CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07

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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.
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A one-sided comparison of two complete norms makes them equivalent

Statement

Assume DC. Let $p, q$ be complete norms on one vector space $V$ . If $q (v) \leq C p (v)$ for all $v$ and some $C > 0$ , then $p$ and $q$ are equivalent norms (Equivalent norms, and the dictionary with equivalent metrics).

Facts & Assumptions

Given: DC, complete norms $p, q$ on $V$ , and $q \leq C p$ .

Proof

technique · direct

1.1given

The identity $I : (V, p) \to (V, q)$ is a bounded linear bijection by the assumed inequality.

2.1step 1.1

Both spaces are Banach (Banach space), so Bounded inverse theorem makes $I^{- 1}$ bounded: $p (v) \leq C^{'} q (v)$ for some $C^{'} > 0$ .

3.1step 2.1∎

The two inequalities are precisely equivalence of $p$ and $q$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 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

Teschl, Topics in Real and Functional Analysis, Theorem 4.6 (standard reference, not scraped)