My question is a general one, whose answer can probably be found in any decent convex analysis book. I unfortunately don't have any at hand right now, so here it is:
Let's consider a "reasonable" Banach space $X$ (say at least reflexive and separable as usual), a convex subset $C\subset X$ and a function $\Phi:C\to\mathbb{R}$ which is continuous for the strong $X$ topology and convex.
What do I need to assume for $\Phi$ to be weakly lower-semicontinuous? Clearly differentiability works with the classical trick $x_n\rightharpoonup x$ and $\Phi(x_k)\geq \Phi(x)+D\Phi(x).(x_k-x)$, but I guess there must be less restrictive conditions than differentiability?
Thank you!
Best Answer
A function is lower semicontinuous (in some topology on $X$) if and only if its epigraph $\{(x,y)\in X\times \mathbb R:y\ge \Phi(x)\}$ is a closed set in the product topology on $X\times \mathbb R$.
In particular, if we use the weak topology on $X$, the product topology on $X\times \mathbb R$ is the weak topology of the product $X\times \mathbb R$, considered as a Banach space of its own.
It is a well-known consequence of Hahn-Banach theorem that a strongly closed convex set is weakly closed, in any normed space.
Combining the above, we conclude: every lower semicontinuous convex function is weakly lower semicontinuous. No further assumptions are needed.