Question

Suppose\[\int\limits_{-\infty}^{\infty} |f(x)|dx < \infty\]\[\int\limits_{-\infty}^{\infty} |g(x)|dx < \infty\]

Prove that the fourier transform of f*g is the fourier transform of f times the fourier transform of g. Where (f * g) is the convolution of f and g. \[\hat{f*g}=\hat{f}\hat{g}\]
You will need:\[(f*g)(x)=\int\limits_{-\infty}^{\infty}f(x-y)g(y)dy\]

Answer 1

lets apply the convolution integral
\[\Large F(f(x).g(x))=F([\int\limits_{-\infty}^{\infty}f(u)g(x-u)du])\]
\[\Large \int\limits_{-\infty}^{\infty}([\int\limits_{-\infty}^{\infty}f(u)g(x-u)du])e^{\iota sx}dx\]

Answer 2

it is difficult to type in Latex :(
wait let me type it somewhere else .

Answer 3

continue from above.
it is more readable !

Answer 4

@abstracted are you ok with this ?

Answer 5

Yes, thanks