Continuous functional calculus under resolvent convergence

If $H=\{0\}$, all operators, resolvents and bounded functions of them are the unique full-domain operator, so all assertions hold. Suppose $H\mathrm{\ne }\{0\}$. For nonreal $w$ put $q_w(\lambda )=(w-\lambda {)}^{-1}$. Both $q_w$ and $\lambda q_w$ are bounded on the real line. The product-domain rule in [A1] shows that $q_w(S)$ maps $H$ into $D(S)$ and $(wI-S)q_w(S)=I$ on $H$, while $q_w(S)(wI-S)=I$ on $D(S)$. Thus $q_w(S)=R_S(w)$ with the convention of [A3]. The bounded calculus agrees with its unbounded truncation definition because the truncations are eventually the same bounded function.

Put $d=z-w$. The scalar identity $(1+d{q}_w)q_z=q_w=q_z(1+d{q}_w)$ and [A2] give these identities for each single operator $S$. Define $D_n(v)=R_{A_n}(v)-R_A(v)$ and $F=I+d{R}_A(w)$, which is independent of $n$. Expanding the product using the preceding identities yields $D_n(w)=(I+d{R}_{A_n}(w))D_n(z)F$: its two terms are $R_{A_n}(w)F$ and $(I+d{R}_{A_n}(w))R_A(w)$, whose mixed terms cancel. The left factor has norm at most $C=1+\mathrm{\mid }d\mathrm{\mid }\mathrm{/}\mathrm{\mid }\mathrm{Im}w\mathrm{\mid }$. Consequently $\mathrm{\parallel }{D}_n(w)x\mathrm{\parallel }\le C\mathrm{\parallel }{D}_n(z)Fx\mathrm{\parallel }\to 0$ in the strong case, since $Fx$ is fixed, and $\mathrm{\parallel }{D}_n(w)\mathrm{\parallel }\le C\mathrm{\parallel }{D}_n(z)\mathrm{\parallel }\mathrm{\parallel }F\mathrm{\parallel }\to 0$ in the norm case. This proves parameter independence without taking a strong limit on a varying vector.

In particular convergence holds at $z$ and $\bar{z}$, and $r(A_n)=-R_{A_n}(z)$, $\bar{r}(A_n)=-R_{A_n}(\bar{z})$, with the analogous formulas for $A$. Both sequences are uniformly bounded. If $B_n\to B$ and $C_n\to C$ strongly and ${\sup }_n\mathrm{\parallel }{B}_n\mathrm{\parallel }\le K$, then $\mathrm{\parallel }(B_n{C}_n-BC)x\mathrm{\parallel }\le K\mathrm{\parallel }(C_n-C)x\mathrm{\parallel }+\mathrm{\parallel }(B_n-B)Cx\mathrm{\parallel }\to 0$. In the norm case the same inequality with operator norms proves convergence of products when the factors are uniformly bounded. Iteration and finite linear combinations, using [A2], therefore prove convergence for every polynomial in $r,\bar{r}$ with zero constant term.

Given $f\in C_0(\mathbb{R})$ and $\delta >0$, choose one such polynomial $p$ with $\mathrm{\parallel }f-p{\mathrm{\parallel }}_{\mathrm{\infty }}<\delta$ by [A4]. The bounded calculus gives $\mathrm{\parallel }(f(A_n)-f(A))x\mathrm{\parallel }\le 2\delta \mathrm{\parallel }x\mathrm{\parallel }+\mathrm{\parallel }(p(A_n)-p(A))x\mathrm{\parallel }$. For fixed $p$ the last term tends to zero by step 3.1; since $\delta$ is arbitrary, strong convergence follows for each $x$ (including $x=0$ directly). The operator-norm inequality is $\mathrm{\parallel }f(A_n)-f(A)\mathrm{\parallel }\le 2\delta +\mathrm{\parallel }p(A_n)-p(A)\mathrm{\parallel }$, proving the norm version as well.

In the norm case, if $f$ has a common finite limit $L$ at both ends, then $g=f-L\in C_0(\mathbb{R})$. Unital linearity gives $f(A_n)-f(A)=g(A_n)-g(A)$, so step 4.1 proves the assertion.

For the strong case let $f$ be any bounded continuous function and set $M=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. For integers $m\ge 1$ define ${\chi }_m(\lambda )=\min (1,\max (0,m+1-\mathrm{\mid }\lambda \mathrm{\mid }))$. These are continuous, compactly supported, between zero and one, equal to one on $[-m,m]$, and converge pointwise to one. Thus $\mathrm{\parallel }(I-{\chi }_m(A))x{\mathrm{\parallel }}^2=\int \mathrm{\mid }1-{\chi }_m{\mathrm{\mid }}^{2}d{E}_x^A\to 0$ by dominated convergence with majorant $1$, integrable against the finite measure of mass $\mathrm{\parallel }x{\mathrm{\parallel }}^2$. For fixed $m$, both ${\chi }_m$ and $f{\chi }_m$ belong to $C_0(\mathbb{R})$, so their calculi converge strongly by step 4.1.

Bounded multiplicativity and linearity give the exact four-term decomposition $f(A_n)-f(A)=f(A_n)(I-{\chi }_m(A))+f(A_n)({\chi }_m(A)-{\chi }_m(A_n))+(f{\chi }_m)(A_n)-(f{\chi }_m)(A)+f(A)({\chi }_m(A)-I)$. Applying it to $x$, its norm is at most $2M\mathrm{\parallel }(I-{\chi }_m(A))x\mathrm{\parallel }+M\mathrm{\parallel }({\chi }_m(A)-{\chi }_m(A_n))x\mathrm{\parallel }+\mathrm{\parallel }((f{\chi }_m)(A_n)-(f{\chi }_m)(A))x\mathrm{\parallel }$. If $M=0$ the claim is immediate. Otherwise choose $m$ so that the first term is less than half a prescribed positive error, using step 5.2, and then $n$ so that the other two together are less than its other half. This proves strong convergence for bounded continuous $f$.

Step 2.1 proves independence of every nonreal parameter, step 5.1 the norm conclusion and step 6.1 the strong conclusion. AC supplies the spectral and countable-choice calculus hypotheses; the cutoffs are explicit. The zero Hilbert space and zero function have been treated in steps 1.1 and 6.1; no limit at either infinity is required in the strong case.