 Solution

The convergence is in the strong operator topology: for each fixed $f\in E$, one must prove $\Vert \lambda R_{\lambda }f-f\Vert \to 0$. First, the contraction semigroup bound givesThus $\lambda R_{\lambda }$ is a contraction. This is the useful scaling of the contraction estimate in the source's hint. Since ${\int }_0^{\infty }\lambda e^{-\lambda t}dt=1$,Fix $\epsilon >0$. By strong continuity, choose $\delta >0$ so that $\Vert P_{t}f-f\Vert <\epsilon$ for $0\le t\le \delta$. Split the integral there and use $\Vert P_{t}f-f\Vert \le 2\Vert f\Vert$ for the tail:Let $\lambda \to \infty$ and then $\epsilon \downarrow 0$. ThereforeThe distinction from operator norm convergence is essential. For example, the multiplication semigroup on ${\ell }^2(\mathbb{N})$ defined by $(P_{t}f{)}_n=e^{-nt}{f}_n$ is a strongly continuous semigroup of contractions: coordinatewise convergence and dominated convergence applied to ${\sum }_n|f_n{|}^2$ prove strong continuity. Its resolvent operator is $(R_{\lambda }f{)}_n=f_n/(\lambda +n)$, andThus the claimed convergence cannot in general be read as convergence in operator norm.