivatives are $ψ_{t} (z_{*}) = ϕ_{t} (z_{*}) + 2 (t_{*} - t_{0})$ and $D ψ (z_{*}) = D ϕ (z_{*}) + 2 (x_{*} - x_{0})$ . Their residual is greater than $θ / 2$ , contradicting the subsolution inequality [F2]. Hence the desired residual at $z_{0}$ is nonpositive.

2.1step 1.1F4∎

General contacts and conclusion. If $w^{*} - ϕ$ merely has a local maximum at $z_{0}$ , fix $r > 0$ with $\overline{B} (z_{0}, r) \subseteq U$ on which the maximum inequality $w^{*} - ϕ \leq w^{*} (z_{0}) - ϕ (z_{0})$ holds and strictify by [F4]: the test $ϕ_{ε} = ϕ + ε ∣ z - z_{0} ∣^{4}$ has the same value and first jet at $z_{0}$ and makes $w^{*} - ϕ_{ε}$ strictly maximised at $z_{0}$ over $\overline{B} (z_{0}, r)$ . Step 1.1 applied to $ϕ_{ε}$ gives $ϕ_{t} (z_{0}) + H (z_{0}, D ϕ (z_{0})) = (ϕ_{ε})_{t} (z_{0}) + H (z_{0}, D ϕ_{ε} (z_{0})) \leq 0$ . Hence $w^{*}$ is a viscosity subsolution of the equation in $U$ ; the selection of the single witness $(w^{'}, y)$ and of the compact maximiser $z_{*}$ involves no choice principle, and the whole argument is pointwise.

Remarks

Where local boundedness above is used. It makes $w^{*}$ real-valued so that the compact-annulus maximum and the test inequality are meaningful; the family is not assumed to consist of locally bounded functions or to be directed, and no member of the family other than the single witness $(w^{'}, y)$ is examined.
Role in Perron's method. This is the load-bearing half of Perron's method for the Cauchy problem: existence between two barriers: the supremum of the admissible subsolutions is made upper semicontinuous by passing to $w^{*}$ , and this theorem says the envelope is still a subsolution.

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Failure of the supersolution test for the lower envelope allows a local bump

Statement

Let $U \subseteq R^{n + 1}$ be open, let $H : U \times R^{n} \to R$ be continuous, and let $w : U \to R$ be an upper semicontinuous viscosity subsolution of $u_{t} + H (x, t, D u) = 0$ in $U$ . Suppose the lower semicontinuous envelope $w_{*}$ fails the supersolution test at $\hat{z} \in U$ in the following precise sense: $w_{*} (\hat{z}) \in R$ and there is $ϕ \in C^{1} (U)$ such that $w_{*} - ϕ$ has a local minimum at $\hat{z}$ and $ϕ_{t} (\hat{z}) + H (\hat{z}, D ϕ (\hat{z})) < 0.$ Then for every sufficiently small $κ > 0$ there is a viscosity subsolution $W_{κ}$ of the same equation in $U$ with $W_{κ} \geq w on U, \sup_{U} (W_{κ} - w) > 0, W_{κ} = w on {z \in U : ∣ z - \hat{z} ∣ \geq κ} .$ Moreover $W_{κ}$ can be taken to be $\max (w, χ)$ on a small ball around $\hat{z}$ and $w$ outside it, where $χ$ is a classical subsolution with $χ (\hat{z}) = w_{*} (\hat{z}) + δ$ for some $δ > 0$ . No choice principle is used.

Facts & Assumptions

Given: Open $U \subseteq R^{n + 1}$ , continuous $H : U \times R^{n} \to R$ , an upper semicontinuous viscosity subsolution $w : U \to R$ , its lower envelope $w_{*}$ , a point $\hat{z} \in U$ with $w_{*} (\hat{z}) \in R$ and a $C^{1}$ test $ϕ$ with $w_{*} - ϕ$ having a local minimum at $\hat{z}$ and $c : = - (ϕ_{t} (\hat{z}) + H (\hat{z}, D ϕ (\hat{z}))) > 0$ .

[F1]

$w_{*} (z) = \sup_{r > 0} \inf {w (y) : ∣ y - z ∣ \leq r}$ is the lower semicontinuous envelope of $w$ , it satisfies $w_{*} \leq w$ pointwise, and for every $η > 0$ there are points $z$ arbitrarily close to $\hat{z}$ with $w (z) < w_{*} (\hat{z}) + η$ (Upper and lower semicontinuous envelopes by local limsup and liminf).

[F2]

A finite maximum of finitely many viscosity subsolutions of the equation in an open set is a viscosity subsolution (Finite maxima of subsolutions and finite minima of supersolutions); a $C^{1}$ function with $χ_{t} + H (z, D χ) \leq 0$ pointwise is a viscosity subsolution of the same equation (Classical solutions are viscosity solutions and differentiable viscosity solutions solve the equation pointwise, Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F3]

A continuous function $f$ with $f (\hat{z}) < 0$ is negative on a neighbourhood of $\hat{z}$ ; here the function in question is $z \mapsto ϕ_{t} (z) + H (z, D ϕ (z))$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem for the smoothness conventions).

Proof

technique · a strict test-function bump plus the finite-maximum rule

1.1F1F2F3algebra

The bump function. Fix $κ > 0$ with $\overline{B} (\hat{z}, κ) \subseteq U$ and choose $r \in (0, κ]$ small. For $γ > 0$ put $\tilde{ϕ} (z) : = ϕ (z) - γ ∣ z - \hat{z} ∣^{4}$ , so that $\tilde{ϕ} \in C^{1} (U)$ , $\tilde{ϕ} (\hat{z}) = ϕ (\hat{z})$ and $D \tilde{ϕ} (\hat{z}) = D ϕ (\hat{z})$ ; shrinking $r$ if necessary and using [F3], we may assume ${\tilde{ϕ}}_{t} (z) + H (z, D \tilde{ϕ} (z)) \leq - c / 2 < 0$ for all $z \in \overline{B} (\hat{z}, r)$ , so every vertical translate of $\tilde{ϕ}$ is a classical, hence viscosity, subsolution there. Since $w_{*} - ϕ$ has a local minimum at $\hat{z}$ , after shrinking $r$ we have $w_{*} (z) - \tilde{ϕ} (z) \geq m + γ ∣ z - \hat{z} ∣^{4}$ for $z \in \overline{B} (\hat{z}, r)$ , where $m : = w_{*} (\hat{z}) - ϕ (\hat{z})$ . Choose $0 < δ < γ (r / 2)^{4}$ and define $χ : = \tilde{ϕ} + m + δ$ , a classical subsolution on $\overline{B} (\hat{z}, r)$ with $χ (\hat{z}) = w_{*} (\hat{z}) + δ$ . On the annulus $r / 2 \leq ∣ z - \hat{z} ∣ \leq r$ we have $w_{*} (z) \geq \tilde{ϕ} (z) + m + γ ∣ z - \hat{z} ∣^{4} \geq χ (z) - δ + γ (r / 2)^{4} > χ (z)$ , and since $w_{*} \leq w$ by [F1] this gives $χ < w$ there. By continuity of $χ$ , choose $r_{0} \in (0, r / 2)$ so that $χ (z) > w_{*} (\hat{z}) + δ / 2$ whenever $∣ z - \hat{z} ∣ < r_{0}$ . The lower-envelope definition [F1] gives a point $z_{0}$ in this ball with $w (z_{0}) < w_{*} (\hat{z}) + δ / 2 < χ (z_{0})$ .

2.1step 1.1F2algebra∎

The bump is a subsolution. Define $W : = \max (w, χ)$ on $B (\hat{z}, r)$ and $W : = w$ on $U ∖ B (\hat{z}, r)$ ; this is well defined because on the sphere $∣ z - \hat{z} ∣ = r$ one has $χ < w$ by step 1.1. Then $W \geq w$ on $U$ , and $\sup_{U} (W - w) > 0$ at the point $z_{0}$ of step 1.1. On the ball $B (\hat{z}, r)$ the function $W$ is the maximum of the viscosity subsolution $w$ and the classical, hence viscosity, subsolution $χ$ , so it is a viscosity subsolution there by [F2]; on the exterior of $\overline{B} (\hat{z}, r)$ it equals the subsolution $w$ ; and near every point of the sphere it equals $w$ , which is a subsolution, so by locality of the definition $W$ is a viscosity subsolution on all of $U$ . Since $χ < w$ on the annulus, $W = w$ outside $B (\hat{z}, r) \subseteq B (\hat{z}, κ)$ , that is $W = w$ on ${z : ∣ z - \hat{z} ∣ \geq κ}$ ; and $W$ is upper semicontinuous as a maximum of the upper semicontinuous $w$ and the continuous $χ$ .

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Time-space barriers enforce the initial trace for the Cauchy problem

Statement

Let $n \geq 1$ , $T > 0$ , $O \subseteq R^{n}$ open, $Z = O \times (0, T)$ , let $H : O \times [0, T] \times R^{n} \to R$ be continuous, and let $u_{0} \in C^{1} (O)$ have bounded gradient. Assume $C_{0} : = \sup_{x \in O, 0 \leq t \leq T} ∣ H (x, t, D u_{0} (x)) ∣ < \infty$ . Define $ϕ_{\pm} (x, t) : = u_{0} (x) \pm C_{0} t$ . Then: (1) $ϕ_{-}$ is a classical subsolution and $ϕ_{+}$ a classical supersolution in $Z$ , each with initial datum $u_{0}$ ; (2) if $w : Z \to R$ is locally bounded with $ϕ_{-} \leq w \leq ϕ_{+}$ on $Z$ , then for every $x \in O$ the relaxed limits satisfy ${lim inf}_{\begin{array}{c} (y, s) \to (x, 0) \\ s > 0 \end{array}} w (y, s) \geq u_{0} (x) \geq {lim sup}_{\begin{array}{c} (y, s) \to (x, 0) \\ s > 0 \end{array}} w (y, s),$ so both equal $u_{0} (x)$ ; (3) consequently $w$ satisfies the relaxed initial condition for the Cauchy problem in both directions, and any continuous extension of $w$ to the initial face takes the value $u_{0}$ pointwise. No choice principle is used.

Facts & Assumptions

Given: Open $O \subseteq R^{n}$ , $T > 0$ , continuous $H : O \times [0, T] \times R^{n} \to R$ , $u_{0} \in C^{1} (O)$ with bounded gradient, $C_{0} = \sup_{O \times [0, T]} ∣ H (x, t, D u_{0} (x)) ∣ < \infty$ , the barriers $ϕ_{\pm} = u_{0} \pm C_{0} t$ , and a locally bounded $w : Z \to R$ with $ϕ_{-} \leq w \leq ϕ_{+}$ .

[F1]

If a $C^{1}$ function satisfies the differential inequality pointwise on the open set $Z$ , then it satisfies the corresponding viscosity test inequality: at a local contact with another $C^{1}$ function, Fermat's theorem makes their first derivatives equal (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem, Fermat's theorem: an interior differentiable local extremum has zero gradient).

[F2]

The functions $(x, t) \mapsto u_{0} (x) \pm C_{0} t$ are $C^{1}$ on $Z$ , with time derivatives $\pm C_{0}$ and spatial gradient $D u_{0} (x)$ ; since $u_{0}$ is continuous on $O$ , they extend continuously to the initial face $O \times {0}$ with value $u_{0}$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F3]

The relaxed initial conditions for a subsolution and a supersolution of the Cauchy problem are stated as limsup and liminf over $(y, s) \to (x, 0)$ with $s > 0$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

Proof

technique · explicit affine-in-time barriers and continuity of the datum

1.1F1F2algebra

The barriers are pointwise classical sub- and supersolutions on $Z$ . By [F2], $ϕ_{\pm}$ are $C^{1}$ there, with $(ϕ_{\pm})_{t} = \pm C_{0}$ and $D ϕ_{\pm} = D u_{0}$ . The definition of $C_{0}$ gives $- C_{0} \leq H (x, t, D u_{0} (x)) \leq C_{0}$ for every $(x, t) \in O \times [0, T]$ , so $(ϕ_{-})_{t} + H (x, t, D ϕ_{-}) = - C_{0} + H (x, t, D u_{0} (x)) \leq 0$ and $(ϕ_{+})_{t} + H (x, t, D ϕ_{+}) = C_{0} + H (x, t, D u_{0} (x)) \geq 0$ pointwise on $Z$ . By [F1] these pointwise inequalities imply the viscosity test inequalities, and [F2] gives the pointwise initial values. No continuity on the lateral boundary $\partial O \times [0, T]$ is needed.

2.1step 1.1algebra

The squeeze at the initial face. Fix $x \in O$ . For $(y, s) \in Z$ the pointwise bounds give $ϕ_{-} (y, s) = u_{0} (y) - C_{0} s \leq w (y, s) \leq u_{0} (y) + C_{0} s = ϕ_{+} (y, s)$ . As $(y, s) \to (x, 0)$ with $s > 0$ we have $s \to 0$ and $y \to x$ , so by continuity of $u_{0}$ at $x$ both $u_{0} (y) - C_{0} s \to u_{0} (x)$ and $u_{0} (y) + C_{0} s \to u_{0} (x)$ ; the squeeze therefore gives $lim inf w \geq u_{0} (x)$ and $lim sup w \leq u_{0} (x)$ , both relaxed limits being taken along $s > 0$ .

3.1step 2.1F3∎

Conclusion. By step 2.1 the two relaxed limits both equal $u_{0} (x)$ , which is exactly the bisided relaxed initial condition of [F3]; in particular a continuous extension of $w$ to $O \times {0}$ must take the value $u_{0}$ there. This is the two-barrier boundary control used by the Perron construction.

Remarks

Sharpness of the hypothesis. The boundedness of $C_{0}$ is what makes the barriers classical; it holds, for example, when $H$ is uniformly bounded on $O \times [0, T] \times {∣ p ∣ \leq ∥ D u_{0} ∥_{\infty}}$ . Boundedness of $O$ or boundedness for each fixed momentum alone does not supply that uniform bound. The barriers are the model two-sided control of the initial face and are used in the Perron existence theorem.

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Perron's method for the Cauchy problem: existence between two barriers

Statement

Let $n \geq 1$ , $T > 0$ , let $H : R^{n} \times [0, T] \times R^{n} \to R$ satisfy the Lipschitz conditions of part (a) of Comparison for first-order Hamilton--Jacobi equations, and let $u_{0} \in C^{1} (R^{n})$ be bounded with bounded gradient. Assume $C_{0} : = \sup_{x \in R^{n}, 0 \leq t \leq T} ∣ H (x, t, D u_{0} (x)) ∣ < \infty$ , and put $ϕ_{\pm} = u_{0} \pm C_{0} t$ as in Time-space barriers enforce the initial trace for the Cauchy problem. Define $W (x, t)$ on $Z = R^{n} \times (0, T)$ as the supremum of all viscosity subsolutions $w$ with $ϕ_{-} \leq w \leq ϕ_{+}$ on $Z$ . Then: (1) $W$ is well defined and $ϕ_{-} \leq W \leq ϕ_{+}$ ; (2) $W^{*}$ is a viscosity subsolution and $W_{*}$ is a viscosity supersolution in $Z$ ; (3) comparison gives $W^{*} \leq W_{*}$ , hence $W = W^{*} = W_{*}$ is continuous, solves the Cauchy problem and carries datum $u_{0}$ in the relaxed sense; (4) $W$ is the unique viscosity solution in the class lying between $ϕ_{-}$ and $ϕ_{+}$ . No choice principle is used.

Facts & Assumptions

Given: The Hamiltonian $H$ with comparison case (a), $u_{0} \in C^{1}$ bounded with bounded gradient, $C_{0} = \sup ∣ H (x, t, D u_{0} (x)) ∣ < \infty$ , the barriers $ϕ_{\pm} = u_{0} \pm C_{0} t$ , and the set $W$ of viscosity subsolutions $w$ of $u_{t} + H (x, t, D u) = 0$ in $Z$ with $ϕ_{-} \leq w \leq ϕ_{+}$ , with $W : = \sup_{w \in W} w$ .

[F1]

$ϕ_{-}$ is a classical subsolution and $ϕ_{+}$ a classical supersolution, each with datum $u_{0}$ , and the barriers control both relaxed initial limits: for every locally bounded $w$ with $ϕ_{-} \leq w \leq ϕ_{+}$ the liminf and limsup at $O \times {0}$ both equal $u_{0}$ (Time-space barriers enforce the initial trace for the Cauchy problem).

[F2]

The upper semicontinuous envelope of a locally bounded-above supremum of a nonempty family of upper semicontinuous viscosity subsolutions is a viscosity subsolution (The upper envelope of a locally bounded supremum of subsolutions is a subsolution); if the lower envelope $w_{*}$ of an upper semicontinuous subsolution $w$ strictly fails the supersolution test at a point, a local bump produces a subsolution $W_{κ} \geq w$ with $W_{κ} > w$ at some point of an arbitrarily small ball about the failure point and $W_{κ} = w$ outside that ball (Failure of the supersolution test for the lower envelope allows a local bump).

[F3]

Comparison case (a) applies to bounded upper semicontinuous subsolutions and bounded lower semicontinuous supersolutions with ordered pointwise initial traces (Comparison for first-order Hamilton--Jacobi equations), and uniqueness in the bounded class follows (Uniqueness and sup-norm contraction for the Cauchy problem).

Proof

technique · envelope subsolution, bump contradiction, comparison

1.1F1F2

Well-definedness and the upper envelope. The barrier $ϕ_{-}$ is itself an admissible subsolution by [F1], so $W$ is nonempty, and every $w \in W$ satisfies $w \leq ϕ_{+}$ , so $W$ is real-valued and bounded above on compact subsets of $Z$ . By [F2] the envelope $W^{*}$ is a viscosity subsolution; moreover $W \leq ϕ_{+}$ gives $W^{*} \leq ϕ_{+}$ because $ϕ_{+}$ is continuous (the limsup defining the envelope of a function bounded above by the continuous $ϕ_{+}$ is at most $ϕ_{+}$ ), and $W^{*} \geq W \geq ϕ_{-}$ . Hence $W^{*}$ is itself an admissible member of $W$ , so $W^{*} \leq W$ by maximality and therefore $W = W^{*}$ is upper semicontinuous.

2.1step 1.1F1F2

The lower envelope is a supersolution. Suppose $W_{*}$ failed the supersolution test strictly at some $\hat{z} \in Z$ : there is $ϕ \in C^{1}$ with $W_{*} - ϕ$ having a local minimum at $\hat{z}$ and $ϕ_{t} (\hat{z}) + H (\hat{z}, D ϕ (\hat{z})) < 0$ . First, $W_{*} (\hat{z}) < ϕ_{+} (\hat{z})$ : otherwise $W_{*} (\hat{z}) = ϕ_{+} (\hat{z})$ and, since $W_{*} \leq ϕ_{+}$ , the function $ϕ_{+} - ϕ$ would have a local minimum at $\hat{z}$ , so the supersolution inequality for the classical supersolution $ϕ_{+}$ would give $ϕ_{t} (\hat{z}) + H (\hat{z}, D ϕ (\hat{z})) \geq 0$ , a contradiction. Choose a small bump supported in $B (\hat{z}, κ)$ ; by [F2] it gives a viscosity subsolution $W_{κ} \geq W$ that exceeds $W$ at some point in that ball and equals $W$ outside it. It is constructed as $\max (W, χ)$ on a smaller ball, where $χ$ is a classical subsolution and is below $W$ on the surrounding annulus. In the construction of the bump lemma, the unshifted smooth part $\tilde{ϕ} + m$ has value $W_{*} (\hat{z}) < ϕ_{+} (\hat{z})$ . First choose its ball radius $r$ small enough that $\tilde{ϕ} + m$ lies strictly below $ϕ_{+}$ throughout the closed ball. Then choose the offset $δ$ smaller than both the positive minimum of $ϕ_{+} - (\tilde{ϕ} + m)$ on that ball and the annular allowance $γ (r / 2)^{4}$ . The resulting $χ = \tilde{ϕ} + m + δ$ stays below $ϕ_{+}$ while all annular gluing inequalities hold; together with $W \leq ϕ_{+}$ this gives $W_{κ} \leq ϕ_{+}$ , while $W_{κ} \geq W \geq ϕ_{-}$ always holds. Hence $W_{κ}$ is squeezed between the barriers, and by the two-sided initial control [F1] it satisfies the relaxed initial condition; so $W_{κ} \in W$ , contradicting maximality because $W_{κ}$ exceeds $W$ at the point supplied by the bump. Therefore $W_{*}$ is a viscosity supersolution.

3.1step 1.1step 2.1F1F3∎

Comparison, continuity and uniqueness. The upper envelope $W^{*} = W$ is a bounded upper semicontinuous subsolution and $W_{*}$ is a bounded lower semicontinuous supersolution; both carry the datum $u_{0}$ in the relaxed sense by [F1] applied to $W$ , which lies between the barriers. Comparison [F3] gives $W^{*} \leq W_{*}$ ; since always $W_{*} \leq W \leq W^{*}$ , all three coincide, so $W$ is continuous and is a viscosity solution of the Cauchy problem with datum $u_{0}$ . For uniqueness, let $V$ be any, possibly discontinuous, viscosity solution with $ϕ_{-} \leq V \leq ϕ_{+}$ . By definition $V^{*}$ is a bounded upper semicontinuous subsolution and $V_{*}$ a bounded lower semicontinuous supersolution, both with datum $u_{0}$ . Continuity of the barriers and $ϕ_{-} \leq V \leq ϕ_{+}$ give $ϕ_{-} \leq V_{*} \leq V \leq V^{*} \leq ϕ_{+}$ , so $V^{*}$ belongs to $W$ and hence $V^{*} \leq W$ by maximality. Comparison between the subsolution $W$ and supersolution $V_{*}$ gives $W \leq V_{*}$ . Thus $W \leq V_{*} \leq V \leq V^{*} \leq W$ , so all are equal.

Remarks

What the barriers do. They provide the nonempty admissible class, keep $W$ locally bounded above, control the initial face in both directions through Time-space barriers enforce the initial trace for the Cauchy problem, and supply the strict inequality $W_{*} (\hat{z}) < ϕ_{+} (\hat{z})$ used to keep the bump below the upper barrier.
Choice. The family is defined by a formula and the supremum is taken in $\overline{R}$ ; no member of the family is selected, and the bump argument uses one compact maximiser at a time.

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The Legendre transform of a finite-valued convex Hamiltonian

Definition

Let $n \geq 1$ and let $H : R^{n} \to R$ be finite-valued. The Legendre transform (convex conjugate) of $H$ is the function $L : R^{n} \to \overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined) defined by $L (v) : = \sup_{p \in R^{n}} (p \cdot v - H (p)), v \in R^{n},$ the supremum being taken in $\overline{R}$ (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ) and the inner product being the Euclidean one (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). The value $L (v) = + \infty$ is possible and is not excluded.

In the biconjugate expressions $p \cdot v - (+ \infty)$ the convention $p \cdot v - (+ \infty) : = - \infty$ is adopted, so that the supremum defining the biconjugate $L^{*} (p) : = \sup_{v \in R^{n}} (p \cdot v - L (v))$ is an ordinary supremum of a set in $R \cup {- \infty}$ when the value $+ \infty$ occurs: every $v$ with $L (v) = + \infty$ contributes the value $- \infty$ and can be discarded.

When $H$ is convex (Convex and strictly convex functions on Euclidean convex sets), $L$ is the convex conjugate used in the Hopf--Lax construction. For every finite-valued $H$ the transform $L$ is convex and lower semicontinuous, being the pointwise supremum of the affine functions $v \mapsto p \cdot v - H (p)$ ; no superlinearity, differentiability, strict convexity, coercivity or smoothness of $H$ is assumed at this point. The notation $H^{*}$ is used interchangeably with $L$ .

Remarks

Why the convention is recorded. The supremum over $v \in R^{n}$ in the biconjugate runs over all of $R^{n}$ even when $L$ takes the value $+ \infty$ ; the convention makes each such term $- \infty$ , so those points neither enlarge nor obstruct the supremum, and $L^{*} (p) = \sup {p \cdot v - L (v) : v \in R^{n}, L (v) < \infty}$ . This is the convention under which the biconjugacy lemma A finite-valued convex Hamiltonian equals its biconjugate is stated, and it is fixed here once for every later use.
Scope. The transform is defined for every finite-valued $H$ ; convexity, lower semicontinuity and the affine-supremum representation are consequences of the definition, not hypotheses. The later Hopf--Lax regime adds superlinearity of $H$ , which is a separate hypothesis and is what makes $L$ real-valued; no choice principle occurs here.

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A finite-valued convex Hamiltonian equals its biconjugate

Statement

Let $n \geq 1$ and let $H : R^{n} \to R$ be convex (Convex and strictly convex functions on Euclidean convex sets), with Legendre transform $L$ as in The Legendre transform of a finite-valued convex Hamiltonian and the convention $p \cdot v - (+ \infty) : = - \infty$ . Then for every $p \in R^{n}$ $H (p) = \sup_{v \in R^{n}} (p \cdot v - L (v)),$ so that $H = L^{*} = H^{* *}$ is the biconjugate of $H$ . Equivalently, the Moreau envelopes $e_{λ} H (p) : = \inf_{q \in R^{n}} {H (q) + \frac{∣ p - q ∣^{2}}{2 λ}} (λ > 0)$ satisfy $e_{λ} H \leq L^{*} \leq H$ for every $λ > 0$ , and $e_{λ} H (p) \to H (p)$ as $λ ↓ 0$ for every $p$ . No superlinearity, differentiability, smoothness, coercivity or strict convexity of $H$ is assumed, and no choice principle is used; in particular the supporting-hyperplane route is not needed.

Facts & Assumptions

Given: An integer $n \geq 1$ , a convex function $H : R^{n} \to R$ , its Legendre transform $L (v) = \sup_{p} (p \cdot v - H (p))$ with the convention $p \cdot v - (+ \infty) : = - \infty$ , the biconjugate $L^{*} (p) = \sup_{v} (p \cdot v - L (v))$ , and the Moreau envelopes $e_{λ} H (p) = \inf_{q} {H (q) + ∣ p - q ∣^{2} / (2 λ)}$ for $λ > 0$ .

[F1]

$L (v) = \sup_{p \in R^{n}} (p \cdot v - H (p))$ is the least upper bound in $\overline{R}$ of the set ${p \cdot v - H (p) : p \in R^{n}}$ , and in the biconjugate the convention $p \cdot v - (+ \infty) : = - \infty$ is adopted (The Legendre transform of a finite-valued convex Hamiltonian).

[F2]

$H$ is convex: $H ((1 - t) x + t y) \leq (1 - t) H (x) + t H (y)$ for all $x, y \in R^{n}$ and $t \in [0, 1]$ (Convex and strictly convex functions on Euclidean convex sets).

[F3]

Every convex function on an open convex set is continuous on it; in particular $H$ is continuous on $R^{n}$ (A convex function on an open convex set is continuous).

[F4]

A nonempty subset of $R^{n}$ is compact if and only if it is closed and bounded, and every continuous real-valued function on a nonempty compact subset attains a maximum and a minimum there (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[F5]

Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , and on a nonempty subset of $R$ bounded in $R$ these agree with the real supremum and infimum (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

[F6]

If $S \subseteq R$ is nonempty and $ℓ = \inf S$ , then $ℓ \leq s$ for every $s \in S$ and $ℓ^{'} \leq ℓ$ for every lower bound $ℓ^{'}$ of $S$ (Greatest lower bound (infimum)).

Proof

technique · conjugate monotonicity plus an attained Moreau minimiser; the supporting inequality is derived from two-point convexity at the minimiser

1.1F1F5algebra

$L^{*} (p) \leq H (p)$ for every $p$ . Fix $p$ and $v$ . By [F1] and the least-upper-bound property of [F5], $p \cdot v - H (p) \leq L (v)$ , that is $p \cdot v - L (v) \leq H (p)$ ; if $L (v) = + \infty$ this reads $- \infty \leq H (p)$ by the convention of [F1], so the inequality holds in all cases. Hence $H (p)$ is an upper bound in $\overline{R}$ of ${p \cdot v - L (v) : v \in R^{n}}$ , and since $L^{*} (p)$ is the least upper bound of that set by [F1] and [F5], $L^{*} (p) \leq H (p)$ .

1.2F2F3F4algebra

A linear-growth lower bound for $H$ . By [F3] the function $H$ is continuous, and the closed unit ball $\overline{B} (0, 1)$ is nonempty, closed and bounded; by [F4] it is compact and $H$ attains on it a minimum $m \in R$ and a maximum $M \in R$ . Put $D : = \max {∣ M - H (0) ∣, ∣ m - H (0) ∣}$ , so that $∣ H (u) - H (0) ∣ \leq D$ for every $∣ u ∣ \leq 1$ . For $∣ q ∣ \geq 1$ put $u : = q / ∣ q ∣$ , so that $u = (1 / ∣ q ∣) q + (1 - 1 / ∣ q ∣) \cdot 0$ is a convex combination of $q$ and $0$ ; convexity [F2] gives $H (u) \leq (1 / ∣ q ∣) H (q) + (1 - 1 / ∣ q ∣) H (0)$ , hence $H (q) \geq ∣ q ∣ (H (u) - H (0)) + H (0) \geq - ∣ q ∣ ∣ H (u) - H (0) ∣ - ∣ H (0) ∣ \geq - (D + ∣ H (0) ∣) ∣ q ∣$ . For $∣ q ∣ \leq 1$ we have $H (q) \geq m \geq - (D + ∣ H (0) ∣)$ . Thus with $C : = D + ∣ H (0) ∣ + 1$ we get $H (q) \geq - C (1 + ∣ q ∣)$ for every $q \in R^{n}$ .

1.3F2F3F6algebra

$e_{λ} H (p) \to H (p)$ as $λ ↓ 0$ , for fixed $p$ . The competitor $q = p$ gives $e_{λ} H (p) \leq H (p)$ for every $λ > 0$ . Fix $ε > 0$ . By continuity of $H$ at $p$ [F3] there is $δ > 0$ with $∣ H (q) - H (p) ∣ \leq ε$ whenever $∣ q - p ∣ \leq δ$ . With $r : = ∣ q - p ∣$ and $φ (q) : = H (q) + r^{2} / (2 λ)$ : if $r \leq δ$ then $φ (q) \geq H (p) - ε$ ; if $r \geq δ$ , put $u : = (q - p) / r$ , so that $p + δ u = (1 - δ / r) p + (δ / r) q$ and convexity [F2] gives $H (p + δ u) \leq (1 - δ / r) H (p) + (δ / r) H (q)$ , that is $H (q) \geq H (p) + (r / δ) (H (p + δ u) - H (p)) \geq H (p) - (ε / δ) r$ . Hence for $r \geq δ$ we have $φ (q) \geq H (p) - (ε / δ) r + r^{2} / (2 λ)$ , and the right-hand side is minimised over $r \geq δ$ at $r = δ$ whenever $λ \leq δ^{2} / ε$ , with value $H (p) - ε + δ^{2} / (2 λ) \geq H (p) - ε / 2$ . Therefore $H (p) - ε$ is a lower bound of the set ${φ (q) : q \in R^{n}}$ whenever $0 < λ \leq δ^{2} / ε$ , and since $e_{λ} H (p)$ is its greatest lower bound [F6] we get $e_{λ} H (p) \geq H (p) - ε$ for all such $λ$ . As $ε > 0$ was arbitrary, ${lim inf}_{λ ↓ 0} e_{λ} H (p) \geq H (p)$ , which with the upper bound gives $e_{λ} H (p) \to H (p)$ .

2.1step 1.2F3F4F6algebra

The Moreau infimum is attained. Fix $p$ and $λ > 0$ and put $φ (q) : = H (q) + ∣ p - q ∣^{2} / (2 λ)$ . By [F3] $φ$ is continuous on $R^{n}$ , and by the bound of step 1.2, $φ (q) \geq - C (1 + ∣ q ∣) + ∣ p - q ∣^{2} / (2 λ)$ , whose right-hand side tends to $+ \infty$ as $∣ q ∣ \to \infty$ because $∣ p - q ∣ \geq ∣ q ∣ - ∣ p ∣$ ; hence there is $R > ∣ p ∣$ with $φ (q) > φ (p)$ for every $∣ q ∣ \geq R$ . The set $S : = {q \in R^{n} : φ (q) \leq φ (p)}$ is then nonempty (it contains $p$ ), contained in the closed ball $\overline{B} (0, R)$ , and closed (it is the preimage under the continuous $φ$ of a closed interval); being closed and bounded it is compact by [F4], so $φ$ attains on $S$ a minimum at some $q_{*} \in S$ by [F4]. For $q \notin S$ we have $φ (q) > φ (p) \geq φ (q_{*})$ , so $q_{*}$ is a global minimiser of $φ$ on $R^{n}$ , and by the definition of the infimum [F6] $φ (q_{*}) = \inf_{q} φ (q) = e_{λ} H (p)$ .

3.1step 2.1F1F2algebra

Supporting inequality and finiteness of $L (v_{*})$ at a minimiser. Keep $p, λ, q_{*}$ and $φ$ from step 2.1, put $w : = p - q_{*}$ , $v_{*} : = w / λ$ , and fix $q \in R^{n}$ with $h : = q - q_{*}$ . For $0 < r \leq 1$ the point $q_{*} + r h$ belongs to $R^{n}$ and minimality of $q_{*}$ gives $φ (q_{*}) \leq φ (q_{*} + r h)$ ; convexity [F2] gives $H (q_{*} + r h) \leq (1 - r) H (q_{*}) + r H (q)$ . Subtracting and using $∣ w - r h ∣^{2} = ∣ w ∣^{2} - 2 r w \cdot h + r^{2} ∣ h ∣^{2}$ , these two inequalities yield $r (H (q) - H (q_{*})) \geq r w \cdot h / λ - r^{2} ∣ h ∣^{2} / (2 λ)$ ; dividing by $r > 0$ and letting $r ↓ 0$ gives $H (q) \geq H (q_{*}) + v_{*} \cdot h$ . Hence $q \cdot v_{*} - H (q) \leq q_{*} \cdot v_{*} - H (q_{*})$ for every $q$ , with equality at $q = q_{*}$ , so the least upper bound $L (v_{*})$ of [F1] equals $q_{*} \cdot v_{*} - H (q_{*}) \in R$ .

4.1step 2.1step 3.1F1F5algebra

$L^{*} (p) \geq e_{λ} H (p)$ . By [F1] and [F5], $L^{*} (p) \geq p \cdot v_{*} - L (v_{*})$ for the vector $v_{*}$ of step 3.1 (the value $L (v_{*})$ is real, so $p \cdot v_{*} - L (v_{*})$ is an ordinary real number and no convention is needed). Substituting $L (v_{*}) = q_{*} \cdot v_{*} - H (q_{*})$ from step 3.1 and $v_{*} = (p - q_{*}) / λ$ , we get $L^{*} (p) \geq (p - q_{*}) \cdot v_{*} + H (q_{*}) = ∣ p - q_{*} ∣^{2} / λ + H (q_{*}) \geq ∣ p - q_{*} ∣^{2} / (2 λ) + H (q_{*}) = e_{λ} H (p)$ , where the last equality is step 2.1.

5.1step 1.1step 1.3step 4.1∎

Conclusion. Steps 1.1 and 4.1 give $e_{λ} H (p) \leq L^{*} (p) \leq H (p)$ for every $λ > 0$ and every $p$ , and step 1.3 gives $e_{λ} H (p) \to H (p)$ as $λ ↓ 0$ ; hence $L^{*} (p) = \sup_{v} (p \cdot v - L (v)) = H (p)$ for every $p$ , that is $H = L^{*}$ .

Remarks

What replaces the subgradient theorem. The only existence input is the attained minimiser of the strictly convex perturbation $q \mapsto H (q) + ∣ p - q ∣^{2} / (2 λ)$ , obtained from continuity, a linear lower bound and compactness. The supporting inequality $H (q) \geq H (q_{*}) + v_{*} \cdot (q - q_{*})$ is then a two-point convexity computation at that minimiser, so no supporting-hyperplane theorem and no choice principle is consumed.
Sharpness of hypotheses. Neither superlinearity nor coercivity of $H$ is used: the quadratic penalty provides the coercivity, and the continuity of $H$ is a consequence of convexity and finite-valuedness by [F3].

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The Hopf--Lax operator and the Hopf--Lax formula

Definition

Let $n \geq 1$ , let $H : R^{n} \to R$ be convex and superlinear, $\lim_{∣ p ∣ \to \infty} \frac{H (p)}{∣ p ∣} = + \infty,$ let $L$ be its Legendre transform (The Legendre transform of a finite-valued convex Hamiltonian), and let $u_{0} : R^{n} \to R$ be bounded and uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point, Lower bound, bounded below, bounded set). For $t > 0$ and $x \in R^{n}$ define $Q_{t} u_{0} (x) : = \inf_{y \in R^{n}} {u_{0} (y) + t L (\frac{x - y}{t})},$ the infimum being computed in $R \cup {+ \infty}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, Greatest lower bound (infimum)) over the extended-real values $u_{0} (y) + t L ((x - y) / t)$ ; the term $t L ((x - y) / t)$ is $+ \infty$ exactly when $L ((x - y) / t) = + \infty$ . Set $Q_{0} u_{0} : = u_{0}$ .

The Hopf--Lax operator with Lagrangian $L$ is the family $(Q_{t})_{t \geq 0}$ , and the function $(x, t) \mapsto Q_{t} u_{0} (x)$ is the Hopf--Lax formula for the Cauchy problem $u_{t} + H (D u) = 0$ , $u (\cdot, 0) = u_{0}$ . The infimum is an extended-real expression at this point: finiteness and the confinement of near-minimisers are proved in Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers, where superlinearity makes $L$ real-valued everywhere.

Remarks

What is fixed and what is postponed. The definition fixes the autonomous Hamiltonian $H : R^{n} \to R$ , convex and superlinear; the datum class, bounded and uniformly continuous $u_{0}$ ; the infimum over all $y \in R^{n}$ of $u_{0} (y) + t L ((x - y) / t)$ for $t > 0$ , read in the extended reals; and the value $Q_{0} u_{0} = u_{0}$ . Neither the attainment of the infimum nor its finiteness is asserted here, and no assertion that $L$ is real-valued is smuggled into the definition; the value $+ \infty$ is kept visible until Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers proves the opposite under superlinearity.
Time scaling. The velocity variable in the formula is $(x - y) / t$ , the average velocity of a straight path from $y$ at time $0$ to $x$ at time $t$ ; the factor $t$ multiplies the Lagrangian density. This normalisation is the one for which the dynamic-programming identity $Q_{t + s} u_{0} = Q_{t} (Q_{s} u_{0})$ of The Hopf--Lax operators form a semigroup (dynamic programming) holds with the coefficient $t + s$ . No choice principle is used in the definition.
Bounded data without continuity. The same pointwise infimum formula defines $Q_{t} f (x)$ for any bounded function $f$ , even when $f$ is not uniformly continuous. Since $L (v) \geq - H (0)$ and the competitor $y = x$ is finite, these values are real. This extension is used for the nonexpansiveness estimate in The Hopf--Lax operator is a contraction in the supremum norm; continuity conclusions such as The Hopf--Lax operator preserves a modulus of continuity retain their stated hypotheses.

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Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers

Statement

Let $H : R^{n} \to R$ be convex and superlinear with Legendre transform $L$ , and let $u_{0} : R^{n} \to R$ be bounded and continuous. Then: (1) $L$ is real-valued on $R^{n}$ , convex, continuous, and superlinear; (2) for every $x \in R^{n}$ and $t > 0$ the infimum defining $Q_{t} u_{0} (x)$ is attained, and for every $δ > 0$ there is a finite radius $ρ = ρ (x, t, δ, ∥ u_{0} ∥_{\infty})$ such that every near-minimiser $y$ with $u_{0} (y) + t L (\frac{x - y}{t}) \leq Q_{t} u_{0} (x) + δ$ satisfies $∣ y - x ∣ \leq ρ$ ; (3) consequently $Q_{t} u_{0}$ is real-valued for every $x$ and every $t \geq 0$ . No choice principle is used.

Facts & Assumptions

Given: A convex superlinear $H : R^{n} \to R$ with $\lim_{∣ p ∣ \to \infty} H (p) / ∣ p ∣ = + \infty$ , its Legendre transform $L$ , a bounded continuous $u_{0} : R^{n} \to R$ , and the operators $Q_{t}$ of The Hopf--Lax operator and the Hopf--Lax formula.

[F1]

For $t > 0$ , $Q_{t} u_{0} (x) = \inf_{y \in R^{n}} {u_{0} (y) + t L ((x - y) / t)}$ , and $Q_{0} u_{0} = u_{0}$ (The Hopf--Lax operator and the Hopf--Lax formula).

[F2]

$L (v) = \sup_{p \in R^{n}} (p \cdot v - H (p))$ for every $v \in R^{n}$ , the supremum being the least upper bound in $\overline{R}$ of the set of real numbers $p \cdot v - H (p)$ (The Legendre transform of a finite-valued convex Hamiltonian).

[F3]

Every convex function on an open convex set is continuous on it; in particular any finite convex function on $R^{n}$ is continuous (A convex function on an open convex set is continuous).

[F4]

A nonempty subset of $R^{n}$ is compact exactly when it is closed and bounded, and every continuous real-valued function on such a set attains a maximum and a minimum (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

Proof

technique · coercivity of the Lagrangian confines near-minimisers to a compact ball, where continuity gives attainment

1.1F2F3F4algebra

Properties of $L$ . Fix $v \in R^{n}$ . Superlinearity of $H$ gives $R > 0$ with $H (p) \geq 2 ∣ v ∣ ∣ p ∣$ for $∣ p ∣ \geq R$ , so $p \cdot v - H (p) \leq - ∣ v ∣ ∣ p ∣ \leq 0$ there; on the closed ball $\overline{B} (0, R)$ , which is closed and bounded and hence compact by [F4], the continuous function $p \mapsto p \cdot v - H (p)$ (continuity of $H$ is [F3]) attains a maximum $M_{v} < \infty$ by [F4]. Hence $p \cdot v - H (p) \leq \max {M_{v}, 0} < \infty$ for every $p$ , while $L (v) \geq - H (0)$ because $p = 0$ is admissible; by the least-upper-bound property of [F2], $L (v)$ is a real number. Convexity of $L$ : for fixed $p$ the map $v \mapsto p \cdot v - H (p)$ is affine, and a pointwise supremum of affine functions is convex; $L$ is real-valued, so it is finite and convex on the open convex set $R^{n}$ and therefore continuous by [F3]. Superlinearity: fix $a > 0$ ; for $v \neq 0$ the admissible test point $p : = a v / ∣ v ∣$ gives $L (v) \geq a ∣ v ∣ - H (a v / ∣ v ∣) \geq a ∣ v ∣ - \max_{∣ q ∣ \leq a} H (q)$ , where the maximum is finite by [F3] and [F4]; dividing by $∣ v ∣$ and letting $∣ v ∣ \to \infty$ gives ${lim inf}_{∣ v ∣ \to \infty} L (v) / ∣ v ∣ \geq a$ , and since $a > 0$ was arbitrary, $L (v) / ∣ v ∣ \to + \infty$ .

2.1step 1.1F1F2F3F4algebra

Attainment and localisation. Fix $x$ , $t > 0$ and $δ > 0$ , and put $φ (y) : = u_{0} (y) + t L ((x - y) / t)$ , so that $Q_{t} u_{0} (x) = \inf φ$ by [F1]. The competitor $y = x$ gives $Q_{t} u_{0} (x) \leq φ (x) = u_{0} (x) + t L (0) \leq ∥ u_{0} ∥_{\infty} + t L (0) = : A < \infty$ , where $L (0) \in R$ . Also every term is at least $\inf u_{0} - t H (0) > - \infty$ by [F2], so $Q_{t} u_{0} (x) \in R$ . Let $y$ satisfy $φ (y) \leq Q_{t} u_{0} (x) + δ$ ; then $t L (\frac{x - y}{t}) \leq A + δ - \inf u_{0} \leq 2 ∥ u_{0} ∥_{\infty} + t L (0) + δ = : C .$ Superlinearity of $L$ from step 1.1 gives $R > 0$ such that $L (w) > C / t$ whenever $∣ w ∣ \geq R$ . If $∣ x - y ∣ / t \geq R$ , then $t L ((x - y) / t) > C$ , contradicting the preceding bound. Hence $∣ y - x ∣ < t R$ , so every $δ$ -near-minimiser lies in the closed ball $\overline{B} (x, ρ)$ with $ρ : = t R < \infty$ ; the radius depends only on $t, δ, ∥ u_{0} ∥_{\infty}$ and the fixed $H$ (and may harmlessly be viewed as a function of $x$ as in the statement). For attainment, let $S : = {y \in R^{n} : φ (y) \leq Q_{t} u_{0} (x) + 1}$ . By the definition of the finite infimum, $S$ is nonempty. It is closed by continuity of $φ$ and bounded by the localisation just proved with $δ = 1$ , so [F4] makes it compact. The continuous function $φ$ attains a minimum on $S$ at some $y_{*}$ . This minimum equals the global infimum: it is at least $Q_{t} u_{0} (x)$ , and for every $0 < ε < 1$ the infimum property gives $y$ with $φ (y) < Q_{t} u_{0} (x) + ε$ , which lies in $S$ , so the minimum is at most $Q_{t} u_{0} (x) + ε$ . Thus $φ (y_{*}) = Q_{t} u_{0} (x)$ and the infimum is attained, without selecting a sequence.

3.1step 1.1step 2.1F1

Real-valuedness of $Q_{t} u_{0}$ . For $t > 0$ each term satisfies $u_{0} (y) + t L ((x - y) / t) \geq \inf_{R^{n}} u_{0} - t H (0) > - \infty$ by the estimate $L \geq - H (0)$ of step 1.1, and the value at $y = x$ is finite; hence $Q_{t} u_{0} (x) \in R$ by [F1]. For $t = 0$ this is $Q_{0} u_{0} = u_{0}$ , real-valued by hypothesis.

4.1step 1.1step 2.1step 3.1∎

Conclusion. Part (1) is step 1.1, part (2) is step 2.1, and part (3) is step 3.1.

Remarks

Dependence of the radius. The radius produced depends on $x$ , $t$ , $δ$ and $∥ u_{0} ∥_{\infty}$ only through the quantifier-free bounds of step 2.1; it is uniform in $x$ on compact $x$ -sets because the estimates are translation invariant. No compactness of the ambient space and no subsequence selection is used, hence no choice principle.

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The Hopf--Lax operator is a contraction in the supremum norm

Statement

Let $H : R^{n} \to R$ be convex and superlinear with Legendre transform $L$ , and let $u_{0}, v_{0} : R^{n} \to R$ be bounded. Then for every $t \geq 0$ and every $x \in R^{n}$ $∣ Q_{t} u_{0} (x) - Q_{t} v_{0} (x) ∣ \leq \sup_{R^{n}} ∣ u_{0} - v_{0} ∣,$ and consequently $\sup_{R^{n}} ∣ Q_{t} u_{0} - Q_{t} v_{0} ∣ \leq ∥ u_{0} - v_{0} ∥_{\infty}$ ; both $Q_{t} u_{0}$ and $Q_{t} v_{0}$ are real-valued (see the proof for the explicit finiteness argument). No uniform continuity of the data is needed for this particular estimate, and no choice principle is used.

Facts & Assumptions

Given: A convex superlinear $H : R^{n} \to R$ , its Legendre transform $L (v) = \sup_{p} (p \cdot v - H (p))$ , bounded data $u_{0}, v_{0} : R^{n} \to R$ , the operators $Q_{t}$ of The Hopf--Lax operator and the Hopf--Lax formula, and $c : = \sup_{R^{n}} ∣ u_{0} - v_{0} ∣ \in [0, \infty)$ .

[F1]

For $t > 0$ and $x \in R^{n}$ , $Q_{t} u_{0} (x) = \inf_{y \in R^{n}} {u_{0} (y) + t L ((x - y) / t)}$ and $Q_{t} v_{0} (x) = \inf_{y \in R^{n}} {v_{0} (y) + t L ((x - y) / t)}$ , with infima computed in $R \cup {+ \infty}$ ; for $t = 0$ , $Q_{0} u_{0} = u_{0}$ and $Q_{0} v_{0} = v_{0}$ (The Hopf--Lax operator and the Hopf--Lax formula).

[F2]

$L (v) = \sup_{p \in R^{n}} (p \cdot v - H (p))$ for every $v$ , so $L (v) \geq - H (0)$ and $L (v)$ is the least upper bound of ${p \cdot v - H (p) : p \in R^{n}}$ (The Legendre transform of a finite-valued convex Hamiltonian).

[F3]

Every convex function on an open convex set is continuous; in particular $H$ is continuous on $R^{n}$ (A convex function on an open convex set is continuous).

[F4]

Proof

technique · compare the two infima termwise, then record the finiteness of the Lagrangian values

1.1F1F2F3F4algebra

The Lagrangian values and Hopf--Lax values are real. Fix $v \in R^{n}$ . The lower bound $L (v) \geq - H (0)$ is [F2]. For the upper bound, superlinearity of $H$ gives $R > 0$ with $H (p) \geq 2 ∣ v ∣ ∣ p ∣$ whenever $∣ p ∣ \geq R$ ; then $p \cdot v - H (p) \leq ∣ v ∣ ∣ p ∣ - 2 ∣ v ∣ ∣ p ∣ \leq 0$ for such $p$ . On the closed ball $\overline{B} (0, R)$ , which is nonempty, closed and bounded and hence compact by [F4], the continuous function $p \mapsto p \cdot v - H (p)$ (continuity of $H$ is [F3]) attains a maximum $M_{v} < \infty$ by [F4]. Therefore $p \cdot v - H (p) \leq \max {M_{v}, 0} < \infty$ for every $p$ , so the least upper bound $L (v)$ of [F2] is real. For $t > 0$ , every Hopf--Lax term is bounded below by $\inf u_{0} - t H (0)$ , and the competitor $y = x$ gives the finite upper bound $Q_{t} u_{0} (x) \leq u_{0} (x) + t L (0)$ ; hence $Q_{t} u_{0} (x) \in R$ , and likewise for $v_{0}$ .

2.1F1F2step 1.1algebra

One-sided comparison for $t > 0$ . Fix $x$ and $t > 0$ , and write $a (y) : = u_{0} (y) + t L ((x - y) / t)$ and $b (y) : = v_{0} (y) + t L ((x - y) / t)$ . By step 1.1 and [F2], $b$ is real-valued, bounded below by $\inf v_{0} - t H (0)$ , and has a finite value at $y = x$ , so $m : = \inf_{y} b (y) = Q_{t} v_{0} (x)$ is real. Since $u_{0} (y) \leq v_{0} (y) + c$ , we have $a (y) \leq b (y) + c$ for every $y$ . For each $ε > 0$ , the infimum property gives a $y$ with $b (y) < m + ε$ , and then $Q_{t} u_{0} (x) = \inf_{y} a (y) \leq a (y) \leq b (y) + c < m + c + ε$ . Letting $ε ↓ 0$ gives $Q_{t} u_{0} (x) \leq Q_{t} v_{0} (x) + c$ .

3.1step 1.1step 2.1F1∎

Conclusion. Fix $t > 0$ and $x$ . Applying step 2.1 to $(u_{0}, v_{0})$ and to $(v_{0}, u_{0})$ , whose value of $c$ is unchanged, gives both one-sided inequalities; the Hopf--Lax values are real by step 1.1, so $∣ Q_{t} u_{0} (x) - Q_{t} v_{0} (x) ∣ \leq c$ . For $t = 0$ this is $∣ u_{0} (x) - v_{0} (x) ∣ \leq c$ by [F1]. Taking the supremum over $x$ gives $\sup ∣ Q_{t} u_{0} - Q_{t} v_{0} ∣ \leq c = ∥ u_{0} - v_{0} ∥_{\infty}$ .

Remarks

Hypotheses actually used. Only convexity, superlinearity, boundedness of the data and the algebraic form of the infimum enter; uniform continuity and the localisation lemma are not needed for this estimate. The argument also shows $L$ is real-valued under superlinearity, a fact used independently in Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers.
Sharpness. The constant $1$ is optimal: for $t > 0$ constant data are preserved by $Q_{t}$ up to the same constant, so the operator is nonexpansive and no smaller universal constant can hold.

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The Hopf--Lax operator preserves a modulus of continuity

Statement

Let $H : R^{n} \to R$ be convex and superlinear with Legendre transform $L$ , and let $u_{0} : R^{n} \to R$ be bounded and uniformly continuous with a nondecreasing modulus of continuity $ω$ satisfying $ω (r) \to 0$ as $r ↓ 0$ and $∣ u_{0} (y) - u_{0} (y^{'}) ∣ \leq ω (∣ y - y^{'} ∣)$ . Then for every $t \geq 0$ and all $x, x^{'} \in R^{n}$ , $∣ Q_{t} u_{0} (x) - Q_{t} u_{0} (x^{'}) ∣ \leq ω (∣ x - x^{'} ∣) .$ Thus every $Q_{t} u_{0}$ has the same modulus of continuity and the family is spatially equicontinuous; this is the equicontinuity input of the initial-trace and vanishing-viscosity arguments. No choice principle is used.

Facts & Assumptions

Given: A convex superlinear $H$ , its Legendre transform $L$ , a bounded uniformly continuous $u_{0}$ with modulus $ω$ as in the statement, the operators $Q_{t}$ of The Hopf--Lax operator and the Hopf--Lax formula, and points $x, x^{'} \in R^{n}$ with $d : = x - x^{'}$ .

[F1]

For $t > 0$ , $Q_{t} u_{0} (x) = \inf_{y \in R^{n}} {u_{0} (y) + t L ((x - y) / t)}$ , with the infimum in $R \cup {+ \infty}$ ; $Q_{0} u_{0} = u_{0}$ (The Hopf--Lax operator and the Hopf--Lax formula).

[F2]

Under the present hypotheses $L$ is real-valued, the infima above are attained, and $Q_{t} u_{0} (x) \in R$ for every $x$ and every $t \geq 0$ (Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers), so all infima compared below are real numbers.

[F3]

The given modulus is a nondecreasing function $ω$ with $ω (r) \to 0$ as $r ↓ 0$ and $∣ u_{0} (y) - u_{0} (y^{'}) ∣ \leq ω (∣ y - y^{'} ∣)$ for all $y, y^{'}$ . This is a hypothesis of the statement; it implies the epsilon--delta uniform continuity of Uniform continuity of a map of metric spaces: one $δ$ serving every point by choosing $δ > 0$ with $ω (δ) < ε$ .

Proof

technique · translate a competitor in the infimum

1.1F1F2F3algebra

Translation of a competitor. Fix $t > 0$ and let $d : = x - x^{'}$ . For every $y^{'} \in R^{n}$ put $y : = y^{'} + d$ . Then $(x - y) / t = (x^{'} - y^{'}) / t$ and, by [F3], $u_{0} (y) \geq u_{0} (y^{'}) - ω (∣ d ∣)$ ; hence $u_{0} (y) + t L ((x - y) / t) \geq u_{0} (y^{'}) + t L ((x^{'} - y^{'}) / t) - ω (∣ d ∣)$ . As $y^{'}$ ranges over $R^{n}$ so does $y$ , so taking the infimum over $y^{'}$ of the right-hand side and using [F1] and [F2] gives $Q_{t} u_{0} (x) \geq Q_{t} u_{0} (x^{'}) - ω (∣ x - x^{'} ∣)$ . Exchanging $x$ and $x^{'}$ gives the reverse inequality $Q_{t} u_{0} (x^{'}) \geq Q_{t} u_{0} (x) - ω (∣ x - x^{'} ∣)$ .

2.1step 1.1F1∎

Conclusion. For $t > 0$ step 1.1 gives $∣ Q_{t} u_{0} (x) - Q_{t} u_{0} (x^{'}) ∣ \leq ω (∣ x - x^{'} ∣)$ , and for $t = 0$ the same inequality is the hypothesis $∣ u_{0} (x) - u_{0} (x^{'}) ∣ \leq ω (∣ x - x^{'} ∣)$ by [F1]. Hence every $Q_{t} u_{0}$ has modulus $ω$ , uniformly in $t$ , and the estimate is translation invariant because $L$ does not depend on the space variable.

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The Hopf--Lax operators form a semigroup (dynamic programming)

Statement

Let $H : R^{n} \to R$ be convex and superlinear with Legendre transform $L$ , and let $u_{0} : R^{n} \to R$ be bounded and uniformly continuous. Then for all $t, s \geq 0$ the Hopf--Lax operators of The Hopf--Lax operator and the Hopf--Lax formula satisfy $Q_{t + s} u_{0} = Q_{t} (Q_{s} u_{0}) = Q_{s} (Q_{t} u_{0}) pointwise on R^{n},$ where the inner operators are applied to the bounded uniformly continuous function $Q_{s} u_{0}$ (or $Q_{t} u_{0}$ ) produced by The Hopf--Lax operator preserves a modulus of continuity. Equivalently, for all $t, s > 0$ and $x \in R^{n}$ the short-time variational principle holds: $Q_{t + s} u_{0} (x) = \inf_{z \in R^{n}} {Q_{s} u_{0} (z) + t L (\frac{x - z}{t})} .$ The family $(Q_{t})_{t \geq 0}$ is therefore a semigroup with $Q_{0} = i d$ on the bounded uniformly continuous data. No choice principle is used.

Facts & Assumptions

Given: A convex superlinear $H$ with Legendre transform $L$ , a bounded uniformly continuous $u_{0}$ , the operators $Q_{t}$ of The Hopf--Lax operator and the Hopf--Lax formula, and $t, s > 0$ .

[F1]

$Q_{t} u_{0} (x) = \inf_{y \in R^{n}} {u_{0} (y) + t L ((x - y) / t)}$ for $t > 0$ , $Q_{0} u_{0} = u_{0}$ , and under the present hypotheses all these infima are real and attained (The Hopf--Lax operator and the Hopf--Lax formula, Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers).

[F2]

$L$ is convex: $L ((1 - λ) w_{1} + λ w_{2}) \leq (1 - λ) L (w_{1}) + λ L (w_{2})$ for all $w_{1}, w_{2}$ and $λ \in [0, 1]$ (Convex and strictly convex functions on Euclidean convex sets).

[F3]

$Q_{s} u_{0}$ is bounded and has the modulus of continuity of $u_{0}$ ; in particular it is bounded and uniformly continuous, so the inner operator $Q_{t}$ is defined on it (The Hopf--Lax operator preserves a modulus of continuity).

Proof

technique · the two-point convexity inequality with weights adapted to $t$ and $s$

1.1F1F2F3algebra

The inequality $Q_{t + s} u_{0} \leq Q_{t} (Q_{s} u_{0})$ . Fix $y, z \in R^{n}$ and write $(x - y) / (t + s) = \frac{t}{t + s} \frac{x - z}{t} + \frac{s}{t + s} \frac{z - y}{s}$ , a convex combination with weights $t / (t + s)$ and $s / (t + s)$ ; by [F2], $L ((x - y) / (t + s)) \leq \frac{t}{t + s} L ((x - z) / t) + \frac{s}{t + s} L ((z - y) / s)$ . Multiplying by $t + s$ and adding $u_{0} (y)$ gives $u_{0} (y) + (t + s) L ((x - y) / (t + s)) \leq [u_{0} (y) + s L ((z - y) / s)] + t L ((x - z) / t)$ . Taking the infimum over $y$ on the left and over $y$ and then $z$ on the right (the double infimum is an infimum over pairs, legitimate for real infima by [F1]) yields $Q_{t + s} u_{0} (x) \leq \inf_{z} {Q_{s} u_{0} (z) + t L ((x - z) / t)} = Q_{t} (Q_{s} u_{0}) (x)$ .

1.2F1F2F3algebra

The reverse inequality. Fix $y \in R^{n}$ and choose the segment point $z : = y + \frac{s}{t + s} (x - y)$ , for which $\frac{z - y}{s} = \frac{x - z}{t} = \frac{x - y}{t + s}$ . Then $u_{0} (y) + s L ((z - y) / s) + t L ((x - z) / t) = u_{0} (y) + (t + s) L ((x - y) / (t + s))$ ; taking the infimum over $y$ gives $Q_{t} (Q_{s} u_{0}) (x) \leq Q_{t + s} u_{0} (x)$ .

2.1step 1.1step 1.2F1F3∎

Conclusion. Steps 1.1 and 1.2 give $Q_{t + s} u_{0} = Q_{t} (Q_{s} u_{0})$ for all $t, s > 0$ . The operator $Q_{s}$ maps the bounded uniformly continuous datum to a bounded uniformly continuous function by [F3], so the composition is well defined; swapping the roles of $t$ and $s$ in the same computation gives $Q_{t + s} u_{0} = Q_{s} (Q_{t} u_{0})$ , and the case $t = 0$ or $s = 0$ is the definition $Q_{0} = i d$ of [F1]. Hence $(Q_{t})_{t \geq 0}$ is a semigroup of operators on the bounded uniformly continuous data, and the displayed short-time variational principle is the identity $Q_{t} (Q_{s} u_{0}) = Q_{t + s} u_{0}$ written out.

Remarks

Choice. The two inequalities are computed by taking infima over explicit sets of reals; the segment point $z$ is given by a formula, so nothing is selected and no choice principle is used.
Why the datum class is preserved. The semigroup statement needs the inner operator to be applied to a bounded uniformly continuous function, which is exactly the content of The Hopf--Lax operator preserves a modulus of continuity together with the boundedness following from $u_{0}$ bounded and $L \geq - H (0)$ .

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A Hopf--Lax minimiser satisfies the characteristic Euler relation at differentiability points

Statement

Let $H : R^{n} \to R$ be convex and superlinear with Legendre transform $L$ , let $u_{0} : R^{n} \to R$ be bounded and continuous, and let $t > 0$ , $x \in R^{n}$ . Let $y$ be a minimiser of $φ (y) : = u_{0} (y) + t L ((x - y) / t)$ , which exists by Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers, put $v : = (x - y) / t$ , and set $u : = Q_{t} u_{0}$ . Then: (1) if $u_{0}$ is differentiable at $y$ and $L$ is differentiable at $v$ , then $D u_{0} (y) = D L (v)$ ; (2) if $L$ is differentiable at $v$ and $u$ is differentiable at $x$ , then $D u (x) = D L (v)$ . No choice principle is used.

Facts & Assumptions

Given: A convex superlinear $H$ with Legendre transform $L$ , bounded continuous $u_{0}$ , $t > 0$ , $x \in R^{n}$ , a minimiser $y$ of $φ (y) = u_{0} (y) + t L ((x - y) / t)$ , $v = (x - y) / t$ , $u = Q_{t} u_{0}$ , and the Euclidean norm $∣ \cdot ∣$ .

[F1]

$Q_{t} u_{0} (x) = \inf_{y \in R^{n}} {u_{0} (y) + t L ((x - y) / t)}$ , and the infimum is attained under the present hypotheses (The Hopf--Lax operator and the Hopf--Lax formula, Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers).

[F2]

$f$ is differentiable at $a$ with derivative $D f (a)$ exactly when $f (a + h) = f (a) + D f (a) h + r (h)$ with $r (h) / ∣ h ∣ \to 0$ as $h \to 0$ ; in that case every directional derivative exists and equals $D f (a) h$ (The total (Fréchet) derivative $D f (a)$ as the linear first-order approximation with $o (∥ h ∥_{2})$ remainder, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

Proof

technique · expand the minimality inequality to first order in the direction of the perturbation

1.1F1F2algebra

First-order condition at a minimiser. Fix $h \in R^{n}$ and $τ \in (0, 1)$ ; minimality of $y$ at the point $x$ gives $u_{0} (y) + t L (v) \leq u_{0} (y + τ h) + t L (v - τ h / t)$ , where $u_{0} (y) + t L (v) = Q_{t} u_{0} (x)$ by [F1]. Assume $u_{0}$ is differentiable at $y$ and $L$ at $v$ . Then $u_{0} (y + τ h) = u_{0} (y) + τ ⟨ D u_{0} (y), h ⟩ + r_{u} (τ h)$ and $L (v - τ h / t) = L (v) - τ ⟨ D L (v), h / t ⟩ + r_{L} (- τ h / t)$ with $r_{u} (τ h) / ∣ τ h ∣ \to 0$ and $r_{L} (- τ h / t) / ∣ τ h ∣ \to 0$ as $τ ↓ 0$ by [F2]. Substituting and cancelling $u_{0} (y) + t L (v)$ gives $τ ⟨ D u_{0} (y), h ⟩ + r_{u} (τ h) \geq τ ⟨ D L (v), h ⟩ - t r_{L} (- τ h / t)$ ; dividing by $τ > 0$ and letting $τ ↓ 0$ gives $⟨ D u_{0} (y), h ⟩ \geq ⟨ D L (v), h ⟩$ .

2.1step 1.1F2algebra∎

The two equalities. The inequality of step 1.1 holds for every $h \in R^{n}$ ; applying it to $- h$ as well gives $⟨ D u_{0} (y) - D L (v), h ⟩ \geq 0$ and $⟨ D u_{0} (y) - D L (v), - h ⟩ \geq 0$ , that is $∣ ⟨ D u_{0} (y) - D L (v), h ⟩ ∣ \leq 0$ for all $h$ . Taking $h = D u_{0} (y) - D L (v)$ gives $∣ D u_{0} (y) - D L (v) ∣^{2} \leq 0$ , hence $D u_{0} (y) = D L (v)$ , which is (1). For (2), minimality of $y$ at the point $x + τ h$ gives $u (x + τ h) = Q_{t} u_{0} (x + τ h) \leq u_{0} (y) + t L (v + τ h / t) = u (x) + t (L (v + τ h / t) - L (v))$ , and if $L$ is differentiable at $v$ and $u$ at $x$ then [F2] gives $u (x + τ h) = u (x) + τ ⟨ D u (x), h ⟩ + r_{1} (τ h)$ and $t (L (v + τ h / t) - L (v)) = τ ⟨ D L (v), h ⟩ + r_{2} (τ h)$ with $r_{i} (τ h) / (τ ∣ h ∣) \to 0$ . Dividing by $τ$ and letting $τ ↓ 0$ gives $⟨ D u (x), h ⟩ \leq ⟨ D L (v), h ⟩$ for every $h$ ; applying this to $- h$ yields $D u (x) = D L (v)$ by the same argument as above, which is (2).

Remarks

Differentiability is assumed only where used. The minimiser exists by the localisation lemma, and the first-order conditions are obtained by perturbing the minimiser in a direction and expanding: no global smoothness of $u_{0}$ , $L$ or $Q_{t} u_{0}$ is asserted, and in the convex-quadratic case $H (p) = ∣ p ∣^{2} / 2$ a minimiser need not be unique when $u_{0}$ is merely continuous.
Direction of the two relations. Part (1) relates the datum to the Lagrangian at the minimiser, part (2) relates the value function to the Lagrangian at the same minimiser; together they identify the slope of the minimising chord with the conjugate momentum.

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Comparison for autonomous convex superlinear Hamiltonians

Statement

Let $n \geq 1$ and let $H : R^{n} \to R$ be finite-valued, continuous, convex and superlinear. Let $T > 0$ . Suppose $u$ and $v$ are bounded uniformly continuous on $R^{n} \times [0, T]$ , their restrictions to $Z = R^{n} \times (0, T)$ are respectively a viscosity subsolution and a viscosity supersolution of $u_{t} + H (D u) = 0$ , and their continuous initial traces satisfy $u (x, 0) \leq v (x, 0)$ for every $x \in R^{n}$ . Then $u \leq v$ on $Z$ . No choice principle is used.

Facts & Assumptions

Given: A finite continuous convex superlinear $H : R^{n} \to R$ , $T > 0$ , bounded uniformly continuous $u, v$ on $R^{n} \times [0, T]$ whose restrictions to $Z$ are a viscosity subsolution and supersolution of $u_{t} + H (D u) = 0$ with $u (\cdot, 0) \leq v (\cdot, 0)$ , and positive parameters $η, η^{'}, ρ, α, P, ε, δ$ .

[F1]

At every local maximum of $w - ϕ$ with $ϕ \in C^{1} (Z)$ a viscosity subsolution $w$ satisfies $ϕ_{t} + H (D ϕ) \leq 0$ , and at every local minimum a viscosity supersolution satisfies $ϕ_{t} + H (D ϕ) \geq 0$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

Uniform continuity of a map on a metric space means: for every $σ > 0$ there is $τ > 0$ such that $∣ f (p) - f (q) ∣ < σ$ whenever $d (p, q) < τ$ ; hence $u$ and $v$ admit bounded time moduli $ω_{u}, ω_{v}$ and their two initial traces admit a bounded common spatial modulus $ω_{0}$ , with $∣ u (x, t) - u (x, t') ∣ \leq ω_{u} (∣ t - t' ∣)$ , $∣ v (y, s) - v (y, s') ∣ \leq ω_{v} (∣ s - s' ∣)$ , and both $∣ u (x, 0) - u (y, 0) ∣, ∣ v (x, 0) - v (y, 0) ∣ \leq ω_{0} (∣ x - y ∣)$ (Uniform continuity of a map of metric spaces: one $δ$ serving every point).

[F3]

A nonempty subset of $R^{n}$ is compact exactly when it is closed and bounded, and a continuous real-valued function on such a set attains its maximum and minimum (For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent).

[F4]

A continuous function on a compact metric space is uniformly continuous there (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous).

[F5]

Every upper semicontinuous real-valued function on a nonempty compact subset of $R^{n}$ is bounded above and attains a maximum (Semicontinuous extreme value theorem on compact Euclidean sets).

Proof

technique · strict time penalties, a bounded-gradient radial spatial penalty with a vanishing coercive weight, and control of the initial faces by the ordered traces and their moduli

1.1F1algebra

Penalisation. Put ${\tilde{u}}_{η} (x, t) = u (x, t) - η / (T - t)$ and ${\tilde{v}}_{η^{'}} (y, s) = v (y, s) + η^{'} / (T - s)$ for $t, s < T$ . If $ϕ$ is a $C^{1}$ upper test for ${\tilde{u}}_{η}$ at an interior point $z_{0}$ , then $ϕ + η / (T - t)$ is a $C^{1}$ upper test for $u$ , so $u_{t}$ evaluation and [F1] give $ϕ_{t} (z_{0}) + η / (T - t_{0})^{2} + H (D ϕ (z_{0})) \leq 0$ , that is $ϕ_{t} + H (D ϕ) \leq - η / (T - t_{0})^{2} \leq - η / T^{2}$ ; dually every $C^{1}$ lower test $ϕ$ for ${\tilde{v}}_{η^{'}}$ satisfies $ϕ_{t} + H (D ϕ) \geq η^{'} / (T - s_{0})^{2} \geq η^{'} / T^{2}$ . Moreover ${\tilde{u}}_{η} (x, t) \leq \sup u - η / (T - t) \to - \infty$ and ${\tilde{v}}_{η^{'}} (y, s) \geq \inf v + η^{'} / (T - s) \to + \infty$ as $t ↑ T$ , respectively $s ↑ T$ , uniformly in the space variable.

1.2assume-contraF2F3F4F5algebra

The doubling function and the initial-face bound. Assume for contradiction that $u (x_{0}, t_{0}) > v (x_{0}, t_{0})$ for some $0 < t_{0} < T$ and put $δ : = \frac{1}{4} (u (x_{0}, t_{0}) - v (x_{0}, t_{0})) > 0$ . Fix $η, η^{'} > 0$ with ${\tilde{u}}_{η} (x_{0}, t_{0}) - {\tilde{v}}_{η^{'}} (x_{0}, t_{0}) > 3 δ$ and put $B : = η + η^{'} > 0$ . Choose $a > 0$ so small that $ω_{0} (a) < B / (8 T)$ , and then $P > 0$ with $P a > \sup_{r \geq 0} ω_{0} (r)$ ; this gives $\sup_{r \geq 0} (ω_{0} (r) - P r) < B / (8 T)$ by splitting at $a$ . then $ε > 0$ with $P ε < B / (8 T)$ , and $ψ (r) : = P (\sqrt{r^{2} + ε^{2}} - ε)$ , so that $∣ P (x - y) / \sqrt{∣ x - y ∣^{2} + ε^{2}} ∣ \leq P$ , $ψ (r) \geq P r - P ε$ and $ψ (0) = 0$ ; put $χ (x) : = \sqrt{1 + ∣ x ∣^{2}}$ . Choose $ρ > 0$ with $ρ \leq 1$ , $2 ρ χ (x_{0}) < δ$ , and such that $∣ H (p) - H (q) ∣ < B / T^{2}$ whenever $∣ p - q ∣ \leq 2 ρ$ and $∣ p ∣, ∣ q ∣ \leq P + 1$ : the last requirement is possible because $H$ is uniformly continuous on the compact set ${∣ p ∣ \leq P + 1}$ by [F3] and [F4]. For $α > 0$ define $Φ_{α} (x, y, t, s) : = {\tilde{u}}_{η} (x, t) - {\tilde{v}}_{η^{'}} (y, s) - ψ (∣ x - y ∣) - \frac{α}{2} ∣ t - s ∣^{2} - ρ (χ (x) + χ (y))$ for $0 \leq t, s < T$ , and set $Φ_{α} : = - \infty$ if $t = T$ or $s = T$ . At the diagonal point $(x_{0}, x_{0}, t_{0}, t_{0})$ we have $Φ_{α} = {\tilde{u}}_{η} (x_{0}, t_{0}) - {\tilde{v}}_{η^{'}} (x_{0}, t_{0}) - 2 ρ χ (x_{0}) > 3 δ - δ = 2 δ$ for every $α$ , while $Φ_{α} \leq \sup u - \inf v < \infty$ . The penalty $- ρ (χ (x) + χ (y))$ makes the superlevel set $S_{α} : = {Φ_{α} \geq 2 δ}$ bounded, and $S_{α}$ is closed because $Φ_{α}$ is upper semicontinuous (it is continuous where $t, s < T$ , and tends to $- \infty$ at the terminal faces, where it is $- \infty$ ) and $2 δ > - \infty$ ; by [F3] $S_{α}$ is compact and it is nonempty by the diagonal estimate. On $S_{α}$ the function $Φ_{α}$ is real-valued, and it is upper semicontinuous as a restriction of an upper semicontinuous function, so it attains on $S_{α}$ a maximum $M_{α} \geq 2 δ$ by [F5], and by [F3] the value $M_{α}$ is finite; a maximum on $S_{α}$ is a global maximum of $Φ_{α}$ because every point outside $S_{α}$ has value $< 2 δ \leq M_{α}$ , and every maximiser lies in $S_{α}$ , hence has $t_{α}, s_{α} < T$ . So for every $α$ there is a maximiser $(x_{α}, y_{α}, t_{α}, s_{α})$ with $Φ_{α} (x_{α}, y_{α}, t_{α}, s_{α}) \geq 2 δ$ and $t_{α}, s_{α} < T$ .

2.1step 1.2F2algebra

Initial faces are excluded for large $α$ . Since ${\tilde{u}}_{η} - {\tilde{v}}_{η^{'}} \leq \sup u - \inf v$ , the inequality $Φ_{α} \geq 2 δ$ gives $\frac{α}{2} ∣ t_{α} - s_{α} ∣^{2} \leq \sup u - \inf v - 2 δ$ , so $∣ t_{α} - s_{α} ∣ \to 0$ as $α \to \infty$ . If $t_{α} = 0$ , then using $u (x_{α}, 0) \leq v (x_{α}, 0)$ , the initial modulus $ω_{0}$ , the time modulus $ω_{v}$ and $ψ (r) \geq P r - P ε$ we get $Φ_{α} \leq ω_{0} (∣ x_{α} - y_{α} ∣) - P ∣ x_{α} - y_{α} ∣ + P ε + ω_{v} (s_{α}) - \frac{B}{T} \leq \frac{B}{8 T} + \frac{B}{8 T} - \frac{B}{T} + ω_{v} (s_{α}) < 0$ for all large $α$ , because $s_{α} \to 0$ by $∣ t_{α} - s_{α} ∣ \to 0$ ; this contradicts $Φ_{α} \geq 2 δ$ . The case $s_{α} = 0$ is identical with $ω_{u}$ in place of $ω_{v}$ . Hence for all sufficiently large $α$ every maximiser has $0 < t_{α}, s_{α} < T$ .

3.1step 1.1step 1.2step 2.1algebradischarge-contradiction∎

Contact inequalities and the contradiction. Fix $α$ large enough that step 2.1 applies and $∣ t_{α} - s_{α} ∣ < T$ . At the maximiser, fixing $(y_{α}, s_{α})$ shows that $ϕ^{U} (x, t) : = ψ (∣ x - y_{α} ∣) + ρ χ (x) + \frac{α}{2} ∣ t - s_{α} ∣^{2}$ is a $C^{1}$ upper test for ${\tilde{u}}_{η}$ at $(x_{α}, t_{α})$ , and fixing $(x_{α}, t_{α})$ shows that $ϕ^{V} (y, s) : = - ψ (∣ x_{α} - y ∣) - ρ χ (y) - \frac{α}{2} ∣ t_{α} - s ∣^{2}$ is a $C^{1}$ lower test for ${\tilde{v}}_{η^{'}}$ at $(y_{α}, s_{α})$ . Their derivatives are $p : = \frac{P (x_{α} - y_{α})}{\sqrt{∣ x_{α} - y_{α} ∣^{2} + ε^{2}}} + \frac{ρ x_{α}}{χ (x_{α})}$ , $q : = \frac{P (x_{α} - y_{α})}{\sqrt{∣ x_{α} - y_{α} ∣^{2} + ε^{2}}} - \frac{ρ y_{α}}{χ (y_{α})}$ and $a : = α (t_{α} - s_{α})$ , with $∣ p ∣, ∣ q ∣ \leq P + ρ \leq P + 1$ and $∣ p - q ∣ = ρ ∣ \frac{x_{α}}{χ (x_{α})} + \frac{y_{α}}{χ (y_{α})} ∣ \leq 2 ρ$ because $∣ z ∣ / χ (z) \leq 1$ for every $z$ . Step 1.1 applied to the two tests gives $a + H (p) \leq - η / T^{2}$ and $a + H (q) \geq η^{'} / T^{2}$ , hence $B / T^{2} = \frac{η + η^{'}}{T^{2}} \leq H (q) - H (p)$ . But $∣ p - q ∣ \leq 2 ρ$ and $∣ p ∣, ∣ q ∣ \leq P + 1$ , so the choice of $ρ$ in step 1.2 gives $∣ H (q) - H (p) ∣ < B / T^{2}$ , a contradiction. Therefore no point with $u > v$ exists in $Z$ , that is $u \leq v$ on $Z$ .

Remarks

Why the radial penalty has bounded gradient. With $ψ (r) = P (\sqrt{r^{2} + ε^{2}} - ε)$ one has $0 \leq ψ^{'} (r) < P$ , so the spatial doubling contributes gradients of modulus at most $P$ and the difference $p - q$ contains exactly the term $ρ (\cdot)$ of the weight. The vanishing of $ε$ is not used as a limit: the estimates hold for a fixed positive $ε$ .
Role of each face. The time penalties give the strict margin $B / T^{2}$ and remove the terminal faces; the weight $ρ (χ (x) + χ (y))$ makes the superlevel sets compact; the initial faces are handled by the pointwise order of the traces and their moduli, so no value-function or semijet machinery beyond the stated hypotheses is needed. The Hilbert-space semijet theorem is not required, which is why $H (p) = ∣ p ∣^{2} / 2$ is covered.

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The Hopf--Lax formula solves the Hamilton--Jacobi Cauchy problem

Statement

Let $H : R^{n} \to R$ be finite-valued, continuous, convex, and superlinear, with Legendre transform $L$ , and let $u_{0} : R^{n} \to R$ be bounded and uniformly continuous. Define $u (x, t) : = Q_{t} u_{0} (x)$ by The Hopf--Lax operator and the Hopf--Lax formula. Then $u$ is a bounded uniformly continuous function on $R^{n} \times [0, T]$ for every $T > 0$ , and: (1) $u$ is a viscosity solution of $u_{t} + H (D u) = 0$ in $R^{n} \times (0, \infty)$ ; (2) $u$ attains the initial datum locally uniformly, $\sup_{x \in K} ∣ u (x, t) - u_{0} (x) ∣ ⟶ 0 (t ↓ 0)$ for every compact $K \subseteq R^{n}$ ; (3) $u$ is the unique bounded uniformly continuous viscosity solution of the Cauchy problem with datum $u_{0}$ ; (4) $u$ satisfies the dynamic-programming relation $Q_{t + s} u_{0} = Q_{t} Q_{s} u_{0}$ of The Hopf--Lax operators form a semigroup (dynamic programming). No choice principle is used.

Facts & Assumptions

Given: A finite continuous convex superlinear $H$ with Legendre transform $L$ , a bounded uniformly continuous datum $u_{0}$ with bounded modulus $ω$ (replace any given modulus by its minimum with $2 ∥ u_{0} ∥_{\infty}$ ), and $u = Q_{t} u_{0}$ .

[F1]

For $t > 0$ , $Q_{t} u_{0} (x) = \inf_{y} {u_{0} (y) + t L ((x - y) / t)}$ , the infimum being attained, and $L$ is real-valued on $R^{n}$ (The Hopf--Lax operator and the Hopf--Lax formula, Finiteness, superlinearity of the Lagrangian and localisation of Hopf--Lax near-minimisers).

[F2]

$Q_{t + s} u_{0} = Q_{t} (Q_{s} u_{0})$ for all $t, s \geq 0$ (The Hopf--Lax operators form a semigroup (dynamic programming)).

[F3]

$H = L^{*}$ , that is $H (p) = \sup_{v} (p \cdot v - L (v))$ (A finite-valued convex Hamiltonian equals its biconjugate).

[F4]

$Q_{t} u_{0}$ has the modulus $ω$ of $u_{0}$ , and $∣ Q_{t} u_{0} (x) - Q_{t} v_{0} (x) ∣ \leq ∥ u_{0} - v_{0} ∥_{\infty}$ (The Hopf--Lax operator preserves a modulus of continuity, The Hopf--Lax operator is a contraction in the supremum norm).

[F5]

Comparison for autonomous convex superlinear Hamiltonians: bounded uniformly continuous subsolutions and supersolutions of $u_{t} + H (D u) = 0$ on $R^{n} \times [0, T]$ with ordered continuous initial traces satisfy the comparison inequality (Comparison for autonomous convex superlinear Hamiltonians).

Proof

technique · the dynamic-programming inequality in both directions, the biconjugacy $H=L^*$, and comparison for uniqueness

1.1F1F2F4algebra

Boundedness and uniform continuity. For every $x$ and $t \geq 0$ we have $\inf u_{0} - t H (0) \leq u (x, t) \leq ∥ u_{0} ∥_{\infty} + t L (0)$ : the upper bound is the competitor $y = x$ in [F1], and the lower bound follows from $L \geq - H (0)$ . By [F4] the map $x \mapsto u (x, t)$ is uniformly continuous with modulus $ω$ uniformly in $t$ . For $t \geq s \geq 0$ , [F2] and [F4] give $∣ u (x, t) - u (x, s) ∣ \leq ∥ Q_{t - s} u_{0} - u_{0} ∥_{\infty}$ ; and $∥ Q_{τ} u_{0} - u_{0} ∥_{\infty} \to 0$ as $τ ↓ 0$ , since $Q_{τ} u_{0} \leq u_{0} + τ L (0)$ and $Q_{τ} u_{0} (x) \geq u_{0} (x) - \sup_{v} {ω (τ ∣ v ∣) - τ L (v)}$ , the supremum tending to $0$ as follows. For $a > 0$ , set $M = ∥ u_{0} ∥_{\infty}$ and choose $A = (2 M + 1) / a$ . Superlinearity and continuity of $L$ give $b \geq 0$ with $L (v) \geq A ∣ v ∣ - b$ everywhere. If $τ ∣ v ∣ \geq a$ , then $ω (τ ∣ v ∣) - τ L (v) \leq - 1 + b τ$ ; if $τ ∣ v ∣ < a$ , then $L (v) \geq - H (0)$ gives the bound $ω (a) + τ H (0)$ . Thus the limsup of the supremum is at most $ω (a)$ , which tends to zero as $a ↓ 0$ ; its liminf is at least zero by the competitor $v = 0$ and $ω (0) = 0$ . Hence $u$ is bounded and uniformly continuous on each strip $R^{n} \times [0, T]$ .

1.2F2F3algebra

The subsolution inequality. Let $ϕ \in C^{1}$ and let $u - ϕ$ have a strict local maximum at $(x_{0}, t_{0})$ with $t_{0} > 0$ . Fix $v \in R^{n}$ and small $h > 0$ , put $y : = x_{0} - h v$ , and use the dynamic-programming identity $u (x_{0}, t_{0}) = Q_{h} (Q_{t_{0} - h} u_{0}) (x_{0}) \leq u (y, t_{0} - h) + h L (v)$ , the inequality coming from the competitor $y$ in the infimum defining $Q_{h}$ . The contact inequality at $(x_{0}, t_{0})$ gives $u (y, t_{0} - h) \leq u (x_{0}, t_{0}) - ϕ (x_{0}, t_{0}) + ϕ (y, t_{0} - h)$ , and combining the two gives $ϕ (x_{0}, t_{0}) - ϕ (x_{0} - h v, t_{0} - h) \leq h L (v)$ . Dividing by $h$ and letting $h ↓ 0$ yields $ϕ_{t} (x_{0}, t_{0}) + ⟨ D ϕ (x_{0}, t_{0}), v ⟩ \leq L (v)$ for every $v$ ; taking the supremum over $v$ and using $H = L^{*}$ of [F3] gives $ϕ_{t} (x_{0}, t_{0}) + H (D ϕ (x_{0}, t_{0})) \leq 0$ . Non-strict maxima are handled by strictification, so $u$ is a viscosity subsolution.

2.1F1F2F3step 1.2algebra

The supersolution inequality. Let $u - ϕ$ have a strict local minimum at $(x_{0}, t_{0})$ with $t_{0} > 0$ . By [F1] there is a minimiser $y$ of $u_{0} (y) + t_{0} L ((x_{0} - y) / t_{0})$ ; put $v : = (x_{0} - y) / t_{0}$ , so that $u (x_{0}, t_{0}) = u_{0} (y) + t_{0} L (v)$ . For $0 < h < t_{0}$ put $z_{h} : = y + \frac{t_{0} - h}{t_{0}} (x_{0} - y)$ , so that $(z_{h} - y) / (t_{0} - h) = v$ and $z_{h} \to x_{0}$ . The dynamic-programming identity at $(z_{h}, t_{0} - h)$ with the competitor $y$ gives $u (z_{h}, t_{0} - h) \leq u_{0} (y) + (t_{0} - h) L (v) = u (x_{0}, t_{0}) - h L (v)$ , hence $u (x_{0}, t_{0}) - u (z_{h}, t_{0} - h) \geq h L (v)$ . The contact inequality at the local minimum gives $u (z_{h}, t_{0} - h) \geq u (x_{0}, t_{0}) - ϕ (x_{0}, t_{0}) + ϕ (z_{h}, t_{0} - h)$ ; combining, $ϕ (x_{0}, t_{0}) - ϕ (z_{h}, t_{0} - h) \geq h L (v)$ . Dividing by $h$ and letting $h ↓ 0$ along $z_{h} \to x_{0}$ gives $ϕ_{t} (x_{0}, t_{0}) + ⟨ D ϕ (x_{0}, t_{0}), v ⟩ \geq L (v)$ , and since $H (D ϕ) = \sup_{w} (⟨ D ϕ, w ⟩ - L (w)) \geq ⟨ D ϕ, v ⟩ - L (v)$ by [F3], we get $ϕ_{t} + H (D ϕ) \geq 0$ . Hence $u$ is a viscosity supersolution, and with step 1.2 it is a viscosity solution of $u_{t} + H (D u) = 0$ .

2.2F1step 1.1algebra

The initial trace. For $τ > 0$ and every $x$ , $Q_{τ} u_{0} (x) \leq u_{0} (x) + τ L (0)$ and $Q_{τ} u_{0} (x) \geq u_{0} (x) - \sup_{v} {ω (τ ∣ v ∣) - τ L (v)}$ , and the latter supremum tends to $0$ by the bounded-modulus estimate in step 1.1; hence $\sup_{R^{n}} ∣ Q_{τ} u_{0} - u_{0} ∣ \to 0$ , which is the stated locally uniform (indeed uniform) attainment of the initial datum.

3.1step 1.2step 2.1F2F5∎

Uniqueness and the semigroup. Any bounded uniformly continuous viscosity solution of the Cauchy problem with datum $u_{0}$ is comparable with $u$ by [F5], in both orders, because both are bounded uniformly continuous and have the same continuous initial trace; hence $u$ is the unique such solution. Property (4) is [F2].

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Vanishing viscosity selects the viscosity solution

Statement

Let $n \geq 1$ , $T > 0$ , $Z = R^{n} \times (0, T)$ , and $Z_{0} = R^{n} \times [0, T)$ . Let $H : R^{n} \times [0, T] \times R^{n} \to R$ satisfy the Lipschitz conditions of part (a) of Comparison for first-order Hamilton--Jacobi equations, let $u_{0} : R^{n} \to R$ be bounded and uniformly continuous, and let $u_{0}^{(ε)} \to u_{0}$ locally uniformly on $R^{n}$ . For each $ε \in (0, 1)$ let $u^{ε} : Z \to R$ be a viscosity solution of $u_{t}^{ε} + H (x, t, D u^{ε}) = ε Δ u^{ε} in Z,$ meaning that for every test function $ϕ \in C^{1, 2} (Z)$ the residual $ϕ_{t} + H (z, D ϕ) - ε Δ ϕ$ is nonpositive at each local maximum of $u^{ε} - ϕ$ and nonnegative at each local minimum. Assume that $u^{ε}$ has initial datum $u_{0}^{(ε)}$ in the relaxed Cauchy sense. Suppose the family is uniformly bounded on $Z$ and locally equicontinuous up to the initial face: there is $M < \infty$ with $∣ u^{ε} (z) ∣ \leq M$ for every $ε \in (0, 1)$ and $z \in Z$ , and for every compact $K \subseteq Z_{0}$ and every $η > 0$ there is $δ > 0$ such that $∣ u^{ε} (z) - u^{ε} (z^{'}) ∣ < η$ for all $ε \in (0, 1)$ and $z, z^{'} \in K \cap Z$ with $∣ z - z^{'} ∣ < δ$ . These estimates give each $u^{ε}$ a continuous trace on the initial face. Then $u^{ε} \to u$ locally uniformly on $Z$ , where $u$ is the unique bounded viscosity solution of $u_{t} + H (x, t, D u) = 0$ with datum $u_{0}$ . Neither existence of the approximants nor a compactness theorem is asserted: the boundedness and equicontinuity estimates are hypotheses. No choice principle is used.

Facts & Assumptions

Given: The Hamiltonian $H$ with the Lipschitz conditions of comparison case (a), bounded uniformly continuous $u_{0}$ , data $u_{0}^{(ε)} \to u_{0}$ locally uniformly, a uniformly bounded family $(u^{ε})$ of viscous solutions, locally equicontinuous up to the initial face, with data $u_{0}^{(ε)}$ in the relaxed sense, and the half-relaxed limits $\overline{u}, \underline{u}$ of the family (Half-relaxed limits of a locally bounded family).

[F1]

For every fixed $ϕ \in C^{1, 2} (Z)$ , at each local maximum of $u^{ε} - ϕ$ one has $ϕ_{t} + H (z, D ϕ) \leq ε Δ ϕ (z)$ , and at each local minimum $ϕ_{t} + H (z, D ϕ) \geq ε Δ ϕ (z)$ . Thus the errors are bounded in absolute value by $c_{ε} (z) : = ε ∣ Δ ϕ (z) ∣$ , which is locally bounded and tends to $0$ locally uniformly by the explicitly assumed second-order test inequalities in the statement; the limit equation is tested in the first-order sense of Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem.

[F2]

The case-(a) finite-cover argument of Half-relaxed limits of sub- and supersolutions with vanishing perturbations proves the subsolution inequality at a strict contact from the inequality for that fixed smooth test and a locally uniformly vanishing error; the dual argument proves the supersolution inequality. The same item proves passage of the relaxed initial datum under local equicontinuity up to that face.

[F3]

Comparison, case (a), applies to the bounded upper semicontinuous subsolution $\overline{u}$ and the bounded lower semicontinuous supersolution $\underline{u}$ when their relaxed initial data agree (Comparison for first-order Hamilton--Jacobi equations); uniqueness in the bounded class is Uniqueness and sup-norm contraction for the Cauchy problem.

[F4]

A compact subset of $Z$ has a finite subcover from every intrinsic open cover (Open cover, subcover, compact metric space, and compact subset of a metric space); every ambient indexed open-ball cover of it also has a finite subcover retaining the indices (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it, clauses 2--3). Upper semicontinuous real-valued functions attain maxima on nonempty compact Euclidean sets (Semicontinuous extreme value theorem on compact Euclidean sets).

[F5]

A nonnegative smooth compactly supported bump equal to $1$ on a smaller ball is supplied by A smooth bump between concentric Euclidean balls. Its integral is finite and positive, so normalization gives a unit-mass bump and the scaled family of The mollifier family generated by a unit-mass smooth bump. Here only compact Riemann integrals are needed: continuous integrands on compact boxes are integrable (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set), and monotonicity and linearity give $∣ \int_{Q} g ∣ \leq vol (Q) \sup_{Q} ∣ g ∣$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ). For a $C^{1}$ function $f$ near a compact ball, first multiply by such a smooth cutoff equal to $1$ on a slightly larger ball and extend by zero, obtaining a globally $C^{1}$ compactly supported function. Its convolution with the fixed bump is smooth: on a fixed integration box every kernel-derivative difference quotient converges uniformly, by the mean value theorem and uniform continuity of the next derivative, so the integral bound passes each derivative through the integral. For the affine changes $y = x - ρ z$ on a compact integration box, Change of variables for an injective $C^{1}$ map on a compact Jordan set applies: the derivative is the invertible matrix $- ρ I$ and the absolute determinant is $ρ^{n + 1}$ . Thus, using the fixed-kernel formula $\int f (x - ρ y) η (y) d y$ gives first derivatives by the same uniform difference-quotient argument. Unit mass then bounds the errors in $f$ and $D_{i} f$ by their moduli of continuity at distance $ρ R_{η}$ , which tend to zero by Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous. These compact-integral arguments use no choice (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

Proof

technique · half-relaxed limits, stability with a vanishing perturbation, comparison, and a compactness-free conversion to local uniform convergence

1.1F1F2F4F5

The relaxed limits are sub- and supersolutions with common initial data. First take a smooth strict upper test $ψ$ for $\overline{u}$ at an interior point. The case-(a) compact finite-cover proof in [F2] applies using the fixed smooth test $ψ + ∣ z - z_{0} ∣^{2}$ at the approximating contacts: its viscosity error is bounded by $ε ∣ Δ_{x} (ψ + ∣ z - z_{0} ∣^{2}) ∣$ , which tends uniformly to zero on the compact contact region, so the half-relaxed limit satisfies $ψ_{t} + H (z, D ψ) \leq 0$ . The dual argument gives the lower-limit supersolution inequality for smooth strict lower tests. To extend these inequalities to an arbitrary $C^{1}$ test $ϕ$ , strictify its contact by adding or subtracting a quartic (Strictification of a viscosity test function by a quartic perturbation). On a closed ball around the contact, convolve $ϕ$ locally with a fixed compactly supported smooth unit-mass bump at scales tending to zero; uniform continuity of $ϕ$ and $D ϕ$ on that ball gives smooth approximants converging in $C^{1}$ . Maximise $\overline{u} - ϕ_{j}$ on the ball for each approximant. The strict contact and uniform convergence imply that the sets of such maximisers are interior for large $j$ and their distance to the original contact tends uniformly to zero. Strictify each smooth test at its maximiser by a quartic and apply the fixed-test argument above. Passing to the limit using $C^{1}$ convergence and continuity of $H$ proves the required inequality for $ϕ$ ; the lower-test argument is dual. Thus $\overline{u}$ is a subsolution and $\underline{u}$ a supersolution for the full $C^{1}$ test definition. Finally, local equicontinuity gives each approximant a continuous initial trace. Its relaxed initial condition makes that trace equal to $u_{0}^{(ε)}$ ; local uniform convergence of these data and the shared boundary modulus then pass the initial trace to both half-relaxed limits.

2.1step 1.1F3

Comparison forces the two limits to agree. The subsolution $\overline{u}$ is bounded and upper semicontinuous and the supersolution $\underline{u}$ is bounded and lower semicontinuous, with the same relaxed initial data $u_{0}$ ; comparison [F3] gives $\overline{u} \leq \underline{u}$ on $Z$ . Since $\underline{u} \leq \overline{u}$ pointwise by the definition of the half-relaxed limits, the two coincide: $\overline{u} = \underline{u} = : u$ , which is therefore continuous; by [F3] it is the unique bounded viscosity solution with datum $u_{0}$ .

3.1step 2.1givenF4algebra∎

Locally uniform convergence. Let $K \subseteq Z$ be compact and $δ > 0$ . For each $z \in K$ , the equalities $\overline{u} (z) = \underline{u} (z) = u (z)$ and the definition of the joint half-relaxed limits in Given give a radius $r_{z} > 0$ such that $∣ u^{ε} (y) - u (z) ∣ < δ / 3$ whenever $0 < ε < r_{z}$ and $y \in Z$ satisfies $∣ y - z ∣ < r_{z}$ . Shrink the radius, if necessary, so that also $∣ u (y) - u (z) ∣ < δ / 3$ there. The family of all such admissible balls covers $K$ ; by [F4] take a finite subcover $B (z_{i}, r_{i})$ and put $ε_{0} : = \min_{i} r_{i} > 0$ . For every $0 < ε < ε_{0}$ and $y \in K$ , one of these balls contains $y$ , so $∣ u^{ε} (y) - u (y) ∣ < 2 δ / 3 < δ$ . This proves uniform convergence on $K$ without selecting a sequence of parameters or points.

Remarks

What is not asserted. No existence of the viscous family is proved and no subsequence is extracted from the family itself; the boundedness and local equicontinuity are hypotheses. The pointwise equality of the two relaxed limits is equivalent to local uniform convergence of the family, which is the content of step 3.1.
Choice. The half-relaxed limits are computed as infima and suprema over sets; the comparison and uniqueness steps are choice-free, and the final conversion uses a finite cover of each compact set.

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Finite speed of dependence for Hamiltonians Lipschitz in momentum

Statement

Let $n \geq 1$ , $T > 0$ , and let $H : R^{n} \times [0, T] \times R^{n} \to R$ be continuous. Suppose there are constants $C > 0$ and $L > 0$ such that, for all $x, y, p, q \in R^{n}$ and $t, s \in [0, T]$ , $∣ H (x, t, p) - H (y, s, p) ∣ \leq C (1 + ∣ p ∣) (∣ x - y ∣ + ∣ t - s ∣), ∣ H (x, t, p) - H (x, t, q) ∣ \leq L ∣ p - q ∣ .$ Let $u, v : R^{n} \times [0, T) \to R$ be bounded, with $u$ upper semicontinuous and $v$ lower semicontinuous; assume that $u$ is a viscosity subsolution and $v$ a viscosity supersolution of $u_{t} + H (x, t, D u) = 0$ on $R^{n} \times (0, T)$ . Fix $x_{0} \in R^{n}$ and $R > 0$ . If $u (x, 0) \leq v (x, 0)$ for every $x \in B (x_{0}, R)$ , then $u (x, t) \leq v (x, t) for 0 \leq t < \min {T, R / L} and ∣ x - x_{0} ∣ < R - L t .$ In particular, if $u$ and $v$ are bounded viscosity solutions with the same initial values on $B (x_{0}, R)$ and are also respectively lower and upper semicontinuous on $R^{n} \times [0, T)$ (so both are continuous there), then $u (x, t) = v (x, t)$ on this open backward cone. The cone is stated with strict spatial inequality because $B (x_{0}, R)$ is open and no continuity of the initial traces is assumed. No choice principle is used.

Facts & Assumptions

Given: Continuous $H$ with the two Lipschitz conditions, bounded $u, v$ on $R^{n} \times [0, T)$ with $u$ upper semicontinuous and $v$ lower semicontinuous, $u$ a subsolution and $v$ a supersolution on $R^{n} \times (0, T)$ , and $u (x, 0) \leq v (x, 0)$ for $∣ x - x_{0} ∣ < R$ .

[F1]

At every $C^{1}$ local maximum of $u - ϕ$ : $ϕ_{t} + H (x, t, D ϕ) \leq 0$ ; at every $C^{1}$ local minimum of $v - ϕ$ : $ϕ_{t} + H (x, t, D ϕ) \geq 0$ (Viscosity subsolutions and supersolutions of a first-order equation and of the Cauchy problem).

[F2]

The difference $u - v$ is upper semicontinuous, since $u$ is upper semicontinuous and $- v$ is upper semicontinuous; adding continuous penalty terms preserves upper semicontinuity (Upper and lower semicontinuity on subsets of $R^{n}$ ).

[F3]

Closed bounded subsets of finite-dimensional Euclidean space are compact, upper semicontinuous real-valued functions attain their maxima on nonempty compact sets, and continuous functions attain their minima there (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, For a nonempty subset of $R^{n}$ with $n \geq 1$ , compactness, closedness and boundedness, pseudocompactness, and attainment of extrema by every continuous real-valued function are equivalent, Semicontinuous extreme value theorem on compact Euclidean sets). In particular, two disjoint compact sets in Euclidean space have positive distance.

[F4]

Comparison case (a) applies to the Hamiltonian $G (x, t, p) = - L ∣ p ∣$ , which has $∣ G (x, t, p) - G (y, s, p) ∣ = 0$ and $∣ G (x, t, p) - G (x, t, q) ∣ \leq L ∣ p - q ∣$ : a bounded upper semicontinuous subsolution and a bounded lower semicontinuous supersolution on the closed slab with ordered initial traces satisfy the comparison inequality (Comparison for first-order Hamilton--Jacobi equations).

Proof

technique · reduce the difference by a localised two-time doubling argument, then compare with a smooth radial cone barrier

1.1F1F2F3algebra

Reduction. Put $w : = u - v$ , which is bounded and upper semicontinuous. Let $ϕ \in C^{1}$ and suppose $w - ϕ$ has a strict local maximum at $z_{*} = (x_{*}, t_{*}) \in Z$ . Choose $T_{*}$ with $t_{*} < T_{*} < T$ and a compact cylinder $Q ⋐ R^{n} \times (0, T_{*})$ around $z_{*}$ on which the maximum is strict. Write $z = (x, t)$ and $z^{'} = (y, s)$ . For fixed $ρ > 0$ set $G_{ρ} (z, z^{'}) : = u (x, t) - v (y, s) - ϕ (x, t) - ρ (∣ x - x_{*} ∣^{2} + ∣ y - x_{*} ∣^{2})$ and $F_{α} (z, z^{'}) : = G_{ρ} (z, z^{'}) - \frac{α}{2} ∣ z - z^{'} ∣^{2}$ . By [F2]--[F3], $F_{α}$ attains a finite maximum $M_{α}$ on $Q \times Q$ , and $M_{α} \geq m_{*} : = (w - ϕ) (z_{*})$ by evaluation at $(z_{*}, z_{*})$ . The values $M_{α}$ decrease with $α$ and are bounded below by $m_{*}$ , so they converge. For every maximiser $(z, z^{'})$ , comparison with $F_{α / 2}$ at that same point gives $α ∣ z - z^{'} ∣^{2} \leq 4 (M_{α / 2} - M_{α}) \to 0$ , uniformly over the maximiser sets; in particular $∣ z - z^{'} ∣ \to 0$ uniformly. Fix any sufficiently small open neighbourhood $V$ of $z_{*}$ with closure in the interior of $Q$ , and put $A : = Q ∖ V$ . Strictness and [F2]--[F3] give $m_{A} : = \max_{z \in A} (w - ϕ) (z) < m_{*}$ . Let $γ : = (m_{*} - m_{A}) / 2 > 0$ and $Δ_{A} : = {(z, z) : z \in A}$ . On $Δ_{A}$ , $G_{ρ} (z, z) \leq m_{A} = m_{*} - 2 γ$ . The compact superlevel set $C : = {(z, z^{'}) \in A \times Q : G_{ρ} (z, z^{'}) \geq m_{*} - γ}$ is disjoint from $Δ_{A}$ . If $C$ is nonempty, [F3] gives a positive distance $d_{*} > 0$ between these compact sets; if it is empty, choose any $d_{*} > 0$ . Thus $G_{ρ} (z, z^{'}) < m_{*} - γ$ whenever $z \in A$ and $∣ z - z^{'} ∣ < d_{*}$ . For all sufficiently large $α$ , every maximiser has $∣ z - z^{'} ∣ < d_{*}$ , so its first slot cannot lie in $A$ , since its value is at least $m_{*}$ . As $V$ was arbitrary, all first slots converge uniformly to $z_{*}$ ; the second slots do also by the diagonal estimate. At each maximiser, fixing one slot gives $C^{1}$ upper and lower contacts for $u$ and $v$ with spatial gradients $p_{α} : = D ϕ (z) + α (x - y) + 2 ρ (x - x_{*})$ and $q_{α} : = α (x - y) - 2 ρ (y - x_{*})$ , and time derivatives $ϕ_{t} (z) + α (t - s)$ and $α (t - s)$ . By [F1] and the two Lipschitz bounds, $ϕ_{t} (z) \leq H (y, s, q_{α}) - H (x, t, p_{α}) \leq L ∣ q_{α} - p_{α} ∣ + 2 C (1 + ∣ p_{α} ∣) ∣ z - z^{'} ∣$ . Since $∣ z - z^{'} ∣ \to 0$ , $α ∣ z - z^{'} ∣^{2} \to 0$ , and $D ϕ$ is bounded on $Q$ , the last error tends to zero uniformly over maximisers. Also $q_{α} - p_{α} = - D ϕ (z) - 2 ρ ((x - x_{*}) + (y - x_{*})) \to - D ϕ (z_{*})$ uniformly for fixed $ρ$ . Passing to these uniform limits gives $ϕ_{t} (z_{*}) - L ∣ D ϕ (z_{*}) ∣ \leq 0$ . The non-strict case follows by Strictification of a viscosity test function by a quartic perturbation; hence $w$ is a viscosity subsolution of $w_{t} - L ∣ D w ∣ = 0$ in $Z$ .

1.2F3algebra

The cone barrier. Let $M : = \max {0, \sup_{R^{n} \times [0, T)} w}$ , and let $h : R \to R$ be the explicit nondecreasing $C^{1}$ cutoff with $h = 0$ on $(- \infty, 0]$ , $h (s) = 3 s^{2} - 2 s^{3}$ for $s \in [0, 1]$ and $h = 1$ on $[1, \infty)$ . For $0 < ε < R$ and $δ > 0$ put $ξ_{ε} (r) : = M h ((r - (R - ε)) / ε)$ and $ψ_{ε, δ} (x, t) : = ξ_{ε} (\sqrt{∣ x - x_{0} ∣^{2} + δ^{2}} + L t)$ . Then $ψ_{ε, δ}$ is $C^{1}$ , bounded and nonnegative, and it is a classical supersolution of $ψ_{t} - L ∣ D ψ ∣ = 0$ on $R^{n} \times (0, T)$ : indeed $∣ D ψ ∣ = ξ_{ε}^{'} \cdot ∣ x - x_{0} ∣ / \sqrt{∣ x - x_{0} ∣^{2} + δ^{2}}$ and $ψ_{t} = L ξ_{ε}^{'}$ , so $ψ_{t} - L ∣ D ψ ∣ = L ξ_{ε}^{'} (1 - ∣ x - x_{0} ∣ / \sqrt{∣ x - x_{0} ∣^{2} + δ^{2}}) \geq 0$ because $ξ_{ε}^{'} \geq 0$ . At $t = 0$ we have $w (x, 0) \leq ψ_{ε, δ} (x, 0)$ for every $x$ : for $∣ x - x_{0} ∣ < R$ this uses $w (x, 0) \leq 0 \leq ψ_{ε, δ} (x, 0)$ , and for $∣ x - x_{0} ∣ \geq R$ it uses $ψ_{ε, δ} (x, 0) = M \geq w (x, 0)$ (the cutoff argument at $t = 0$ is $(\sqrt{∣ x - x_{0} ∣^{2} + δ^{2}} - (R - ε)) / ε \geq 1$ ).

2.1step 1.1step 1.2F3F4∎

Comparison with the barrier and conclusion. By step 1.1 the difference $w$ is a bounded upper semicontinuous subsolution of $w_{t} + G (x, t, D w) = 0$ and by step 1.2 the barrier is a bounded continuous supersolution of the same equation with ordered initial traces; comparison [F4] gives $w \leq ψ_{ε, δ}$ on $R^{n} \times (0, T)$ . At $t = 0$ the desired inequality is the assumed initial order. Now fix $0 < t < \min {T, R / L}$ and $∣ x - x_{0} ∣ < R - L t$ . Choose $ε > 0$ with $∣ x - x_{0} ∣ + L t < R - ε$ and then $δ > 0$ with $\sqrt{∣ x - x_{0} ∣^{2} + δ^{2}} + L t \leq R - ε$ ; for these parameters the cutoff argument is at most $0$ , so $ψ_{ε, δ} (x, t) = 0$ and comparison gives $w (x, t) \leq 0$ , that is $u (x, t) \leq v (x, t)$ . Under the additional semicontinuity assumptions in the equality clause, $v$ is an upper semicontinuous subsolution and $u$ a lower semicontinuous supersolution on the same half-closed slab. Applying the same conclusion to $(v, u)$ with the initial agreement then gives the reverse inequality and hence equality on the cone.

Remarks

Why the strict cone. The initial agreement is assumed only on the open ball and the initial traces need not be continuous; the barrier is built with $R - ε$ and the limiting argument therefore produces the strict inequality $∣ x - x_{0} ∣ < R - L t$ .
The reduction is not the comparison theorem for $u - v$ directly. The reduction uses the two-sided doubling contacts and the momentum-Lipschitz bound, so the difference satisfies the Hamilton--Jacobi equation with the Hamiltonian $- L ∣ p ∣$ , to which comparison case (a) applies.

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Value functions and the Hamilton--Jacobi--Bellman equation: orientation only

Remarks

For a controlled dynamical system $y^{'} (\cdot) = b (y, v)$ with running cost $L (y, v)$ and initial cost $u_{0}$ , consider the value function $u (x, t) = \inf {\int_{0}^{t} L (γ (s), v (s)) d s + u_{0} (γ (0)) : γ (t) = x, γ^{'} = b (γ, v)}$ over admissible controls. When hypotheses make this value finite and continuous, ensure the dynamic programming principle, and give the viscosity characterization, $u$ is a viscosity solution of the Hamilton--Jacobi--Bellman equation $u_{t} + H (x, D u) = 0$ with $H (x, p) = \sup_{v} {p \cdot b (x, v) - L (x, v)} .$ To express the Legendre duality precisely, define the effective velocity cost $ℓ (x, ξ) : = \inf {L (x, v) : b (x, v) = ξ},$ with value $+ \infty$ when the fiber is empty. Then $H (x, \cdot)$ is the convex conjugate of $ℓ (x, \cdot)$ ; when $ℓ (x, \cdot)$ is proper, lower semicontinuous and convex, Fenchel--Moreau gives $ℓ (x, ξ) = \sup_{p} {p \cdot ξ - H (x, p)}$ (Clason, Theorem 5.1(iii), recorded here as source-only orientation; The Legendre transform of a finite-valued convex Hamiltonian fixes the conjugate notation but does not prove this extended-valued result). If comparison holds in the chosen solution class, the viscosity solution is unique.

The sign convention is the one of this page: velocities are integrated forward from time $0$ to time $t$ , the running cost is accumulated forward, the initial cost is paid at time $0$ , and $H$ is convex in $p$ because it is a supremum of affine functions of $p$ , irrespective of convexity of the control or velocity set. With this convention the Hopf--Lax operator of The Hopf--Lax operator and the Hopf--Lax formula is the special case of the formula in which the infimum over paths has been reduced to a single infimum over the starting point, $u (x, t) = \inf_{y} {u_{0} (y) + t L ((x - y) / t)}$ , for the autonomous convex superlinear case.

This remark records the interpretation only: no admissible-control framework, no measurable selection, no existence of optimal controls and no dynamic programming theorem for control systems is developed on this page, The stationary specialization of the displayed evolution equation is $H (x, D u) = 0$ ; a separately discounted formulation leads to equations such as $λ u + H (x, D u) = f$ . The only dynamic-programming content of this page is the semigroup law for the Hopf--Lax operator The Hopf--Lax operators form a semigroup (dynamic programming), which is proved directly from the convexity of the Lagrangian. No choice principle is used; this orientation is recorded so that the Cauchy problems of The Hamilton--Jacobi Cauchy problem and its classical solutions can be read against their control origin.

5 · Examples, counterexamples and false statements

None yet.