TITLE: A smooth sequence of function converging in the L^2 norm implies uniform convergence.
QUESTION [1 upvotes]: I am stuck on the following.
Let $f_n$ be a sequence of functions on the circle, and suppose we know the following about its Fourier coefficients:

For every $n\in\mathbb{N}, $$|\hat{f}_n(k)|\leq \frac{C}{1+|k|^N}$, for every $N\in\mathbb{N}$. That is, the $f_n$ is sequence of $C^{\infty}$-functions.
$\lim_{n\rightarrow \infty} \sum_{k\in\mathbb{Z}}|\hat{f}_n(k)|^2 = 0$, i.e.  $f_n$ converges to zero in the $L^2$-norm.

Does this imply uniform convergence to zero:
$$ ||f_n||_{\infty} := \sup_{x\in[0,2\pi)} |f(x)|  \rightarrow 0,\quad \mbox{as }n\rightarrow\infty\mbox{ ?} $$

REPLY [0 votes]: Proof by contradiction.
Suppose $f_n$ does not converge uniformly to zero, then there would exist an $\epsilon>0$ and a subsequence $n_k$ such that $|f_{n_k}(t_k)|\geq \epsilon$ for all $k\in \mathbb{N}$. However, $f_n$ is an equicontinous family. That is, for every $\epsilon>0$, there exists $\delta>0$ such that for all $n\in\mathbb{N}$:
$$ d(f_n(x),f_n(y)) < \epsilon,\quad \mbox{whenever }d(x,y)<\delta.$$
This can be seen from the fact that the derivatives $f_n'$ are uniformly bounded:
$$ f_n'(x) = \sum_{k\in\mathbb{Z}} i k \hat{f}_n(k) e^{ikx} \quad \Rightarrow \quad  |f_n'(x)| \leq \sum_{k\in\mathbb{Z}} |k| |\hat{f}_n(k)| \leq \sum_{k\in\mathbb{Z}} |k|\frac{C}{1+|k|^N}< \infty \mbox{ for some }N\geq3.$$
Now, pick a $\delta>0$ such that $d(f_n(x),f_n(y)) < \epsilon/2$ whenever $d(x,y)<\delta$. We see that:
$$ || {f}_{n_{k}} ||_2 \geq \epsilon/2 \sqrt{\delta}.  $$
But ${f}_{n}$ converges to zero in the $L^2$-norm, which is contradiction. Hence,  $f_n$ must converge uniformly to zero.
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