We say $f \in L^2([-\pi, \pi])$ is an element of the Sobolev space $H^s(\mathbb{T})$ of order $s \geq 0$ if $$\sum_{n \in \mathbb{Z}} |\hat{f}(n)|^2 (1 + |n|^2)^s < \infty.$$
I am trying to prove the following:
Let $s > \frac{1}{2}$. If $f \in H^s(\mathbb{T})$ then $\exists g \in C([-\pi, \pi])$ such that $f = g$ a.e.
I have already shown that $H^s(\mathbb{T})$ is a Hilbert space with Hermitian inner product $$\langle f, g \rangle_{H^s(\mathbb{T})} := \sum_{n \in \mathbb{Z}} \hat{f}(n) \overline{\hat{g}(n)} (1 + |n|^2)^s,$$ and I believe I can also show that $\exists C(s) > 0$ such that $$||f||_{\infty} \leq C(s) ||f||_{H^s(\mathbb{T})},$$ with an added hint that the Weierstrass M-test may be useful, but I am overall not sure how to proceed.
What Giuseppe said. By Cauchy-Schwarz, for $s>\frac 1 2$ $$\sum_{n\in\mathbb Z} |\widehat f(n)| \leq \sqrt{\sum_{n\in\mathbb Z} \frac{1}{(1+n^2)^s}}\sqrt{\sum_{n\in\mathbb Z} |\widehat f(n)|^2(1+n^2)^s}<+\infty$$ Thus the Fourier series converges uniformly (in fact it is normally convergent).
For every $x$, the Fourier series converges pointwise to a number. Call it $g(x)$. And by uniform convergence $g$ is continuous. Finally, since we know the Fourier series converges to $f$ almost everywhere, it follows that $f=g$ almost everywhere.