Discuss compactness of the set of $L^2$ bounded functions

Question

Discuss weak and strong compactness of the following subsets of $L^2(0,1)$:

$A=\{u\in L^2(0,1):||u||_{L^2}\le1\}.$

I know some theorems which might be helpful, but I don't know if I applied them correctly.

The set $A$ is a closed unit ball of the Banach space $L^2(0,1)$, which is an infinite dimensional normed space, so by Riesz theorem $A$ is not strongly compact. Moreover, since $L^2(0,1)$ is reflexive, by Banach-Alaoglu theorem its closed unit ball $A$ is weakly compact.

Is this correct?

Answer 1

Concerning point A, you solved it correctly via "abstract nonsense". However, I think that it is always better to use concrete examples when possible.

In this case, a concrete example of a sequence contained in $A$ that has no converging subsequences is $$ f_n(x)=\sqrt n f(nx), $$ where $f$ is any nonzero element of $A$.

Here I omitted an important piece of information. The function $f$ is defined on $[0, 1]$, but we implicitly consider that $$f(x)=0,\qquad \text{if }x\notin [0,1].$$ In particular, for all $x\ne 0$, it holds that $nx>1$ for all sufficiently large $n$, and so $f(nx)=0$ eventually. Therefore $$ \lim_{n\to \infty} f_n(x)= 0,$$ and so $f_n\to 0$ pointwise almost everywhere.

Now the change of variable formula for integrals yields $$\tag{1}\|f_n\|_{L^2}=\|f\|_{L^2},\qquad \forall n\ge 1;$$ so $f_n\in A$ for all $n$. Now suppose for a contradiction that there exists $g\in L^2$ and a subsequence $f_{k_n}$ such that $$\|f_{k(n)}-g\|_{L^2}\to 0.$$ By (1), it must be that $\|g\|_{L^2}=\|f\|_{L^2}\ne 0$; thus, $$g\ne 0.$$ However, any sequence that converges in $L^2$ has a subsequence that converges pointwise almost everywhere, so there exists a sub-sub-sequence $f_{k(h(n))}$ such that $$ f_{k(h(n))}\to g,\qquad \text{almost everywhere.}$$ And this is a contradiction, for $f_{k(h(n))}\to 0$ almost everywhere by the remark in the colored box, and so it would imply that $g=0$.