# Fourier series and convolution

Let $f$ and $g$ be $2\pi$-periodic, piece-wise smooth functions having Fourier series $f(x)=\sum_n\alpha_ne^{inx}$ and $g(x)=\sum_n\beta_ne^{inx}$, and define the convolution of $f$ and $g$ to be $f\star g(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)g(x-t)dt$. Show that the complex form of the Fourier series for $f\star g$ is $$f\star g(x)=\sum_{n=-\infty}^{\infty}\alpha_n\beta_ne^{inx}.$$

I have been trying different approaches to this for a while but I haven't been able to figure it out. I think that I am supposed to use the fact that $$\frac{1}{2\pi}\langle f,g\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\overline{g(t)}dt=\sum_{n=-\infty}^{\infty}\alpha_n\overline{\beta_n}$$ but I can't figure out how to manipulate the algebra to show the correct result. I may be forgetting some key property of the inner product or something.

• Try convolving the Fourier series expressions instead. Note that the convolution operator is linear, so you just have to convolve two exponentials. – Cameron Williams Feb 2 '16 at 2:39
• I have, but that gets me $\frac{1}{2\pi}\int_{-\pi}^{\pi}\sum_n\alpha_ne^{int}\sum_n\beta_ne^{in(x-t)}dt$ and I don't know how to simplify that – Matt G Feb 2 '16 at 2:40
• Bring out the sums (and use different indices on the sums!). – Cameron Williams Feb 2 '16 at 2:41
• Bring them outside the integral? – Matt G Feb 2 '16 at 2:53
• Matt. You need only express one of the functions as a Fourier series. I posted a way forward that avoids the "messiness" of the double sum and ensuing manipulation thereof. - Mark – Mark Viola Feb 2 '16 at 3:55

Let the Fourier Series representation for $f$ and $g$ be written respectively by

$$f(x)=\sum_{n=-\infty}^\infty \alpha_ne^{inx}$$

$$g(x)=\sum_{n=-\infty}^\infty \beta_ne^{inx}$$

with coefficients $\alpha_n$ and $\beta_n$ given by

$$\alpha_n=\frac1{2\pi}\int_{-\pi}^\pi f(t)e^{-int}\,dt$$

$$\beta_n=\frac1{2\pi}\int_{-\pi}^\pi g(t)e^{-int}\,dt$$

The convolution of $f$ and $g$ is defined as

\begin{align} (f*g)(x)&\equiv \frac1{2\pi}\int_{-\pi}^\pi f(t)g(x-t)\,dt\\\\ &=\frac1{2\pi}\int_{-\pi}^\pi f(t)\left(\sum_{n=-\infty}^\infty \beta_n e^{in(x-t)}\right)\,dt\\\\ &=\sum_{n=-\infty}^\infty \beta_n e^{inx}\left(\frac1{2\pi}\,\int_{-\pi}^\pi f(t)e^{-int}\,dt\right)\\\\ &=\sum_{n=-\infty}^\infty \alpha_n\,\beta_n e^{inx} \end{align}

and we are done!

• Nice answer :) I haven't seen this approach before. – Cameron Williams Feb 2 '16 at 4:56
• @CameronWilliams Thank you! And glad to see you're back Cam. - Mark – Mark Viola Feb 2 '16 at 5:24
• Yeah I've been in a bit of a rut lately so I've been pretty scarce around here. But it has gotten better now :) – Cameron Williams Feb 2 '16 at 6:02

Since Dr. MV already posted an answer, I'll post the one I alluded to in my comment:

Convolving $f$ with $g$ gives

\begin{align} (f*g)(x) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \left(\sum_m \alpha_m e^{imt}\right) \left(\sum_n \beta_n e^{in(x-t)}\right)\,dt \\ &= \frac{1}{2\pi} \sum_m\sum_n \alpha_m\beta_n e^{inx} \int_{-\pi}^{\pi} e^{i(m-n)t}\,dt. \end{align}

This integral evaluates to $2\pi$ if $m=n$ and $0$ otherwise. (Check this yourself by breaking it into these two cases.) Meaning that the double sum is zero if $m\neq n$ so really you can drop down to one sum to get

$$(f*g)(x) = \sum_n \alpha_n \beta_n e^{inx}.$$

• How can you combine the two sums like that? – Matt G Feb 2 '16 at 17:53
• Try working it out with finite sums to see why it works. – Cameron Williams Feb 2 '16 at 17:58