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

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Fredholm index is locally constant

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field. Then the Fredholm operators $X \to Y$ (Fredholm operator cokernel and index) form an open subset of the space $B (X, Y)$ of bounded linear operators with the operator norm (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum): for every Fredholm $T$ there is a real $δ > 0$ such that every bounded $A : X \to Y$ with $∥ A - T ∥ < δ$ is Fredholm, and then $ind A = ind T$ .

Facts & Assumptions

[A1]

A Fredholm $T$ admits bounded projections splitting $X = \ker T \oplus X_{1}$ and $Y = ran T \oplus Y_{0}$ , with $N : = \ker T$ and $Y_{0}$ finite dimensional, $\dim Y_{0} = \dim coker T$ , and with $T_{1} : = T ∣_{X_{1}} : X_{1} \to ran T$ a bounded isomorphism whose inverse $T_{1}^{- 1}$ is bounded (Fredholm splitting and parametrix).

[A2]

Neumann: if $∥ T_{1}^{- 1} ∥ ∥ C ∥ < 1$ then $T_{1} + C$ is invertible with bounded inverse (Neumann series and small perturbations of bounded inverses, The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A3]

Fredholm operators are closed under composition between Banach spaces and the index is additive, $ind (U T) = ind U + ind T$ (Fredholm index is additive, Fredholm operator cokernel and index); an invertible bounded operator is Fredholm with index $0$ , its kernel and cokernel being ${0}$ .

[A4]

A linear map defined on a finite-dimensional normed space is bounded (A linear map from a finite-dimensional normed space is bounded); rank-nullity (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); and for a block-diagonal operator $diag (A_{11}, S)$ on $ran T \oplus Y_{0}$ the kernel is $\ker A_{11} \oplus \ker S$ and the cokernel is isomorphic to $coker A_{11} \oplus coker S$ (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: $A C$ , Banach spaces $X, Y$ over one scalar field, a Fredholm $T : X \to Y$ , and the splitting $X = N \oplus X_{1}$ , $Y = ran T \oplus Y_{0}$ of [A1].

1.1A1

The projections $P_{N}$ , $P_{X_{1}}$ , $P_{ran T}$ , $P_{Y_{0}}$ of the two splittings are bounded; write $κ : = \max (1, ∥ P_{ran T} ∥, ∥ P_{Y_{0}} ∥)$ .

1.2A1A4

If $X_{1} = {0}$ , then $ran T = {0}$ , so $X = N$ and $Y = Y_{0}$ are finite dimensional, $T = 0$ , and for every bounded $A : X \to Y$ rank-nullity gives $ind A = \dim \ker A - \dim coker A = \dim X - \dim Y = ind T$ ; so the claim holds with any $δ > 0$ in this case.

2.1step 1.1A1A2algebra

Assume $X_{1} \neq {0}$ , so $ran T \neq {0}$ and $∥ T_{1}^{- 1} ∥ > 0$ , and put $δ : = 1 / (2 κ ∥ T_{1}^{- 1} ∥) > 0$ . For every bounded $A$ with $∥ A - T ∥ < δ$ , writing $A_{11} = P_{ran T} A ∣_{X_{1}}$ and $T_{11} = T_{1}$ one has $∥ A_{11} - T_{1} ∥ \leq κ ∥ A - T ∥$ , so $∥ T_{1}^{- 1} (A_{11} - T_{1}) ∥ < 1 / 2 < 1$ and $A_{11} = T_{1} (I + T_{1}^{- 1} (A_{11} - T_{1}))$ is invertible with bounded inverse by [A2].

3.1step 1.1step 2.1A1algebra

Under the hypothesis of [step 2.1], reorder the domain splitting as $X = X_{1} \oplus N$ and keep the codomain splitting $Y = ran T \oplus Y_{0}$ . Let $U : Y \to Y$ and $V : X \to X$ be the bounded operators whose block matrices in these stated orders are $U = (\begin{array}{cc} I_{ran T} & 0 \\ - A_{21} A_{11}^{- 1} & I_{Y_{0}} \end{array})$ and $V = (\begin{array}{cc} I_{X_{1}} & - A_{11}^{- 1} A_{12} \\ 0 & I_{N} \end{array})$ , where $A_{12} = P_{ran T} A ∣_{N}$ , $A_{21} = P_{Y_{0}} A ∣_{X_{1}}$ and $A_{22} = P_{Y_{0}} A ∣_{N}$ ; then $U A V = diag (A_{11}, S)$ with $S : = A_{22} - A_{21} A_{11}^{- 1} A_{12}$ , and $U, V$ are invertible with bounded inverses given by the same matrices with the off-diagonal signs reversed.

4.1step 3.1A4

Under the hypothesis of [step 2.1], $U A V = diag (A_{11}, S)$ is Fredholm with index $ind (U A V) = ind A_{11} + ind S = ind S$ , because $A_{11}$ is an isomorphism of $X_{1}$ onto $ran T$ and $S$ maps the finite-dimensional space $N$ boundedly into the finite-dimensional space $Y_{0}$ ; by [A4] its index is $ind S = \dim \ker S - \dim coker S = \dim \ker S + \dim im S - \dim Y_{0} = \dim N - \dim Y_{0}$ .

5.1step 3.1step 4.1A3

Under the hypothesis of [step 2.1], $A$ is Fredholm with $ind A = \dim N - \dim Y_{0} = ind T$ : since $U, V$ and their inverses are invertible hence Fredholm of index $0$ , [A3] gives first that $A = U^{- 1} (U A V) V^{- 1}$ is Fredholm, and then $ind A = ind (U^{- 1}) + ind (U A V) + ind (V^{- 1}) = ind (U A V)$ .

6.1step 1.2step 5.1∎

In the case of [step 1.2] and in the case of [step 5.1] every bounded $A$ with $∥ A - T ∥$ below the corresponding $δ$ (any positive number in the first case, the $δ$ of [step 2.1] in the second) is Fredholm of index $ind T$ , so the Fredholm operators are open in $B (X, Y)$ and the index is locally constant at $T$ .

Fredholm index is locally constant

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources