Holder Continuity

Measuring Continuity by a Power

Uniform continuity says that inputs sufficiently close together have outputs close together, with one choice of input tolerance working throughout the domain. Sometimes we can say more: the output difference is bounded by a fixed power of the input difference. This quantitative control is called Holder continuity.

A Lipschitz function has output differences bounded by a constant times the input distance. Holder continuity allows a general positive power of that distance. Exponents between zero and one permit more variation at small scales than a Lipschitz bound does; the square-root estimate established earlier in this course is one example. An exponent equal to one gives precisely the Lipschitz condition.

Definition: Let $E\subseteq\mathbb{R}$, let $f:E\to\mathbb{R}$, and let $\alpha>0$. The function $f$ is Holder continuous with exponent $\alpha$, or $\alpha$-Holder, on $E$ if there is a constant $C\geq0$ such that, for every $x,y\in E$, $$ |f(x)-f(y)|\leq C|x-y|^\alpha. $$ The constant $C$ is called a Holder constant for $f$ with exponent $\alpha$.

The definition allows $C=0$, in which case $f$ is constant on $E$. When $\alpha=1$, the condition is exactly that $f$ is Lipschitz. When $0<\alpha<1$, it is often possible for the difference quotient $|f(x)-f(y)|/|x-y|$ to become large as $x$ and $y$ approach one another, even though the Holder bound remains valid.

As established in “Examples of Uniformly Continuous Functions,” a bound by a positive power of $|x-y|$ implies uniform continuity. Thus every Holder function is uniformly continuous; Holder continuity adds a specific rate of control, rather than a different notion of closeness.

How the Exponent Depends on the Domain

On a bounded domain, a Holder estimate with a larger exponent also gives estimates with smaller positive exponents. This lets us describe one function using several exponents, although the associated constants may differ. Boundedness of the domain is essential to this comparison.

Theorem: Let $E\subseteq\mathbb{R}$ be bounded, and suppose $f:E\to\mathbb{R}$ is $\alpha$-Holder with constant $C$, where $\alpha>0$. If $0<\beta\leq\alpha$, then $f$ is $\beta$-Holder on $E$.

Proof. If $E$ contains at most one point, every difference $f(x)-f(y)$ is zero, so the conclusion holds with constant zero. Otherwise let $D=\sup\{|x-y|:x,y\in E\}$, which is finite and positive because $E$ is bounded and contains distinct points. For $x,y\in E$, set $d=|x-y|$. If $d=0$, the desired inequality holds because both sides are zero. If $d>0$, then $d\leq D$, and $\alpha-\beta\geq0$, so

|f(x)-f(y)| \leq C d^\alpha =C d^\beta d^{\alpha-\beta} \leq C D^{\alpha-\beta}d^\beta.

Thus $C D^{\alpha-\beta}$ is a Holder constant with exponent $\beta$. When $\alpha=\beta$, the factor $D^0$ is $1$, so the original estimate is unchanged. $\square$

This theorem explains why a Holder exponent on a bounded domain is not always unique. If a function has exponent $\alpha$, it has every smaller positive exponent as well. It does not say that the smaller exponent is the best one, nor does it guarantee an exponent larger than $\alpha$. On an unbounded domain the argument can fail: there need not be a finite upper bound $D$ for the input distances.

Worked Example: A Polynomial with Several Holder Exponents

Consider $f(x)=x^2$ on $[0,2]$. For $x,y\in[0,2]$,

|x^2-y^2|=|x-y||x+y|\leq4|x-y|,

because $0\leq x+y\leq4$. Therefore $f$ is Lipschitz, hence $1$-Holder, with constant $4$. The diameter of $[0,2]$ is $2$. The exponent-comparison theorem gives, for every $0<\beta\leq1$,

|x^2-y^2| \leq4\cdot2^{1-\beta}|x-y|^\beta.

In particular, $x^2$ is $1/2$-Holder on this interval with constant $4\sqrt{2}$. The original Lipschitz estimate is stronger at small distances, but the weaker power estimate is still valid.

Exponents Above One on an Interval

The flexibility to lower an exponent on a bounded domain does not mean that exponents can be increased at will. In fact, an exponent greater than one imposes a severe restriction whenever the domain contains every point between any two of its points—that is, whenever the domain is an interval.

Theorem: Let $I\subseteq\mathbb{R}$ be an interval, and let $f:I\to\mathbb{R}$ be $\alpha$-Holder for some $\alpha>1$. Then $f$ is constant on $I$.

Proof. If $I$ contains at most one point, the conclusion holds immediately. Otherwise fix any $a,b\in I$ with $a<b$. Because $I$ is an interval, each of the $n+1$ points

x_k=a+\frac{k(b-a)}{n},\qquad k=0,1,\ldots,n,

belongs to $I$. Let $C$ be a Holder constant for $f$ with exponent $\alpha$. By telescoping the differences between consecutive points and applying the triangle inequality and the Holder bound,

\begin{aligned} |f(b)-f(a)| &\leq\sum_{k=1}^{n}|f(x_k)-f(x_{k-1})|\\ &\leq\sum_{k=1}^{n}C\left(\frac{b-a}{n}\right)^\alpha\\ &=C(b-a)^\alpha n^{1-\alpha}. \end{aligned}

Since $\alpha>1$, the exponent $1-\alpha$ is negative, so $n^{1-\alpha}\to0$ as $n\to\infty$. The left side is nonnegative and bounded above by a quantity tending to zero; hence $|f(b)-f(a)|=0$. We have proved $f(a)=f(b)$ for every $a<b$ in $I$. Reversing the order covers any pair of distinct points, so $f$ is constant on $I$. $\square$

The interval hypothesis matters. For a domain with gaps, the subdivision points used in the proof might not belong to the domain. For example, on $E=\{0,1\}$, the function $f(0)=0$, $f(1)=1$ satisfies a Holder bound with exponent $2$ and constant $1$: the only nonzero input distance is $1$, and $1\leq1\cdot1^2$. It is not constant. Thus the theorem is a statement about intervals, not arbitrary subsets of the real line.

Power Functions and Sharp Exponents

For $0<\alpha\leq1$, power functions on nonnegative inputs give a useful family of Holder examples. The key estimate is that raising nonnegative numbers to this power cannot increase their difference beyond the same power of the original difference.

To see the estimate, first note that $(s+t)^\alpha\leq s^\alpha+t^\alpha$ for $s,t\geq0$. If $s=0$, equality holds. If $s>0$, set $r=t/s$. For $r>0$, the function $q(r)=(1+r)^\alpha-1-r^\alpha$ has derivative

q'(r)=\alpha\big((1+r)^{\alpha-1}-r^{\alpha-1}\big)\leq0,

because $\alpha-1\leq0$ and $1+r\geq r>0$. Since $q(r)\to0$ as $r\to0^+$, it follows that $q(r)\leq0$. Rescaling by $s^\alpha$ proves the inequality. Now if $u\geq v\geq0$, write $u=v+(u-v)$. The inequality just proved gives

u^\alpha\leq v^\alpha+(u-v)^\alpha, \qquad\text{so}\qquad u^\alpha-v^\alpha\leq(u-v)^\alpha.

Switching $u$ and $v$ when necessary yields $|u^\alpha-v^\alpha|\leq|u-v|^\alpha$.

Worked Example: A Sharp Exponent for $x^{2/3}$

Let $f(x)=x^{2/3}$ on $[0,1]$. Applying the power estimate with $\alpha=2/3$, $u=x$, and $v=y$ gives

|x^{2/3}-y^{2/3}|\leq|x-y|^{2/3}.

Thus $f$ is $2/3$-Holder with constant $1$. This exponent is sharp: for any $\beta>2/3$, a $\beta$-Holder estimate with finite constant $C$ would, by taking $y=0$ and $0<x\leq1$, require

x^{2/3}\leq Cx^\beta, \qquad\text{and hence}\qquad x^{2/3-\beta}\leq C.

But $2/3-\beta<0$, so $x^{2/3-\beta}$ becomes unbounded as $x\to0^+$. No such finite $C$ exists. The function is Holder with exponent $2/3$, but with no larger exponent on this domain.

Holder Exponents Under Composition

A Holder estimate can also be carried through a composition. The exponents multiply: the inner function first turns an input distance into a power, and the outer function then raises that bound to its own power.

Theorem: Let $E,F\subseteq\mathbb{R}$, let $f:E\to F$ be $\alpha$-Holder with constant $C_f$, and let $g:F\to\mathbb{R}$ be $\beta$-Holder with constant $C_g$, where $\alpha,\beta>0$. Then $g\circ f$ is $\alpha\beta$-Holder on $E$, with constant $C_g C_f^\beta$.

Proof. For any $x,y\in E$, apply the Holder estimate for $g$ to the points $f(x),f(y)\in F$, and then apply the estimate for $f$:

\begin{aligned} |g(f(x))-g(f(y))| &\leq C_g|f(x)-f(y)|^\beta\\ &\leq C_g\big(C_f|x-y|^\alpha\big)^\beta\\ &=C_g C_f^\beta |x-y|^{\alpha\beta}. \end{aligned}

This is the required Holder estimate. The proof also covers $C_f=0$: then $f$ is constant, so the composition is constant and the displayed bound remains valid. $\square$

Worked Example: Composing Two Power Functions

On $[0,1]$, define $f(x)=x^{1/3}$ and $g(t)=t^{1/2}$. The power estimate gives

|f(x)-f(y)|\leq|x-y|^{1/3}, \qquad |g(s)-g(t)|\leq|s-t|^{1/2}

for $x,y,s,t\in[0,1]$. The composition theorem therefore gives exponent $(1/3)(1/2)=1/6$ and constant $1$. Directly, $g(f(x))=x^{1/6}$, and the resulting estimate is

|x^{1/6}-y^{1/6}|\leq|x-y|^{1/6}.

The multiplication of exponents records how the two stages of the composition affect small input differences.

What Holder Continuity Tells Us

A Holder estimate gives more information than uniform continuity alone: it specifies a power rate at which output differences are controlled. It also clarifies why uniform continuity and Lipschitz continuity should not be treated as synonyms. A function can be uniformly continuous while failing every Lipschitz estimate, yet satisfy a Holder estimate with exponent below one.

When applying the definition, keep track of both the exponent and the domain. On a bounded domain, a larger Holder exponent implies every smaller positive exponent. On an interval, an exponent greater than one forces the function to be constant. On an arbitrary domain that last conclusion need not hold, because the domain may not contain the subdivision points needed to compare values along the space between two inputs.

Finally, a Holder exponent need not be the largest possible one. A Lipschitz function on a bounded interval also has every exponent between zero and one, while the example $x^{2/3}$ shows that some functions have a genuine upper limit on their possible exponents. Testing the estimate near points where the function changes most rapidly is often the simplest way to identify that limit.

Check Your Understanding

Use the definition and results above to answer the following questions.

What does it mean for a function to be $\alpha$-Holder, and what does the definition become when $\alpha=1$?
Why does a Holder estimate with exponent $\alpha$ on a bounded set also give an estimate with every exponent $0<\beta\leq\alpha$?
Where does the interval hypothesis enter the proof that an exponent greater than one forces constancy?
Why is $x^{2/3}$ on $[0,1]$ not Holder with any exponent greater than $2/3$?
If the inner function has Holder exponent $\alpha$ and the outer function has exponent $\beta$, what exponent does their composition have?

What You'll Learn

Measuring Continuity by a Power

How the Exponent Depends on the Domain

Worked Example: A Polynomial with Several Holder Exponents

Exponents Above One on an Interval

Power Functions and Sharp Exponents

Worked Example: A Sharp Exponent for \(x^{2/3}\)

Holder Exponents Under Composition

Worked Example: Composing Two Power Functions

What Holder Continuity Tells Us

Check Your Understanding

Holder Continuity

What You'll Learn

Measuring Continuity by a Power

How the Exponent Depends on the Domain

Worked Example: A Polynomial with Several Holder Exponents

Exponents Above One on an Interval

Power Functions and Sharp Exponents

Worked Example: A Sharp Exponent for \(x^{2/3}\)

Holder Exponents Under Composition

Worked Example: Composing Two Power Functions

What Holder Continuity Tells Us

Check Your Understanding

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