Evaluating a Power Series With Scalar Coefficients with an Operator Argument

Question

If I have a power series/polynomial $$P(x) = \sum_{i=0}^\infty a_n x^n$$ with all scalar coefficients, $a_n$, and I want to evaluate it with a operator argument $A$. The operator is Hermitian and admits an eigendecomposition $$A = V^T \Lambda V$$ where $\Lambda$ is diagonal and the operators $V$ are unitary. If the spectrum of $\Lambda$ falls within the domain of convergence of $P(x)$ is that a sufficient condition to say that: $$P(A) = V^T P(\Lambda) V?$$ Or does the series have to converge everywhere in order to make this statement?

Answer 1

Suppose $P(z)=\sum_{n=0}^{\infty}a_n z^n$ converges absolutely for $|z| \le R$. If $\Lambda$ is a bounded selfadjoint operator with $\sigma(\Lambda) \subseteq \{ z : |z|\le R \}$, then $P(\Lambda)=\sum_{n=0}^{\infty}a_n \Lambda^n$ converges absolutely in the operator norm because $\|\Lambda^n\|=\|\Lambda\|^n$ for a seladjoint $\Lambda$, and $\|\Lambda\|=r_{\sigma}(\Lambda)$ is the spectral radius, which is assumed to be bounded by $R$.