Let $0 \in \mathbf{N}$. Let $P_m(x): [0,1] \to \mathbf{C}$ be bounded analytic functions for every $m\in \mathbf{N}$. Formally, define $$ f(x) = \sum_{m\in \mathbf{N}}c_m P_m(x)\overline{P_m}(x), $$ where $ \overline{P_m}(x) $ is the complex conjugate of $P_m(x)$.
Conjecture. There exist coefficients $c_m > 0$ such that $f$ is also analytic in $[0,1]$. Furthermore, $$ f^{(n)}(x) = \sum_{m\in \mathbf{N}}c_m \frac{d^n}{dx^n}\left(P_m(x)\overline{P_m}(x)\right). $$
Ideas. It is my understanding that since $P_m$ is analytic, then $\overline{P_m}$ is also analytic. Additionally, the product of two analytic functions is also analytic. Hence, $P_m(x)\overline{P_m}(x)$ is also analytic. So, it suffices to prove that an absolutely convergent series of analytic functions is itself analytic. For sake of notation, let $q_m(x) = P_m(x)\overline{P_m}(x)$. As $q_m$ is analytic, we have that $$ q_m(x) = \sum_{n\in \mathbf{N}}d_{mn} x^n $$ for some coefficients $d_{mm}$. So, we have $$ f(x) = \sum_{m\in \mathbf{N}}c_m P_m(x)\overline{P_m}(x) = \sum_{m\in \mathbf{N}}c_m \sum_{n\in \mathbf{N}}d_{mn} x^n. $$ Therefore, if $$ \sum_{m\in \mathbf{N}}\sum_{n\in \mathbf{N}}|c_md_{mn}| < \infty, $$ then $f$ would be analytic as that would allow us to change the order of summation. As $q_m$ is analytic, its power series is absolutely convergent. Hence, $$ \sum_{n\in \mathbf{N}}|d_{mn}| < M_{m} < \infty. $$ Therefore, let $$ c_m = \frac{1}{2^m M_m}. $$ Thus, $$ \sum_{m\in \mathbf{N}}\sum_{n\in \mathbf{N}}|c_md_{mn}| < \infty, $$ and $$ f(x) = \sum_{m\in \mathbf{N}}c_m P_m(x)\overline{P_m}(x) = \sum_{m\in \mathbf{N}}c_m \sum_{n\in \mathbf{N}}d_{mn} x^n = \sum_{n\in \mathbf{N}}\left(\sum_{m\in \mathbf{N}}c_md_{mn}\right)x^n, $$ implying $f$ is analytic.
Questions. Does the prior argument work? Did I make any glaring mistakes? Also, I'm quite stuck on showing that the summation and differentiation commute in this instance. I believe it suffices in showing that $$ \sum_{m\in \mathbf{N}}c_m \frac{d^n}{dx^n}\left(P_m(x)\overline{P_m}(x)\right) $$ converges uniformly. Does this immediately follow from the assumptions that the $P_m$ functions are analytic hence continuous. And from the fact that $$ \sum_{m\in \mathbf{N}}c_m \left|\frac{d^n}{dx^n}\left(P_m(x)\overline{P_m}(x)\right)\right| $$ is continous for all $n$. Or does that not hold. Any advice or counterexamples would be appreciated.