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

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Fredholm index is stable under compact perturbations

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $X$ and $Y$ be Banach spaces over the same scalar field, let $T : X \to Y$ be a Fredholm operator and let $K : X \to Y$ be compact (Fredholm operator cokernel and index, Compact linear operator, A bounded linear operator between normed spaces). Then $T + K$ is Fredholm and $ind (T + K) = ind T$ .

Facts & Assumptions

[A1]

By Atkinson's theorem a bounded operator is Fredholm exactly when it has a bounded parametrix modulo compact operators (Atkinson); compact operators are closed under sums, scalar multiples and composition with bounded operators (Linear combinations of compact operators are compact, Compositions with a compact operator are compact, Fredholm operator cokernel and index).

[A2]

The Fredholm operators form an open subset of $B (X, Y)$ and the index is locally constant (Fredholm index is locally constant, The operator norm as the least bound and as the unit-sphere or unit-ball supremum): for each Fredholm $A$ there is $δ_{A} > 0$ such that every bounded $B$ with $∥ B - A ∥ < δ_{A}$ is Fredholm with $ind B = ind A$ .

[A3]

The interval $[0, 1]$ is a connected subset of $R$ (A subset of $R$ is connected if and only if it is order-convex, that is, an interval, Separated sets, disconnection, and connected subset of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). A real point lies in the closure of a set exactly when each of its neighbourhoods meets that set (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points). Convergence in operator norm is metric convergence (Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Open ball, closed ball and sphere in a metric space).

Proof

technique · direct

Given: $A C$ , Banach spaces $X, Y$ over one scalar field, a Fredholm $T : X \to Y$ , a compact $K : X \to Y$ , and a parametrix $S$ for $T$ with $S T - I_{X}$ and $T S - I_{Y}$ compact.

1.1A1

For every $t \in [0, 1]$ the operator $T + t K$ is Fredholm: $S$ is a parametrix for it modulo compact operators, because $S (T + t K) - I_{X} = (S T - I_{X}) + t S K$ and $(T + t K) S - I_{Y} = (T S - I_{Y}) + t K S$ are compact by [A1], so Atkinson applies.

1.2A1A3algebra

The path $t \mapsto T + t K$ is continuous for the operator norm: $∥ (T + s K) - (T + t K) ∥ \leq ∣ s - t ∣ ∥ K ∥$ for all real $s, t$ .

2.1step 1.2A2

Put $U : = {t \in [0, 1] : ind (T + t K) = ind T}$ . For every $t \in U$ , local constancy [A2] gives $δ > 0$ with all operators within $δ$ of $T + t K$ having the same index. With $η : = δ / (1 + ∥ K ∥) > 0$ , every $s \in [0, 1]$ satisfying $∣ s - t ∣ < η$ has $∥ (T + s K) - (T + t K) ∥ \leq ∣ s - t ∣ ∥ K ∥ < δ$ , hence lies in $U$ .

2.2step 1.2A2

Put $V : = [0, 1] ∖ U$ . For every $t \in V$ , the same argument gives $η > 0$ such that every $s \in [0, 1]$ with $∣ s - t ∣ < η$ has index $ind (T + t K) \neq ind T$ and hence lies in $V$ .

3.1step 2.1step 2.2A3

Hence $U = [0, 1]$ . Indeed $0 \in U$ . If $V$ were nonempty, then $U \cup V = [0, 1]$ would be a disconnection: if $t \in V$ , step 2.2 supplies a neighbourhood of $t$ whose intersection with $[0, 1]$ is contained in $V$ , so this neighbourhood misses $U$ and [A3] gives $t \notin \overline{U}$ ; hence $\overline{U} \cap V = \emptyset$ . Similarly step 2.1 gives $U \cap \overline{V} = \emptyset$ . Thus the two nonempty sets would be separated, contradicting connectedness of $[0, 1]$ in [A3].

4.1step 1.1step 3.1∎

In particular $1 \in U$ , so $T + K$ is Fredholm with $ind (T + K) = ind T$ , as claimed.

Fredholm index is stable under compact perturbations

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources