# Identical Fourier coefficients of continuous $f,\varphi\Rightarrow f=\varphi$

1. Nov 17, 2014

### DavideGenoa

Hi, friends! Let $f:[a,b]\to\mathbb{C}$ be an http://librarum.org/book/10022/173 [Broken] periodic function and let its derivative be Lebesgue square-integrable $f'\in L^2[a,b]$. I have read a proof (p. 413 here) by Kolmogorov and Fomin of the fact that its Fourier series uniformly converges to a continuous function $\varphi$ whose Fourier coefficients are the same as the Fourier coefficients of $f$.

I read in the same proof that, since $\varphi$ has the same Fourier coefficients of $f$, because of the continuity of the two functions we get $f=\varphi$. I do not understand why continuity guarantees the equality. Could anybody explain that?

I $\infty$-ly thank you!!!

Last edited by a moderator: May 7, 2017
2. Nov 17, 2014

### mathwonk

3. Nov 17, 2014

### DavideGenoa

I suspect that the answer is there: $(f-\varphi)(x)=0$ almost everywhere, if I correctly understand...
Thank you so much!!!