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.
Convexity and strong lower semicontinuity imply weak lower semicontinuity
convex-analysisfunctional-analysishilbert-spacesweak-convergence
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b) My Google responses on "weakly sequentially lower semicontinuous" query yield something about convexity. For instance, if $X$ is a normed space, and $f:X\to\overline{\mathbb R}$ is quasi-convex and (strongly) lower semicontinuous then $f$ is also weakly sequentially lower semicontinuous [1, Ex.3]. You can try to google for more such facts.
[1] Peter Holthe Hansen, 01716 Advanced Topics in Applied Functional Analysis, Assignment 10.
a) It seems the following.
An example of a continuous (and, hence, a lower semicontinuous) function $f:\ell_2\to\mathbb R$ such that $f$ is not weakly sequentially lower semicontinuous. For each point $x_0\in\ell_2$ define a function $f_{x_0}:\ell_2\to\mathbb R$ by putting $f_{x_0}(x)=\min\{\|x-x_0\|-1/3,0\}$ for each $x\in\ell_2$. For each natural number $n$ let $e_n$ be the standard $n$-th ort of the space $\ell_2$. Put $f=\sum_{n=1}^\infty f_{e_n}$. Then $f$ is continuous as a sum of a family of continuous functions with the locally finite family of the supports. From the other side, $f$ is not weakly sequentially lower semicontinuous, because $f(0)=0$, but the sequence $\{e_n\}$ weakly converges to $0$ and $f(e_n)=-1/3$ for each $n$.
An example of a weakly sequentially lower semicontinuous function $f:\ell_2\to\mathbb R$ such that $f$ is not weakly lower semicontinuous.
Maybe we should first to construct a subset $A$ of the space $\ell_2$ such that $0$ belongs to the weak closure of $A$, but $0$ does not belong to the weak sequential closure of $A$.
I found such a set $A=\{x_{nm}\}$ in Bill Johnson’s answer to Question “A point in the weak closure but not in the weak sequential closure”. The zero is the unique not weakly isolated point of the set $A$. So we may define a function $f:\ell_2\to\mathbb R$ by putting $f(0)=-1$, $f|A\setminus\{0\}=-2$ and $f|\ell_2\setminus A=0$.
Consider $\ell^p$, the space of real sequences for which $\sum_j|x_j|^p$ is finite with natural norm. If $p\neq 2$, this space is not a Hilbert space. If $1\lt p\lt \infty$, take $e_{j}=(0,\dots,0,1,0,\dots)$, where the $1$ is at the $j$-th position. This sequence converges weakly to $0$ in $\ell^p$ but not strongly. In order to see the weak convergence, we can use the fact that $f$ can be represented as $f\left(\left(x_n\right)_{n\geqslant 1}\right)=\sum_{n\geqslant 1}f_n x_n$ where $\left(f_n\right)_{n\geqslant1}\in \ell^{p'}$ and $p'$ is the conjugate exponent of $p$, that is,$1/p+1/p'=1$. Then $f\left(e_n\right)=f_n$ and $\left\lvert f_n\right\rvert^{p'}\to 0$ because of the convergence of $\sum_{n\geqslant 1}\left\lvert f_n\right\rvert^{p'}$.
The case $p=1$ is different. If weak convergence and convergence in norm in $(X,\lVert\cdot\rVert)$ are equivalent, we say that $(X,\lVert\cdot\rVert)$ has the Schur property.
Best Answer
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.