TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Contraction implies Lipschitz implies uniformly continuous implies continuous; every Hölder map is uniformly continuous, and a Lipschitz map on a bounded space is Hölder for every exponent

Statement

Let $(X, d_{X})$ and $(Y, d_{Y})$ be metric spaces (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $f : X \to Y$ be a function, with the three regularity conditions as in Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction. Then:

If $f$ is a contraction, it is Lipschitz.
If $f$ is Lipschitz, it is uniformly continuous (Uniform continuity of a map of metric spaces: one $δ$ serving every point).
If $f$ is uniformly continuous, it is continuous (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).
If $f$ is $α$ -Hölder for some rational $α$ with $0 < α \leq 1$ , it is uniformly continuous.
Suppose $X$ is nonempty and bounded, and put $R : = diam (X) + 1$ , a real with $R > 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). If $f$ is Lipschitz with constant $L$ , then for every rational $α$ with $0 < α \leq 1$ the map $f$ is $α$ -Hölder with constant $L R^{1 - α}$ .

Claim 5 carries its boundedness hypothesis for a reason, and no unconditional "Lipschitz implies Hölder" is asserted anywhere here. On an unbounded space the implication is false; the witness and its verification are in the first remark below.

Strictness is not claimed. The five implications are asserted and nothing more; that none of them reverses is witnessed elsewhere, and those witnesses are not prerequisites of this theorem. See the second remark.

Facts & Assumptions

Given: Metric spaces $(X, d_{X})$ , $(Y, d_{Y})$ , a function $f : X \to Y$ , a real $ε > 0$ , and a rational $α$ with $0 < α \leq 1$ .

[A1]

$f$ is a contraction with constant $q$ : $f$ is Lipschitz with constant $q$ and $0 \leq q < 1$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[A2]

$f$ is Lipschitz with constant $L \geq 0$ : $d_{Y} (f (x), f (x^{'})) \leq L d_{X} (x, x^{'})$ for all $x, x^{'} \in X$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[A3]

$f$ is $α$ -Hölder with constant $C \geq 0$ : $d_{Y} (f (x), f (x^{'})) \leq C d_{X} (x, x^{'})^{α}$ for all $x, x^{'} \in X$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[A4]

$X$ is nonempty and bounded, so $diam (X)$ exists and $d_{X} (x, x^{'}) \leq diam (X)$ for all $x, x^{'} \in X$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Complete ordered field (least-upper-bound property)).

[L1]

Uniform continuity: one $δ > 0$ per $ε > 0$ serving every pair (Uniform continuity of a map of metric spaces: one $δ$ serving every point); continuity at a point allows $δ$ to depend on the point as well (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

[L2]

Rational powers of a positive base, with $a^{1} = a$ , $a^{0} = 1$ , and the supplementary clause $0^{r} = 0$ for rational $r > 0$ (Rational powers $a^{r}$ of a positive base, Order on the rationals).

[L3]

Exponent laws for positive bases: $a^{r + s} = a^{r} a^{s}$ , $(a^{r})^{s} = a^{r s}$ , and $a^{r} > 0$ (Laws of rational exponents).

[L4]

Monotonicity in the base: for rational $r > 0$ and $0 < a < b$ one has $a^{r} < b^{r}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L5]

A metric is nonnegative, and $d (x, x^{'}) = 0$ forces $x = x^{'}$ (Nonnegativity of a metric is a consequence of the other axioms, not an axiom, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L6]

Positivity of inverses and multiplication of inequalities by positives (Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

Proof

technique · direct

1.1

Claim 1 is immediate from the definitions: a contraction with constant $q$ is Lipschitz with the constant $q$ , and $q \geq 0$ .

A1A2

1.2

Claim 2: assume [A2] and put $δ : = ε / (L + 1)$ , a positive real since $L + 1 > 0$ . If $d_{X} (x, x^{'}) < δ$ then, using $d_{X} (x, x^{'}) \geq 0$ , $d_{Y} (f (x), f (x^{'})) \leq L d_{X} (x, x^{'}) \leq (L + 1) d_{X} (x, x^{'}) < (L + 1) δ = ε$ ; the same $δ$ served every pair, so $f$ is uniformly continuous.

A2L1L5L6

1.3

Claim 3: assume $f$ uniformly continuous and let $a \in X$ ; the $δ$ belonging to $ε$ satisfies $d_{X} (x, a) < δ \Rightarrow d_{Y} (f (x), f (a)) < ε$ for all $x$ , which is continuity at $a$ , and $a$ was arbitrary.

1.4

Claim 4, the case of equal points: assume [A3]; if $d_{X} (x, x^{'}) = 0$ then $x = x^{'}$ , so $d_{Y} (f (x), f (x^{'})) = 0 < ε$ whatever $δ$ is.

A3L5

1.5

Claim 4, the main case: put $η : = ε / (C + 1) > 0$ and $δ : = η^{1 / α}$ , which is a positive real because $1 / α$ is a positive rational and $η > 0$ .

L2L3L6choose

1.6

Claim 5: assume [A2] and [A4], put $R : = diam (X) + 1$ and $C : = L R^{1 - α}$ . Since $diam (X) \geq 0$ we have $R \geq 1 > 0$ , so $R^{1 - α}$ is defined and positive and $C \geq 0$ .

A2A4L2L3

2.1

Let $t : = d_{X} (x, x^{'})$ with $0 < t < δ$ . Then $t^{α} < δ^{α} = (η^{1 / α})^{α} = η^{(1 / α) α} = η^{1} = η$ , so $d_{Y} (f (x), f (x^{'})) \leq C t^{α} \leq (C + 1) t^{α} < (C + 1) η = ε$ .

step 1.5A3L2L3L4L6

2.2

Let $x, x^{'} \in X$ and $t : = d_{X} (x, x^{'})$ ; then $0 \leq t \leq diam (X) < R$ . If $t = 0$ then $d_{Y} (f (x), f (x^{'})) \leq L \cdot 0 = 0 = C \cdot 0^{α} = C t^{α}$ , because $α > 0$ .

step 1.6A2A4L2L5

2.3

If $t > 0$ then $t = t^{α + (1 - α)} = t^{α} t^{1 - α}$ and $t^{1 - α} \leq R^{1 - α}$ : for $α < 1$ the exponent $1 - α$ is a positive rational and $0 < t < R$ , and for $α = 1$ both sides are $1$ . Multiplying by $L t^{α} \geq 0$ gives $L t = L t^{α} t^{1 - α} \leq L R^{1 - α} t^{α} = C t^{α}$ , and hence $d_{Y} (f (x), f (x^{'})) \leq L t \leq C t^{α}$ .

step 1.6A2L2L3L4L6

3.1

Steps 1.4 and 2.1 cover every pair with $d_{X} (x, x^{'}) < δ$ , and $δ$ did not depend on the pair, so $f$ is uniformly continuous: claim 4 holds.

step 1.4step 2.1L1

4.1

Steps 2.2 and 2.3 give $d_{Y} (f (x), f (x^{'})) \leq C d_{X} (x, x^{'})^{α}$ for every pair, so $f$ is $α$ -Hölder with constant $C = L R^{1 - α}$ , which is claim 5; claims 1 to 4 are steps 1.1, 1.2, 1.3 and 3.1.

step 1.1step 1.2step 1.3step 3.1step 2.2step 2.3∎

Remarks

Boundedness in claim 5 cannot be dropped, and here is the witness. Take $X = Y = R$ with the usual metric and $f (x) = x$ , which is Lipschitz with constant $1$ . Suppose $f$ were $α$ -Hölder with constant $C \geq 0$ for some rational $α$ with $0 < α < 1$ , so that $∣ x - y ∣ \leq C ∣ x - y ∣^{α}$ for all reals $x, y$ . Taking $y = 0$ and $x = t$ with $t > 0$ , and writing $t = t^{α} t^{1 - α}$ (Laws of rational exponents), division by $t^{α} > 0$ gives $t^{1 - α} \leq C$ for every real $t > 0$ . At $t = 1$ this reads $1 \leq C$ , so $C > 0$ ; and then choosing a natural $n$ with $n > C^{1 / (1 - α)}$ , which exists by the Archimedean property (Every complete ordered field is Archimedean), and raising to the positive rational power $1 - α$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ , Laws of rational exponents) gives $n^{1 - α} > (C^{1 / (1 - α)})^{1 - α} = C$ , contradicting $t^{1 - α} \leq C$ at $t = n$ . So the identity of $R$ is Lipschitz and $α$ -Hölder for no exponent $α < 1$ . A chain reading "Lipschitz implies Hölder implies uniformly continuous" is therefore false as stated, which is why claims 4 and 5 are separated here and why claim 5 carries a hypothesis.
No implication reverses, and two of the witnesses are on the companion page. The square root on $[0, \infty)$ is $1 / 2$ -Hölder and not Lipschitz ( $\sqrt{\cdot}$ on $[0, \infty)$ is uniformly continuous and exactly $1 / 2$ -Hölder, and is not Lipschitz ↗), and $x \mapsto 1 / x$ on $(0, 1)$ is continuous and not uniformly continuous ( $x \mapsto 1 / x$ is continuous on $(0, 1)$ and sends the Cauchy sequence $(1 / (k + 2))_{k \geq 0}$ to an unbounded one ↗). Both are read here as orientation only: this theorem does not depend on them and claims nothing about strictness. That a Lipschitz map need not be a contraction is visible already in the identity map, whose only Lipschitz constants are the reals $\geq 1$ .
The constant $L + 1$ and the constant $C + 1$ are there to avoid a case split. Dividing by $L$ or by $C$ would require them to be nonzero, and a Lipschitz constant or a Hölder constant is allowed to be $0$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction). Enlarging the constant by $1$ is harmless, since the defining inequalities are preserved by enlarging the constant.
What claim 5 costs. The constant it produces, $L R^{1 - α}$ , grows with the diameter of $X$ , and that dependence is exactly what the unbounded counterexample above exploits: as $R \to \infty$ no single Hölder constant survives.

Depends on

Used by

Dependency tree · two levels

43 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

Lipschitz continuity (Wikipedia) (standard reference, not scraped)
Hölder condition (Wikipedia) (standard reference, not scraped)
Uniform continuity (Wikipedia) (standard reference, not scraped)