Nyquist frequency , sampling distance I have few questions I tried to solve regarding nyquist theorem, and I would like to see your opinion if I'm doing it correctly?(one I know the answer second one not sure).

1.Let $f(x)$ and $g(x)$ be functions with max frequencys $B_1$ and $B_2$, what is the max sampling distance allowed according to nyquist to the function $h(x)=f(x)g(x)$.


My answer: according to the convolution theorem $H(x)=F(x)*G(x)$(* for convolution), so $H(x)=\int_{-B_1}^{B_1}F(u)G(x-u)du$ so $u \leq B_1$ now we knowthat $x-u \leq B_2$ so we get $x\leq B_2+B_1$ and this $x$ is our max frequency, now I can say that our sample rate is $f_s\geq 2(B_1+B_2)$ so our sample max distance denote $T_s$ is $T_s \leq \frac{1}{2(B_1+B_2)}$
is that correct?


2.lets mark function $f(x,y)=sinx*cosy$ what is the max sampling rate?


to be honest here I'm really not sure, I think its zero by intuition but not really sure if I'm correct and if so why?


Please note: those are the notation I saw when I found those questions, thanks
Your answer to the first question is right, provided that the signals involved are real and lowpass (baseband) nature.

The second question is tricky, assuming ideal infinite length sinusoidals, your intuition is correct in the sense that their Fourier transforms multiplied would yield zero (unless their frequencies were the same), and has zero bandwidth. Also note that the convolution was divergent in the first place, if that's a problem for you.

However, any practical implementation of that algorithm would include windows and therefore result of Frequency domain multiplication will be nonzero and the result would be non bandlimited, hence cannot be sampled without anti-aliasing filtering.