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        riddles >> hard >> fourier transform  of distributions
        
(Message started by: trusure on Mar 16^th, 2010, 1:57pm)

Title: fourier transform of distributions
Post by trusure on Mar 16^th, 2010, 1:57pm

the Fourier transform of a derivative is:
(d/dx(f))^ = i s (f^)(s)

Now I have the function
f(t)={ sin(t) , if -pi/2 <=t<=pi/2
0 , else
}

how to prove that
(d/dx(f))^ = i s (f^)(s)

in distribution sense; I mean applying the notation
<f^, phi> = <f, (phi)^>
(or any other rules like : <f ', phi> = - <f, (phi)'> )

for any test function phi

thanks

Title: Re: fourier transform of distributions
Post by MonicaMath on Mar 17^th, 2010, 8:40am

Did you find the FT for your function? I will give it to you

f^(s)=sqrt(2/pi) i s cos(s.pi/2)/(s^2 - 1)

use this now.

Title: Re: fourier transform of distributions
Post by Obob on Mar 17^th, 2010, 10:50am

Rather than work with this particular function, you can do it for any (tempered) distribution f as follows (writing D for d/ds and p = phi for the test function to save space):

<i s f^, p> = <f^, i s p> = <f, (i s p)^> = - <f, D(p^)> = <Df, p^> = <(Df)^, p>,

which says that (Df)^ = i s f^. The crucial step here is <f, (i s p)^> = - <f, D(p^)>, which follows from (i s p)^ = - D(p^). This is just a statement about test functions; you can prove it by writing out the definition of the Fourier transform and differentiating under the integral.