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

  1. Nov 17, 2014 #1
    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
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  3. Nov 17, 2014 #2

    mathwonk

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  4. Nov 17, 2014 #3
    I suspect that the answer is there: ##(f-\varphi)(x)=0## almost everywhere, if I correctly understand...
    Thank you so much!!!
     
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