TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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If $|f(x) - f(y)| \le C|x-y|^{\alpha}$ on an interval for some rational $\alpha > 1$ then $f$ is constant

Statement

Let $I \subseteq \mathbb{R}$ be order-convex (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length), let $f : I \to \mathbb{R}$ , let $C \in \mathbb{R}$ with $C \ge 0$ , and let $\alpha \in \mathbb{Q}$ with $\alpha > 1$ (Order on the rationals). Suppose

$|f(x) - f(y)| \;\le\; C\,|x - y|^{\alpha} \qquad \text{for all } x, y \in I ,$

the power being the rational power of a nonnegative base (Rational powers $a^r$ of a positive base, with the convention $0^{\alpha} = 0$ for $\alpha > 0$ ). Then $f$ is constant on $I$ : $f(x) = f(y)$ for all $x, y \in I$ .

The hypothesis is written out, and not expressed through Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction, because it cannot be. That definition introduces the $\alpha$ -Hölder condition for rational $\alpha$ with $0 < \alpha \le 1$ only, and says explicitly that no claim is made about an exponent above $1$ . The displayed inequality is the natural extension of the formula to $\alpha > 1$ , and this theorem is what that extension is worth: for rational $0 < \alpha \le 1$ the same inequality is the $\alpha$ -Hölder condition of Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction instantiated at $I \subseteq \mathbb{R}$ by Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace, clause 4, and then it makes $f$ uniformly continuous, hence continuous (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, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point); above $1$ it makes $f$ constant, which is why the definition stops at $1$ .

Order-convexity is essential. On a domain that is not order-convex the conclusion fails: on $I = \{0\} \cup \{1\}$ the function $f(0) = 0$ , $f(1) = 1$ satisfies the inequality with $C = 1$ and any $\alpha$ , and is not constant. What the proof uses is that the whole segment between two points of $I$ lies in $I$ , so that the distance between them can be subdivided.

Facts & Assumptions

Given: An order-convex $I \subseteq \mathbb{R}$ , a function $f : I \to \mathbb{R}$ , a real $C \ge 0$ and a rational $\alpha > 1$ with $|f(x) - f(y)| \le C|x-y|^{\alpha}$ for all $x, y \in I$ . Natural numbers are identified with their canonical images in $\mathbb{R}$ , as elsewhere in this library.

[L1]

Order-convexity: $x, y \in I$ with $x \le z \le y$ gives $z \in I$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

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

[L3]

Laws of rational exponents for $a, b > 0$ and $r, s \in \mathbb{Q}$ : $a^{r} > 0$ ; $a^{r+s} = a^{r}a^{s}$ ; $(ab)^{r} = a^{r}b^{r}$ ; $a^{-r} = 1/a^{r}$ ; $(a^{r})^{s} = a^{rs}$ (Laws of rational exponents).

[L4]

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

[L5]

Finite sums: $\sum_{k<n}(c_{k+1} - c_k) = c_n - c_0$ ; $\sum_{k<n} \lambda = n\lambda$ ; and $\bigl|\sum_{k<n} a_k\bigr| \le \sum_{k<n} |a_k|$ (Laws of finite sums and finite products, Finite sums and finite products, by recursion, Triangle inequality for finite sums).

[L6]

Archimedean property: for every real $t$ there is a natural $m \ge 1$ with $t < m$ ; and for every real $\eta > 0$ there is a natural $q \ge 1$ with $1/q < \eta$ (Every complete ordered field is Archimedean, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L7]

Reciprocals: $0 < s \le t$ implies $0 < 1/t \le 1/s$ , and $0 < s < t$ implies $1/t < 1/s$ (Inverses of positives are positive, and reciprocation reverses order).

[L8]

Absolute value and ordered-field arithmetic: $|u| \ge 0$ ; $|u| = 0$ exactly when $u = 0$ ; $|u| = u$ for $u \ge 0$ ; the order is total; a real that is $\ge 0$ and smaller than every positive real is $0$ (Basic properties of the absolute value, Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1

Normalisations. Since $|x-y|^{\alpha} \ge 0$ by [L2] and [L3], the hypothesis with the constant $C$ implies the same inequality with the constant $C + 1 > 0$ ; so we may and do assume $C > 0$ . Also, the hypothesis and the conclusion are symmetric in $x$ and $y$ and are trivial when $x = y$ , so it suffices to prove $f(x) = f(y)$ for $x, y \in I$ with $x < y$ ; fix such a pair and put $A := C\,(y-x)^{\alpha}$ , a real with $A > 0$ by [L3].

L2L3L8suffices: prove it for C positive and x strictly less than y

1.2

The exponent gap. Put $\beta := \alpha - 1$ , a rational with $\beta > 0$ . By [L6] fix a natural $q \ge 1$ with $1/q < \beta$ .

L6choose

1.3

Subdividing. Let $n \in \mathbb{N}$ with $n \ge 1$ and put $h := (y-x)/n > 0$ and $t_k := x + k h$ for $k \in \mathbb{N}$ . For $k \le n$ one has $x \le t_k \le y$ , so $t_k \in I$ by [L1]; and $|t_{k+1} - t_k| = h$ for $k < n$ . Define the sequence $c : \mathbb{N} \to \mathbb{R}$ by $c_k := f\bigl(t_{\min\{k,n\}}\bigr)$ , so that $c_0 = f(x)$ , $c_n = f(y)$ , and $c_{k+1} - c_k = f(t_{k+1}) - f(t_k)$ for every $k < n$ .

L1L8

2.1

The telescoped estimate. By [L5], $f(y) - f(x) = c_n - c_0 = \sum_{k<n}(c_{k+1} - c_k)$ , hence $|f(y) - f(x)| \le \sum_{k<n} |c_{k+1} - c_k| = \sum_{k<n} |f(t_{k+1}) - f(t_k)| \le \sum_{k<n} C h^{\alpha} = n\,C\,h^{\alpha}$ , the middle inequality being the hypothesis applied to the pair $t_k, t_{k+1}$ of points of $I$ and the last equality being the constant-sum rule of [L5].

step 1.3L5

2.2

The bound can be made arbitrarily small. Let a real $\eta > 0$ be given and put $R := A/\eta > 0$ . By [L6] fix a natural $N$ with $R^{q} < N$ ; then $N \ge 1$ , since $R^{q} > 0$ by [L3] and [L2]. By [L4] applied with the rational exponent $1/q > 0$ to the bases $R^{q} < N$ , and by [L2] and [L3] which give $(R^{q})^{1/q} = R^{q \cdot (1/q)} = R^{1} = R$ , we get $N^{1/q} > R$ .

step 1.1step 1.2L2L3L4L6choose

3.1

Rewriting the bound. By [L3], $h^{\alpha} = \bigl((y-x)\cdot n^{-1}\bigr)^{\alpha} = (y-x)^{\alpha}\,(n^{-1})^{\alpha} = (y-x)^{\alpha}\,n^{-\alpha}$ , and $n\,n^{-\alpha} = n^{1}n^{-\alpha} = n^{1-\alpha} = n^{-\beta} = 1/n^{\beta}$ . Hence $n\,C\,h^{\alpha} = A/n^{\beta}$ , and step 2.1 gives $|f(y) - f(x)| \le A/n^{\beta}$ for every natural $n \ge 1$ .

step 1.1step 1.2step 2.1L2L3

3.2

By [L4], $N^{1/q} \le N^{\beta}$ : this is an equality if $N = 1$ , since then both sides are $1$ by [L3], and it is the strict inequality of [L4] for the base $N > 1$ and the exponents $1/q < \beta$ . Hence $N^{\beta} \ge N^{1/q} > R > 0$ , so $1/N^{\beta} < 1/R$ by [L7] and therefore $A/N^{\beta} < A/R = \eta$ .

step 1.2step 2.2L3L4L7

4.1

Combining steps 3.1 and 3.2, $|f(y) - f(x)| \le A/N^{\beta} < \eta$ . The real $\eta > 0$ was arbitrary and $|f(y) - f(x)| \ge 0$ , so $|f(y) - f(x)| = 0$ by [L8], that is $f(y) = f(x)$ . Since $x < y$ in $I$ were arbitrary, and by the reduction of step 1.1, $f$ is constant on $I$ .

step 1.1step 3.1step 3.2L8∎

Remarks

The mechanism in one line. Splitting $[x,y]$ into $n$ equal pieces costs $n$ applications of the hypothesis, each of size $C(|x-y|/n)^{\alpha}$ , for a total of $C|x-y|^{\alpha}n^{1-\alpha}$ . For $\alpha = 1$ the factor $n^{1-\alpha}$ is $1$ and the estimate says nothing new; for $\alpha < 1$ it grows and the estimate is useless; only for $\alpha > 1$ does it tend to $0$ , and then it forces the increment to vanish.
Why the vanishing of $n^{1-\alpha}$ is proved rather than asserted. Neither real exponents nor a general theorem of the form $n^{-\beta}\to0$ is available at this point in the reading order, so the proof supplies the one rational instance it needs. The general real-power theory is developed later in Real powers for positive bases, with the zero-base positive-exponent convention ↗. Steps 2.2 and 3.2 supply the one instance that is needed, by reducing to the exponent $1/q$ with $q$ a natural number, where the $q$ -th root of Existence and uniqueness of $n$ -th roots: a unique $a^{1/n} \ge 0$ with $(a^{1/n})^n = a$ and the Archimedean property do the work.
The boundary case $\alpha = 1$ is exactly the Lipschitz condition, which does not force constancy: the identity is $1$ -Lipschitz and not constant. So the theorem is sharp at the endpoint of the range that Lipschitz map, $\alpha$ -Hölder map for rational $0 < \alpha \le 1$ , and contraction admits, and the strict nesting of the classes below $1$ is witnessed on the companion page by On $[0,1]$ the function $x^{\beta}$ is $\beta$ -Hölder and is $\alpha$ -Hölder for no rational $\alpha > \beta$ , so the Hölder classes are strictly nested ↗.

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Sources

Hölder condition (Wikipedia) (standard reference, not scraped)
Lipschitz continuity (Wikipedia) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.4 (standard reference, not scraped)
Cornell numerical methods notes: Calculus (standard reference, not scraped)

If ∣f(x)−f(y)∣≤C∣x−y∣α|f(x) - f(y)| \le C|x-y|^{\alpha}∣f(x)−f(y)∣≤C∣x−y∣α on an interval for some rational α>1\alpha > 1α>1 then fff is constant