Modulus of Continuity

From a Continuity Condition to a Function of Distance

Holder continuity measures output differences by a fixed power of the input distance. The modulus of continuity gives a broader description: for each allowed input distance, it records the largest output difference that can occur. This makes it possible to study how the amount of output variation changes as the input scale changes.

The modulus depends on both the function and its domain. In particular, a supremum is taken only over pairs of points in that domain. If the function is unbounded, the supremum at a positive distance may be infinite; we allow that possibility so the definition applies without an extra boundedness assumption.

Definition: Let $E\subseteq\mathbb{R}$ be nonempty, and let $f:E\to\mathbb{R}$. For $t\geq0$, the modulus of continuity of $f$ on $E$ is $$ \omega_f(t)=\sup\{|f(x)-f(y)|:x,y\in E,\ |x-y|\leq t\}. $$ The value $\omega_f(t)$ is allowed to be $+\infty$. In particular, $\omega_f(0)=0$.

The set in this supremum is nonempty: for any $x\in E$, the pair $x,y=x$ is allowed. At $t=0$, every allowed pair has $x=y$, so every output difference is zero. If $f$ is bounded, say $|f(x)|\leq M$ on $E$, then $\omega_f(t)\leq2M$ for every $t$, and all its values are finite.

The modulus is nondecreasing: if $0\leq s\leq t$, every pair counted in $\omega_f(s)$ is also counted in $\omega_f(t)$. Thus $\omega_f(s)\leq\omega_f(t)$. Its value at a particular scale is a uniform bound: whenever $x,y\in E$ and $|x-y|\leq t$,

|f(x)-f(y)|\leq\omega_f(t).

This inequality follows directly from the definition of supremum, even if the supremum is not attained by any pair.

Worked Examples of Moduli

Worked Example: A Truncated Linear Function

Define $f:[0,\infty)\to\mathbb{R}$ by $f(x)=\min\{x,1\}$. For any $x,y\geq0$, assume first that $x\leq y$. If $y\leq1$, then $f(y)-f(x)=y-x$. If $x\leq1\leq y$, then $f(y)-f(x)=1-x\leq y-x$. If $1\leq x\leq y$, the difference is zero. Reversing the roles of $x$ and $y$ when necessary proves

|f(x)-f(y)|\leq|x-y|.

Also, $f$ takes values in $[0,1]$, so $|f(x)-f(y)|\leq1$. These facts give $\omega_f(t)\leq\min\{t,1\}$. If $0\leq t\leq1$, the pair $x=0,y=t$ gives $|f(t)-f(0)|=t$. If $t\geq1$, the pair $x=0,y=1$ gives $|f(1)-f(0)|=1$. Therefore the upper bounds are attained, and

\omega_f(t)=\min\{t,1\}.

The modulus is linear for small distances, then becomes constant because the function's entire range has length one.

Worked Example: The Cubic Function on a Bounded Interval

Let $f(x)=x^3$ on $[0,1]$. Fix $0\leq d\leq1$. For points $y$ and $y+d$ in this interval, $0\leq y\leq1-d$, and

$$ (y+d)^3-y^3=3y^2d+3yd^2+d^3. $$

For fixed $d\geq0$, the right side is nondecreasing as $y$ increases over nonnegative values: each term $3y^2d$ and $3yd^2$ is nondecreasing, and the last term is constant in $y$. Hence among pairs at distance $d$, the largest difference occurs at $y=1-d$. That difference is

$$ 1-(1-d)^3. $$

This expression is nondecreasing in $d\in[0,1]$, since $1-d$ decreases from $1$ to $0$, and its cube decreases as well. Thus, among distances $d\leq t$, the largest difference occurs at $d=t$ when $0\leq t\leq1$. For $t\geq1$, the largest possible difference is the full range $1-0=1$, attained by the pair $0,1$. We conclude that

\omega_f(t)= \begin{cases} 1-(1-t)^3,&0\leq t\leq1,\\ 1,&t\geq1. \end{cases}

For small $t$, the first expression equals $3t-3t^2+t^3$. The exact formula therefore records more than just the fact that $f$ is uniformly continuous: it describes the maximal change at each scale.

Worked Example: Distance to the Integers

Let $E=\mathbb{R}$, and define $f(x)=\inf\{|x-n|:n\in\mathbb{Z}\}$, the distance from $x$ to the integers. Its values lie in $[0,1/2]$, because every real number is within distance $1/2$ of an integer. For any $x,y\in\mathbb{R}$ and any integer $n$, the triangle inequality gives $|x-n|\leq|x-y|+|y-n|$. Taking the infimum over $n$ yields $f(x)\leq|x-y|+f(y)$; reversing $x,y$ then gives $|f(x)-f(y)|\leq|x-y|$. Together with the range bound, this implies $\omega_f(t)\leq\min\{t,1/2\}$.

If $0\leq t\leq1/2$, the pair $x=0,y=t$ has $f(0)=0$ and $f(t)=t$, so its output difference is $t$. If $t\geq1/2$, the pair $x=0,y=1/2$ has output difference $1/2$. Thus

\omega_f(t)=\min\{t,1/2\}.

The function repeats its pattern across the real line, but its modulus is determined by its local slope and its bounded range.

The Modulus Characterizes Uniform Continuity

Uniform continuity asks for one input tolerance that works for all pairs of points in the domain. The modulus packages the output differences from all such pairs into a single quantity, so the definition becomes a limit condition at zero.

Theorem: Let $E\subseteq\mathbb{R}$ be nonempty, and let $f:E\to\mathbb{R}$. Then $f$ is uniformly continuous on $E$ if and only if $$ \lim_{t\to0^+}\omega_f(t)=0. $$

Proof. Suppose first that $f$ is uniformly continuous. Let $\varepsilon>0$. By the definition of uniform continuity, there exists $\delta>0$ such that $x,y\in E$ and $|x-y|<\delta$ imply $|f(x)-f(y)|<\varepsilon$. If $0\leq t<\delta$, every pair with $|x-y|\leq t$ has $|x-y|<\delta$. Consequently all output differences in the supremum defining $\omega_f(t)$ are at most $\varepsilon$, and $\omega_f(t)\leq\varepsilon$. To obtain the usual strict bound for a prescribed $\varepsilon$, apply this argument first with $\varepsilon/2$; then $\omega_f(t)\leq\varepsilon/2<\varepsilon$ for all sufficiently small $t$. This proves the stated limit.

Conversely, suppose $\omega_f(t)\to0$ as $t\to0^+$. Given $\varepsilon>0$, choose $\delta>0$ such that $\omega_f(\delta)<\varepsilon$. Whenever $x,y\in E$ satisfy $|x-y|<\delta$, the definition of the modulus gives

|f(x)-f(y)|\leq\omega_f(\delta)<\varepsilon.

The same $\delta$ works for every pair in $E$, so $f$ is uniformly continuous. $\square$

The theorem gives a useful distinction between continuity at each point and uniform continuity. Pointwise continuity can allow the required tolerance to depend on the center point. The modulus takes a supremum over all centers at once. If that supremum does not become small as the allowed distance shrinks, uniform continuity fails.

Worked Example: A Discontinuous Indicator Function

Let $f:\mathbb{R}\to\mathbb{R}$ equal $1$ on the rational numbers and $0$ on the irrational numbers. For every $t>0$, there are rational and irrational numbers at distance at most $t$. For example, choose a rational number $q$, and then choose an irrational number $r$ with $0<|q-r|<t$; such irrationals exist in every nonempty open interval. This pair has $|f(q)-f(r)|=1$. Since all values of $f$ are either zero or one, no output difference exceeds one. Hence

\omega_f(0)=0,\qquad \omega_f(t)=1\quad(t>0).

The modulus does not tend to zero at the origin, agreeing with the theorem that $f$ is not uniformly continuous.

Hölder Bounds in Modulus Form

The modulus also expresses Holder continuity without referring separately to every pair of points. The equivalence below is simply the definition organized by input scale; it is useful when the modulus can be calculated or estimated directly.

Theorem: Let $E\subseteq\mathbb{R}$ be nonempty, let $f:E\to\mathbb{R}$, and let $\alpha>0$. The function $f$ is $\alpha$-Holder with constant $C\geq0$ on $E$ if and only if $$ \omega_f(t)\leq Ct^\alpha\qquad\text{for every }t>0. $$

Proof. If $f$ is $\alpha$-Holder with constant $C$, then for any pair with $|x-y|\leq t$,

|f(x)-f(y)|\leq C|x-y|^\alpha\leq Ct^\alpha.

Taking the supremum over those pairs gives $\omega_f(t)\leq Ct^\alpha$. Conversely, suppose the modulus bound holds for every $t>0$. For distinct $x,y\in E$, set $t=|x-y|>0$. Then

|f(x)-f(y)|\leq\omega_f(|x-y|)\leq C|x-y|^\alpha.

For $x=y$, the Holder inequality is $0\leq0$, so it holds as well. This proves the equivalence. $\square$

In particular, an estimate of the form $\omega_f(t)\leq Ct$ is a Lipschitz estimate, and a bound by $Ct^\alpha$ for any positive exponent implies that the modulus tends to zero. This recovers the fact established earlier in the course that Holder continuity implies uniform continuity.

A Domain-Sensitive Property: Subadditivity on Intervals

On an interval, a distance can be split into two shorter distances using an intermediate point that still belongs to the domain. This leads to a useful inequality for the modulus. The interval condition cannot be omitted for an arbitrary subset of the real line.

Theorem: Let $I\subseteq\mathbb{R}$ be an interval, and let $f:I\to\mathbb{R}$. For $s,t\geq0$, $$ \omega_f(s+t)\leq\omega_f(s)+\omega_f(t), $$ where the inequality is understood in the extended nonnegative reals.

Proof. Take $x,y\in I$ with $|x-y|\leq s+t$, and suppose first that $x\leq y$. Write $d=y-x$, and set $z=x+\min\{s,d\}$. Since $x\leq z\leq y$ and $I$ is an interval, $z\in I$. Also, $z-x=\min\{s,d\}\leq s$. If $d\leq s$, then $y-z=0\leq t$; if $d>s$, then $y-z=d-s\leq t$, because $d\leq s+t$. The triangle inequality and the definition of the modulus now give

\begin{aligned} |f(y)-f(x)| &\leq |f(y)-f(z)|+|f(z)-f(x)|\\ &\leq\omega_f(t)+\omega_f(s). \end{aligned}

If $y<x$, interchange the roles of $x$ and $y$ and apply the same construction. Thus every pair counted in $\omega_f(s+t)$ has output difference at most $\omega_f(s)+\omega_f(t)$. Taking the supremum proves the inequality. $\square$

For a domain with gaps, an intermediate point may not exist. For instance, take $E=\{0,2\}$, with $f(0)=0$ and $f(2)=1$. Then $\omega_f(1)=0$, because distinct points of $E$ are two units apart, while $\omega_f(2)=1$. Thus $\omega_f(2)\leq\omega_f(1)+\omega_f(1)$ would say $1\leq0$, which is false. The scale-splitting proof depends on the interval hypothesis, not merely on the fact that the domain is a subset of $\mathbb{R}$.

How to Use a Modulus Carefully

A modulus is a global summary, not necessarily a value achieved by some pair of points. For example, on an unbounded domain a supremum can be infinite even when the function is continuous at every point. The square function on the real line illustrates why one should not infer uniform behavior from pointwise continuity: for any fixed $t>0$, pairs $x$ and $x+t$ have output difference $|2tx+t^2|$, which is unbounded as $x$ varies. Therefore its modulus is infinite at every positive $t$, consistent with the fact that it is not uniformly continuous there.

When calculating a modulus, first establish an upper bound for every admissible pair. Then look for pairs that attain that bound, or approach it arbitrarily closely. Keeping those two tasks separate prevents a common error: finding a pair with a large output difference proves only a lower bound for the modulus, not an exact formula. The domain matters at both stages, since it determines which pairs are available and whether an intermediate point can be used in an estimate.

Check Your Understanding

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

Why is $\omega_f(0)=0$, and why can $\omega_f(t)$ be infinite for $t>0$?
State the condition on the modulus that is equivalent to uniform continuity.
How does an $\alpha$-Holder estimate translate into an upper bound for $\omega_f(t)$?
Where does the interval hypothesis enter the proof of subadditivity?
For the truncated linear function in the first worked example, why does its modulus become constant for $t\geq1$?

What You'll Learn