LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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Riesz Schauder ascent and descent stabilize

Statement

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $X$ be a Banach space over $R$ or $C$ , let $K : X \to X$ be a compact operator (Compact linear operator) and put $A : = I - K$ . Then there is $m_{0} \in N$ such that for every $n \geq m_{0}$

$\ker A^{n} = \ker A^{m_{0}}, ran A^{n} = ran A^{m_{0}},$

and for every such $m$ the following hold:

$X = \ker A^{m} \oplus ran A^{m}$ as a direct sum of linear subspaces;
$\ker A^{m}$ is finite dimensional (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) and $ran A^{m}$ is closed in $X$ ;
$A$ maps $ran A^{m}$ bijectively onto itself, and the restricted map $ran A^{m} \to ran A^{m}$ , $y \mapsto A y$ , is a bounded linear isomorphism with bounded inverse (Bounded inverse theorem).

Facts & Assumptions

[A1]

For every $n \geq 1$ there is a compact $K_{n}$ with $A^{n} = I - K_{n}$ : $A = I - K$ , and if $A^{n} = I - K_{n}$ with $K_{n}$ compact then $A^{n + 1} = (I - K_{n}) A = I - K_{n} - K + K_{n} K$ , where $K_{n} + K - K_{n} K$ is compact by the ideal and linear-subspace properties (Compositions with a compact operator are compact, Linear combinations of compact operators are compact, Compact linear operator, A bounded linear operator between normed spaces).

[A2]

For compact $C$ the kernel $\ker (I - C)$ is finite dimensional (Kernel of identity minus compact is finite dimensional) and $ran (I - C)$ is closed, with $dist (x, \ker (I - C)) \leq C_{0} ∥ (I - C) x ∥$ for some real $C_{0} > 0$ (Range of identity minus compact is closed, The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[A3]

Riesz lemma: for a proper closed subspace $M$ of a normed space and $0 < α < 1$ there is $x$ with $∥ x ∥ = 1$ and $dist (x, M) > α$ (Riesz lemma, The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M)), A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[A4]

Under DC, Countable Choice is available, and a compact operator sends bounded sequences to sequences with convergent subsequences (Dependent choice implies countable choice, Sequential characterization of compact operators, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ); a subsequence of a bounded sequence is bounded.

[A5]

A closed linear subspace of a Banach space is a Banach space (A closed subspace of a Banach space is Banach), and a bounded bijection between Banach spaces has a bounded inverse (Bounded inverse theorem, Banach space).

Proof

technique · direct

Given: $D C$ , a Banach space $X$ over $R$ or $C$ , a compact $K : X \to X$ , and $A = I - K$ .

1.1

For every $n \geq 1$ the operator $A^{n}$ equals $I - K_{n}$ with $K_{n}$ compact.

2.1

For every $n \geq 1$ the kernel $\ker A^{n}$ is finite dimensional with $\ker A^{n} \subseteq \ker A^{n + 1}$ , and the range $ran A^{n}$ is closed with $ran A^{n + 1} \subseteq ran A^{n}$ .

step 1.1A2

3.1

For every $n$ with $\ker A^{n} \neq \ker A^{n + 1}$ , the finite-dimensional subspace $\ker A^{n}$ is closed in the normed space $\ker A^{n + 1}$ , so Riesz's lemma gives $x \in \ker A^{n + 1}$ with $∥ x ∥ = 1$ and $dist (x, \ker A^{n}) > 1 / 2$ (A finite-dimensional normed subspace is closed).

step 2.1A3

3.2

For every $n$ with $ran A^{n} \neq ran A^{n + 1}$ there is $x \in ran A^{n}$ with $∥ x ∥ = 1$ and $dist (x, ran A^{n + 1}) > 1 / 2$ , the distance being computed in $X$ .

step 2.1A3

4.1

The chain $\ker A^{n}$ stabilizes: if $\ker A^{n} \neq \ker A^{n + 1}$ for every $n$ , Countable Choice in [A4] applied to [step 3.1] gives unit vectors $x_{n} \in \ker A^{n + 1}$ with $dist (x_{n}, \ker A^{n}) > 1 / 2$ ; for $m < n$ one has $x_{m} \in \ker A^{n}$ , $A x_{m} \in \ker A^{n}$ and $A x_{n} \in \ker A^{n}$ , so $K x_{n} - K x_{m} = x_{n} - (x_{m} + A x_{n} - A x_{m})$ is at distance $> 1 / 2$ from $0$ , and the bounded sequence $(K x_{n})$ has no convergent subsequence, contradicting [A4].

step 3.1A4

4.2

The chain $ran A^{n}$ stabilizes: if $ran A^{n} \neq ran A^{n + 1}$ for every $n$ , Countable Choice in [A4] applied to [step 3.2] gives unit vectors $x_{n} \in ran A^{n}$ with $dist (x_{n}, ran A^{n + 1}) > 1 / 2$ ; for $m > n$ one has $x_{m} \in ran A^{n + 1}$ , $A x_{n} \in ran A^{n + 1}$ and $A x_{m} \in ran A^{n + 1}$ , so $K x_{n} - K x_{m} = x_{n} - (x_{m} + A x_{n} - A x_{m})$ has norm $> 1 / 2$ , and the bounded sequence $(K x_{n})$ has no convergent subsequence, contradicting [A4].

step 3.2A4

5.1

Choose $m$ so that both chains are constant from $m$ onward by [step 4.1] and [step 4.2], and put $N : = \ker A^{m}$ , $Y : = ran A^{m}$ . Then $X = N + Y$ : for $x \in X$ the element $A^{m} x$ lies in $Y = ran A^{2 m}$ , so $A^{m} x = A^{2 m} y$ for some $y \in X$ , and $x - A^{m} y \in N$ .

step 4.1step 4.2

6.1

Under the choice of [step 5.1] one has $N \cap Y = {0}$ : if $x \in N \cap Y$ , say $x = A^{m} y$ with $A^{m} x = 0$ , then $A^{2 m} y = 0$ , so $y \in \ker A^{2 m} = \ker A^{m}$ by [step 4.1] and $x = A^{m} y = 0$ .

step 5.1

6.2

Under the choice of [step 5.1], $A (Y) = Y$ : $A (Y) = A (A^{m} X) = A^{m + 1} X = Y$ by the stabilization of [step 4.2].

step 5.1

7.1

Under the choice of [step 5.1], $X = N \oplus Y$ , and $N$ is finite dimensional and $Y$ is closed by [step 2.1].

step 5.1step 6.1step 2.1

8.1

Under the choice of [step 5.1], the restriction $A ∣_{Y}$ is injective: if $A y = 0$ with $y = A^{m} z \in Y$ , then $A^{m + 1} z = 0$ , so $z \in \ker A^{m + 1} = \ker A^{m}$ by [step 4.1] and $y = A^{m} z = 0$ .

step 7.1

9.1

Under the choice of [step 5.1], $A ∣_{Y} : Y \to Y$ is a bounded bijection by [step 6.2] and [step 8.1], and $Y$ is a Banach space by [step 7.1] and [A5], so the inverse $(A ∣_{Y})^{- 1}$ is bounded by [A5].

step 7.1step 6.2step 8.1A5

10.1

Taking $m_{0} : = m$ from [step 5.1] gives the stabilization, and [step 7.1], [step 6.2] and [step 9.1] give the decomposition, the finite-dimensional kernel, the closed range and the bounded isomorphism on that range; every larger $m$ works as well because the chains are constant from $m_{0}$ onward.

step 5.1step 7.1step 6.2step 9.1∎

Depends on

Used by

Fredholm alternative for identity minus compact Theorem
Riesz schauder spectrum of a compact operator Theorem

Dependency tree · two levels

90 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 — §6.6 pp.191–192, Lemmas 6.32–6.33 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis — §5.2.3 p.226, Remark 5.23, algebraic part (standard reference, not scraped)