TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Continuous functional calculus under resolvent convergence

Statement

Assume the Axiom of Choice. Let $A_{n}, A$ be self-adjoint operators on a complex Hilbert space $H$ and suppose $A_{n} \to A$ in the strong resolvent sense. Then $f (A_{n}) x \to f (A) x$ for every bounded continuous $f : R \to C$ and every $x \in H$ . If $A_{n} \to A$ in the norm resolvent sense, then $∥ f (A_{n}) - f (A) ∥ \to 0$ for every bounded continuous $f$ with $\lim_{t \to + \infty} f (t) = \lim_{t \to - \infty} f (t)$ . In both cases the conclusion does not depend on the nonreal parameter used in the definition of convergence.

Facts & Assumptions

[A1]

On nonzero complex $H$ , AC supplies the spectral PVM of each self-adjoint $S$ , representing $S$ as the integral of the identity on its squared-integrability domain. The unbounded calculus has the exact product domain $D (g (S)) \cap D ((f g) (S))$ , and sums and products agree with their pointwise counterparts on their domains (Spectral theorem for unbounded self-adjoint operators (PVM form), Unbounded Borel functional calculus: domains, products, spectral mapping).

[A2]

The bounded PVM calculus is linear, unital, multiplicative and conjugation preserving, satisfies $∥ h (S) ∥ \leq ∥ h ∥_{\infty}$ and $∥ h (S) x ∥^{2} = \int ∣ h ∣^{2} d E_{x}^{S}$ , and $E_{x}^{S} (R) = ∥ x ∥^{2}$ . Its Countable Choice assumptions are supplied by AC (Bounded borel pvm integral, Pvm integral is a star homomorphism, Scalar and complex measures from a pvm, The Axiom of Choice).

[A3]

$R_{S} (z) = (z I - S)^{- 1}$ exists for nonreal $z$ with $∥ R_{S} (z) ∥ \leq 1 / ∣ Im z ∣$ . Convergence is initially assumed at one fixed nonreal parameter only (Resolvent of a self-adjoint operator: nonreal resolvents and the estimate, Resolvent and spectrum of an unbounded operator, Norm and strong resolvent convergence).

[A4]

For fixed nonreal $z$ , the linear span of $r^{k} {\overline{r}}^{j}$ , $k + j \geq 1$ , $r (λ) = (λ - z)^{- 1}$ , is uniformly dense in $C_{0} (R)$ (The resolvent star algebra is dense in C_0(R)). Only its function-algebra density assertion is used; parameter independence is proved below.

[A5]

Scalar dominated convergence holds with an integrable majorant (Dominated convergence). Operator norm bounds give $∥ B x ∥ \leq ∥ B ∥ ∥ x ∥$ and, by applying this twice, $∥ B C ∥ \leq ∥ B ∥ ∥ C ∥$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

Proof

technique · direct

Given: AC, the self-adjoint operators and convergence at a fixed nonreal $z$ in the indicated mode.

1.1A1A2A3given

If $H = {0}$ , all operators, resolvents and bounded functions of them are the unique full-domain operator, so all assertions hold. Suppose $H \neq {0}$ . For nonreal $w$ put $q_{w} (λ) = (w - λ)^{- 1}$ . Both $q_{w}$ and $λ q_{w}$ are bounded on the real line. The product-domain rule in [A1] shows that $q_{w} (S)$ maps $H$ into $D (S)$ and $(w I - S) q_{w} (S) = I$ on $H$ , while $q_{w} (S) (w I - S) = I$ 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.

2.1A2A3A5step 1.1

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 $C = 1 + ∣ d ∣ / ∣ Im w ∣$ . Consequently $∥ D_{n} (w) x ∥ \leq C ∥ D_{n} (z) F x ∥ \to 0$ in the strong case, since $F x$ is fixed, and $∥ D_{n} (w) ∥ \leq C ∥ D_{n} (z) ∥ ∥ F ∥ \to 0$ in the norm case. This proves parameter independence without taking a strong limit on a varying vector.

3.1A2A3A5step 1.1step 2.1

In particular convergence holds at $z$ and $\overline{z}$ , and $r (A_{n}) = - R_{A_{n}} (z)$ , $\overline{r} (A_{n}) = - R_{A_{n}} (\overline{z})$ , with the analogous formulas for $A$ . Both sequences are uniformly bounded. If $B_{n} \to B$ and $C_{n} \to C$ strongly and $\sup_{n} ∥ B_{n} ∥ \leq K$ , then $∥ (B_{n} C_{n} - B C) x ∥ \leq K ∥ (C_{n} - C) x ∥ + ∥ (B_{n} - B) C x ∥ \to 0$ . 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 $r, \overline{r}$ with zero constant term.

4.1A2A4A5step 3.1

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

5.1A2step 4.1

In the norm case, if $f$ has a common finite limit $L$ at both ends, then $g = f - L \in C_{0} (R)$ . Unital linearity gives $f (A_{n}) - f (A) = g (A_{n}) - g (A)$ , so step 4.1 proves the assertion.

5.2A2A5step 4.1

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

6.1A2A5step 5.2

Bounded multiplicativity and linearity give the exact four-term decomposition $f (A_{n}) - f (A) = f (A_{n}) (I - χ_{m} (A)) + f (A_{n}) (χ_{m} (A) - χ_{m} (A_{n})) + (f χ_{m}) (A_{n}) - (f χ_{m}) (A) + f (A) (χ_{m} (A) - I)$ . Applying it to $x$ , its norm is at most $2 M ∥ (I - χ_{m} (A)) x ∥ + M ∥ (χ_{m} (A) - χ_{m} (A_{n})) x ∥ + ∥ ((f χ_{m}) (A_{n}) - (f χ_{m}) (A)) x ∥$ . 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 $f$ .

7.1A1A2A3step 1.1step 2.1step 5.1step 6.1∎

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.

Source notes

Teschl, Theorem 6.31 and its proof, pp.179-180, gives the polynomial approximation and four-term cutoff route, with Corollary 6.32 giving parameter independence. Here the latter is proved first using a resolvent-difference factorization with a fixed right factor. The library convention is $(z I - S)^{- 1}$ , so $r (S) = - R_{S} (z)$ . No general continuity of adjoints for strongly convergent bounded operators is assumed.

Depends on

Used by

Unitary groups converge under strong resolvent convergence Corollary

Dependency tree · two levels

71 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Mathematical Methods in Quantum Mechanics, second edition (standard reference, not scraped)