# convexity and lower semi-continuity for weak convergence

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!

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.
• Very elegant answer, thank you. This clearly works if $\Phi$ is defined on the whole space $X$. However, I'm a little concerned with the case when $\Phi$ is only defined on a convex subset $C\subset X$: if I'm not mistaken your cute argument would require $C$ to be strongly closed, which I'm not assuming here... What do you think? – leo monsaingeon Jul 9 '13 at 21:36
• @leomonsaingeon Since the epigraph $\{(x,y):y\ge \Phi(x),x\in C\}$ is convex, so is its closure (in the norm topology of $X\times \mathbb R$). This closure is the epigraph of a convex lsc function defined on $X$ (which takes on the value $+\infty$ outside of the closure of $C$). – 40 votes Jul 9 '13 at 21:55