The key assumption is that the signal is band-limited. This is a frequency-domain assumption. Any sensible proof must go through the frequency domain.
Same goes for the proof being an approximation argument. Any real proof must be an approximation argument, including the one you alluded to, or its rigorous version.
Here is another Hilbert space argument that, in the end, gives us approximation in the topology of uniform convergence (much better than $L^1$ or $L^2$):
Let $H$ be the set of band-limited elements of $L^2(\mathbb{R})$ (no extra assumptions). By unitarity of the Fourier transform $\mathcal{F}$, $H$ is a Hilbert subspace of $L^2$. Since $L^2$ elements of compact support lie in $L^1$, the Fourier inversion theorem implies that elements of $H$ are in fact continuous almost everywhere. So band-limited assumption implies (for now) continuity and therefore sampling make sense.
$\mathcal{F}(H)$ has orthonormal basis $\{ e^{- 2 \pi i k\frac{\xi}{2T}} \}_{k \in \mathbb{Z}}$. So now it's natural to compute the Hilbert space expansion of $\hat{f}$ in this basis then apply $\mathcal{F}^{-1}$. By unitarity
$$
\langle \hat{f}, e^{- 2 \pi i k\frac{\xi}{2T}} \rangle = \langle f, \delta_{\frac{k}{2T}}\rangle = f(\frac{k}{2T}).
$$
Strictly speaking, one needs a rigged Hilbert space that includes distributions to make sense of inner products with delta functions but everything works out. On the other hand, the inverse Fourier transform of the basis $\{ e^{- 2 \pi i k\frac{\xi}{2T}} \}_{k \in \mathbb{Z}}$ are just shifts of the $\mbox{sinc}$ function. So we have that Shannon's sampling formula holds in the $L^2$-sense.
To strengthen the convergence, notice $L^2$-convergence (in the frequency domain) implies $L^1$-convergence by the band-limited assumption. By property of $\mathcal{F}^{-1}$, back in the time domain we have uniform convergence.
Since $\mbox{sinc}$ functions and its shifts are all smooth, we can actually conclude that a band-limited $L^2$ function is in fact smooth almost everywhere.
Hint: let me ask you something: How do the plots look when $k$ is an integer multiple of the fundamental frequency, compared to when it is not? Have you tried varying the sampling rate (say, 100 -> 50) to see what the effect on the plots is?
My guess is that you will see very sharp peaks when $k$ is an integer multiple; you will see slightly wider peaks when $k$ is not; and these peaks will increase in wideness as the sampling rate decreases.
Can you figure out what is going on?
Best Answer
Sampling rate ν is also in Hz. It is the number of observations per second - the rate at which you are sampling data.
Both sides of the equation are in the same units. Hertz.
And that’s a Greek letter ν (nu), not an English letter v.