Q:

For any real sequence converging to $0$ , is there a $2\pi$ periodic continuous function whose Fourier coefficients are larger than the sequence?

Let $\{a_n\}_{n\in \mathbb Z}$ be a sequence of positive real numbers such that $a_n \to 0$ as $|n| \to \infty$ . Does there exist a continuous $2\pi$ periodic function $f: \mathbb R \to \mathbb C$ such that $|\hat f(n)| \ge a_n , \forall n \in \mathbb Z$  ? Or at least , does there exist a continuous $2\pi$ periodic function $f: \mathbb R \to \mathbb C$ such that $|\hat f(n)| \ge a_n$ , for infinitely many $n \in \mathbb Z$ ? 
Here $\hat f(n):=\dfrac 1 {2\pi} \int_0^{2\pi} f(x)e^{-inx} dx$

A:

As Daniel Fischer noted, the inequality $|\hat f(n)|\ge a_n$ cannot hold for all $n$ if $(a_n)$ is not square-summable, since $\sum |\hat f(n)|^2<\infty$ for every function in $L^2([0,2\pi])$, in particular for every continuous function on $[0,2\pi]$. 
But, as reuns said, it's possible to find continuous $f$ such that $|\hat f(n)|\ge a_n$ infinitely often. Indeed, there is a subsequence $a_{n_k}$ such that $|a_{n_k}|\le 2^{-k}$ for every $k$. Let 
$$
f(x) = \sum_{k} a_{n_k}\exp(n_k i x)
$$
By the Weierstrass M-test, the series converges uniformly, hence $f$ is continuous. It has the desired Fourier coefficients by construction.

