If $\underbrace{f*f*\ldots*f}_{n\text{ times}}\to f$ uniformly, then the continuous $2\pi$-periodic function $f$ is a trigonometric polynomial 
Let $f$ be a $2 \pi$-periodic continuous function. Given $$g_1 = f, \qquad g_2 = f * f, \quad \cdots \quad g_{n} = \underbrace{f * f * \cdots * f}_{n \text{ times}} $$ where $*$ denotes convolution, and assume $ g_{n} \xrightarrow{n\to\infty} f $ uniformly.
Prove that $f $ is a trigonometric polynomial.

Im not sure where to start, I have proved in previous parts that if $f$ is Dirichlet kernel $$\sum_{n=-N}^{N}e^{i n x}$$ then for any $n\in \mathbb{N} $ we have $g_n=f$, but I am not sure how to use it, if it is relevant at all. Any help would be appreciated. Thanks in advance.

By the convolution theorem, we have
$$c_k[g_n] = c_k[f]^n \qquad \forall n \ge 0, \ \forall k \in \mathbb{Z},$$
where $c_k$ denotes the $k$-th Fourier coefficient.
Furthermore, we have $c_k[g_n] \to c_k[f]$ for all $k \in \mathbb{Z}$ due to the uniform convergence $g_n \rightrightarrows f$ (can you show this?).
But $c_k[f]^n \xrightarrow{n \to \infty} c_k[f]$ implies that $c_k[f] \in \{ 0, 1 \}$ for every $k \in \mathbb{Z}$.
But in this case $f(x) = \sum_{k \in \mathbb{Z}} c_k[f] e^{i k x}$ diverges in $L^2([0, 2 \pi])$ unless there is a $m \in \mathbb{N}$ such that $c_k[f] = 0$ for all $k \in \mathbb{Z}$ with $| k | > m$.
Hence $f$ is a trigonometric polynomial of degree at most $m$(, whose coefficients are either 0 or 1).
