How do I let Fourier Transform know that it is a linear operator?

Question

I have the following issue with FourierTransform, Fourier transform doesn't seem to know it is a linear operator.

ClearAll["Global`*"]
aa = FourierTransform[s1[t],t,omega];
bb = FourierTransform[s1'[t],t,omega];

cc = FourierTransform[s1[t] + s1'[t],t,omega];

Expand[aa+bb] == Expand[cc]

This yields false (more precisely, it doesn't yield true). I would expect the Fourier transform to know it is a linear operator. Do I have to do something like the following?

fourierTransformSubs = 
{
FourierTransform[a1_ f_[t_],t_,omega_] :>  a1 FourierTransform[f[t],t,omega],
FourierTransform[f_[t_]+g_[t_] ,t_,omega_] :> FourierTransform[f[t],t,omega] + 
                                              FourierTransform[g[t],t,omega]
}

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Are you asking how to show the identity with s[t] defined or undefined? — bbgodfrey, Sep 26 '15 at 01:36
s[t] is undefined. I want to do things symbolically for now. — fred, Sep 26 '15 at 01:49
Well, it's an interesting question what should happen for the symbolic case. It's possible that a sum has a Fourier transform but its terms don't. — John Doty, Sep 26 '15 at 02:59
Hmmm, I have to think about that. However is it possible that the Fourier transform of a function times a nonzero scalar exists, but the Fourier transform for the functions doesn't? — fred, Sep 26 '15 at 03:07

How do I let Fourier Transform know that it is a linear operator?

0 Answers0