Lower Semicontinuity Concepts

Solution 1:

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 $f|\ell_2\setminus A=0$.