Norm and strong resolvent convergence

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $A_n$ $(n\in \mathbb{N})$ and $A$ be self-adjoint operators on the same Hilbert space $H$ and fix a nonreal $z_0$. One writes $A_n\to A$ in the norm resolvent sense when $\mathrm{\parallel }{R}_{A_n}(z_0)-R_A(z_0)\mathrm{\parallel }⟶0$ in operator norm, and in the strong resolvent sense when $R_{A_n}(z_0)x⟶R_A(z_0)x\text{for every }x\in H,$ that is, strong operator convergence (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces, Weak convergence of nets and sequences).

Both notions are well posed for every choice of nonreal $z_0$: by Resolvent of a self-adjoint operator: nonreal resolvents and the estimate each nonreal number belongs to $\rho (A_n)\cap \rho (A)$, so all resolvents occurring are bounded with $\mathrm{\parallel }{R}_{A_n}(z_0)\mathrm{\parallel }\le 1\mathrm{/}\mathrm{\mid }\mathrm{Im}{z}_0\mathrm{\mid }$. The definition deliberately does not assert independence of the parameter $z_0$: that independence is a theorem, proved for the norm case by the resolvent-star-algebra density lemma below and used in the continuous-calculus-under-resolvent-convergence theorem below. Norm resolvent convergence implies strong resolvent convergence, and both are notions about the resolvents rather than about the operators: no convergence of the operators themselves is asserted or implied.