Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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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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The reverse triangle inequality in a normed space

Statement

Let V be a normed space. For all x,y∈V, ∣∥x∥−∥y∥∣≤∥x−y∥. In particular, the norm map is 1-Lipschitz for the norm metric.

Facts & Assumptions

Given: A normed space V and vectors x,y∈V.

[L1]

A norm satisfies the triangle inequality and absolute homogeneity (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Proof

technique · direct
1.1L1

Since x=(x−y)+y, the triangle inequality in [L1] gives ∥x∥≤∥x−y∥+∥y∥.

1.2L1algebra

Since y=(y−x)+x and ∥y−x∥=∥x−y∥ by absolute homogeneity at the scalar −1, [L1] also gives ∥y∥≤∥x−y∥+∥x∥.

2.1step 1.1step 1.2algebra

Step 1.1 yields ∥x∥−∥y∥≤∥x−y∥, and step 1.2 yields ∥y∥−∥x∥≤∥x−y∥; together these are exactly ∣∥x∥−∥y∥∣≤∥x−y∥.

3.1step 2.1∎

The displayed inequality says precisely that the map x↦∥x∥ is 1-Lipschitz for the metric d(u,v)=∥u−v∥.

Depends on

Used by

Dependency tree · two levels

14 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