LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-22

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Compositions with a compact operator are compact

Statement

Let $W$ , $X$ , $Y$ and $Z$ be normed spaces over the same scalar field. If $T : X \to Y$ is compact (Compact linear operator) and $A : W \to X$ and $B : Y \to Z$ are bounded linear operators (A bounded linear operator between normed spaces), then the composites $T A : W \to Y$ and $B T : X \to Z$ are compact.

Facts & Assumptions

[A1]

A bounded linear operator is continuous and satisfies $∥ S w ∥ \leq ∥ S ∥ ∥ w ∥$ for all $w$ (For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, The operator norm as the least bound and as the unit-sphere or unit-ball supremum); $T$ is compact exactly when $\overline{T (E)}$ is compact for every bounded $E \subseteq X$ , in particular for $E = {\overline{B}}_{X} = {x : ∥ x ∥ \leq 1}$ (Compact linear operator).

[A2]

A subset $E$ of a metric space is bounded when $E = \emptyset$ or $E \subseteq B (x_{0}, r)$ for some point $x_{0}$ and real $r > 0$ ; a subset of a bounded set is bounded (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[A3]

A continuous image of a compact subset 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); scalar multiplication by a fixed scalar is continuous (Vector addition and scalar multiplication are continuous in a normed space); 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 $W, X, Y, Z$ over one scalar field, a compact $T : X \to Y$ , and bounded linear $A : W \to X$ , $B : Y \to Z$ .

1.1

If $E \subseteq W$ is bounded and nonempty, say $E \subseteq B (w_{0}, r)$ , then $∥ A w ∥ \leq ∥ A ∥ (∥ w_{0} ∥ + r)$ for every $w \in E$ by [A1], so $A (E) \subseteq B (0, ∥ A ∥ (∥ w_{0} ∥ + r) + 1)$ is bounded; and $A (\emptyset) = \emptyset$ is bounded, so $A$ carries bounded sets to bounded sets.

A1A2algebra

1.2

If $E = \emptyset$ , take $R = 1$ , so $E \subseteq R {\overline{B}}_{X}$ holds immediately. If $E$ is a bounded subset of $X$ with $E \neq \emptyset$ and $E \subseteq B (x_{0}, R_{0})$ , then $∥ x ∥ \leq ∥ x_{0} ∥ + R_{0} = : R$ for every $x \in E$ by [A2] and the triangle inequality, so $E \subseteq R {\overline{B}}_{X}$ ; the same holds in any normed space.

A2algebra

1.3

The set $B (\overline{T ({\overline{B}}_{X})})$ is compact: $\overline{T ({\overline{B}}_{X})}$ is compact by [A1], and $B$ is continuous by [A1], so the image under $B$ is compact by [A3].

A1A3

2.1

For every bounded $E \subseteq W$ the image $A (E)$ is bounded by [step 1.1], so $\overline{T (A (E))}$ is compact by [A1]; hence $T A$ is compact.

step 1.1A1

2.2

For every bounded $E \subseteq X$ , [step 1.2] gives $E \subseteq R {\overline{B}}_{X}$ for some real $R \geq 0$ , so $T (E) \subseteq R \overline{T ({\overline{B}}_{X})}$ and hence $B (T (E)) \subseteq R B (\overline{T ({\overline{B}}_{X})})$ , which is compact by [step 1.3] and [A3] and therefore closed; thus $\overline{B (T (E))} \subseteq R B (\overline{T ({\overline{B}}_{X})})$ is a closed subset of a compact set, hence compact, and $B T$ is compact.

step 1.2step 1.3A1A3

3.1

Both composites $T A$ and $B T$ are therefore compact.

step 2.1step 2.2∎

Depends on

Used by

Absolute value and singular values of a compact operator Definition
Calkin algebra Definition
Relative compactness with respect to an operator Definition
Linear combinations of compact operators are compact Lemma
Riesz Schauder ascent and descent stabilize Lemma
Cyclicity of the trace Theorem
Fredholm index is additive Theorem
Fredholm index is stable under compact perturbations Theorem
Riesz schauder spectrum of a compact operator Theorem
Schauder compact adjoint theorem Theorem
Spectral theorem for compact self adjoint operators Theorem
Trace class iff product of two Hilbert Schmidt operators Theorem
Weyl's theorem: invariance of the essential spectrum Theorem

Dependency tree · two levels

57 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 pp.69–70, Theorem 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)