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

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Resolvent identity

Statement

Let $A$ be a unital complex Banach algebra and let $a, b \in A$ . With the resolvent $R (z, a) = (z 1 - a)^{- 1}$ of Spectrum and resolvent set in a Banach algebra:

for all $z, w \in ρ_{A} (a)$ , $R (z, a) - R (w, a) = (w - z) R (z, a) R (w, a);$ in particular $R (z, a)$ and $R (w, a)$ commute;
for all $z \in ρ_{A} (a) \cap ρ_{A} (b)$ , $R (z, a) - R (z, b) = R (z, a) (a - b) R (z, b) .$

Both identities are equalities of two-sided products; no commutativity of $A$ is assumed, and the factor order shown is the one that is used later.

Facts & Assumptions

Given: A unital complex Banach algebra $A$ , elements $a, b \in A$ , complex numbers $z, w$ with $z, w \in ρ_{A} (a)$ and $z \in ρ_{A} (b)$ .

[L1]

The norm is submultiplicative, $∥ 1 ∥ = 1$ , and multiplication is associative, bilinear, and satisfies $1 u = u 1 = u$ for all $u \in A$ (Unital Banach algebra).

[L2]

For $z \in ρ_{A} (a)$ the resolvent $R (z, a)$ is the unique element of $A$ with $(z 1 - a) R (z, a) = R (z, a) (z 1 - a) = 1$ , and similarly for $b$ ; if $u, v$ are invertible then $(u v)^{- 1} = v^{- 1} u^{- 1}$ (Spectrum and resolvent set in a Banach algebra, Invertible element and general linear group of a Banach algebra).

Proof

technique · direct

1.1L1L2algebra

Each resolvent is a two-sided inverse of its own argument: $R (z, a) (z 1 - a) = (z 1 - a) R (z, a) = 1$ , $R (w, a) (w 1 - a) = (w 1 - a) R (w, a) = 1$ and $R (z, b) (z 1 - b) = (z 1 - b) R (z, b) = 1$ by [L2]; and the scalar identities $(z 1 - a) - (w 1 - a) = (z - w) 1$ and $(z 1 - a) - (z 1 - b) = b - a$ are immediate. No commutativity between $a$ and $b$ , and no commutativity between $R (z, a)$ and $R (z, b)$ , is claimed or needed: the two computations below multiply each resolvent against its own argument only.

2.1step 1.1L2algebra

Multiplying the identity $(z 1 - a) - (w 1 - a) = (z - w) 1$ on the left by $R (z, a)$ and on the right by $R (w, a)$ yields $R (z, a) (z 1 - a) R (w, a) - R (z, a) (w 1 - a) R (w, a) = (z - w) R (z, a) R (w, a)$ ; the two terms on the left equal $R (w, a)$ and $R (z, a)$ respectively, so $R (w, a) - R (z, a) = (z - w) R (z, a) R (w, a)$ , which is claim 1.

2.2step 1.1L2algebra

Multiplying the identity $(z 1 - a) - (z 1 - b) = b - a$ on the left by $R (z, a)$ and on the right by $R (z, b)$ yields $R (z, a) (z 1 - a) R (z, b) - R (z, a) (z 1 - b) R (z, b) = R (z, a) (b - a) R (z, b)$ ; the two terms on the left equal $R (z, b)$ and $R (z, a)$ respectively, so $R (z, b) - R (z, a) = R (z, a) (b - a) R (z, b)$ , which is claim 2 in the stated form after moving the term and reversing the sign: $R (z, a) - R (z, b) = R (z, a) (a - b) R (z, b)$ .

3.1step 2.1step 2.2∎

Claim 1 and claim 2 are exactly the two displayed identities of the statement, so the lemma is proved.

Resolvent identity

Statement

Facts & Assumptions

Proof

Depends on

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