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. 
 A: 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.
