Why is the infimum in the Hopf-Lax formula a minimum?

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:

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.

• 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 $\delta$?) – Jack Apr 6 '15 at 20:27
• @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. – Matt Rosenzweig Apr 6 '15 at 20:34
• Fair enough. It does not affect the proof at all. Just for clarification. Thank you! – Jack Apr 6 '15 at 21:10
• @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. – Xianjin Yang Oct 20 '16 at 19:50
• @XianjinYang: I think the first claim in my answer addresses your question. – Matt Rosenzweig Oct 20 '16 at 21:30