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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.

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)