Convolutions, Fourier coefficients

[Incand]

Homework Statement

When ##f## and ##g## are ##2\pi##-periodic Riemann integrable functions define their convolution by
##(f*g)(x) = \frac{1}{2\pi} \int_0^{2\pi} f(y)g(x-y)dy##
Denoting Fourier coefficients by ##c_n(f)## show that ##c_n(f * g) = c_n(f)c_n(g)##.

Homework Equations

##c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-inx}dx##

The Attempt at a Solution

We note that in both ##c_n(f*g)## and ##c_n(f)c_n(g)## get a factor of ##\frac{1}{4\pi^2}## before the integrals so all we have to show is that
##\int_{-\pi}^\pi e^{-inx}\int_0^{2\pi} f(y)g(x-y)dydx = \int_{-\pi}^\pi f(x)e^{-inx}dx \int_{-\pi}^\pi g(y)e^{-iny}dy##
Starting with the left side and changing the order of integration we get
##\int_{-\pi}^\pi e^{-inx}\int_0^{2\pi} f(y)g(x-y)dydx = \int_0^{2\pi} f(y) \int_{-\pi}^\pi g(x-y)e^{-inx}dx.##
Making a change of variables ##z=x-y##
##\int_0^{2\pi} f(y)e^{-iny} \int_{-\pi-y}^{\pi-y} g(z)e^{-inz}dzdy##
which seems somewhat promising except I have the wrong integration interval. Any hints on what I should do to show this? I also tried things like partial integration with derivation under the integral sign but that didn't seem to help me any.

 

[Svein]

First: ##c_{n}(f)\cdot c_{n}(g)=\int_{-\pi}^{\pi}f(u)e^{-inu}du\cdot \int_{-\pi}^{\pi}g(v)e^{-inv}dv=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(u)g(v)e^{-in(u+v)}dudv##. Then let ##v=x-u##:
##c_{n}(f)\cdot c_{n}(g)=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(u)g(v)e^{-in(u+v)}dudv=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(u)g(x-u)e^{-inx}dudx=\int_{-\pi}^{\pi}e^{-inx}(\int_{-\pi}^{\pi}f(u)g(x-u)du)dx##. Some justifications are needed...

 

[Incand]

First: ##c_{n}(f)\cdot c_{n}(g)=\int_{-\pi}^{\pi}f(u)e^{-inu}du\cdot \int_{-\pi}^{\pi}g(v)e^{-inv}dv=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(u)g(v)e^{-in(u+v)}dudv##. Then let ##v=x-u##:
##c_{n}(f)\cdot c_{n}(g)=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(u)g(v)e^{-in(u+v)}dudv=\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}f(u)g(x-u)e^{-inx}dudx=\int_{-\pi}^{\pi}e^{-inx}(\int_{-\pi}^{\pi}f(u)g(x-u)du)dx##. Some justifications are needed...

Thanks for responding! You seem to be doing a similar approach to mine except the other way around. I'm having trouble understanding how you keep the same integration interval when changing variables in the next last step. If you set ##v=x-u##
shouldn't ##x=\pi+u## when ##v=\pi## for example? Perhaps this is one of the justifications I need to discover. I know the change of integration order should be fine since the functions are integrable (i.e. the integrals are convergent).

The other problem is that the convolution integral was defined from ##0\to 2\pi##. Having ##e^{-in\pi} = e^{in\pi}## could possibly help I guess but I don't really get anywhere here either.

Edit: Since ##f## and ##g## is ##2\pi##-periodic perhaps I'm allowed to switch the interval, have to think on this a bit.

 

[Svein]

Since ff and gg is 2π2\pi-periodic perhaps I'm allowed to switch the interval, have to think on this a bit.

Exactly!

The justifications are needed elsewhere.

 

[Incand]

Exactly!

The justifications are needed elsewhere.

I think I get how to change the integration inteval in the last integral now.
The inner integral is
##\int_{-\pi}^\pi f(u)g(x-u)du = \int_{-\pi}^0 f(u)g(x-u)du + \int_0^\pi f(u)g(x-u)du = \int_\pi^{2\pi} f(u)g(x-u)du + \int_0^\pi f(u)g(x-u)du = \int_0^{2\pi} f(u)g(x-u)du## since ##f## and ##g## are two periodic and then so is ##fg##.

And the variable substitution similary
##\int_{-\pi}^\pi f(u) \int_{-\pi}^\pi g(v) e^{-in(u+v)}dvdu = \int_{-\pi}^\pi f(u) \int_{-\pi + u}^{\pi+\pi} g(x-u)e^{-inx}dxdu##
Lets take a closer look at the inner integral
##\int_{-\pi + u}^{\pi+u} g(x-u)e^{-inx}dx = \int_{-\pi}^{\pi} g(x-u)e^{-inx}dx##
since ##f,g## and ##e^{ix}## are all ##2\pi##-periodic. In fact I could've done both these calculations in one step knowing this.
This are all the justifications needed right or is there something else i missed? Thanks for helping!

 

[Svein]

This are all the justifications needed right or is there something else i missed?

You need to quote the Fubini theorem when manipulating double integrals. In this case the theorem is trivially satisfied.