Continuity and Lipschitz Property of Infimal Convolution

Question

If $f$ is convex and Lipschitz-continuous on a real Hilbert space $H$ and $g$ is lsc and convex then is the infimal convolution $f\square g$ Lipschitz?

Answer 1

$\newcommand{\R}{\mathbb{R}}\newcommand{\Rbar}{\overline{\R}}$This is partially answered by Theorem 10.56 in Rockafellar and Wets' "Variational Analysis". The theorem states that if

1. $f=g \# h$, where $g:\R^n\to\Rbar$ is a proper, lsc function and $h:\R^n\to\R$ and
2. for every $a\in\R$, the mapping $x \mapsto \{w: g(w) + h(x-w) \leq a\}$ is locally bounded and
3. $h$ is strictly continuous,

then, $f$ is strictly continuous with strict continuity modulus

$$ \operatorname{lip} f(x) \leq \max_{w\in P(x)} \operatorname{lip} h(x-w), $$

where $P(x) = \mathrm{argmin}_{w\in\R^n} \{g(w) + h(x-w)\}$.

This means that a condition for local Lipscthiz continuity at $\bar{x}$ is that there is an $M\geq 0$ so that $\operatorname{lip} h(\bar x-w) \leq M$ for all $w\in P(\bar x)$.

You assume that $g$ is convex, so the proximal operator will be single-valued. I don't think that the assumption of convexity can be used to weaken conditions 2 and 3.

A special case is when $h$ is the function $h(x) = \tfrac{1}{2}\|x\|^2$ for which $\operatorname{lip} h(x) = \|x\|$. Then $P(x) = \operatorname{prox}_g (x)$, which in general may be multi-valued. We have

$$ \operatorname{lip} f(x) \leq \max_{w\in\operatorname{prox}_g (x)}\|w-x\|^2. $$

This upper bound is finite if $\operatorname{prox}_g (x)$ is compact. This will be the case if $\phi(x, w) = g(x) + \tfrac{1}{2}\|x-w\|^2$ is level-bounded in $x$, locally uniformly in $w$ (by Theorem 1.17 in the same book).

If $g$ is convex, then

$$ \operatorname{lip} f(x) \leq \|\operatorname{prox}_g (x)-x\|^2. $$

Note that the condition of global Lipschitz continuity (i.e., $\sup_x \operatorname{lip} f(x)$ is finite) is rather strong (so we need additional assumptions).