It is said in Evan's Partial Differential Equations that the Hopf-Lax formula $$ u(x,t)=\inf_{y\in{\Bbb R}^n}\left\{tL(\frac{x-y}{t})+g(y)\right\} $$ is actually a minimum where $g$ is assumed to be Lipschitz continuous and the Lagragian $L$ is convex and $$ \lim_{|v|\to\infty}\frac{L(v)}{|v|}=\infty. $$ Question: why is this true?

A quick search for "Hopf-Lax formula" in Google returns the following proof, which looks promising:

enter image description here

But it seems to me that it proves nothing since the formula underscored is true for all $y\in{\Bbb R}^n$ by the definition of the infimum.


Fix $x,t$. Observe that the function

$$\varphi(y):=tL\left(\dfrac{x-y}{t}\right)+g(y),\ y\in\mathbb{R}^{n}$$

is continuous, being the sum of the two continuous functions. Using the hypothesis that $g$ is Lipschitz, we have the estimate

$$\varphi(y)\geq tL\left(\dfrac{x-y}{t}\right)-\left\|g\right\|_{Lip}\left|y-x\right|-\left|g(x)\right|=\left|x-y\right|\left[\dfrac{L\left(\frac{x-y}{t}\right)}{\frac{\left|x-y\right|}{t}}-\left\|g\right\|_{Lip}-\dfrac{\left|g(x)\right|}{\left|x-y\right|}\right]$$

I claim that the expression in brackets tends to $+\infty$ as $\left|y\right|\rightarrow+\infty$. Indeed, it is clear that $\lim\frac{\left|g(x)\right|}{\left|x-y\right|}=0$ and $\lim \dfrac{L(\frac{x-y}{t})}{\frac{\left|x-y\right|}{t}}=+\infty$ by hypothesis. Whence, there exists an $R>0$ such that $\left|y\right|>R$ implies that $\varphi(y)>u(x,t)+\delta$, for fixed $\delta>0$.

By Weierstrass' extreme value theorem, $\varphi$ attains its minimum on the closed ball $\overline{B}(0,R)$ at some point $y_{*}$. And by definition of infimum, there exists $y'\in\overline{B}(0,R)$ such that $\varphi(y')<u(x,t)+\delta/2$. So

$$\inf_{\left|y\right|>R}\varphi(y)\geq u(x,t)+\delta>u(x,t)+\delta/2>\varphi(y')\geq\varphi(y_{*})$$

and by construction $\inf_{\left|y\right|\leq R}\varphi(y)\geq\varphi(y_{*})$. These two results yield that $u(x,t)\geq\varphi(y_{*})$, from which equality is immediate.

  • $\begingroup$ The estimate $|y|>R$ implies that $\varphi(y)>u(x,t)+\delta$ for fixed $\delta>0$ is little confusing for me. Are you just using the proved fact that $$\lim_{|y|\to\infty}\varphi(y)=\infty?$$ (Does $R$ depend on $\delta$?) $\endgroup$ – Jack Apr 6 '15 at 20:27
  • $\begingroup$ @Jack: Yes, I'm just using the proved fact that $\lim_{\left|y\right|\rightarrow\infty}\varphi(y)=\infty$; basically unraveling the definition of a limit. $R$ may depend on the quantity $u(x,t)+\delta$, but I don't see how that affects the proof. $\endgroup$ – Matt Rosenzweig Apr 6 '15 at 20:34
  • $\begingroup$ Fair enough. It does not affect the proof at all. Just for clarification. Thank you! $\endgroup$ – Jack Apr 6 '15 at 21:10
  • $\begingroup$ @MattRosenzweig We first need to prove that $u(x,t)\not=-\infty$ right? Otherwise, $u(x,t)+\delta/2>\varphi(y')$ may not be true. $\endgroup$ – Xianjin Yang Oct 20 '16 at 19:50
  • $\begingroup$ @XianjinYang: I think the first claim in my answer addresses your question. $\endgroup$ – Matt Rosenzweig Oct 20 '16 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.