Lower Semi Continuous w.r.t weak Topology

Question

I have a question regarding (weakly) lower continuity of some Functionals.

Let $\mathbb{H}$ be a reflexive Hilbert space and $A\subseteq \mathbb{H}$ be a closed set in the weak topology of $\mathbb{H}$. I have shown that a map $f:A\to \mathbb{R}$ is uniformly continuous as a map between metric spaces. Is there a chance that $f$ is already weakly lower semi-continuous ? Or do I have to show first that $f$ is convex?

Thanks a lot for your help!

Answer 1

If $f$ is continuous and convex then $\{x:f(x)<\lambda\}$ is closed and convex for all $\lambda$. Such sets are weakly closed. Therefore such functions are wlsc.

If we drop the assumption on convexity then we can consider the function $f(x)=-\|x\|$. Then the set $\{x:f(x)>-1\}=B(0,1)$ isn't weakly open. Therefore $f$ isn't wlsc.