from my understanding Riemann-Lebesgue Lemma says that every continuous and 2π periodic function, the Fourier coefficient go to zero as n approach to infinity
$\lim_{|n|\to+\infty}\hat{f}(n)=0,$
but i occur problems such as this: (convolution) If f*g=g then g is a trigonometric polynomial
that to prove a function is trigonometric polynomial i should assume the function is continuous and 2π periodic and prove:
$\lim_{|n|\to+\infty}\hat{f}(n)=0,$
so my confusion is how do i prove a function is trigonometric polynomial if i know it's continuous and 2π periodic do i miss any other condition? or i didn't understand Riemann-Lebesgue Lemma?
There's a difference between (i) $\lim_{|n| \to \infty} \hat f(n) = 0$ and (ii) $\hat f(n) = 0$ for all $|n| > N$ for some $N \in \Bbb N$. The former is the Riemann-Lebesgue lemma, but the latter comes from the definition of trigonometric polynomials. In particular, if you have a trigonometric polynomial $$p(t) := \sum_{n=-N}^N a_n e^{2\pi int}$$ then $\hat p(n) = 0$ for all $n$ satisfying $|n| > N$ (clearly.) This is stronger than what the Riemann-Lebesgue lemma tells us - for trigonometric polynomials, not only does $\hat f(n)$ approach $0$ as $|n|\to\infty$, but also $\hat f(n) = 0$ for all but finitely many $n\in \Bbb Z$.