Fourier Transforms - The Convolution Theorem.

[binbagsss]

Ok so I've seen the convolution theorem written as:

F(h(x)\otimesg(x))=H(k)G(k)

(And this is how it appears when I have a quick google).

My book then does a problem in which is uses:

F(h(x)g(x))=H(k)\otimesG(k)

Where H(k)=F(h(x)) and similarly G(k)=F(g(x)),
and F represents a Fourier transform


My question
- I can't see how these are equivalent at all?

Many Thanks to anyone who can help shed some light !

 

[SteamKing]

binbagsss:
Please post your math questions in one of the Math HW forums. This helps to organize the questions so that posters can receive appropriate responses to questions.

 

[LCKurtz]

Ok so I've seen the convolution theorem written as:

F(h(x)\otimesg(x))=H(k)G(k)

(And this is how it appears when I have a quick google).

My book then does a problem in which is uses:

F(h(x)g(x))=H(k)\otimesG(k)

Where H(k)=F(h(x)) and similarly G(k)=F(g(x)),
and F represents a Fourier transform


My question
- I can't see how these are equivalent at all?

Many Thanks to anyone who can help shed some light !

Take a look at:
http://ugastro.berkeley.edu/infrared09/PDF-2009/convolution2.pdf

It gives an argument for the formula on the last page.