Concerning the question about Nyquist theorem (assuming equally time-spaced sampling points):
You are always stuck to the Nyquist theorem, which means that you cannot directly measure the amplitude $B_\omega$ at a specific angular frequency $\omega$, but will always evaluate $B_{\omega,\textrm{meas}} = \sum_{i = 0}^{\infty}B_{\omega+i \omega_s}$, where $\omega_s$ denotes the angular frequency of your sampling process.
This theorem is independent of the algorithm evaluating $B_\omega$.
But you can use this to reduce the sampling rate to at least two times the bandwidth of your signal.
Usually one lowpass filters the signal before sampling to ensure there is no signal for $i>0$. But of course you can also bandpass filter at a higher frequency to only measure a signal for some $i>0$. Then you will sample the alias of your real signal.
Example:
Considering you want to analyse the spectrum from $\omega_0 = 2\pi\, 100\, \mathrm{M} \mathrm{rad}^{-1}$ to $\omega_1 = 2\pi\, 101\, \mathrm{M} \mathrm{rad}^{-1}$ you could sample with $\omega_s = 2\pi\, 2\,\mathrm{M} \mathrm{rad}^{-1}$ and you would get signal for $B_\omega$ with $\omega \leq 2\pi\, 1\, \mathrm{M} \mathrm{rad}^{-1}$ (assuming the analogue parts of your hardware are capable of handling $\omega_1$).
(In real world, as the bandpass filter has to filter from $\omega_0 = 2\pi\, 100\, \mathrm{M} \mathrm{rad}^{-1}$ to $\omega_1 = 2\pi\, 101\, \mathrm{M} \mathrm{rad}^{-1}$, you would have to use a slight higher sampling rate as $\omega_s = 2\pi\, 2\,\mathrm{M} \mathrm{rad}^{-1}$, as there is no perfect rectangular bandpass filter.)
