Definition of Uniform Continuity

The Shared Scale in the Definition

In “Why Uniform Continuity Is Different,” the central distinction was that one input radius must work for every pair of points in the domain. Here we make that definition precise as a tool: we examine exactly what may depend on what, and derive useful consequences without changing the condition. This careful reading matters because uniform continuity concerns a function together with its domain, not just a formula.

Definition (recall): Let $E\subseteq\mathbb{R}$ and $f:E\to\mathbb{R}$. The function $f$ is uniformly continuous on $E$ if for every $\varepsilon>0$, there exists a $\delta>0$ such that, for all $x,y\in E$, $$ |x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<\varepsilon. $$ For a fixed $\varepsilon$, the same $\delta$ must work for every pair $x,y$. It may depend on $\varepsilon$, the function, and its domain, but not on the particular pair.

The order of the choices is important. First an output tolerance $\varepsilon$ is specified; then one radius $\delta$ is chosen; finally, the implication must hold for all domain points $x,y$. Choosing a new radius after seeing the pair would not meet the definition. Nor does the definition require one radius to work for every possible $\varepsilon$: smaller requested output tolerances may require smaller input radii.

The implication uses strict inequalities. If $|x-y|<\delta$, the required conclusion is $|f(x)-f(y)|<\varepsilon$. A proof can often establish an estimate with a non-strict inequality, such as $|f(x)-f(y)|\leq C|x-y|$; the strict input inequality then supplies the strict output conclusion when $\delta$ is chosen appropriately. Keeping these two roles clear prevents a common mistake: concluding only that the output difference is at most $\varepsilon$ when the definition asks for it to be less than $\varepsilon$.

Measuring Variation at a Given Input Scale

For a nonempty domain, the definition can be viewed through the largest output difference among pairs whose input distance is below a chosen scale. This gives a compact way to describe how much the function can vary at small distances. The supremum need not be finite, so we allow it to take the value $+\infty$.

Definition (oscillation at scale): Let $E\subseteq\mathbb{R}$ be nonempty, let $f:E\to\mathbb{R}$, and let $r>0$. Define $$ \omega_f(r)=\sup\{|f(x)-f(y)|:x,y\in E,\ |x-y|<r\}, $$ where the supremum is taken in $[0,+\infty]$. The quantity $\omega_f(r)$ records the greatest output variation possible between inputs less than $r$ apart.

The set in this supremum is nonempty: for any $x\in E$, the pair $x,x$ qualifies and gives output difference zero. Also, increasing $r$ can only add eligible pairs, so $\omega_f(r)$ cannot decrease as $r$ increases. Uniform continuity says that these possible output differences become arbitrarily small as the input scale shrinks.

Theorem (Oscillation Criterion for Uniform Continuity): Let $E\subseteq\mathbb{R}$ be nonempty and $f:E\to\mathbb{R}$. Then $f$ is uniformly continuous on $E$ if and only if $\omega_f(r)\to0$ as $r\downarrow0$.

Proof. Suppose first that $f$ is uniformly continuous. Let $\eta>0$. By the definition, there is a $\delta>0$ such that whenever $x,y\in E$ satisfy $|x-y|<\delta$, we have $|f(x)-f(y)|<\eta$. Thus, for every $r$ with $0<r\leq\delta$, every pair used to define $\omega_f(r)$ also satisfies $|x-y|<\delta$. Each corresponding output difference is less than $\eta$, so

\omega_f(r)\leq\eta \qquad (0<r\leq\delta).

To verify the limit is zero, let $\varepsilon>0$ and apply this reasoning with $\eta=\varepsilon/2$. For all sufficiently small positive $r$, we obtain $\omega_f(r)\leq\varepsilon/2<\varepsilon$. This is precisely $\omega_f(r)\to0$ as $r\downarrow0$.

Conversely, suppose $\omega_f(r)\to0$ as $r\downarrow0$, and let $\varepsilon>0$. By the definition of this limit, there is a $\delta>0$ such that $\omega_f(\delta)<\varepsilon$. If $x,y\in E$ and $|x-y|<\delta$, then their output difference is one of the quantities whose supremum defines $\omega_f(\delta)$. Consequently,

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

The definition of uniform continuity is satisfied. Therefore the two conditions are equivalent. $\square$

The use of a limit, rather than just the assertion that $\omega_f(r)$ is finite for some $r$, is essential. A function may have bounded output variation at a particular input scale without that variation becoming arbitrarily small at finer scales. The theorem captures exactly the shrinking-scale requirement.

Worked Example: The Square-Root Function on Its Whole Domain

Let $f(x)=\sqrt{x}$ on $[0,\infty)$. Take any $x,y\in[0,\infty)$, and suppose first that $x\geq y$. Since $\sqrt{x}\geq\sqrt{y}\geq0$,

(\sqrt{x}-\sqrt{y})^2 \leq(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) =x-y.

The inequality holds because $\sqrt{x}+\sqrt{y}\geq\sqrt{x}-\sqrt{y}$, and both factors being compared are nonnegative. Taking nonnegative square roots gives $\sqrt{x}-\sqrt{y}\leq\sqrt{x-y}$. If $y\geq x$, the same argument with $x$ and $y$ interchanged gives the corresponding bound. Hence, in either case,

|\sqrt{x}-\sqrt{y}|\leq\sqrt{|x-y|}.

Given $\varepsilon>0$, choose $\delta=\varepsilon^2$. Whenever $|x-y|<\delta$, the estimate yields

|\sqrt{x}-\sqrt{y}| \leq\sqrt{|x-y|} <\sqrt{\delta} =\varepsilon.

This one choice works for every pair in $[0,\infty)$, including pairs near zero. Therefore the square-root function is uniformly continuous on its entire domain.

What Happens When the Domain Is Changed?

Uniform continuity on a set automatically gives uniform continuity on every smaller set. This follows directly from the quantifiers: a radius that works for all pairs in a larger set still works for the fewer pairs in a subset. The converse need not hold, because a subset may omit the pairs or regions responsible for failure.

Theorem (Restriction Preserves Uniform Continuity): Let $E\subseteq\mathbb{R}$, let $f:E\to\mathbb{R}$ be uniformly continuous on $E$, and let $A\subseteq E$. Then the restriction $f|_A:A\to\mathbb{R}$ is uniformly continuous on $A$.

Proof. Fix $\varepsilon>0$. Since $f$ is uniformly continuous on $E$, there is a $\delta>0$ such that for all $x,y\in E$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. Now take any $x,y\in A$ with $|x-y|<\delta$. Because $A\subseteq E$, both points also belong to $E$, so the same implication gives $|f(x)-f(y)|<\varepsilon$. Thus $\delta$ works for every pair in $A$, proving that $f|_A$ is uniformly continuous. $\square$

This proof also covers $A=\varnothing$: there are no pairs in $A$ that could violate the defining implication. More generally, the empty function is uniformly continuous vacuously. On a one-point domain, every function is uniformly continuous as well, since every pair of domain points has output difference zero.

Worked Example: Any Function on a Finite Domain

Consider $E=\{-2,1,4\}$, and define $f(-2)=7$, $f(1)=-3$, and $f(4)=10$. The output values differ substantially, but uniform continuity concerns output differences only when the inputs are sufficiently close. The distances between distinct points of $E$ are

|{-2}-1|=3,\qquad |1-4|=3,\qquad |{-2}-4|=6.

Given any $\varepsilon>0$, choose $\delta=3$. If $x,y\in E$ and $|x-y|<3$, then $x=y$, because every pair of distinct points has distance at least $3$. It follows that

|f(x)-f(y)|=|f(x)-f(x)|=0<\varepsilon.

Thus this particular function is uniformly continuous. The same argument proves a general fact: every function on a finite subset of $\mathbb{R}$ is uniformly continuous. If the finite set has at least two points, there is a smallest positive distance between distinct points; choosing that distance as $\delta$ forces any qualifying pair to be identical. If the set has zero or one point, uniform continuity holds by the empty-domain or one-point reasoning above.

Worked Example: A Bounded Rational Formula on the Nonnegative Half-Line

Define $g(x)=x/(1+x)$ for $x\in[0,\infty)$. For $x,y\geq0$, a common denominator gives the exact identity

\left|\frac{x}{1+x}-\frac{y}{1+y}\right| =\frac{|x(1+y)-y(1+x)|}{(1+x)(1+y)} =\frac{|x-y|}{(1+x)(1+y)}.

Since $1+x\geq1$ and $1+y\geq1$, the denominator is at least $1$. Therefore

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

Given $\varepsilon>0$, take $\delta=\varepsilon$. If $x,y\in[0,\infty)$ satisfy $|x-y|<\delta$, then

|g(x)-g(y)|\leq|x-y|<\delta=\varepsilon.

The formula therefore defines a uniformly continuous function on this unbounded domain. The proof does not need the domain to be bounded; it relies on a direct estimate that holds for every pair.

A Consequence for Cauchy Sequences

Uniform continuity also controls sequences whose terms need not approach any point of the domain. Recall that a sequence is Cauchy when its terms eventually become arbitrarily close to one another. Applying the shared input scale to all sufficiently late pairs shows that a uniformly continuous function preserves this property.

Theorem (Uniform Continuity Preserves Cauchy Sequences): Let $E\subseteq\mathbb{R}$, and let $f:E\to\mathbb{R}$ be uniformly continuous. If $(x_n)$ is a Cauchy sequence in $E$, then $(f(x_n))$ is a Cauchy sequence in $\mathbb{R}$.

Proof. Let $\varepsilon>0$. Uniform continuity supplies a $\delta>0$ such that $x,y\in E$ and $|x-y|<\delta$ imply $|f(x)-f(y)|<\varepsilon$. Since $(x_n)$ is Cauchy, there is an index $N$ such that whenever $m,n\geq N$,

|x_m-x_n|<\delta.

Both $x_m$ and $x_n$ belong to $E$, so the uniform-continuity implication applies to this pair. Thus for all $m,n\geq N$,

|f(x_m)-f(x_n)|<\varepsilon.

This is the Cauchy condition for $(f(x_n))$. Therefore the image sequence is Cauchy. $\square$

The shared scale is what makes this argument work: the index $N$ can be chosen from the Cauchy condition after fixing the single $\delta$ supplied by uniform continuity. Pointwise continuity gives control near each specified point, but a Cauchy sequence in a domain need not converge to a point of that domain. The theorem is therefore a genuinely useful consequence of uniform control across the whole set.

In practice, the definition can be approached in three complementary ways: track the quantifiers directly, estimate output differences for arbitrary domain pairs, or study the oscillation $\omega_f(r)$ as the input scale shrinks. Always keep the domain in view. A formula can be uniformly continuous on one set and fail to be uniformly continuous on a larger set, while a finite domain makes every function uniformly continuous because distinct inputs cannot be arbitrarily close.

Check Your Understanding

Use the definition and the results proved here to answer the following questions.

For a fixed $\varepsilon$, which choices may the radius $\delta$ depend on, and which choices must it be independent of?
Why does $\omega_f(r)\to0$ imply the defining inequality with a strict output bound?
Explain why restricting a uniformly continuous function to a subset does not require finding a new radius.
Why is every function on a finite subset of $\mathbb{R}$ uniformly continuous, regardless of its output values?
In the Cauchy-sequence theorem, where is the uniformity of $f$ used?

Definition of Uniform Continuity

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