Operator norm convergence in functional calculus

Question

Let $X$ be a complex Banach space. Suppose that $A:X \to X$ is a bounded linear operator and that $(F_n)_{n \in \mathbb{N}}$ is a sequence of analytic functions in a fixed neighbourhood $D \subset \mathbb{C}$ of the spectrum $\sigma(A)$ of $A$. Moreover, assume that $F_n \to F$ uniformly on $D$ as $n \to \infty$, so that in particular $F$ is also an analytic function on $D$. I want to show that $F_n(A) \to F(A)$ in the operator norm, where $F_n(A)$ and $F(A)$ are understood in the sense of functional calculus. I would be grateful for any help.

Answer 1

Hint. If $\mathcal C$ is a closed curve, with the spectrum of $A$ in its interior, and $\mathrm{Ind(\mathcal C,z_0)=1}$, for all $z_0\in \sigma(A)$, then $$ f(A)=\frac{1}{2\pi i}\int_{\mathcal C}\frac{f(z)\,dz}{zI-A}, $$ for all $f$ analytic in a neighbourhood of the spectrum.