Given a complex Banach algebra $\mathcal{X}$, and a complex power series $f(z)=\sum_{n=0}^\infty a_nz^n$ with radius of convergence $R>0$, it is not difficult to see that for an $x\in \mathcal{X}$ such that $\limsup_{n}\lVert x^n\rVert^{1/n}<R$, we have that $f(x)=\sum_{n=0}^\infty a_nx^n$ converges. Indeed, on that case, there would exist $0\leq r< R$ and an integer $n_0$ such that $\lVert x^n\rVert^{1/n}<r$ for every $n\geq n_0$. Therefore $f(x)=\sum_{n=0}^\infty a_nx^n$ would converge absolutely: $$ \begin{align} \sum_{n=0}^\infty |a_n|\left\lVert x^n\right\rVert &=\sum_{n=0}^{n_0-1}|a_n|\left\lVert x^n\right\rVert +\sum_{n=n_0}^{\infty}|a_n|\left\lVert x^n\right\rVert\\ &\leq \sum_{n=0}^{n_0-1}|a_n|\left\lVert x^n\right\rVert +\sum_{n=n_0}^{\infty}|a_n|r^n\\ &=\sum_{n=0}^{n_0-1}|a_n|\left\lVert x^n\right\rVert +f(r)-\sum_{n=0}^{n_0-1}|a_n|r^n\\ &<+\infty, \end{align} $$ and so, since $\mathcal{X}$ is a Banach space, $f(x)$ would converge. In particular, any $x\in \mathcal{X}$ such that $\lVert x\rVert<R$ also satisfies $\limsup_{n}\lVert x^n\rVert^{1/n}<R$.
Now, I am interested in proving the following fact:
Proposition. Suppose $f(z)=\sum_{n=0}^\infty a_nz^n$ and $g(z)=\sum_{n=0}^\infty b_nz^n$ are complex power series of radius of convergence $R>0$ and $+\infty$, respectively. Suppose also that $g(f(z))=z$ for all $z\in\mathbb{C}$ such that $|z|<R$. Then, if $\mathcal{X}$ is a Banach algebra and $x\in \mathcal{X}$ is such that $\limsup_{n}\lVert x^n\rVert^{1/n}<R$, we have that $$g(f(x))=x.$$
Note that under the hypothesis of the proposition and by the comments above, $f(x)=\sum_{n=0}^\infty a_nx^n$ converges and, hence, also $g(f(x))=\sum_{n=0}^\infty b_nf(x)^n$ converges, so that $g(f(x))$ is well-defined.
In particular, I am interested in the canonical example in which $\mathcal{X}$ is the Banach algebra of linear bounded operators on a Banach space, $\mathcal{X}=L(X)$, for $X$ a Banach space. If some characteristic of this concrete example would help to prove the proposition, I would be content with that proof, but I suspect that a direct proof would be general for arbitrary Banach algebras.
So far, I've been thinking of trying to expand the series $$ g(f(z))=\sum_{n=0}^\infty b_n \left( \sum_{m=0}^\infty a_m z^m \right)^n $$ to get an expression of the form $g(f(z))=\sum_{n=0}^\infty c_nz^n$, and obtaining and explicit expression of $c_n$ in terms of the $a_n$'s and the $b_n$'s (and then, setting $c_1=1$ and $c_n=0$ for every $n\neq 1$). By continuity of products and sums in a Banach algebra, we could do the analogous expansion for $g(f(x))$, for $x\in\mathcal{X}$ such that $\limsup_{n}\lVert x^n\rVert^{1/n}<R$, and conclude the result. However, this proof has the downsides that (i) one would need to learn about explicit computations of composition of power series, and by which what is said in some MSE posts, it does look like a pain; and (ii) this proof would be much more general, showing actually the more general result that “the induced power series of the composition is the composition of the power series,” and we would not be exploiting the property that our composition yields the identity, and perhaps there is a simpler proof exploiting this fact.
For the case $\mathcal{X}=L(X)$, I've been indicated the following
Hint: Show that if $R_k(z):=\sum_{n=0}^{N_k}c_{kn}z^n$ is a sequence of polynomials that converges to the polynomial $z$ uniformly on the closed ball of radius $\rho$ centered at the origin, then $c_{k1}\to 1$ and $\sup\limits_{2\leq n\leq N_k}|c_{kn}|\rho^n\leq c_k\to 0$ as $k\to\infty$, for certain bounds $c_k$. (Recall Cauchy's integral formula.)
The problem is that, even if I were to suppose as true this fact, I don't know how to utilize it to prove the proposition.
Any thoughts or ideas will be appreciated :)
