Let $f$ be analytical with power series $\sum_{n=0}^\infty a_n z^n $ and convergence radius $R>0$. Let $X$ be a Banach space and $T\in B(X)$ (bounded from $X$ to itself) such that $||T||<R$. Prove that $f(T):=\lim_{N\to \infty}\sum_{n=0}^N a_nT^n$ is a bounded operator and find $||f(T)||$.
In the proof, it was first proven that $f(T)$ is well defined and bounded. Only then the author went on to computing $||f(T)||$. When I attempted to solve this myself, I concluded that it is already well defined since the hypothesis includes the fact that series converges ($||T||<R$ so it's in the convergence radius, no?), and went directly to computing $||f(T)||$ by claiming the following:
$$ ||f(T)|| = \lim_{N\to \infty}|| \sum_{n=0}^N a_n T^n|| \leq \sum_{n=0}^\infty |a_n| ||T||^n$$
where the first transition is by continuity of the norm and the fact that the series converges for this T, and the second is the triangle inequality and again the fact that $T$ is within the radius of convergence (absolute convergence in the norm).
What is wrong with my reasoning? Why would I first have to claim that $f(T)$ is well defined (converges) if I already know that...? Is it perhaps a case of the order of convergence that could be different? (I do not recall much about analytic functions and their power series etc., so I must be missing something in this respect).
Thanks.
In a Banach space $Y,$ if the series $\sum y_n$ is absolutely convergent, i.e. $\sum \|y_n\|<\infty$, then it is convergent. In your case $Y=B(X)$ and $y_n=a_nT^n.$ We have $$\sum \|y_n\|=\sum |a_n|\,\|T^n\|\le \sum |a_n|\,\|T\|^n<\infty $$ as $\|T\|<R.$ Hence the series $f(T):=\sum a_n T^n$ is convergent.
Remark I do not think it is possible to find a norm $f(T)$ explicitly in terms of $f$ and $\|T\|.$ In particular there is no explicit expression for the norm $\|T^2\|$ in terms of $\|T\|.$ It is possible to find the spectral radius of $f(T)$ in terms of $f$ and $\sigma(T).$ Namely $$r(f(T))=\max\{|f(z)|\,:\, z\in \sigma(T)\}$$