# Fourier Series of Product of Continuous Functions

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.

• What is $f \cdot g$ here? I don't think that you take the product of two functions! Maybe you take the convolution on $[0,1]$? – p4sch Oct 7 '18 at 14:16
• Double checked the problem, and it is indeed the product of two functions, unless the problem itself has a typo. – Stackman Oct 7 '18 at 17:08
• Yes, you are right! The identity should be true! – p4sch Oct 8 '18 at 7:49

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}