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

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Linear combinations of compact operators are compact

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field. Then the compact operators $X \to Y$ (Compact linear operator) form a linear subspace of $B (X, Y)$ (A bounded linear operator between normed spaces): the zero operator is compact, and if $S, T : X \to Y$ are compact and $λ$ is a scalar, then $S + T$ and $λ S$ are compact.

Consequently, if $K : X \to X$ is a compact endomorphism, if $m \geq 1$ is a natural number and $a_{1}, \dots, a_{m}$ are scalars, then the polynomial $\sum_{k = 1}^{m} a_{k} K^{k} = a_{1} K + a_{2} K^{2} + \dots + a_{m} K^{m}$ is compact.

Facts & Assumptions

[A1]

$S$ is compact exactly when $\overline{S (E)}$ is compact for every bounded $E$ ; in particular $\overline{S ({\overline{B}}_{X})}$ is compact for the closed unit ball (Compact linear operator), and a bounded linear operator such as $K$ satisfies $∥ K w ∥ \leq ∥ K ∥ ∥ w ∥$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

[A2]

The zero operator has range ${0}$ , and the space ${0}$ admits the empty ordered basis of finite length, so the zero operator is compact (Bounded finite rank operators are compact, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[A3]

If $T$ is compact and $C$ is bounded linear, then $T C$ and $C T$ are compact (Compositions with a compact operator are compact).

[A4]

Addition $+ : Y \times Y \to Y$ and scalar multiplication are continuous (Vector addition and scalar multiplication are continuous in a normed space); a finite product of compact spaces is compact (A product of finitely many compact spaces is compact in the product topology); a continuous image of a compact set is compact (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism); a compact subset of a metric space is closed, and a closed subset of a compact metric space is compact (A compact subset of a metric space is closed and bounded, A closed subset of a compact metric space is compact).

Proof

technique · direct

Given: Normed spaces $X, Y$ over one scalar field, compact operators $S, T : X \to Y$ , a scalar $λ$ , and a compact endomorphism $K : X \to X$ .

1.1

The zero operator is compact by [A2].

1.2

For bounded $E \subseteq X$ the sets $\overline{S (E)}$ and $\overline{T (E)}$ are compact by [A1], so their product is compact in $Y \times Y$ by [A4] and its image under the continuous addition map is compact by [A4]; since $(S + T) (E) \subseteq \overline{S (E)} + \overline{T (E)}$ , its closure is a closed subset of that compact image, hence compact, and $S + T$ is compact.

A1A4algebra

1.3

For bounded $E \subseteq X$ the set $λ \overline{S (E)}$ is the image of the compact set $\overline{S (E)}$ under the continuous map $y \mapsto λ y$ , hence compact by [A4]; since $(λ S) (E) \subseteq λ \overline{S (E)}$ , its closure is compact by [A4], so $λ S$ is compact.

A1A4

1.4

For every natural $j \geq 1$ the power $K^{j}$ is compact: $K^{1} = K$ is compact, and if $K^{j}$ is compact then $K^{j + 1} = K \circ K^{j}$ is compact by [A3].

2.1

By [step 1.1], [step 1.2] and [step 1.3] the compact operators $X \to Y$ contain the zero operator and are closed under addition and scalar multiplication, so they form a linear subspace of $B (X, Y)$ .

step 1.1step 1.2step 1.3

3.1

Let $K$ be compact and $a_{1}, \dots, a_{m}$ scalars with $m \geq 1$ . Each $K^{k}$ with $1 \leq k \leq m$ is compact by [step 1.4]; by induction on $m$ using [step 2.1], a finite sum of scalar multiples of compact operators is compact, so $\sum_{k = 1}^{m} a_{k} K^{k}$ is compact.

step 1.4step 2.1∎

Depends on

Used by

Lambda identity minus compact has index zero Corollary
Calkin algebra Definition
Relative compactness with respect to an operator Definition
Riesz Schauder ascent and descent stabilize Lemma
Fredholm index is additive Theorem
Fredholm index is stable under compact perturbations Theorem
Riesz schauder spectrum of a compact operator Theorem

Dependency tree · two levels

74 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 — §3.1 Theorem 3.1 and Problem 3.1 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis — §4.2 p.185, Theorem 4.28(i) (standard reference, not scraped)