Convexity and strong lower semicontinuity imply weak lower semicontinuity

I have seen that if a set $$K$$ on an Hilbert space $$H$$ is convex and strongly sequentially-closed, it is weakly closed. The teacher said that if you take a convex and weakly lower semicontinuous functional $$F$$, using the fact that the sets $$F^{-1}(-\infty, \lambda]$$ are convex and that closure implies weak closure, it is easy to conclude that convexity and strong lower semicontinuity imply weak lower semicontinuity. I do not see how to do that though. I would like to see a proof not involving weak topologies etc. The way he said it menat it was supposed to be done using only the definitions, or little more.

• when you say that a functional $F$ is closed or weakly closed, what is your definition for that? – supinf Oct 25 '18 at 15:39
• With successions. Strongly closed means that if a succession $x_n$ of elements of this set converges to some $x$, then $x$ is also an element of the set. Weakly closed is the same, but with weak convergence – tommy1996q Oct 25 '18 at 16:01
• Oh sorry I made a mistake. I’ll edit that – tommy1996q Oct 25 '18 at 16:03
• "Succession" = sequence. It is called sequential closure. – A.Γ. Oct 25 '18 at 17:31
• math.stackexchange.com/questions/2355100/… – A.Γ. Oct 25 '18 at 17:38

Fact 1. Let $$(X,\mathcal{T})$$ be a topological space and let $$f \colon (X,\mathcal{T}) \to \left[{-}\infty,{+}\infty\right]$$. Then $$f$$ is lower semicontinuous if and only if, for every $$\xi \in \mathbb{R}$$, the lower level set $$f^{-1}(\left[{-}\infty,\xi\right])$$ is closed. Here, by lower semicontinuity, I mean: for every $$x \in X$$ and for every $$\xi \in \left]-\infty,f(x)\right[$$, there exists a neighborhood $$V$$ of $$x$$ in such that $$(\forall y \in V)\; f(y) > \xi$$.
Remark Lower semicontinuity goes with the topology on the domain of $$f$$. In particular, in your question, lower semicontinuous means "$$f$$ is lower semicontunuous wrt to the strong topology" whereas "weakly lower semicontinuous" means "$$f$$ is lower semicontinuous wrt to the weak topology on $$H$$." So I guess there is no way to avoid weak topology in the proof as it directly relates to the topologies on the domain.
Fact 2. Let $$C$$ be a convex subset of $$H$$ (in your question). Then $$C$$ is closed in the topology induced by the hilbertian norm of $$H$$ if and only if $$C$$ is closed in the weak topology.
Returning to your question and assume that $$f$$ is lower semicontinuous w.r.t the strong topology (induced by the norm of $$H$$) and that $$f$$ is convex. We must show that $$f$$ is weakly lower semicontinuous, i.e., $$f$$ is continuous when $$H$$ is equipped with the weak topology. Let us use Fact 1 to do this, i.e., take $$\xi \in \mathbb{R}$$ and show that $$f^{-1}(\left[{-}\infty,\xi\right])$$ is weakly closed. Since $$f$$ is convex, the set $$f^{-1}(\left[{-}\infty,\xi\right])$$ is convex. On the other hand, since $$f$$ is lsc w.r.t to the strong topology, the set $$f^{-1}(\left[{-}\infty,\xi\right])$$ is closed in the strong topology by Fact 1. Altogether, Fact 2 implies that it is indeed weakly closed.
So, we have shown that, for every $$\xi \in \mathbb{R}$$, the set $$f^{-1}(\left[{-}\infty,\xi\right])$$ is closed in the weak topology. In view of Fact 1, we conclude that $$f$$ is weakly lsc, i.e., lower semicontinuous when $$H$$ is equipped with the weak topology.