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Proof of the convolution theorem

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Homework Statement


With the Fourier transform of f(x) defined as F(k)=1/√(2π)∫-∞dxf(x)e-ikx and a convolution of g(x) and h(x) defined by f(x)=[g*h](x)=∫-∞h(x-y)g(y)dy show that the Fourier transform of f(x) equals √(2π)H(k)G(k).

Homework Equations


In problem

The Attempt at a Solution


So I understand the general idea, but I'm stuck on a few little issues. I'm going to drop the integral limits from now on.

F(k)=1/√(2π)∫dx{∫h(x-y)g(y)dy}e-ikx (we can put e-ikx into the integral over y, allowing the next step)

Now the idea is to let z=x-y (we want to get h(z) which will lead us to the form of a Fourier transform for h). However all proofs seems to say this means dx=dz to give

F(k)=1/√(2π)∫dz{∫h(z)g(y)e-ikye-ikzdy}

The question is, why isn't dz=d(x-y)=dx-dy used (given no sources actually discuss this I gather I am being very silly). I also seem

Then
F(k)=√(2π)[1/√(2π)∫h(z)e-ikzdz][1/√(2π)∫g(y)e-ikydy]
=√(2π)H(k)G(k)

If anybody could explain that it would be great, thanks :)
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


With the Fourier transform of f(x) defined as F(k)=1/√(2π)∫-∞dxf(x)e-ikx and a convolution of g(x) and h(x) defined by f(x)=[g*h](x)=∫-∞h(x-y)g(y)dy show that the Fourier transform of f(x) equals √(2π)H(k)G(k).

Homework Equations


In problem

The Attempt at a Solution


So I understand the general idea, but I'm stuck on a few little issues. I'm going to drop the integral limits from now on.

F(k)=1/√(2π)∫dx{∫h(x-y)g(y)dy}e-ikx (we can put e-ikx into the integral over y, allowing the next step)

Now the idea is to let z=x-y (we want to get h(z) which will lead us to the form of a Fourier transform for h). However all proofs seems to say this means dx=dz to give

F(k)=1/√(2π)∫dz{∫h(z)g(y)e-ikye-ikzdy}

The question is, why isn't dz=d(x-y)=dx-dy used (given no sources actually discuss this I gather I am being very silly). I also seem

Then
F(k)=√(2π)[1/√(2π)∫h(z)e-ikzdz][1/√(2π)∫g(y)e-ikydy]
=√(2π)H(k)G(k)

If anybody could explain that it would be great, thanks :)
Use the standard two-dimensional change-of-variable formula from ##(x,y)## to ##(u,v)##, where ##u = y, v = x-y##:
[tex] dx \, dy = \frac{\partial(x,y)}{\partial(u,v)} du \, dv [/tex]
Here, ##\partial(x,y)/\partial(u,v)## denotes the Jacobian of the "transformation" ##x(u,v), \, y(u,v)##.
 
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  • #3
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Use the standard two-dimensional change-of-variable formula from ##(x,y)## to ##(u,v)##, where ##u = y, v = x-y##:
[tex] dx \, dy = \frac{\partial(x,y)}{\partial(u,v)} du \, dv [/tex]
Here, ##\partial(x,y)/\partial(u,v)## denotes the Jacobian of the "transformation" ##x(u,v), \, y(u,v)##.
Ahh, silly me, never thought about it as a 2D variable change!

So we have
F(k)=1/√(2π)∫{∫h(x-y)g(y)e-ikxdy}dx

(x,y)->(w,z) where w=y, z=x-y (so that x=w+z, y=w)
Then the Jacobian |∂(x,y)/∂(w,z)|=|(∂x/∂w)(∂y/∂z)-(∂x/∂z)(∂y/∂w)|=|(1)(0)-(1)(1)|=1
Then dxdy=dwdz=dydz and so we're sorted - thanks :)
 
  • #4
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Use the standard two-dimensional change-of-variable formula from ##(x,y)## to ##(u,v)##, where ##u = y, v = x-y##:
[tex] dx \, dy = \frac{\partial(x,y)}{\partial(u,v)} du \, dv [/tex]
Here, ##\partial(x,y)/\partial(u,v)## denotes the Jacobian of the "transformation" ##x(u,v), \, y(u,v)##.
Sorry but I am encountering some trouble with this again:

My notes say that we can show that a convolution
f(x)=∫h(x-y)g(y)dy
is symmetric by writing z=x-y so then (dz=-dy, limits swap)
f(x)=∫h(z)g(x-z)dz
and noting that z is just a dummy variable so we could call it y again if we really wanted so
f(x)=∫h(y)g(x-y)dy.

However how does dy=-dz in the variable change here? We don't have a 2D integral to solve the issue here so I'm confused...

Edit: I gather f(x) is f evaluated at some fixed point x, so that x is fixed and dx=0 here, unlike in the double integral... Correct?
 
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  • #5
Ray Vickson
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Sorry but I am encountering some trouble with this again:

My notes say that we can show that a convolution
f(x)=∫h(x-y)g(y)dy
is symmetric by writing z=x-y so then (dz=-dy, limits swap)
f(x)=∫h(z)g(x-z)dz
and noting that z is just a dummy variable so we could call it y again if we really wanted so
f(x)=∫h(y)g(x-y)dy.

However how does dy=-dz in the variable change here? We don't have a 2D integral to solve the issue here so I'm confused...

Edit: I gather f(x) is f evaluated at some fixed point x, so that x is fixed and dx=0 here, unlike in the double integral... Correct?
In your first integral, ##y## goes from ##-\infty## to ##+\infty##, while in the second, ##z## goes from ##+\infty## to ##-\infty##. When you re-write that as ##z## going from ##-\infty## to ##+\infty##, your ##-dz## will become a ##+dz##.
 
  • #6
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In your first integral, ##y## goes from ##-\infty## to ##+\infty##, while in the second, ##z## goes from ##+\infty## to ##-\infty##. When you re-write that as ##z## going from ##-\infty## to ##+\infty##, your ##-dz## will become a ##+dz##.
In your first integral, ##y## goes from ##-\infty## to ##+\infty##, while in the second, ##z## goes from ##+\infty## to ##-\infty##. When you re-write that as ##z## going from ##-\infty## to ##+\infty##, your ##-dz## will become a ##+dz##.
The question was, why is dx zero? But I think I may have answered it myself...
 
  • #7
Ray Vickson
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The question was, why is dx zero? But I think I may have answered it myself...
Here was your question: "However how does dy=-dz in the variable change here? We don't have a 2D integral to solve the issue here so I'm confused..."
 
  • #8
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Here was your question: "However how does dy=-dz in the variable change here? We don't have a 2D integral to solve the issue here so I'm confused..."
Sorry I guess it wasn't that clear - but z=x-y, dz=dx-dy, so how does dy=-dz (i.e how is dx=0)...
 
  • #9
Ray Vickson
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Sorry I guess it wasn't that clear - but z=x-y, dz=dx-dy, so how does dy=-dz (i.e how is dx=0)...
Have you forgotten that you are trying to equate a product of FTs to the FT of the convolution? That means that you will be integrating over ##x## as well (times ##e^{\pm i kx}##, of course), but that ##x## lies outside the ##y## or ##z## integral, so is like a constant when you are looking at the convolution for fixed ##x##.
 

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