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

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

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

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 $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$, 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 $\epsilon$-$\delta$ 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.

• The mechanism in one line. Splitting $[x,y]$ into $n$ equal pieces costs $n$ applications of the hypothesis, each of size $C(\mathrm{\mid }x-y\mathrm{\mid }\mathrm{/}n{)}^{\alpha }$, for a total of $C\mathrm{\mid }x-y{\mathrm{\mid }}^{\alpha }{n}^{1-\alpha }$. 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 $n^{-\beta }\to 0$ 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 $1\mathrm{/}q$ with $q$ a natural number, where the $q$-th root of Existence and uniqueness of $n$-th roots: a unique $a^{1\mathrm{/}n}\ge 0$ with $(a^{1\mathrm{/}n}{)}^n=a$ 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 ↗.