Let $E$ be a t.v.s. and $f, g:E \to (-\infty, +\infty]$ be convex l.s.c. such that $f,g \not \equiv +\infty$. We define the infimal-convolution $f \square g:E \to [-\infty, +\infty]$ by
$$
(f \square g)(x) := \inf_{y\in E} [f(x-y) + g(y)], \quad \forall x\in E.
$$
From Wikipedia, $f \square g$ is convex l.s.c. if $E =\mathbb R^n$. Does this property still hold in case $E$ is a t.v.s. or a n.v.s?
Update:
- It's written in Robert Phelps's texbook Convex functions, monotone operators, and differentiability that
- In below attempt, I'm stuck proving the existence of such $y_0$.
It's clear that $(x,y) \mapsto f(x-y)$ and $(x,y) \mapsto g(y)$ are l.s.c., so is the sum $\phi:(x,y) \mapsto f(x-y)+g(y)$. It follows that $\operatorname{epi} \phi := \{(x, y, \lambda) \mid \phi (x,y) \le \lambda\}$ is closed. Our goal is to show that $\operatorname{epi} (f \square g)$ is closed. Fix $\epsilon>0$. Assume $(x_n, \lambda_n)$ is a sequence in $\operatorname{epi} (f \square g)$ such that $x_n \to x\in E$ and $\lambda_n \to \lambda \in \mathbb R$. We want to show that $(x,\lambda) \in \operatorname{epi} (f \square g)$.
For each $n$, there is $y_n \in E$ such that $\phi(x_n, y_n) \le \lambda_n +\epsilon$. If there exist $\color{blue}{y_0}\in E$ and a subsequence $(y_{n_m})_m$ such that $y_{n_m} \to y_0$ as $m \to \infty$, then by l.s.c. of $\phi$ we get $\phi(x,y_0) \le \lambda+\varepsilon$. The claims then follows.
Best Answer
The result is wrong.
Suppose that $f$ and $g$ are indicator functions of sets $A$ and $B$ respectively.
Then $f\Box g$ is the indicator function of the set $$A+B $$
Now assume that $A$ and $B$ are convex closed nonempty.
Then $f$ and $g$ are convex, lower semicontinuous, and proper.
However, $A+B$ is convex and nonempty, but not necessarily closed:
Indeed, if $X=\mathbb{R}^2$, then let $A$ and $B$ be the epigraphs of $\exp(x)$ and $\exp(-x)$ respectively.
Then $A+B$ is the open upper halfplane in $\mathbb{R}^2$.
In particular, $A+B$ is not closed.
Hence the infimal convolution is not lower semicontinuous.
Note: For some conditions guaranteeing that $f\Box g$ is lower semicontinuous, see Proposition 15.7 in the book by Bauschke-Combettes. For instance, $f$ is coercive and $g$ bounded below would to. Or the domains of $f^*$ and $g^*$ overlap sufficiently.