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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.

Depends on

Used by

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Michael G. Crandall, Hitoshi Ishii and Pierre-Louis Lions, User's guide to viscosity solutions of second order partial differential equations, Bulletin of the American Mathematical Society 27 (1992), 1--67 (complete article) (standard reference, not scraped)
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)