I have seen in this thread a nice answer where it is shown that Thread
that the adjoint operator of a compact operator is compact by using the Arzela Ascoli theorem. Unfortunately, there is one thing I do not understand: It is not shown that $(f_n)_n$ is closed, so why is this sequence compact then? It is only shown that it is equicontinuous and bounded but not closed and since we derive from Arzela Ascoli that there is a convergent subsequence we probably need compactness.
I should point how we formulated Arzela Ascoli: Let $M \subset C(S)$ where $S$ is a compact set, then $M$ is compact iff: $M$ is bounded, closed and equicontinuous.
Thank you in advance for every helpful comment.