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Neumann series and small perturbations of bounded inverses

Statement

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

If $R \in B (X)$ 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 $\sum_{n < N} R^{n}$ , the norm limit being taken in $B (X)$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators), and $∥ \sum_{n = 0}^{\infty} R^{n} ∥ \leq \frac{1}{1 - ∥ R ∥} .$
If $A \in B (X, Y)$ is invertible with $A^{- 1} \in B (Y, X)$ and $E \in B (X, Y)$ satisfies $∥ A^{- 1} E ∥ < 1$ , then $A + E$ is invertible with $(A + E)^{- 1} = (I + A^{- 1} E)^{- 1} A^{- 1} \in B (Y, X)$ .

Facts & Assumptions

[A1]

$∥ R^{n} ∥ \leq ∥ R ∥^{n}$ for every $n$ , by induction from $∥ S T ∥ \leq ∥ S ∥ ∥ T ∥$ (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 $\sum r^{k}$ converges with sum $1 / (1 - r)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges); in particular $∥ R ∥ < 1$ makes $\sum_{n} ∥ R ∥^{n}$ 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: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

Proof

technique · direct

Given: Banach spaces $X, Y$ over one scalar field, $R \in B (X)$ with $∥ R ∥ < 1$ , and the partial sums $S_{N} : = \sum_{n < N} R^{n}$ .

1.1

For every $n$ one has $∥ R^{n} ∥ \leq ∥ R ∥^{n}$ , and $\sum_{n} ∥ R ∥^{n}$ converges to $1 / (1 - ∥ R ∥)$ .

A1A2

2.1

The space $B (X)$ is Banach, so the absolutely convergent series $\sum_{n} R^{n}$ converges in operator norm to some $S \in B (X)$ . For every $N$ , the finite triangle inequality and [step 1.1] give $∥ S_{N} ∥ \leq \sum_{n < N} ∥ R^{n} ∥ \leq \sum_{n = 0}^{\infty} ∥ R ∥^{n} = \frac{1}{1 - ∥ R ∥} .$ Since $S_{N} \to S$ and the norm is continuous, taking the limit yields $∥ S ∥ \leq 1 / (1 - ∥ R ∥)$ .

step 1.1A2A3A4

2.2

For every $N$ one has $(I - R) S_{N} = S_{N} (I - R) = I - R^{N}$ , and $∥ R^{N} ∥ \leq ∥ R ∥^{N} \to 0$ , so $R^{N} \to 0$ .

step 1.1A1algebra

3.1

From [step 2.1] and [step 2.2], $(I - R) S = \lim_{N} (I - R) S_{N} = \lim_{N} (I - R^{N}) = I$ and likewise $S (I - R) = I$ : the first limit holds because $∥ (I - R) (S - S_{N}) ∥ \leq (1 + ∥ R ∥) ∥ S - S_{N} ∥ \to 0$ .

step 2.1step 2.2A1A4

4.1

Hence $I - R$ is invertible with inverse $S = \sum_{n \geq 0} R^{n}$ and $∥ (I - R)^{- 1} ∥ \leq 1 / (1 - ∥ R ∥)$ , which is claim 1.

step 2.1step 3.1

4.2

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

step 3.1algebra

5.1

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

step 4.1step 4.2∎

Depends on

Used by

The Banach inverse theorem for a small Lipschitz perturbation of the identity Example
Fredholm index is locally constant Theorem
Implicit function theorem for Banach spaces Theorem
Inverse function theorem for Banach spaces Theorem
Riesz schauder spectrum of a compact operator Theorem

Dependency tree · two levels

42 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, Topics in Real and Functional Analysis, version November 17, 2017 — §6.1 p.163, Lemma 6.1 and p.164, Corollary 6.2 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis — §4.1, Neumann series for bounded inverses (standard reference, not scraped)