mfw you're trying to take the Fourier Transform of a distribution
I don't get it, and the world's quickest and last comprehensive DDG search didn't find an explanation either.
It's a reach, but the Fourier transformation of a Schwarz (rapidly decaying) function is also a Schwarz function. Compact support is a strictly stronger condition than Schwarz (the function must eventually decay to 0) but doesn't have this nice property with respect to Fourier transforms, i.e. the FT of a compactly supported function is Schwarz but not necessarily compactly supported
