- #1
DavideGenoa
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Hi, friends! Let ##f:[a,b]\to\mathbb{C}## be an http://librarum.org/book/10022/173 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!
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!
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