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Define:

$ \left(f*g\right)\left(x\right)=\frac{1}{2\pi}\intop_{-\pi}^{\pi}f\left(t\right)g\left(x-t\right)dt $

I want to prove that $ f*\left(g*h\right)=\left(f*g\right)*h $

I havent really learned double integrals so I used some information I found online and Im not sure that I've change variables correctly. here's what I've tried:

By definition:

$ f*\left(g*h\right)\left(x\right)=\frac{1}{2\pi}\intop_{-\pi}^{\pi}f\left(t\right)\left(g*h\right)\left(x-t\right)dt=\frac{1}{2\pi}\intop_{-\pi}^{\pi}f\left(t\right)\left(\frac{1}{2\pi}\intop_{-\pi}^{\pi}g\left(s\right)h\left(x-t-s\right)ds\right)dt $

$ =\frac{1}{2\pi}\cdot\frac{1}{2\pi}\intop_{-\pi}^{\pi}\intop_{-\pi}^{\pi}f\left(t\right)g\left(s\right)h\left(x-t-s\right)dsdt $

Now change variables :

$ s=u-v,t=v $

so $ dsdt=\begin{vmatrix}\frac{\partial s}{\partial u} & \frac{\partial s}{\partial v}\\ \frac{\partial t}{\partial u} & \frac{\partial t}{\partial v} \end{vmatrix}dudv$

Thus

$ dsdt=\begin{vmatrix}1 & -1\\ 0 & 1 \end{vmatrix}dudv=dudv $

My problem is with the bounderies of the new integral.

Is this correct ? :

$ f*\left(g*h\right)\left(x\right)=\frac{1}{2\pi}\cdot\frac{1}{2\pi}\intop_{-\pi}^{\pi}\intop_{v-\pi}^{v+\pi}f\left(v\right)g\left(u-v\right)h\left(x-u\right)dudv $

$ =\frac{1}{2\pi}\intop_{v+\pi}^{v-\pi}\left(\frac{1}{2\pi}\intop_{-\pi}^{\pi}f\left(v\right)g\left(u-v\right)dv\right)h\left(x-u\right)du $

Feels a bit of a lie because Im integrating over the variable $ v $ while v still appear outside of the $ dv $ integral (in the bounderies of the outside intergral )

If its not correct I'll be glad if someon can show me how to change the variables correctly (namely what to do with the bounderies)

if its true it will solve that problem because $ (f*g) $ is periodic with period $ 2\pi $ according to former assumptions in this discussion, so if I could treat the $ v $ in the integral as a regular constant, that wouldnt bother me since the length of the segment is $ 2\pi $.

Thanks in advance.

FreeZe
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  • @b00nheT I just want to know if the variables change is correct regardless to the whole proof. – FreeZe Nov 18 '20 at 10:00
  • It's not a lie, but it's not true either. Draw the region $-\pi < v < \pi \land -\pi < u - v < \pi$, you'll see that if $v$ is the inner variable, the iterated integral should be $$\left( \int_{-2 \pi}^0 \int_{-\pi}^{u + \pi} + \int_0^{2 \pi} \int_{u - \pi}^\pi \right) F(u, v) , dv du.$$ – Maxim Nov 18 '20 at 17:33
  • @Maxim but t=v exactly ao how come the bouderies of the dv integral are not $ -\pi $ and $ \pi $ – FreeZe Nov 18 '20 at 17:58

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