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Suppose $f, g$ are two continuous $1$-periodic functions. Prove that $$\widehat{f \cdot g}(n)=\sum_{m\in \mathbb{Z}}\hat{f}(n-m)\hat{g}(m),$$ where $\widehat{f\cdot g}$ is the Fourier coefficient of $f\cdot g$, and $\hat{f}$ and $\hat{g}$ are the Fourier coefficients of $f$ and $g$, respectively.

Could you provide a small hint to set me on the right direction? Any help is much appreciated.

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  • $\begingroup$ What is $f \cdot g$ here? I don't think that you take the product of two functions! Maybe you take the convolution on $[0,1]$? $\endgroup$ – p4sch Oct 7 '18 at 14:16
  • $\begingroup$ Double checked the problem, and it is indeed the product of two functions, unless the problem itself has a typo. $\endgroup$ – Stackman Oct 7 '18 at 17:08
  • $\begingroup$ Yes, you are right! The identity should be true! $\endgroup$ – p4sch Oct 8 '18 at 7:49
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We can write $$g(x) = \sum_{m \in \mathbb{Z}} \widehat{g}(m) \mathrm{e}^{2\pi x m}$$ and the convergence is in $L^2$. Using the Cauchy-Schwartz inequality, we see that $$g(x) f(x) =\sum_{m \in \mathbb{Z}} \widehat{g}(m) f(x) \mathrm{e}^{2\pi \mathrm{i} x m} $$ converges in $L^1$. Thus, we get by interchanging the sum and integration \begin{align} \widehat{f \cdot g}(n)= \int_0^1 \exp(-2 \pi \mathrm{i} n x) g(x) f(x) \, \mathrm{d} x &= \sum_{m \in \mathbb{Z}} \widehat{g} (m) \int_0^1 \exp(2 \pi \mathrm{i} (m-n) x) f(x) \, \mathrm{d} x \\ &= \sum_{m \in \mathbb{Z}} \widehat{g}(m) \widehat{f}(m-n). \end{align}

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