I'm reading through a paper that states at some point something like
A set of $N$ coefficients, $y[n]$, are computed by convolving with a kernel $h(x)$ and uniformly sampling the output with sample spacing $\Delta_x = \frac{2\pi}{N}$: $$ y[n] = \int_{0}^{2\pi} dx h(n \Delta_x - x) f(x), \;\; n \in \left\{0,1,\ldots,N-1 \right\}$$
I cannot manage to see actually why.
If $f(x)$ is in $L^2([-T/2,T/2])$ the fuorier coefficients can be computed as
$$ y[n] = \frac{1}{T} \int_{-T/2}^{T/2} f(x)e^{-i2\pi\frac{nx}{T}} dx $$
and the only clue I have is trying to manipulate
$$ y[n+k] = \frac{1}{T} \int_{-T/2}^{T/2} f(x)e^{-i2\pi\frac{(n+k)x}{T}} dx $$
but I cannot see any translation, can anyone give me a clue of what's happening?
$$L^{p}(\Gamma) \subset L^{q}(\Gamma), \quad p > q$$ whenever $\Gamma$ is a set of finite measure. Therefore the Fourier coefficients in your question are well defined for $f \in L^{2}(-T/2,T/2)$ because using the fact I mentioned above $f \in L^{2}(-T/2,T/2)$ $\implies$ $f \in L^{1}(-T/2,T/2)$.
Therefore,
$$| y[n] | = \frac{1}{T}\Big| \int_{-T/2}^{T/2}f(x) e^{2i\pi nx/T}\Big| \le \int_{-T/2}^{T/2}|f(x)| < \infty$$