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Second resolvent identity for a closed perturbation

Statement

Assume Dependent Choice. Let $A$ and $C$ be closed operators with $D (C) = D (A)$ , put $B : = C - A$ , and assume $B$ is bounded for the graph norm of $A$ (Relative boundedness with respect to an operator). Then for every $z \in ρ (A) \cap ρ (C)$ $R_{C} (z) - R_{A} (z) = R_{A} (z) B R_{C} (z) = R_{C} (z) B R_{A} (z),$ and every product here is defined on all of $H$ and is bounded.

Facts & Assumptions

[A1]

For $z \in ρ (A)$ the operator $R_{A} (z)$ maps $H$ bijectively onto $D (A)$ and $A R_{A} (z) = z R_{A} (z) - I$ on $H$ , since $A = z - (z - A)$ (Resolvent and spectrum of an unbounded operator).

[A2]

$B$ is bounded for the graph norm of $A$ : there are $a, b \geq 0$ with $∥ B x ∥ \leq a ∥ A x ∥ + b ∥ x ∥$ for $x \in D (A)$ ; each $B R_{A} (z)$ is therefore everywhere defined and bounded, since $∥ B R_{A} (z) y ∥ \leq a ∥ A R_{A} (z) y ∥ + b ∥ R_{A} (z) y ∥$ and both terms are bounded in $y$ (Relative boundedness with respect to an operator, [A1]).

[A3]

$(z - C) R_{C} (z) = I$ and $(z - A) R_{A} (z) = I$ on $H$ , so $R_{A} (z) (C - z) R_{C} (z) = - R_{A} (z)$ and $R_{C} (z) (A - z) R_{A} (z) = - R_{C} (z)$ , because $R_{C} (z)$ has range $D (C) = D (A)$ (Resolvent and spectrum of an unbounded operator).

[A4]

The graph norms of two closed operators with the same domain are equivalent: both domains are Banach, and the identity map from the $C$ -graph norm to the $A$ -graph norm has closed graph, hence is bounded by the closed graph theorem (Closed graph theorem, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain).

Proof

technique · direct

Given: Closed $A, C$ with $D (A) = D (C)$ , $B = C - A$ graph-norm bounded for $A$ , and $z \in ρ (A) \cap ρ (C)$ .

1.1

The operator $B R_{A} (z)$ is bounded by [A1] and [A2]. By [A4] the $A$ -graph norm is bounded by a constant times the $C$ -graph norm, so the same relative bound makes $B$ bounded for the $C$ -graph norm; applying [A1] with $C$ in place of $A$ shows that $B R_{C} (z)$ is bounded as well.

A1A2A4

2.1

$- R_{C} (z) + R_{A} (z) B R_{C} (z) = - R_{A} (z)$ , that is $R_{C} (z) - R_{A} (z) = R_{A} (z) B R_{C} (z)$ : by [A3], $R_{A} (z) (C - z) R_{C} (z) = - R_{A} (z)$ , and expanding $C = A + B$ gives $R_{A} (z) (C - z) R_{C} (z) = R_{A} (z) (A - z) R_{C} (z) + R_{A} (z) B R_{C} (z) = - R_{C} (z) + R_{A} (z) B R_{C} (z)$ , because $R_{A} (z) (A - z)$ is minus the identity on $D (A) = D (C)$ and $R_{C} (z)$ takes values in $D (C)$ .

A2A3step 1.1

2.2

By the same computation with the roles of $A$ and $C$ interchanged (so that the perturbation is $- B$ ), $- R_{A} (z) - R_{C} (z) B R_{A} (z) = - R_{C} (z)$ , that is $R_{C} (z) - R_{A} (z) = R_{C} (z) B R_{A} (z)$ .

A2A3step 1.1

3.1

Rearranging steps 2.1 and 2.2 gives $R_{C} (z) - R_{A} (z) = R_{A} (z) B R_{C} (z)$ and $R_{C} (z) - R_{A} (z) = R_{C} (z) B R_{A} (z)$ , which is the stated identity; all products are bounded by step 1.1. ∎

Depends on

Used by

Weyl's theorem: invariance of the essential spectrum Theorem

Dependency tree · two levels

25 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)