Alphabeta Math
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∈A. With the resolvent R(z,a)=(z1−a)−1 of Spectrum and resolvent set in a Banach algebra:

  1. for all z,w∈ρA(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;
  2. for all z∈ρA(a)∩ρA(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∈A, complex numbers z,w with z,w∈ρA(a) and z∈ρA(b).

[L1]

The norm is submultiplicative, ∥1∥=1, and multiplication is associative, bilinear, and satisfies 1u=u1=u for all u∈A (Unital Banach algebra).

[L2]

For z∈ρA(a) the resolvent R(z,a) is the unique element of A with (z1−a)R(z,a)=R(z,a)(z1−a)=1, and similarly for b; if u,v are invertible then (uv)−1=v−1u−1 (Spectrum and resolvent set in a Banach algebra, Invertible element and general linear group of a Banach algebra).

Proof

technique · direct
1.1

Each resolvent is a two-sided inverse of its own argument: R(z,a)(z1−a)=(z1−a)R(z,a)=1, R(w,a)(w1−a)=(w1−a)R(w,a)=1 and R(z,b)(z1−b)=(z1−b)R(z,b)=1 by [L2]; and the scalar identities (z1−a)−(w1−a)=(z−w)1 and (z1−a)−(z1−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.

L1L2algebra
2.1

Multiplying the identity (z1−a)−(w1−a)=(z−w)1 on the left by R(z,a) and on the right by R(w,a) yields R(z,a)(z1−a)R(w,a)−R(z,a)(w1−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.

step 1.1L2algebra
2.2

Multiplying the identity (z1−a)−(z1−b)=b−a on the left by R(z,a) and on the right by R(z,b) yields R(z,a)(z1−a)R(z,b)−R(z,a)(z1−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).

step 1.1L2algebra
3.1

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

step 2.1step 2.2∎

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