Property of the convolution product

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SUMMARY

The convolution property of the Fourier transform states that the Fourier transform of the convolution of two functions is the product of their individual Fourier transforms: \(\mathcal{F}\{f \ast g\} = \mathcal{F}\{f\} \mathcal{F}\{g\}\). This discussion explores whether this property holds for the inverse Fourier transform as well, proposing that \(\mathcal{F}^{-1}\{F \ast G\} = \mathcal{F}^{-1}\{F\} \ast \mathcal{F}^{-1}\{G\}\) should be valid due to the similarity in the exponential kernel. The conclusion suggests that there is no apparent reason for the convolution property to not apply to the inverse transform, despite a lack of formal affirmation in existing literature.

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Jhenrique
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It is known that:\mathcal{F}\{f\ast g\}=\mathcal{F}\{f\}\mathcal{F}\{g\}\mathcal{F}\{f g\}=\mathcal{F}\{f\}\mathcal\ast{F}\{g\}

But this property is valid for inverse tranform too?\mathcal{F}^{-1}\{F\ast G\}=\mathcal{F}^{-1}\{F\} \mathcal{F}^{-1}\{G\}\mathcal{F}^{-1}\{F G\}=\mathcal{F}^{-1}\{F\}\ast \mathcal{F}^{-1}\{G\}
 
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Think about this for a moment...
 
Ben Niehoff said:
Think about this for a moment...

The only difference between ##\mathcal{F}## and ##\mathcal{F}^{-1}## is the sign in the exponential in the kernel, thus, appears there isn't reason for the property of the convolution not be valid for inverse transform, I think. Although I never saw this affirmation/property for inverse transform in anywhere.
 
You don't need to worry about the integral definitions. Just use this:

Jhenrique said:
It is known that:\mathcal{F}\{f\ast g\}=\mathcal{F}\{f\}\mathcal{F}\{g\}\mathcal{F}\{f g\}=\mathcal{F}\{f\}\ast\mathcal{F}\{g\}

and apply ##\mathcal F^{-1}## to each line.
 
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Oh yeah! Thank you!
 

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