$ x\in \mathbb Z $
$$(f*g)(x)= \sum_{k=0}^{N-1} f(k) g(x-k).$$
How can I prove the commutativity of the convolution product with this expression and without integrals?
I tried with the substitution $j= x-k$ but I have
$$ (f*g)(x)= \sum_{k=0}^{N-1} f(k) g(x-k) = \sum_{j=x-N +1 }^x f(x-j) g(j) $$
and I don't know what to do next.
For the discrete convolution to be commutative, I think you'd need either the bounds of summation to be $-\infty,+\infty$ or suppose some sort of periodicity. As you're summing between $0$ and $N-1$, let's say that $f$ and $g$ are $N$-periodic.
The sum you obtained on the right is perfectly all right, the only problem is that you sum between $x-N+1$ and $x$. But if you look careful, there are $N$ terms in there, so it's exactly like summing between $0$ and $N-1$ by periodicity.