Let $A$ be a bounded self-adjoint operator on a separable Hilbert space. Assume that $f_n:\mathbb{R}\to\mathbb{C}$ is a sequence of continuous bounded functions such that $f_n \to f$ point wise. Then $f_n(A)\to f(A)$ strongly (Reed & Simon Thm. VII.2 (d)). If we further assume that $f_n \to f$ uniformly, then $f_n(A)\to f(A)$ in operator norm, right?
Are there weaker conditions on $f_n \to f$ so that $f_n(A)\to f(A)$ in operator norm?
Functional calculus $f\mapsto f(A)$ is an isometry from $C(\sigma(A))$ to $B(H)$. Thus $f_n\to f$ uniformly on $\sigma(A)$ if and only if $f_n(A)\to f(A)$ in operator norm.