# Operator norm convergence in functional calculus

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.

• what have you tried? what tools of the functional calculus do you have available? – supinf Nov 5 '18 at 10:56

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.