Convergence of a subsequence in $(C(\mathbb{T}), \|\cdot\|_2)$ Problem: Define, $ \mathbb{T} := \mathbb{R}/{2\pi\mathbb{Z}} $.
Consider a sequence of functions $(g_n)_{n\in \mathbb{N}} \in C^4(\mathbb{T})$ such that, $ \sup_{n \in \mathbb{N}}(\| g_n \|_2 + \| g'_n \|_2  + \| g''_n \|_2) < \infty $. Prove that $(g_n)_{n\in \mathbb{N}}$ has a subsequence converging in $(C(\mathbb{T}), \|\cdot\|_2)$, where, $\|\cdot\|_2$ is the norm,
$\|f\|_2 = (\frac{1}{2\pi}\int_0^{2\pi}|f(t)|^2dt)^{1/2}$.
Background: First course in analysis. This part of the course covers Fourier series, Fejér's theorem, covergence in the $L^\infty$ and $L^2$ norms. I know that the $C^4$ condition on the sequence of functions guarantees that the fourier series approximations of each function $g_n$ and its derivatives converge in both norms.
I don't really see how to approach the problem, unless there's some way to show that the space is compact?
Thanks.
 A: Yes, you need to prove compactness. This is an instance of Rellich–Kondrachov theorem, but presumably the point of the problem is to show compactness directly. To get uniform convergence, it suffices to have Fourier coefficients converging in $\ell^1(\mathbb{Z})$, so let's look at those. 
Let $g_n^{(k)}$ be the $k$th coefficient of $g_n$. Then $g_n''$ has coefficients $-k^2 g_n^{(k)}$. So, the assumption that both $g_n$ and $g_n''$ are   bounded in $L^2$ norm implies the existence of a constant $M$ such that 
$$\sum_{k\in\mathbb{Z}}(k^2+1)^2|g_n^{(k)}|^2\le M$$
Hence, $|g_n^{(k)}|\le M/(k^2+1)$. This is the main step of the proof: we got a summable dominating sequence $a_k=M/(k^2+1)$. For the rest, only this fact matters.
There are various ways to finish. One could use diagonalization to get coordinate-wise convergent subsequence and apply the dominated convergence theorem. Or one could prove the compactness of the set $A=\{c\in \ell^1: |c_k|\le a_k\ \forall k\}$ directly. Indeed, it is obviously closed in $\ell^1$, hence complete. To show total boundedness: given $\epsilon>0$, pick $N$ such that $\sum_{|k|>N}a_k<\epsilon/2$, and observe that $A$ is contained in the $\epsilon/2$-neighborhood of the set 
$$A_N=\{c\in \ell^1: |c_k|\le a_k\ \forall k,\text{ and } c_k=0 \text{ for }|k|>N\}$$
Since $A_N$ is compact, it admits a finite $\epsilon/2$ net. This net serves as an $\epsilon$-net of $A$.
