Can someone argue why the inequality must be true for all $2\pi$-periodic integrable functions:
$$ |f(x)| \le \sum_{n\in\mathbb{Z}}|\hat{f(n)}|$$
for almost all $x\in\mathbb{R}$.
I have an intuitive feeling that this statement is true, I als don't think that this statement remains true for all $x \in \mathbb{R}$. Could someone give me a hint on how to proof this? Thank you!
If $\sum_{k\in\mathbb Z}|\hat f(k)|=\infty $, the result is obviously true. Suppose $\sum_{k\in\mathbb Z}|\hat f(k)|<\infty $.
Define $g(x)=\sum_{m\in\mathbb Z}\hat f(m)e^{2i\pi mx}$. Notice that $g$ is well defined and continuous because the series converges normally (and thus uniformly). You have $$\hat g(k)=\sum_{m\in\mathbb Z}\hat f(m)\int_0^{2\pi}e^{2i\pi mx}e^{-2i\pi kx}=\hat f(k).$$ Therefore, $f=g$ a.e., and thus $$f(x)=\sum_{k\in\mathbb Z}f(k)e^{2i\pi x}\quad a.e.$$ Therefore $$|f(x)|\leq \sum_{k\in\mathbb Z}|\hat f(k)|.$$