Some properties of Yosida-Moreau transform

Let $f(x)$ be a continuous function on $\mathbb{R}^n$, $f(x) \geqslant 0$ for any $x$. Define $$ f_{\alpha}(x) = \inf\limits_{y}\left( f(y) +\frac{|x-y|^2}{2\alpha} \right) $$ where $\alpha > 0$. How to show that $f_{\alpha}(x)$ is locally Lipschitz continuous and $f_{\alpha} \to f$ when $\alpha \to 0+0$ uniformly on any compact $K$ in $\mathbb{R}^n$?

Solution 1:

Let us restrict attention to $|x|\le R$. Let $M=\sup_{|x|\le R}f$. The main observation is that the infimum can be taken only over $y$ such that $|x-y|\le \sqrt{2\alpha M}$. Indeed, when $|x-y|>\sqrt{2\alpha M}$, the second term makes the sum greater than $M$, hence greater than $f(x)$.

1) Let $R'=R+\sqrt{2\alpha M}$. For any fixed $y$ such that $|y|\le R'$, the function $x\mapsto f(y)+|x-y|^2/(2\alpha)$ is Lipschitz with a constant $L=L(R',\alpha)$. It is easy to see that the infimum of any family of $L$-Lipschitz functions is $L$-Lipschitz (or $\equiv -\infty$, which is what we rule out with a lower bound for $f$).

2) is now easy: if $\alpha$ is small, $\sqrt{2\alpha M}$ is small, so the infimum will not be far away from $f(x)$ by the uniform continuity of $f$ on $\{|x|\le R'\}$.