Does absolute and uniform convergence imply normal convergence? Let $f_n$ be a sequence of holomorphic functions defined on some compact set $K$ in $\mathbb C$ (i.e. each function in the sequence is holomorphic on some nbd of $K$).
If $\sum f_n$ converges uniformly and absolutely on $K$ then can we say it coverges normally on $K$ (i.e $\sum \Vert f_n\Vert$ is finite)?
Note- It’s very easy to construct counterexamples for smooth functions using bump functions.
I am curious about this because in J. B. Conway’s Complex Analysis book, in one lemma (Lemma 5.8, Chapter 7) needed for Weierstrass factorization theorem, he states uniform and absolute convergence in the hypothesis but uses normal convergence in the proof.
 A: We can also construct counterexamples using holomorphic functions. Let $K$ be the closed unit disk, define $\zeta_n = \exp(i/n)$ for $n \in \mathbb{N}\setminus \{0\}$, and $h(z) = \frac{1}{2}(1 + z)$. Choose positive integers $m_n$ such that
$$\lvert h(z)\rvert^{m_n} \leqslant 2^{-n}$$
for all $z \in K$ with $\lvert z-1\rvert \geqslant \frac{1}{10n^2}$ and put
$$f_n(z) = \frac{1}{n}h(\zeta_n z)^{m_n}\,.$$
Then $\lVert f_n\rVert = \frac{1}{n}$, so the series is not normally convergent on $K$. But it is absolutely and uniformly convergent on $K$. For every $z \in K$, we have $\lvert f_n(z)\rvert \leqslant 2^{-n}$ for all but at most one $n$, so
$$\sum_{n = m}^{\infty} \lvert f_n(z)\rvert \leqslant \frac{1}{m} + \sum_{n = m}^{\infty} 2^{-n} = \frac{1}{m} + 2^{1-m}\,.$$
A: This is not as good an answer as Daniel Fischer's, but I include it because it's not too much different from the $C^\infty$ bump function idea that you mentioned.
Choose pairwise disjoint closed intervals $[a_n,b_n],n=1,2,\dots$ contained in $[0,1].$ For each $n,$ choose a continuous $f_n$ on $[0,1],$ with support $[a_n,b_n],$ such that $\|f_n\|=1/n.$ Then $\sum f_n$ converges uniformly on $[0,1].$
By the Weierstrass approximation theorem, there exist polynomials $p_n$ such that $\|p_n-f_n\|<1/2^n.$ Claim: $\sum_{n=1}^{\infty}p_n$ has the desired property on $K=[0,1].$
Proof of claim: For each $x\in [0,1],$
$$\sum_{n=1}^{\infty}|p_n(x)| \le \sum_{n=1}^{\infty}|p_n(x)-f_n(x)| + \sum_{n=1}^{\infty}|f_n(x)| <\infty.$$
So $\sum p_n$ converges absolutely on $[0,1].$ Given $M<N,$ we have
$$|\sum_{M}^{N}p_n(x)| \le |\sum_{M}^{N}(p_n(x)-f_n(x))|+|\sum_{M}^{N}f_n(x)|,\,\, x\in [0,1].$$
This implies the partial sums of $\sum p_n$ are uniformly Cauchy on $[0,1],$ hence $\sum p_n$ converges uniformly on $[0,1].$ Since $\sum \|f_n\| = \infty,$ the claim is proved.
A: Maybe this is relevant for you: In Remmert's excellent book Theory of Complex Functions, he says on page 106 in the section on normal convergence:

If $X$ is a locally compact metric space and $\sum \| f_\nu \|_K < \infty $ for every compactum $K$ in $X$, then $\sum f_\nu$ is normally convergent in $X$.

