Let f be a periodic $ 2\pi $ and continuous function. given $$g_1=f,g_2=f*f,\dots g_{n}=\underbrace{f*f*f*f}_{n\,times} $$ 1.assuming $ f(x)=D_N=\sum_{n=-N}^{N}e^{inx} $ prove that $f=g_n $
- assuming $ g_{n}\xrightarrow[n\to\infty]{}f $ uniformally.
for 1: Using induction for case $n=1 $ $ f=f $ is easy. We know the Fourier coefficients $$ \hat{f}\left(m\right)=\begin{cases} 1 & \left|m\right|\le N\\ 0 & else \end{cases} $$
Assuming $g_n =f $, applying $*f $ to both sides we get that $$g_{n+1}=\underbrace{f*f*\dots*f}_{n+1\,times}=f*f $$ Because of convolution theorem : $$ \left(\hat{f}\left(m\right)\right)^{n+1}=\begin{cases} 1\cdot1=1 & \left|m\right|\le N\\ 0\cdot0 & else \end{cases}=\hat{f}\left(m\right) $$ Hence the fourier coefficients for $\left(\hat{f}\left(m\right)\right)^{n}=\hat{f}\left(m\right) $ we can infer that $ \hat{g}_{n}\left(m\right)=\left(\hat{f}\left(m\right)\right)^{n}=\hat{f}\left(m\right) $ hence $ f=g_n $
For part 2: had a bit of trouble here. my line of thought was that if we contradict Riemann-Lebesgue theorem we can prove it. Lets assume f is not a trigonometric polynomial.
Because f is continuous Riemann Lebesgue applies to it so we know that $\hat{f}\left(m\right)\xrightarrow{m\to\infty}0 $
Because of convolution theorem we once again get$$\lim_{n\to\infty}\left(\hat{f}\left(m\right)\right)^{n}=\hat{f}\left(m\right) $$ for all m. assuming for some large m $ \hat{f}\left(m\right)\ne0 $ we can divide both sides by $ \hat{f}(m) $ We can get $ \lim_{n\to\infty}\hat{f}\left(m\right)=1 $ therefore $\hat{f}(m)$ is a root of unity, but not 0. which is a contradiction to Riemann-Lebesgue.