Continuity tells us that a function does not jump, but it says nothing about how quickly the values are allowed to change as the input scale shrinks. That missing rate matters whenever a limiting process has to survive differentiation, integration, approximation, or compactness: a [continuous function](/page/Continuous%20Function) can oscillate more and more violently at small scales while still obeying the epsilon-delta definition, so continuity alone is often too coarse for analysis.

The first useful refinement is not differentiability. Differentiability asks for a linear approximation, which is too rigid for many boundary-value problems, singular integrals, and variational limits. Hölder continuity sits between continuity and Lipschitz continuity: it asks for a power-law control

\begin{align*} |f(x)-f(y)| \le C|x-y|^\gamma \end{align*}

with an exponent $0 < \gamma \le 1$. The exponent measures the roughness: larger $\gamma$ means stronger control, while smaller $\gamma$ permits sharper cusps.

[example: A Continuous Function with a Precise Cusp Rate] Let $f:[-1,1]\to\mathbb R$ be given by \begin{align*} f(x)=\sqrt{|x|}. \end{align*} We show first that $f$ has the square-root modulus $r\mapsto r^{1/2}$. For $a,b\ge0$ with $a\ge b$, set $d=a-b$. If $d=0$, then $|\sqrt a-\sqrt b|=0$. If $d>0$, then \begin{align*} \sqrt a-\sqrt b=\frac{a-b}{\sqrt a+\sqrt b}=\frac{d}{\sqrt a+\sqrt b}. \end{align*} Since $d=a-b\le a$, we have $\sqrt d\le \sqrt a\le \sqrt a+\sqrt b$, and therefore \begin{align*} \frac{d}{\sqrt a+\sqrt b}\le \frac{d}{\sqrt d}=\sqrt d. \end{align*} Thus $|\sqrt a-\sqrt b|\le \sqrt{|a-b|}$ for all $a,b\ge0$. Applying this with $a=|x|$ and $b=|y|$ gives \begin{align*} |\sqrt{|x|}-\sqrt{|y|}|\le \sqrt{||x|-|y||}. \end{align*} Also, the triangle inequality gives $|x|\le |x-y|+|y|$, so $|x|-|y|\le |x-y|$; swapping $x$ and $y$ gives $|y|-|x|\le |x-y|$. Hence \begin{align*} ||x|-|y||\le |x-y|. \end{align*} Combining the two estimates, \begin{align*} |\sqrt{|x|}-\sqrt{|y|}|\le |x-y|^{1/2} \end{align*} for all $x,y\in[-1,1]$. In particular, $f$ is Hölder continuous with exponent $1/2$, hence continuous. The same function is not Lipschitz on $[-1,1]$. For $0<x\le1$, \begin{align*} \frac{|f(x)-f(0)|}{|x-0|}=\frac{|\sqrt x-0|}{x}=\frac{\sqrt x}{x}=x^{-1/2}. \end{align*} If a Lipschitz constant $L$ existed, then this quotient would satisfy $x^{-1/2}\le L$ for every $0<x\le1$, but choosing $0<x<1/L^2$ when $L>0$ gives $x^{-1/2}>L$, and $L=0$ is already impossible from $x=1$. Thus $f$ is continuous with a precise square-root rate, but its cusp at $0$ is too sharp for any linear distance bound. [/example]

This example is the basic reason Hölder continuity deserves its own page rather than being treated as a minor variant of continuity. The square-root cusp is not a pathology to discard; it is a stable regularity class that appears naturally in elliptic equations, harmonic functions near rough boundaries, and compactness arguments.

## Definition

The parent notion is [continuity](/page/Continuity): small changes in input force small changes in output. Hölder continuity strengthens this by prescribing a specific power of the distance that controls the output change. The definition is usually made on metric spaces, because the only structure it needs is a notion of distance in the domain and codomain.

[definition: Hölder Continuity] Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces, let $0<\gamma\le 1$, and let $f:X\to Y$ be a function. The function $f$ is Hölder continuous with exponent $\gamma$ if there exists a constant $C\ge 0$ such that \begin{align*} d_Y(f(x),f(y)) \le C d_X(x,y)^\gamma \end{align*} for all $x,y\in X$. [/definition]

The parameter $\gamma$ is part of the assertion. Saying that a function is Hölder continuous without naming the exponent usually means that there exists at least one exponent $\gamma\in(0,1]$ for which the condition holds. In Euclidean spaces, the definition becomes the familiar estimate $|f(x)-f(y)|\le C|x-y|^\gamma$.

The rest of the page develops the machinery that makes this estimate usable: moduli that compare different rates of continuity, seminorms that measure the best constant, local variants for interior regularity, and Hölder spaces that support compactness and PDE estimates.

## From Continuity to Quantitative Continuity

### Global rate estimates

Continuity at a point allows the permissible input radius to depend on a tolerance in an unspecified way. Hölder continuity replaces that dependence by an explicit formula. This gives estimates that can be transported through inequalities, compactness arguments, and approximation schemes.

Before using Hölder continuity as a regularity assumption, we need to check that it really is a strengthening of continuity. The power-law bound supplies a uniform modulus of continuity on the whole domain.

[quotetheorem:8226]

This theorem explains why Hölder continuity belongs under the continuity family in the definition graph. It is not a different kind of mapping condition; it is a quantified strengthening of the same idea. The next question is whether the exact exponent is rigid, or whether controlling a sharper power also controls rougher powers. On bounded sets, the diameter provides the missing scale conversion.

[quotetheorem:8227]

The boundedness hypothesis is not cosmetic. When distances can be arbitrarily large, the inequality $|x-y|^\gamma\le C|x-y|^\alpha$ fails for large $|x-y|$ if $\alpha<\gamma$. Local estimates avoid this issue by working on compact subsets.

[example: Why Boundedness Matters for Exponent Comparison] Let $f:\mathbb R\to\mathbb R$ be given by $f(x)=x$. For any $x,y\in\mathbb R$, \begin{align*} |f(x)-f(y)|=|x-y|. \end{align*} Thus $f$ is Lipschitz with constant $1$, and therefore satisfies the Hölder estimate with exponent $1$. Now fix $0<\alpha<1$. We show that no constant can make $f$ Hölder continuous with exponent $\alpha$ on all of $\mathbb R$. If such a constant $C\ge0$ existed, then for every $x>0$, applying the estimate to the pair $(x,0)$ would give \begin{align*} |f(x)-f(0)|\le C|x-0|^\alpha. \end{align*} Since $f(x)=x$ and $f(0)=0$, this becomes \begin{align*} x\le Cx^\alpha. \end{align*} Because $x>0$, dividing both sides by $x^\alpha$ gives \begin{align*} x^{1-\alpha}\le C. \end{align*} This inequality would have to hold for all $x>0$. If $C=0$, taking $x=1$ gives $1\le0$, impossible. If $C>0$, choose $x>C^{1/(1-\alpha)}$; then raising both sides to the positive power $1-\alpha$ gives \begin{align*} x^{1-\alpha}>C. \end{align*} This contradicts $x^{1-\alpha}\le C$. Hence $f$ is Lipschitz on the unbounded domain $\mathbb R$, but it is not globally Hölder continuous there with any exponent $\alpha<1$; the boundedness assumption in exponent comparison supplies the missing large-scale control. [/example]