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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]

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).

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.

Depends on

Used by

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Sources

Hung Vinh Tran, Hamilton--Jacobi Equations: Theory and Applications, 2020 preliminary author manuscript of AMS Graduate Studies in Mathematics 213 (complete text) (standard reference, not scraped)