Let $f\in\mathcal C^{2}(\mathbb{T})$. Prove that the Fourier series of $f$ converges uniformly to $f$, that is, $$f(x)=\sum_{n=-\infty}^{\infty} \hat{f}(n)e^{inx}$$ converges uniformly on $\mathbb{R}$.
Proof: Since $f\in\mathcal C^{2}(\mathbb{T})$, if we integrate by parts we have the following
$$\hat{f'}(n)=in\hat{f}(n)$$
and therefore
$$\hat{f''}(n)=in\hat{f'}(n)$$ So $$\hat{f''}(n)=-n\hat{f}(n)$$
Now check Weierstrass M-test hypothesis
$$\lvert \hat{f}(n)e^{inx} \rvert = \lvert \hat{f}(n) \rvert = \bigg\lvert \frac{\hat{f''}(n)}{-n^2} \bigg\rvert = \frac{\lvert \hat{f''}(n) \rvert}{n^2} \leq \frac{\|f''\|_1}{n^2}$$
Since $f\in\mathcal C^{2}(\mathbb{T})$, we have that $f'' \in L_1(\mathbb{T})$ and, as a consequence, $\|f''\|_{1} < \infty$.
Thus $\sum_{n=0}^{\infty} \frac{\|f''\|_1}{n^2}$ converges and by Weierstrass M-test the series $\sum_{n=0}^{\infty} \hat{f}(n)e^{inx}$ converges uniformly. Repeating for $\sum_{n=-\infty}^{-1} \frac{\|f''\|_1}{n^2}$ we have that the series $\sum_{n=-\infty}^{-1} \hat{f}(n)e^{inx}$ converges uniformly. We conclude that the series $\sum_{n=-\infty}^{\infty} \hat{f}(n)e^{inx}$ converges uniformly.