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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Bounded finite rank operators are compact

Statement

Let $X$ and $Y$ be normed spaces over the same scalar field and let $T : X \to Y$ be a bounded linear operator (A bounded linear operator between normed spaces) whose range $T (X)$ admits an ordered basis of finite length (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis). Then $T$ is compact (Compact linear operator).

Facts & Assumptions

[A1]

$T$ is compact exactly when $\overline{T ({\overline{B}}_{X})}$ is compact, where ${\overline{B}}_{X} = {x \in X : ∥ x ∥ \leq 1}$ (Compact linear operator); the operator norm satisfies $∥ T x ∥ \leq ∥ T ∥ ∥ x ∥$ for every $x$ (The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

[A2]

A subspace $W$ of a normed space $V$ admitting an ordered basis of finite length is a closed subset of $V$ (A finite-dimensional normed subspace is closed, Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis), and its closed unit ball ${w \in W : ∥ w ∥ \leq 1}$ is compact (The closed unit ball is compact if and only if the normed space is finite-dimensional).

[A3]

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), and scalar multiplication is continuous on a normed space (Vector addition and scalar multiplication are continuous in a normed space).

[A4]

A compact subset of a metric space is closed and bounded (A compact subset of a metric space is closed and bounded); a closed subset of a compact topological space is a compact subset (A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact).

Proof

technique · direct

Given: Normed spaces $X, Y$ over one scalar field and a bounded linear $T : X \to Y$ whose range $R : = T (X)$ admits an ordered basis of finite length.

1.1

Every point of $T ({\overline{B}}_{X})$ lies in $R$ , and $∥ T x ∥ \leq ∥ T ∥$ for every $x \in {\overline{B}}_{X}$ by [A1]; writing $s : = ∥ T ∥$ , this says $T ({\overline{B}}_{X}) \subseteq s {\overline{B}}_{R}$ , where ${\overline{B}}_{R} = {v \in R : ∥ v ∥ \leq 1}$ .

A1algebra

1.2

By [A2] the set ${\overline{B}}_{R}$ is compact and $R$ is closed in $Y$ .

2.1

The set $s {\overline{B}}_{R}$ is the image of the compact set ${\overline{B}}_{R}$ under the continuous map $v \mapsto s v$ , so it is a compact subset of $Y$ by [A3]; it is therefore closed in $Y$ by [A4].

step 1.2A3A4

3.1

Since $T ({\overline{B}}_{X}) \subseteq s {\overline{B}}_{R}$ by [step 1.1], the closure $\overline{T ({\overline{B}}_{X})}$ is contained in the closed set $s {\overline{B}}_{R}$ ; being a closed subset of the compact space $s {\overline{B}}_{R}$ , it is compact by [A4].

step 1.1step 2.1A4

4.1

By [A1] compactness of $\overline{T ({\overline{B}}_{X})}$ is exactly compactness of $T$ , so $T$ is compact.

step 3.1A1∎

Depends on

Used by

Compact operator iff approximation numbers tend to zero Corollary
Finite rank operators are norm dense in compact Hilbert space operators Corollary
A compact operator can have nondense range Counterexample
Compactness is not preserved by strong operator limits Counterexample
Approximable operator Definition
Calkin algebra Definition
Trace class operator Definition
Adjoint, norm and trace of an operator of rank at most one Example
Diagonal operator on ell p is compact iff diagonal tends to zero Example
Diagonal Schatten class criteria on ell two Example
Finite-rank truncations of a square-integrable kernel Example
Linear combinations of compact operators are compact Lemma
Positive square root of a compact positive operator Lemma
Atkinson Theorem
Trace class is a two sided Banach operator ideal Theorem

Dependency tree · two levels

67 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

Theo Bühler and Dietmar Salamon, Functional Analysis — §4.2 p.184, Example 4.23 (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.1 p.69, Theorem 3.1 (standard reference, not scraped)