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Resolvent identity for closed operators

Statement

Let $A : D (A) \subseteq X \to X$ be a closed linear operator on a Banach space $X$ , with resolvent $R (λ, A) = (λ I - A)^{- 1}$ on $ρ (A)$ (Resolvent and spectrum of a closed operator on a Banach space). For all $λ, μ \in ρ (A)$ , $R (λ, A) - R (μ, A) = (μ - λ) R (λ, A) R (μ, A),$ and consequently $R (λ, A) R (μ, A) = R (μ, A) R (λ, A)$ .

Facts & Assumptions

Given: A closed linear operator $A : D (A) \subseteq X \to X$ on a Banach space $X$ , scalars $λ, μ \in ρ (A)$ , and the resolvents $R (λ, A), R (μ, A) \in B (X)$ (Resolvent and spectrum of a closed operator on a Banach space).

[F1]

$R (λ, A) = (λ I - A)^{- 1}$ satisfies $R (λ, A) X = D (A)$ , $R (λ, A) (λ I - A) y = y$ for $y \in D (A)$ and $(λ I - A) R (λ, A) x = x$ for $x \in X$ ; the same holds with $μ$ in place of $λ$ (Resolvent and spectrum of a closed operator on a Banach space). Ranges of resolvents lie in $D (A)$ , and composition with the bounded maps $R (λ, A), R (μ, A)$ is associative and bilinear wherever defined (A bounded linear operator between normed spaces, Unbounded linear operators: domain, graph and extension).

Proof

technique · direct, factoring the difference of the two inverse shifts through the common domain $D(A)$

1.1F1algebra

For $x \in X$ the vector $R (μ, A) x$ lies in $D (A)$ by [F1], so $(λ I - A) R (μ, A) x$ and $(μ I - A) R (μ, A) x$ are defined and differ by $(λ - μ) R (μ, A) x$ ; applying the bounded linear map $R (λ, A)$ and using linearity gives $(λ - μ) R (λ, A) R (μ, A) x = R (λ, A) (λ I - A) R (μ, A) x - R (λ, A) (μ I - A) R (μ, A) x$ .

2.1F1step 1.1

The first term on the right equals $R (μ, A) x$ , because $R (λ, A) (λ I - A) y = y$ for $y \in D (A)$ and $R (μ, A) x \in D (A)$ ; the second equals $R (λ, A) x$ , because $(μ I - A) R (μ, A) = I$ on $X$ . Hence $(λ - μ) R (λ, A) R (μ, A) x = R (μ, A) x - R (λ, A) x$ , that is $R (λ, A) x - R (μ, A) x = (μ - λ) R (λ, A) R (μ, A) x$ .

3.1step 2.1

Since $x \in X$ was arbitrary, $R (λ, A) - R (μ, A) = (μ - λ) R (λ, A) R (μ, A)$ .

4.1step 3.1algebra∎

Interchanging $λ$ and $μ$ gives $R (μ, A) - R (λ, A) = (λ - μ) R (μ, A) R (λ, A)$ ; adding the two identities yields $0 = (μ - λ) [R (λ, A) R (μ, A) - R (μ, A) R (λ, A)]$ . For $λ \neq μ$ this gives $R (λ, A) R (μ, A) = R (μ, A) R (λ, A)$ , and for $λ = μ$ the equality is trivial.

Depends on

Used by

Resolvent power estimates for semigroup generators Corollary
Yosida approximants Definition
Lumer-Phillips generation theorem Theorem

Dependency tree · two levels

17 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

Klaus-Jochen Engel and Rainer Nagel, One-Parameter Semigroups for Linear Evolution Equations, Graduate Texts in Mathematics 194 (complete author-hosted monograph) (standard reference, not scraped)
Roland Schnaubelt, Evolution Equations, Karlsruhe Institute of Technology (2023/24 course, complete lecture notes) (standard reference, not scraped)