ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27↗ rests on later material

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On $[0, 1]$ the function $x^{β}$ is $β$ -Hölder and is $α$ -Hölder for no rational $α > β$ , so the Hölder classes are strictly nested

Example

Let $β \in Q$ with $0 < β \leq 1$ (Order on the rationals) and let

$f_{β} : [0, 1] \to R, f_{β} (x) : = x^{β}$

be the rational power of a nonnegative base (Rational powers $a^{r}$ of a positive base, with the convention $0^{β} = 0$ ), on the closed bounded interval $[0, 1]$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). Hölder conditions for a real function on $[0, 1]$ are the metric ones instantiated, by Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace, clause 4: $g$ is $γ$ -Hölder with constant $C$ when $∣ g (x) - g (y) ∣ \leq C ∣ x - y ∣^{γ}$ for all $x, y \in [0, 1]$ (Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction). Then:

$f_{β}$ is $β$ -Hölder with constant $1$ : $∣ x^{β} - y^{β} ∣ \leq ∣ x - y ∣^{β} for all x, y \in [0, 1] .$
$f_{β}$ is $α$ -Hölder for no rational $α$ with $β < α \leq 1$ : for such an $α$ there is no real $C \geq 0$ with $∣ x^{β} - y^{β} ∣ \leq C ∣ x - y ∣^{α}$ throughout $[0, 1]$ .
The classes are nested: if $0 < β < α \leq 1$ are rational and $g : [0, 1] \to R$ is $α$ -Hölder with constant $C$ , then $g$ is $β$ -Hölder with the same constant $C$ .
Hence the nesting is strict, at every pair of rational exponents $0 < β < α \leq 1$ : the $α$ -Hölder functions on $[0, 1]$ form a proper subclass of the $β$ -Hölder ones, $f_{β}$ lying in the second and not the first. Taking $α = 1$ : for rational $0 < β < 1$ the function $f_{β}$ is uniformly continuous on $[0, 1]$ (Uniform continuity of $f : A \to R$ : one $δ$ serving every pair of points of $A$ ) and is not Lipschitz.

What this witnesses. 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 asserts Lipschitz $\Rightarrow$ uniformly continuous $\Rightarrow$ continuous and $α$ -Hölder $\Rightarrow$ uniformly continuous, and claims no converse; it says so explicitly. This item supplies the missing witnesses on the real line, and it is one of the two named in the remarks of Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace. The other is $x \mapsto 1 / x$ is continuous on $(0, 1)$ and not uniformly continuous there, the pairs $1 / (k + 2)$ and $1 / (k + 3)$ defeating every $δ$ , which separates continuity from uniform continuity.

Why the exponents are rational. Rational powers $a^{r}$ of a positive base is the exponent theory available at this page's position in the reading order, so the example is stated for rational exponents. The later Real powers for positive bases, with the zero-base positive-exponent convention ↗ supplies real exponents; the restriction here belongs to the local toolkit, not to the Hölder notion. Exponents above $1$ are excluded there for a reason of substance: they force constancy (If $∣ f (x) - f (y) ∣ \leq C ∣ x - y ∣^{α}$ on an interval for some rational $α > 1$ then $f$ is constant).

Facts & Assumptions

Given: A rational $β$ with $0 < β \leq 1$ , the interval $[0, 1]$ , and $f_{β} (x) = x^{β}$ . Naturals are identified with their canonical images in $R$ .

[L1]

Rational powers: $a^{r}$ is defined for $a > 0$ and $r \in Q$ , with $a^{1} = a$ and $a^{q / 1} = a^{q}$ agreeing with the integer power; $0^{r} = 0$ for rational $r > 0$ ; and $1^{r} = 1$ (Rational powers $a^{r}$ of a positive base, Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ , Integer powers $a^{m}$ , Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L2]

Laws of rational exponents for $a, b > 0$ and $r, s \in Q$ : $a^{r} > 0$ ; $a^{r + s} = a^{r} a^{s}$ ; $(a b)^{r} = a^{r} b^{r}$ ; $a^{- r} = 1 / a^{r}$ ; $(a^{r})^{s} = a^{r s}$ . The product law persists for $a, b \geq 0$ when $r > 0$ (Laws of rational exponents).

[L3]

Monotonicity: for $0 < a < 1$ and rationals $r < s$ one has $a^{r} > a^{s}$ ; for $a > 1$ and $r < s$ one has $a^{r} < a^{s}$ ; for $a = 1$ all powers are $1$ ; and for rational $r > 0$ and $0 < a < b$ one has $a^{r} < b^{r}$ (Monotonicity of $r \mapsto a^{r}$ and of $a \mapsto a^{r}$ ).

[L4]

Hölder conditions for real functions on $[0, 1]$ are $∣ g (x) - g (y) ∣ \leq C ∣ x - y ∣^{γ}$ for all $x, y$ , and an $α$ -Hölder real function with rational $0 < α \leq 1$ is uniformly continuous, hence continuous; "Lipschitz" is the case $γ = 1$ (Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction, 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, Uniform continuity of $f : A \to R$ : one $δ$ serving every pair of points of $A$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L5]

Archimedean property: for every real $η > 0$ there is a natural $q \geq 1$ with $1 / q < η$ , and for every real $t$ a natural $n$ with $t < n$ ; and $0 < s < t$ implies $0 < 1 / t < 1 / s$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L6]

Absolute value and order in $R$ : $∣ u ∣ \geq 0$ ; $∣ u ∣ = u$ for $u \geq 0$ ; the order is total, so two points of $[0, 1]$ may be named so that one is $\leq$ the other; and $[0, 1] = {x : 0 \leq x \leq 1}$ (Basic properties of the absolute value, Ordered field, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1

For $0 \leq t \leq 1$ one has $t^{β} \geq t$ . If $t = 0$ then $t^{β} = 0 = t$ by [L1]; if $t = 1$ then $t^{β} = 1 = t$ by [L1]. If $0 < t < 1$ then, when $β < 1$ , [L3] with $r : = β < s : = 1$ gives $t^{β} > t^{1} = t$ , and when $β = 1$ it is an equality.

L1L3L6

1.2

Claim 3. Let $0 < β < α \leq 1$ be rational and let $g$ satisfy $∣ g (x) - g (y) ∣ \leq C ∣ x - y ∣^{α}$ on $[0, 1]$ . For $x, y \in [0, 1]$ put $a : = ∣ x - y ∣$ , so $0 \leq a \leq 1$ by [L6]. If $a = 0$ then $a^{α} = a^{β} = 0$ by [L1]; if $a = 1$ then both are $1$ by [L1]; and if $0 < a < 1$ then [L3] with $r : = β < s : = α$ gives $a^{α} < a^{β}$ . In every case $a^{α} \leq a^{β}$ , so $∣ g (x) - g (y) ∣ \leq C a^{α} \leq C a^{β}$ and $g$ is $β$ -Hölder with the same constant.

L1L3L4L6

1.3

Claim 2, the setup. Let $α \in Q$ with $β < α \leq 1$ and suppose, for contradiction, that some real $C \geq 0$ satisfies $∣ x^{β} - y^{β} ∣ \leq C ∣ x - y ∣^{α}$ for all $x, y \in [0, 1]$ . Taking $x : = 1$ and $y : = 0$ gives $1 = ∣ 1 - 0 ∣ \leq C \cdot 1 = C$ by [L1], so $C \geq 1 > 0$ . Taking $y : = 0$ and an arbitrary $x$ with $0 < x \leq 1$ gives $x^{β} \leq C x^{α}$ .

L1L6

2.1

Subadditivity: $(u + v)^{β} \leq u^{β} + v^{β}$ for all reals $u, v \geq 0$ . If $u + v = 0$ then $u = v = 0$ and both sides are $0$ by [L1]. Otherwise put $s : = u + v > 0$ , $p : = u / s$ and $q : = v / s$ , so $p, q \geq 0$ and $p + q = 1$ , whence $0 \leq p \leq 1$ and $0 \leq q \leq 1$ . By step 1.1, $p^{β} \geq p$ and $q^{β} \geq q$ , so $p^{β} + q^{β} \geq p + q = 1$ . By the product law of [L2], valid for nonnegative bases since $β > 0$ , $u^{β} = (p s)^{β} = p^{β} s^{β}$ and $v^{β} = q^{β} s^{β}$ ; hence $u^{β} + v^{β} = (p^{β} + q^{β}) s^{β} \geq s^{β} = (u + v)^{β}$ , using $s^{β} > 0$ from [L2].

step 1.1L1L2L6

2.2

Claim 2, the estimate. Put $γ : = α - β$ , a rational with $γ > 0$ . For $0 < x \leq 1$ , dividing the inequality of step 1.3 by $x^{α} > 0$ and using [L2] gives $x^{β - α} = x^{β} x^{- α} \leq C$ , that is $1 / x^{γ} \leq C$ and hence $x^{γ} \geq 1 / C > 0$ by [L5]. Applying this at $x : = 1 / n$ for a natural $n \geq 1$ , and using $(1 / n)^{γ} = 1 / n^{γ}$ from [L2], gives $n^{γ} \leq C$ for every natural $n \geq 1$ .

step 1.3L2L5

3.1

Claim 1. Let $x, y \in [0, 1]$ ; by [L6] name them so that $y \leq x$ . Put $u : = y \geq 0$ and $v : = x - y \geq 0$ , so $x = u + v$ . By step 2.1, $x^{β} \leq y^{β} + (x - y)^{β}$ , that is $x^{β} - y^{β} \leq (x - y)^{β} = ∣ x - y ∣^{β}$ . Also $y^{β} \leq x^{β}$ : for $y = 0$ this reads $0 \leq x^{β}$ by [L1] and [L2], and for $0 < y \leq x$ it is [L3] with the exponent $β > 0$ , together with equality when $y = x$ . Hence $∣ x^{β} - y^{β} ∣ = x^{β} - y^{β} \leq ∣ x - y ∣^{β}$ , so $f_{β}$ is $β$ -Hölder with constant $1$ .

step 2.1L1L2L3L4L6

3.2

Claim 2, the contradiction. By [L5] fix a natural $q \geq 1$ with $1 / q < γ$ , and then a natural $n$ with $C^{q} < n$ ; since $C^{q} > 0$ we have $n \geq 1$ , and since $C \geq 1$ we have $C^{q} \geq 1$ and so $n > 1$ . By [L3] with the exponent $1 / q > 0$ applied to the bases $C^{q} < n$ , and by $(C^{q})^{1 / q} = C^{q \cdot (1 / q)} = C^{1} = C$ from [L1] and [L2], we get $n^{1 / q} > C$ ; and by [L3] with the base $n > 1$ and the exponents $1 / q < γ$ we get $n^{γ} > n^{1 / q} > C$ . That contradicts step 2.2, so no such $C$ exists and claim 2 holds.

step 2.2L1L2L3L5

4.1

Claim 4. Let $0 < β < α \leq 1$ be rational. Every $α$ -Hölder function on $[0, 1]$ is $β$ -Hölder by step 1.2, and $f_{β}$ is $β$ -Hölder by step 3.1 and not $α$ -Hölder by step 3.2; so the inclusion of classes is proper. With $α : = 1$ and $0 < β < 1$ : $f_{β}$ is $β$ -Hölder, hence uniformly continuous on $[0, 1]$ by [L4], and it is not $1$ -Hölder, that is not Lipschitz.

step 3.1step 1.2step 3.2L4∎

Remarks

The witness is as concrete as it can be here. For $β = 1 / 2$ the function is $\sqrt{x}$ , and the failure of the Lipschitz condition is the familiar one: $∣ \sqrt{x} - \sqrt{0} ∣ = \sqrt{x}$ exceeds $C x$ for every $C$ once $x < 1 / C^{2}$ . Step 2.2 is that computation written for a general rational exponent.
Subadditivity is the whole of claim 1, and it is proved by normalising to $p + q = 1$ and using $t^{β} \geq t$ on $[0, 1]$ . No derivative and no convexity argument is used; neither is available at this point in the reading order.
What happens at the two ends of the range. At $β = 1$ the function is the identity, Lipschitz and not $α$ -Hölder for any rational $α > 1$ — indeed no nonconstant function is, by If $∣ f (x) - f (y) ∣ \leq C ∣ x - y ∣^{α}$ on an interval for some rational $α > 1$ then $f$ is constant, which is why Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction stops at $1$ . As $β$ decreases the class grows, and claim 4 says it grows strictly at every rational step.

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Sources

Hölder condition (Wikipedia) (standard reference, not scraped)
Lipschitz continuity (Wikipedia) (standard reference, not scraped)
Uniform continuity (Wikipedia) (standard reference, not scraped)
University of Zaragoza thesis on Hölder continuity (standard reference, not scraped)
University of Wisconsin Math 521 exercises (standard reference, not scraped)

On [0,1][0,1] the function xβxβ is ββ-Hölder and is αα-Hölder for no rational α>βα>β, so the Hölder classes are strictly nested