# $\mathcal{F}$ convex and lower continous $\Rightarrow$ $\mathcal{F}$ weakly lower continous

I'm having troubles with one part of a problem consisting out of several subquestions and hope some of you can help me!

Let $$X$$ be a Banach-space and let $$\mathcal{F} : X \rightarrow (-\infty,\infty]$$ a convex and lower continous functional. I have to show, that $$\mathcal{F}$$ is weakly lower continous too.

We defined these types of continuity in the following way:

$$\mathcal{F}$$ is called lower continous if $$u_k \rightarrow u$$ in X $$\Rightarrow \mathcal{F}(u) \leq \liminf_{k\rightarrow \infty}\mathcal{F}(u_k)$$

$$\mathcal{F}$$ is called weakly lower continous if the same holds for $$u_k \rightharpoonup u$$.

I guess, that it might be a good idea to use one of the last statements we got in our lectur, which states that a convex subset $$C$$ of a Banach space $$X$$ is closed in strong topology if and only if $$C$$ is closed in the weak topology, but I even wasn't able to prove it using this lemma.

I would be grateful, if someone could help me! :)

PS: I've already looked for similar questions on stackexchange and found this one and this one but both use some different definition of lower contionus, which wasn't introduced in our lecture. So I would appreciate if someone could help solving this task using the definitions I mentioned above, due to I don't only want to solve this problem, but also want to improve my understanding of things introduced in our lecture.

If a set is convex and closed it is also weakly closed. The reasoning here is that the weak topology and is generated by linear functionals/halfspaces, and convex sets can be written as intersections of halfspaces.

With this in mind the proof works by considering the epigraph of $$f$$ (which is in fact a common approach in convex analysis). Note further that a convex function is lower semicontinuous if and only if its epigraph is closed. Since $$f$$ is assumed to be convex and lower semicontinuous its epigraph is convex and closed (in the regular topology) which makes it closed in the weak topology and thus weakly closed, i.e. weakly lower semicontinuous.

• Thanks for your help! – pcalc May 23 at 5:02

You may proceed as follows. Define the lower level sets at height $$\xi\in\mathbb{R}$$ of $$\mathcal{F}$$ as $$L_\xi=\left\{x\in X:\mathcal{F}(x)\leq\xi\right\}.$$

Step 1: Show that, if $$\mathcal{F}$$ is convex, then $$(\forall\xi\in\mathbb{R})\;L_\xi$$ is a convex subset of $$\mathbb{R}$$.

Step 2: Show that $$\mathcal{F}$$ is lower semicontinuous in the strong topology if and only if, for every $$\xi\in\mathbb{R}$$, $$L_\xi$$ is closed in the strong topology.

Step 3: Show that $$\mathcal{F}$$ is weakly lower semicontinuous if and only if, for every $$\xi\in\mathbb{R}$$, $$L_\xi$$ is closed in the weak topology. To do this, you just need to use the definition that you mentioned. Hint: use the sequential characterization of closed sets. :)

Step 4: Combine the above steps and the "last statement in your lecture."

Would be happy to provide more details if needed.

PS: I think we should refer to the "weak lower semicontinuity" in your question as "weakly sequentially lower semicontinuous." Although these guys coincide in the case of convex functions, it is not true in general. We need to use the notion of "nets" to deal with the weak topology.