I'm trying to understand exactly what restrictions a finite duration time domain signal has on its Fourier transform.
I found a helpful table in The Fourier Transform and its Applications by Ronald Bracewell:
Where the comb symbol is defined by the usual $\mathrm{III}(x) = \Sigma_n \delta(x - n) $.
I know that a time domain signal sampled at rate $M$, by the normal Nyquist limit, will only have useful frequency content up to $M/2$. But this table seems to imply that the Fourier transform of such a signal will have frequency content up to $M$?
Similarly I'm confused about the Fourier domain equivalent of this statement (is there a name for this result? I haven't been able to find much searching "Fourier domain equivalent of Shannon-Nyquist"). I read the table as stating that a signal with duration $2T=D$ in the time domain will have a minimum resolvable frequency of $1/T = 2/D$ in the frequency domain. Is there a factor of 2 missing here as above, for useful information? Is there a Fourier-domain equivalent of Shannon-Nyquist?