CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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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.

Any two completions of a normed space are uniquely linearly isometric

Statement

Let $X$ be a normed space, and let $(Y, j)$ and $(Z, k)$ be two completions of $X$ . Then there is a unique linear isometric isomorphism $U : Y \to Z$ such that $U \circ j = k .$

Facts & Assumptions

Given: A normed space $X$ and two completions $(Y, j)$ and $(Z, k)$ of $X$ .

[L1]

Any two metric completions are related by a unique isometry commuting with the dense embeddings (A completion is unique up to a unique isometry fixing the original space, and uniformly continuous maps into complete spaces extend through it).

[L2]

Bounded linear maps extend uniquely across a completion (Bounded linear maps extend uniquely across the completion).

[L3]

A linear isometric isomorphism is a bijective linear isometry (Linear isometries and isometric isomorphisms, Completion of a normed space).

Proof

technique · direct

1.1

By [L1], there is a unique isometry $U : Y \to Z$ with $U \circ j = k$ .

1.2

On the dense subspace $j [X] \subseteq Y$ , the map $j (x) \mapsto k (x)$ is linear and norm-preserving. Applying [L2] to this dense linear isometry extends it to a bounded linear map $\tilde{U} : Y \to Z$ with $\tilde{U} \circ j = k$ .

L2L3

2.1

Both $U$ and $\tilde{U}$ are continuous maps $Y \to Z$ extending the same map on $j [X]$ , so the uniqueness in step 1.1 forces $U = \tilde{U}$ . Hence $U$ is linear.

step 1.1step 1.2L1

3.1

Since $U$ is already an isometry by step 1.1, it is a linear isometric isomorphism. Uniqueness of such a map is again the uniqueness from [L1].

step 2.1L3∎

Depends on

Used by

Dependency tree · two levels

33 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 (standard reference, not scraped)
Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)