2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.