Suppose $f(T)$ and $g(T)$ are formal power series such that the constant coefficient of $g$ is $0$. We write:
$$f(T)=\sum_{n=0}^{\infty}a_nT^n$$ $$g(T)=\sum_{n=1}^{\infty}b_nT^n$$
One defines the composite formal series of $f$ and $g$ by being the series $$(f\circ g)(T)=\sum_{n=1}^{\infty}c_nT^n$$
where $c_n$ is the $n$-th coefficient of the series $a_0+a_1g(T)+\dots+a_ng(T)^n$.
I'm trying to understand the proof of the following Theorem from Lang's Complex Analysis:
Let $$f(z)=\sum_{n\geq 0}a_n z^n$$ and $$h(z)=\sum_{n\geq 1}b_n z^n$$ be convergent power series, and assume that the constant term of $h $ is $0$. Assume that $f(z)$ is absolutely convergent for $|z|\leq r$ with $r>0$, and that $s>0$ is a number such that:
$$\sum|b_n|s^n\leq r.$$
Let $g=f(h)$ be the formal power series obtained by composition. Then $g$ converges absolutely for $|z|\leq s$, and for such $z$, $g(z)=f(h(z))$.
I'm having trouble on understanding several steps.
Proof: Let $g(T)=\sum c_n T^n$. Then $g(T)$ is dominated by the formal series obtained by composing: $$\sum_{n=0}^\infty|a_n|\left(\sum_{k=1}^\infty |b_k| T^k\right)^n$$
By hypothesis, the series on the right converges absolutely for $|z|\leq s$, so $g(z)$ converges absolutely for $|z|\leq s$.
What is stated above is intuitive, however, I'm having trouble on seeing why this is true. I have managed to prove that this new series really dominates $g$, however, I'm having trouble on seeing why does this second series converge for $|z|\leq s$. I think I can't just say that this happens just because the inner sum is $\leq s$ since the definition of composite series is kinda confusing.
Let $$f_N(T)=a_0+a_1T+\dots+a_{N-1}T^{n-1}$$ be the polynomial of degree $\leq N-1$ beginning the power series $f$. Then the composite series given by $f(h(T))-f_N(h(T))$ is dominated by the series $\sum_{n\geq N}|a_n|\left(\sum|b_k|T^k\right)^n$
Again, I'm having trouble on understanding this step.
I can try to help whoever comes to this question to understand a possible demonstration.
As requirements we already need to understand that:
Given a sequence of complex numbers $\lbrace a_n \rbrace_{n\in \mathbb{N}_0}$, and a complex number $z_0$ there exists a unique number $l\in [ 0; \infty ]$such that the series $ \sum_{n=0}^\infty a_n\cdot (z-z_0)^n$ converges if $|z-z_0|<l$ and diverges if $|z-z_0|>l$ and $l=\frac{1}{\limsup_n |a_n|}$
If I have a power series centered at $z_0$: $ \sum_{n=0}^\infty a_n (z-z_0)^n $ with positive radius $\sigma_0>0$ and another power series centered at $z_0$ $ \sum_{n=0}^\infty b_n (z-z_0)^n $ with positive radius $\sigma_1>0$ then it is valid that the product of these two power series in a neighborhood of distance $\sigma_2 = \min\lbrace \sigma_0, \sigma_1 \rbrace$ is a power series: $$ \left(\sum_{n=0}^\infty a_n (z-z_0)^n \right) \cdot \left(\sum_{n=0}^\infty b_n (z-z_0)^n \right) =\sum_{n=0}^\infty \left( \sum_{s=0}^n(a_s\cdot b_{n-s}) \right)\cdot (z-z_0)^n$$
A bit of summation theory (Fubbini): If we have a sequence $a:\mathbb{N}_0^2 \to \mathbb C $ such that exists a positive constant $C\in \mathbb{R}_{>0}$ such that for every finite set $F\subseteq\mathbb{N}_0^2$ $$ \sum_{(n,m)\in F} |a_{n,m}| < C $$ then all the sums that appear in the following line are convergent and the equality holds $$ \sum_{n=0}^\infty \left( \sum_{m=0}^\infty a_{n,m} \right) = \sum_{m=0}^\infty \left( \sum_{n=0}^\infty a_{n,m} \right) $$
Power series are continuous functions within their radius of convergence.
The idea is that I want to prove that if $ f(z)= \sum_{n=0}^\infty a_n (z-z_0)^n$ is a power series of positive radius $\sigma_f>0$ and $g(z)= \sum_{n=0}^\infty b_n (z-f(z_0))^n$ is a power series of positive radius $\sigma_g>0$ centered in $f(z_0)=a_0$ then exists $\sigma>0$ such that the composition $$g\circ f : B_{\sigma}(z_0) \to \mathbb{C}$$ $$ z \mapsto g(f(z))$$ is well defined and is a power series.
Let's start the proof. Let us consider the power series $$\nu: B_{\sigma_f}(0)\to \mathbb{C} $$ $$ \nu(z)= \sum_{n=1}^\infty |a_n| z^n$$
By observation 1 this series has the same radius of convergence as f. And And, since it is continuous at $z=0$, there exists a number $\sigma>0$ such that $\sigma < \sigma_f$ and $$ \forall z \in C |z|<\sigma \implies |\nu(z)|<\sigma_g$$. Another important property of this number $\sigma<\sigma_f$ is that
$$ \forall z\in \mathbb{C} |z-z_0| <\sigma \implies |f(z)-f(z_0)| < \sigma_g$$
This happens because if $|z-z_0|<\sigma$, $|f(z)-f(z_0)| = |\sum_{n=1}^\infty a_n (z-z_0)^n|$ and $|\sum_{n=1}^\infty a_n (z-z_0)^n| \leq \sum_{n=1}^\infty |a_n| |z-z_0|^n = \nu(|z-z_0|) \leq \nu(\sigma) < \sigma_g$
What this tells us is that we can compose f with g in $B_\sigma(z_0)$.
Let's take $z\in B_\sigma(z_0)$ and calculate:
$$ g(f(z))= \sum_{m=0}^\infty b_m (f(z)-f(z_0))^m $$ $$ =\sum_{m=0}^\infty b_m \left( \sum_{n=0}^\infty a_n (z-z_0)^n -f(z_0) \right)^m$$ but taking into account that $f(z_0)= a_0$ $$ =\sum_{m=0}^\infty b_m \left( \sum_{n=1}^\infty a_n (z-z_0)^n \right)^m=...$$
From observation 2 (observation that deserves its proof apart) used m times, we know that $ \left( \sum_{n=1}^\infty a_n (z-z_0)^n \right)^m $ is actually the power series $ \sum_{n=m}^\infty \left( \sum_{k_1 + k_2 +...+k_m = n} a_{k_1} \cdot a_{k_2} \cdot ... \cdot a_{k_m} \right) \cdot (z-z_0)^n $ So, we have
$$...= \sum_{m=0}^\infty b_m \left( \sum_{n=m}^\infty \left( \sum_{k_1 + k_2 +...+k_m = n} a_{k_1} \cdot a_{k_2} \cdot ... \cdot a_{k_m} \right) \cdot (z-z_0)^n \right) = $$
$$ ...=\sum_{m=0}^\infty \sum_{n=m}^\infty b_m\left( \sum_{k_1 + k_2 +...+k_m = n} a_{k_1} \cdot a_{k_2} \cdot ... \cdot a_{k_m}\right) \cdot (z-z_0)^n =... $$
As a matter of convenience, since we want to swap the summations, we would like the second summation to start from 0 and not from m. Let's use the convention that $\sum_{k_1 + k_2 +...+k_m = n} a_{k_1} \cdot a_{k_2} \cdot ... \cdot a_{k_m}=0 $ if $n<m$ and we write
$$ ...= \sum_{m=0}^\infty \sum_{n=0}^\infty b_m\left( \sum_{k_1 + k_2 +...+k_m = n} a_{k_1}\cdot a_{k_2} \cdot ... \cdot a_{k_m} \right) \cdot (z-z_0)^n $$
Now for a moment we have to detach ourselves from this chain of accounts. We want to see that we can interchange the summations with Fubbini. For that, we need to see that the partial sums of the doubly indexed terms are bounded. (see observation 3) Let us consider the power series $$ r(z)=\sum_{m=0}^\infty |b_m| z^m $$ The radius of convergence of this series is $\sigma_g$. Since $\nu(|z-z_0|)<\sigma_g$ because $|z-z_0|<\sigma$, then $$\sum_{m=0}^\infty |b_m| \nu(|z-z_0|)^m < \infty$$ That is, we have to
$$\sum_{m=0}^\infty |b_m| \left( \sum_{n=1}^\infty |a_n|\cdot|z-z_0|^n \right)^m < \infty $$
Now, applying observation 2 again, we can say that: $$\sum_{m=0}^\infty |b_m| \sum_{n=0}^\infty\left( \sum_{k_1 + k_2 +...+k_m = n} |a_{k_1}|\cdot |a_{k_2}| \cdot ... \cdot |a_{k_m}| \right) \cdot (z-z_0)^n < \infty $$
And
$$\infty >\sum_{m=0}^\infty \sum_{n=0}^\infty |b_m| \left( \sum_{k_1 + k_2 +...+k_m = n} |a_{k_1}|\cdot |a_{k_2}| \cdot ... \cdot |a_{k_m}| \right) \cdot |z-z_0|^n>... $$ $$...>\sum_{m=0}^\infty \sum_{n=0}^\infty |b_m| \left| \ \sum_{k_1 + k_2 +...+k_m = n} a_{k_1}\cdot a_{k_2} \cdot ... \cdot a_{k_m} \right| \cdot |z-z_0|^n $$
Then $ \sum_{m=0}^\infty \sum_{n=0}^\infty |b_m| \left| \ \sum_{k_1 + k_2 +...+k_m = n} a_{k_1}\cdot a_{k_2} \cdot ... \cdot a_{k_m} \right| \cdot |z-z_0|^n< \infty$ and this is the condition we need to invert the sum...
$$ ...= \sum_{m=0}^\infty \sum_{n=0}^\infty b_m\left( \sum_{k_1 + k_2 +...+k_m = n} a_{k_1}\cdot a_{k_2} \cdot ... \cdot a_{k_m} \right) \cdot (z-z_0)^n =...$$
$$...= \sum_{n=0}^\infty \left[ \sum_{m=0}^\infty b_m\left( \sum_{k_1 + k_2 +...+k_m = n} a_{k_1}\cdot a_{k_2} \cdot ... \cdot a_{k_m} \right) \right] \cdot (z-z_0)^n $$
This equality finishes proving that g composed with f is a power series ():
$$g(f(z))= \sum_{n=0}^\infty \left[ \sum_{m=0}^\infty b_m\left( \sum_{k_1 + k_2 +...+k_m = n} a_{k_1}\cdot a_{k_2} \cdot ... \cdot a_{k_m} \right) \right] \cdot (z-z_0)^n $$
(The magic of proving that the power series we are proposing is well defined and convergent was left to Fubbini's theorem.)