I have been struggling to understand the relation between $z$-transform and the study of analysis but there is something that puzzles me.
Stability of a discrete time system is happens if and only if the Region of Convergence includes the unit circle. In particular, Fourier Analysis and Applications by C. Gasket and P. Witomski (by the way, excellent book) states:
"Assume that the filter $D$ is a convolution system with transfer function $H(z)$ that converges in a nonempty annulus $A$:
i) $D$ is stable if and only if the unit circle $z=1$ is in $A$"
They show a proof but I do not understand it really proofs "if and only if". They state that a system is stable if and only if $\sum_{n=-\infty}^{\infty}|h_n|<+\infty$ and I agree with them.
They state that the previous condition is equivalent to saying that the series:
$H(z)=\sum_{n=-\infty}^{\infty}h_nz^{-n}<+\infty$
is absolutely convergent for $|z|=1$ (I also agree) so that proves i)... and I do not understand that very last point. In my (maybe wrong) opinion, that proves:
$\sum_{n=-\infty}^{\infty}|h_n|<+\infty \Rightarrow |H(z)|_{z=e^{-j\omega}}<+\infty$
but only in one way.
In fact, we know that square summable sequences can have Fourier transform, which does not converge uniformly, so in those cases $|H(z)|_{z=e^{-j\omega}}<+\infty$ but the system is not stable (in continuous time, sinc function is an example but I am not sure if that example works in discrete time). In other words, I do not see that the existence of $z$-transform is linked to absolute convergence.