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.

  • $\begingroup$ when you say that a functional $F$ is closed or weakly closed, what is your definition for that? $\endgroup$ – supinf Oct 25 '18 at 15:39
  • $\begingroup$ 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 $\endgroup$ – tommy1996q Oct 25 '18 at 16:01
  • $\begingroup$ Oh sorry I made a mistake. I’ll edit that $\endgroup$ – tommy1996q Oct 25 '18 at 16:03
  • $\begingroup$ "Succession" = sequence. It is called sequential closure. $\endgroup$ – A.Γ. Oct 25 '18 at 17:31
  • $\begingroup$ math.stackexchange.com/questions/2355100/… $\endgroup$ – A.Γ. Oct 25 '18 at 17:38

Let us start with two facts and a remark.

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.


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