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I saw this theorem in Wikipedia, and didn't see it in other place. I tried to proof it, but failed.

How to prove that for $2\pi-$periodic functions one has The convolution theorem for fourier series.: $$ \widehat{f*g}(x) =2π\hat{g}(x)\cdot\hat{f}(x) $$ Note: this is convolution theorem for fourier series, not transforms

see here

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  • $\begingroup$ Is there a question you wanted us to answer? $\endgroup$ – John Hughes Jan 16 '18 at 18:48
  • $\begingroup$ A proof of this is in nearly every book / lecture notes on fourier analysis $\endgroup$ – TheWaveLad Jan 16 '18 at 18:49
  • $\begingroup$ note, this theorem is for fourier series, not fourier transform. I tried to find a proof on the web, and I didn't find even the theorem itself on other sources. $\endgroup$ – Mr.OY Jan 16 '18 at 19:22
  • $\begingroup$ @JohnHughes It will be nice to see a proof to this theorem. $\endgroup$ – Mr.OY Jan 16 '18 at 19:23
  • $\begingroup$ Of what theorem? You've shown us a proof of a theorem that isn't the one you care about (apparently). If you could carefully state the theorem whose proof you'd like to see, perhaps we could help you. As @TheWaveLad says, it's almost certainly in any book you pick up, but we could actually tell you the theorem NUMBER in a book or two ... $\endgroup$ – John Hughes Jan 16 '18 at 19:44
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The proof use the same token as in the case of Fourier transform. Indeed, $f$ and $g$ be $2\pi$-periodic functions. But since the function $y\mapsto e^{-iyx}g(y)$ is $2π-$periodic, then using this we have , $$\int_{0}^{2\pi} e^{-ix(t-u)}g(t-u)dt\overset{y=t-u}{=} \color{blue}{\int_{-u}^{2\pi-u} e^{-ixy}g(y)dy =\int_{0}^{2\pi} e^{-ixy}g(y)dy}= 2π\hat{g}(x) $$

Inserting this in the following relation we get

$$ \color{blue}{\widehat{f*g}(x) =} \frac{1}{2\pi}\int_{0}^{2\pi}e^{-ixt}f*g(t)dt = \frac{1}{2\pi}\int_{0}^{2\pi}e^{-ixt}\int_{0}^{2\pi} f(u)g(t-u)dudt\\= \frac{1}{2\pi}\int_{0}^{2\pi}\left(e^{-iux}f(u)\int_{0}^{2\pi} e^{-ix(t-u)}g(t-u)dt\right)du\\= 2π\hat{g}(x) \frac{1}{2\pi}\int_{0}^{2\pi}e^{-iux}f(u)du= \color{red}{2π\widehat{g}(x)\cdot\widehat{f}(x)} $$

we used this An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$.

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