Closed $L^2$ unit ball in a closed subspace of $C[0,1]$ is compact

Question

Let $S$ be a closed subspace of $C[0,1]$. I want to show that the closed unit ball of $S$ is compact with respect to the $L2$-.norm

My idea is that we can show weak convergence and then show that this implies convergence in norm. I'm unsure of how to work with this however.

Answer 1

This is definitely not true. Define $$ S=\{x \in C([0,1]): \ x(0)=0\}. $$ This is a closed subspace of $C([0,1])$. Now take the functions $$ f_n(x) := \sin(2n \pi x). $$ It is well-known that $f_n\rightharpoonup 0$ in $L^2(0,1)$, but $\|f\|_{L^2(\Omega)} = \frac1{\sqrt2}$, hence $f_n \not\to0$ in $L^2(0,1)$.

Since $(f_n)$ is a bounded sequence in $L^2(0,1)$ and $C([0,1])$, the desired result is not true no matter which unit ball you choose.

Answer 2

Just to finish the question: Take the case $S= C[0,1].$ Let $B$ be the closed unit ball of $C[0,1].$ For $n>2$ let $f_n$ be the element of $C[0,1]$ whose graph connects the points $(0,0),(1/2,0),(1/2+ 1/n, 1), (1,1)$ with line segments. Then each $f_n\in B.$ However, as you can verify, $f_n \to \chi_{[1/2,1]}$ in $L^2.$ Since $\chi_{[1/2,1]}\notin C[0,1],$ we have shown $B$ is not even closed in $L^2,$ much less compact in $L^2.$