Prove or disprove a set $F$ is closed.

Question

This is an example in my book that talks about $F$ being precompact;

Let $F$ be the subset of $C([0,1])$ that consists of functions $f$ of the form $$f(x) = \sum_{n=1}^{\infty}a_n\sin(n\pi x) \hspace{.5cm} \text{ with } \hspace{.5 cm} \sum_{n=1}^{\infty} n|a_n|\leq 1.$$

I know $F$ is bounded and equicontinuous so it is precompact. I am interested in knowing if $F$ is also closed or not? So I could use the Arzela-Ascoli theorem so say that $F$ is either compact or not compact.

Answer 1

If $(f_k,k\geqslant 1)$ is a sequence in $F$ which converges uniformly to $f\in C[0,1]$, then $((a_{n,k})_n,k\geqslant 1)$ is Cauchy in $\ell^2$ (using Parseval's equality), so for each $n$, $a_{n,k}\to b_n$ as $k\to \infty$. Now we have to show that $f\in F$. We have $\sum_{n= 1}^Nn|b_n|=\lim_{k\to \infty}\sum_{n=1}^Nn|a_{n,k}|\leqslant 1$ hence $\sum_n n|b_n|\leqslant 1$ so it's enough to show that $f(x)=\sum_nb_n\sin(nx)$. To this aim we can compute the Fourier series (we have normal convergence).