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

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Fredholm alternative for identity minus compact

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ be a Banach space over a fixed field $F \in {R, C}$ , let $K : X \to X$ be a compact operator (Compact linear operator) and put $A : = I - K$ . Then:

$A$ is injective if and only if it is surjective, and in that case $A$ is boundedly invertible;
for every $y \in X$ the equation $A x = y$ has a solution if and only if $φ (y) = 0$ for every $φ \in \ker A^{*}$ , the transpose $A^{*} = I - K^{*}$ acting on $X^{*}$ (The transpose of a bounded operator);
$\ker A$ and the cokernel $X / ran A$ (The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)) are finite dimensional with equal dimensions over $F$ (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Facts & Assumptions

[A1]

The compact $K$ is bounded; if $C$ is a bound for $K$ , then $∥ A x ∥ \leq (1 + C) ∥ x ∥$ , so $A$ is bounded and linear (Compact linear operator, A bounded linear operator between normed spaces). Under DC there is $m_{0}$ from which the kernel and range chains stabilize; since every larger exponent has the same properties, take $m \geq \max {m_{0}, 1}$ . Then $\ker A^{m} = \ker A^{n}$ and $ran A^{m} = ran A^{n}$ for all $n \geq m$ , and with $N : = \ker A^{m}$ , $Y : = ran A^{m}$ one has $X = N \oplus Y$ , $N$ finite dimensional, $Y$ closed, $A (Y) = Y$ and $A ∣_{Y}$ a bounded isomorphism (Riesz Schauder ascent and descent stabilize, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); $A C$ supplies DC (AC supplies the countable and dependent choices used in Banach integration).

[A2]

$A^{*} = I - K^{*}$ : the transpose of the identity is the identity and the transpose is additive (Transposition reverses composition, The transpose of a bounded operator); under DC the range of $I - K$ is closed (Range of identity minus compact is closed). For a bounded linear $T$ with closed range one has $(ran T)^{⊥} = \ker T^{*}$ and $\overline{ran T} =^{⊥} (\ker T^{*})$ (Elementary kernel and range annihilator identities).

[A3]

A bounded bijection between Banach spaces has a bounded inverse (Bounded inverse theorem, Banach space). For a linear map $T : V \to W$ over $F$ with $V$ finite dimensional, $\dim_{F} V = \dim_{F} \ker T + \dim_{F} ran T$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis); for a linear subspace $U \leq V$ the quotient $V / U$ is a vector space and a surjective linear map induces a linear isomorphism $V / \ker T ≅ ran T$ (First isomorphism theorem for vector spaces: $V / \ker T$ is isomorphic to $im T$ , The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M), Linear subspace of a vector space).

Proof

technique · direct

Given: $A C$ , a Banach space $X$ over the fixed field $F \in {R, C}$ , a compact $K : X \to X$ , $A = I - K$ ; and $m$ , $N = \ker A^{m}$ , $Y = ran A^{m}$ as in [A1].

1.1

$N$ is finite dimensional, $Y$ is closed, $A ∣_{Y}$ is a bounded isomorphism of $Y$ onto $Y$ , and $X = N \oplus Y$ . Moreover $A (N) \subseteq N$ , because $A^{m} (A n) = A (A^{m} n) = 0$ for $n \in N$ . All subspaces, quotients and dimensions below are over $F$ .

2.1

$\ker A \subseteq N$ and $\ker A = \ker (A ∣_{N})$ : if $A x = 0$ then $A^{m} x = 0$ , so $x \in N$ ; conversely $x \in N$ with $A x = 0$ means $x \in \ker A$ .

step 1.1

2.2

$ran A = A (N) + Y$ : because $A$ is linear, $X = N + Y$ and $A (Y) = Y$ .

step 1.1

2.3

$A (N) \neq N$ whenever $N \neq {0}$ : $A^{m}$ vanishes on $N$ , and if $A (N) = N$ then $A ∣_{N}$ is a surjective linear endomorphism of the finite-dimensional space $N$ , hence also injective, so $A^{m} ∣_{N} = A ∣_{N}^{m}$ would be injective while $A^{m} ∣_{N} = 0$ and $N \neq {0}$ , a contradiction.

step 1.1A3

3.1

$A^{*} = I - K^{*}$ , and $ran A$ is closed by [A2], so $(ran A)^{⊥} = \ker A^{*}$ and $\overline{ran A} =^{⊥} (\ker A^{*})$ with the closure equal to $ran A$ itself.

step 2.2A2

3.2

The inclusion $N ↪ X$ induces a linear isomorphism $N / A (N) ≅ X / ran A$ : the map $φ (n) = n + ran A$ has kernel $N \cap ran A = N \cap (A (N) + Y) = A (N) + (N \cap Y) = A (N)$ , and it is surjective because every coset $x + ran A$ with $x = n + y$ equals $n + ran A$ .

step 1.1step 2.2A3

3.3

The following are equivalent: $A$ injective, $\ker A = {0}$ , $N = {0}$ , $A$ surjective. Indeed $\ker A = {0}$ is $A$ injective; $N = {0}$ gives $\ker A = {0}$ by [step 2.1], and conversely $N \neq {0}$ makes $A ∣_{N}$ non-injective by [step 2.3], so $\ker A \neq {0}$ ; finally $N = {0}$ gives $X = Y = ran A$ so $A$ is surjective, while if $N \neq {0}$ and $A$ is surjective then $N = N \cap ran A = A (N) + (N \cap Y) = A (N)$ , contradicting [step 2.3].

step 1.1step 2.1step 2.2step 2.3A3

4.1

If $A$ is injective, hence bijective by [step 3.3], then $A^{- 1}$ is bounded by [A3].

step 3.3A3

4.2

For $y \in X$ the equation $A x = y$ is solvable if and only if $y \in ran A$ if and only if $φ (y) = 0$ for every $φ \in \ker A^{*}$ , by [step 3.1].

step 3.1A2

4.3

Rank–nullity for $A ∣_{N} : N \to N$ gives $\dim_{F} N = \dim_{F} \ker A + \dim_{F} A (N)$ . The surjective quotient map $N \to N / A (N)$ has kernel $A (N)$ , so rank–nullity also gives $\dim_{F} N = \dim_{F} A (N) + \dim_{F} (N / A (N))$ . Both its image and the first map's kernel are finite dimensional by rank–nullity. Cancelling the common natural summand and using the isomorphism in step 3.2 yields $\dim_{F} \ker A = \dim_{F} (X / ran A)$ , with both spaces finite dimensional.

step 2.1step 3.2A3algebra

5.1

Collecting: [step 3.3] and [step 4.1] give claim 1, [step 4.2] gives claim 2, and [step 4.3] gives finite dimensionality and equality of the dimensions of kernel and cokernel, claim 3.

step 3.3step 4.1step 4.2step 4.3∎

Depends on

Used by

Fredholm alternative for an integral equation Example
Riesz schauder spectrum of a compact operator Theorem

Dependency tree · two levels

76 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 — §6.5 p.189, Theorem 6.30 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis — §4.4 p.198, Remark 4.42 (standard reference, not scraped)