# How do I let Fourier Transform know that it is a linear operator? [duplicate]

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 1:36
• s[t] is undefined. I want to do things symbolically for now. – fred Sep 26 '15 at 1:49
• 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 2:59
• – xzczd Sep 26 '15 at 8:46