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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.

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    Are you sure that it is correct? The whole space $S=C[0,1]$ is obviously closed. But the closed unit ball isn't compact. – 2017-01-24
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    Riesz's Lemma shows that infinite dimensional normed spaces never have a compact unit ball. – 2017-01-24
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    I think you meant that the closed unit ball of $C[0,1]$ is compact with respect to the $L^2$ norm. Please edit the question to clarify. Right now it does not even parse as English ("in $S$ in $S$ is compact") – 2017-01-24
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    You could see that the closure of the unit ball of $C[0,1]$ in $L^2[0,1]$ is compact by embedding $L^2[0,1]$ into $L^2(\mathbb R)$ and applying the Kolmogorov Riesz theorem (cf theorem 5 in [this link](https://www.math.ntnu.no/conservation/2009/037.pdf) ). Then the closure of any ball in a subspace is compact since it is a closed subset of a a compact set. – 2017-01-26
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    @s.harp: Not true. The uniform estimate on the translations will fail, see my answer. – 2017-01-26
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    @daw Yes, I misread the conditions to be not-uniform – 2017-01-26

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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.

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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 $L^2.$