Property of the convolution product

[Jhenrique]

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?$$\mathcal{F}^{-1}\{F\ast G\}=\mathcal{F}^{-1}\{F\} \mathcal{F}^{-1}\{G\}$$$$\mathcal{F}^{-1}\{F G\}=\mathcal{F}^{-1}\{F\}\ast \mathcal{F}^{-1}\{G\}$$

 

[Ben Niehoff]

Think about this for a moment...

 

[Jhenrique]

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.

 

[Ben Niehoff]

You don't need to worry about the integral definitions. Just use this:

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.

 

[Jhenrique]

Oh yeah! Thank you!