Corollary

Let $\mathscr{T}_s$ be the topology of simple convergence on $\mathscr{L}(E)$. Let $f\in \mathscr{C}_b(\mathbf{R})$. For every sequence $(u_n)_{n\in\mathbf{N}}$ in $\mathscr{A}(E)$ which converges to $u$ for the $\mathscr{T}_s$-resolvent topology, the sequence $(f(u_n))_{n\in\mathbf{N}}$ converges to $f(u)$ in $\mathscr{L}(E)$ endowed with the topology $\mathscr{T}_s$.

Let $(u_n)$ be a sequence in $\mathscr{A}(E)$ which converges to $u$ for the $\mathscr{T}_s$-resolvent topology. Let $x\in E$ and let $\varepsilon >0$.

For every integer $N\in \mathbf{N}$, let $\varphi_N\in \mathscr{K}(\mathbf{R})$ be a function with support contained in $[-(N + 1),N + 1]$ such that $0\leqslant \varphi_N\leqslant 1$ and $\varphi_N(t) = 1$ for every $t\in [-N,N]$. The functions $\varphi_N$ converge simply to 1 as N tends to infinity, and satisfy $|\varphi_N|\leqslant 1$, hence prop. 6 of IV, p. 276 implies that there exists $N\in \mathbf{N}$ such that $\|\varphi_N(u)x-x\|\leqslant \varepsilon$. Put $f_N=f \varphi_N$.

For every $n\in \mathbf{N}$, one has

For every $v\in \mathscr{A}$ (E), one has

(prop. 5, c) of IV, p. 275). Since $\varphi_N\in \mathscr{C}_0(\mathbf{R})$, assertion a) of prop. 11 applied to the $\mathscr{T}_s$-resolvent topology implies that $\varphi_N(u_n)x$ converges to $\varphi_N(u)x$ when $n\rightarrow +\infty$. Consequently, for every sufficiently large $n$, one has

Inequality (15) then becomes

for every sufficiently large $n$. The proof is concluded by means of assertion a) of loc. cit., applied to the function $f_N\in \mathscr{C}_0(\mathbf{R})$ and to the $\mathscr{T}_s$-resolvent topology.