Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 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.

Neumann series and small perturbations of bounded inverses

Statement

Let X and Y be Banach spaces over the same scalar field (Banach space).

  1. If R∈B(X) satisfies ∥R∥<1 (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum), then I−R is invertible with inverse the operator-norm limit of the partial sums ∑n<NRn, the norm limit being taken in B(X) (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators), and ∥∑n=0∞Rn∥≤11−∥R∥.
  2. If A∈B(X,Y) is invertible with A−1∈B(Y,X) and E∈B(X,Y) satisfies ∥A−1E∥<1, then A+E is invertible with (A+E)−1=(I+A−1E)−1A−1∈B(Y,X).

Facts & Assumptions

[A1]

∥Rn∥≤∥R∥n for every n, by induction from ∥ST∥≤∥S∥ ∥T∥ (Composition satisfies |ST|\le|S|,|T|, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A2]

For ∣r∣<1 the scalar series ∑rk converges with sum 1/(1−r) (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges); in particular ∥R∥<1 makes ∑n∥R∥n converge to the real number 1/(1−∥R∥).

[A3]

If Y is Banach then B(X,Y) is Banach for the operator norm (If (Y) is Banach then (\mathcal B(X,Y)) is Banach, The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators, Banach space); a series in a Banach space that converges absolutely converges (Series criterion for Banach spaces).

[A4]

Addition and scalar multiplication are continuous on a normed space (Vector addition and scalar multiplication are continuous in a normed space), and the reverse triangle inequality makes every norm continuous with respect to norm convergence (The reverse triangle inequality in a normed space, Convergence of a sequence in a metric space: xk→x iff d(xk,x)→0 in R).

Proof

technique · direct

Given: Banach spaces X,Y over one scalar field, R∈B(X) with ∥R∥<1, and the partial sums SN:=∑n<NRn.

1.1A1A2

For every n one has ∥Rn∥≤∥R∥n, and ∑n∥R∥n converges to 1/(1−∥R∥).

2.1step 1.1A2A3A4

The space B(X) is Banach, so the absolutely convergent series ∑nRn converges in operator norm to some S∈B(X). For every N, the finite triangle inequality and [step 1.1] give ∥SN∥≤∑n<N∥Rn∥≤∑n=0∞∥R∥n=11−∥R∥. Since SN→S and the norm is continuous, taking the limit yields ∥S∥≤1/(1−∥R∥).

2.2step 1.1A1algebra

For every N one has (I−R)SN=SN(I−R)=I−RN, and ∥RN∥≤∥R∥N→0, so RN→0.

3.1step 2.1step 2.2A1A4

From [step 2.1] and [step 2.2], (I−R)S=lim⁡N(I−R)SN=lim⁡N(I−RN)=I and likewise S(I−R)=I: the first limit holds because ∥(I−R)(S−SN)∥≤(1+∥R∥)∥S−SN∥→0.

4.1step 2.1step 3.1

Hence I−R is invertible with inverse S=∑n≥0Rn and ∥(I−R)−1∥≤1/(1−∥R∥), which is claim 1.

4.2step 3.1algebra

For the perturbation, A+E=A(I+A−1E) and ∥−A−1E∥=∥A−1E∥<1, so by [step 3.1] applied to R:=−A−1E∈B(X) the operator I+A−1E is invertible with bounded inverse, and therefore (A+E)−1=(I+A−1E)−1A−1∈B(Y,X), which is claim 2.

5.1step 4.1step 4.2∎

Claims 1 and 2 are exactly the two parts of the statement.

Depends on

Dependency tree · 2 levels

Sources