LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Bounded local displacement on a geodesic space implies coarse Lipschitz control

Statement

Let $X$ be a geodesic metric space and $Y$ a metric space. Suppose $f : X \to Y$ satisfies $d_{Y} (f (x), f (y)) \leq C whenever d_{X} (x, y) \leq 1,$ for some real $C \geq 0$ . Then $f$ is coarse Lipschitz; more precisely, $d_{Y} (f (x), f (y)) \leq C d_{X} (x, y) + C for all x, y \in X .$

Facts & Assumptions

Given: A geodesic metric space $X$ , a metric space $Y$ , a map $f : X \to Y$ , and a real $C \geq 0$ such that $d_{Y} (f (x), f (y)) \leq C$ whenever $d_{X} (x, y) \leq 1$ .

[L1]

In a geodesic metric space, every two points $x, y$ are joined by a geodesic segment $γ : [0, ℓ] \to X$ of length $ℓ = d_{X} (x, y)$ (Geodesics and geodesic metric spaces).

[L2]

A map is coarse Lipschitz when there are reals $A, B \geq 0$ with $d_{Y} (f (x), f (y)) \leq A d_{X} (x, y) + B$ for all $x, y$ (Coarse Lipschitz maps and quasi-isometric embeddings).

[L3]

The Archimedean property says that for every real $t$ there is a natural number $m$ with $t < m$ (Every complete ordered field is Archimedean).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

Proof

technique · direct

1.1

Fix $x, y \in X$ . If $d_{X} (x, y) \leq 1$ , the displayed hypothesis already gives $d_{Y} (f (x), f (y)) \leq C \leq C d_{X} (x, y) + C$ .

givenalgebra

1.2

Suppose $d_{X} (x, y) > 1$ . By [L1], choose a geodesic $γ : [0, ℓ] \to X$ from $x$ to $y$ with $ℓ = d_{X} (x, y)$ . By [L3] and [L4], let $m$ be the least natural number with $ℓ \leq m$ . Then $m - 1 < ℓ \leq m$ , so $m \leq ℓ + 1$ .

L1L3L4choose

2.1

Put $x_{i} : = γ (i ℓ / m)$ for $0 \leq i \leq m$ . Consecutive points satisfy $d_{X} (x_{i - 1}, x_{i}) = ℓ / m \leq 1$ , so the hypothesis gives $d_{Y} (f (x_{i - 1}), f (x_{i})) \leq C$ for every $i$ . Summing along the chain yields $d_{Y} (f (x), f (y)) \leq m C \leq C ℓ + C = C d_{X} (x, y) + C$ .

givenstep 1.2algebra

3.1

Steps 1.1 and 2.1 give the displayed global bound for all $x, y$ , so [L2] makes $f$ coarse Lipschitz.

L2step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

C. Löh, Geometric Group Theory, Sections 4.4 and 5.1 (standard reference, not scraped)
C. Drutu and M. Kapovich, Lectures on Geometric Group Theory, Chapter 5 (standard reference, not scraped)