Epsilon-Delta Versus Uniform Continuity

Where the Radius May Depend

Continuity at a point and uniform continuity both use an epsilon-delta implication. The distinction is not the shape of the implication, but where the radius is allowed to depend. For continuity at a fixed point $a$, the radius may depend on $a$. For uniform continuity, one radius must work across the entire domain. This difference in quantifiers can determine whether a function behaves consistently across a large or unbounded set.

Definition (recall): A function $f:E\to\mathbb{R}$ is continuous at $a\in E$ if for every $\varepsilon>0$, there exists a $\delta>0$ such that for every $x\in E$, $$ |x-a|<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon. $$ The radius may depend on $\varepsilon$ and $a$. The function is uniformly continuous on $E$ if for every $\varepsilon>0$, there exists a $\delta>0$ such that for every $x,y\in E$, $$ |x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<\varepsilon. $$ Here the radius may depend on $\varepsilon$, but not on either input point.

For continuity, the base point $a$ is fixed before the radius is chosen. A different point may require a different radius. Uniform continuity makes a stronger demand: for each requested output tolerance, the choice must work regardless of which pair of inputs is considered. In symbols, the distinction is

\text{continuity: }\quad \text{for every }a\in E\text{ and }\varepsilon>0,\text{ there exists }\delta>0;

\text{uniform continuity: }\quad \text{for every }\varepsilon>0,\text{ there exists }\delta>0 \text{ that works for every }x,y\in E.

The order matters. In the first condition, the radius can be selected after the point $a$ is specified. In the second, one radius must be selected before the input pair is known. Theorem (Uniform Continuity Implies Continuity), proved earlier in “Why Uniform Continuity Is Different,” gives one direction: uniform continuity implies continuity at every point. The converse is the question this comparison makes precise.

A Common-Center Test

There is a useful way to compare the two definitions while keeping a fixed base point in each implication. For a given $\varepsilon$, ordinary continuity at every point gives a radius for each center $a$. Uniform continuity holds exactly when those centered implications can all use one common radius.

Theorem (Common-Center Characterization): Let $E\subseteq\mathbb{R}$ and $f:E\to\mathbb{R}$. The function $f$ is uniformly continuous on $E$ if and only if for every $\varepsilon>0$, there exists a $\delta>0$ such that for every center $a\in E$ and every $x\in E$, $$ |x-a|<\delta \quad\Longrightarrow\quad |f(x)-f(a)|<\varepsilon. $$ The same $\delta$ must work for all centers $a$.

Proof. Suppose first that $f$ is uniformly continuous. Fix $\varepsilon>0$ and choose a uniform-continuity radius $\delta>0$. For any $a,x\in E$ with $|x-a|<\delta$, apply uniform continuity to the pair $x,a$. It gives $|f(x)-f(a)|<\varepsilon$. Thus this one $\delta$ works for every center.

Conversely, suppose that for each $\varepsilon>0$ there is a common radius $\delta>0$ for all centers. Take any $x,y\in E$ with $|x-y|<\delta$. Use the centered condition with center $a=x$ and point $y$. Then

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

Since the choice of $x,y$ was arbitrary, the defining implication for uniform continuity holds for every pair. Therefore $f$ is uniformly continuous. $\square$

This characterization pinpoints the issue: continuity at each point supplies centered radii, but it does not guarantee that those radii have a positive lower bound that works throughout the domain. A proof of uniform continuity must control the radii across all centers, not merely produce them one at a time.

Worked Example: The Square Function Is Continuous at Every Point

Let $f(x)=x^2$ on $\mathbb{R}$. Fix a point $a\in\mathbb{R}$ and a tolerance $\varepsilon>0$. The algebraic identity

$$ |x^2-a^2|=|x-a||x+a| $$

shows why a radius can be chosen using the location of $a$. Set

\delta=\min\left(1,\frac{\varepsilon}{2|a|+1}\right).

If $|x-a|<\delta$, then $|x-a|<1$, and the triangle inequality gives

|x+a|\leq|x-a|+2|a|<1+2|a|.

Consequently,

|x^2-a^2| =|x-a||x+a| <\delta(2|a|+1) \leq\varepsilon.

The last inequality follows from $\delta\leq\varepsilon/(2|a|+1)$. Thus $f$ is continuous at $a$, and since $a$ was arbitrary, it is continuous everywhere. Notice, however, that this choice of $\delta$ depends on $a$. As $|a|$ increases, this particular radius can shrink. The pointwise proof alone does not establish a common radius for all centers.

Pointwise Continuity Need Not Be Uniform

The square function provides a direct demonstration that continuity at every point is not enough. To disprove uniform continuity, it is useful to fix one output tolerance and find pairs whose input distances become arbitrarily small while their output differences remain too large. This approach directly contradicts the requirement that a single radius work for every pair.

Theorem (The Square Function Is Not Uniformly Continuous on $\mathbb{R}$): The function $f(x)=x^2$ is continuous at every real number but is not uniformly continuous on $\mathbb{R}$.

Proof. Continuity at every point was proved in the preceding example. To show failure of uniform continuity, fix $\varepsilon_0=1$. Let $\delta>0$ be any proposed uniform radius. Choose a positive integer $n$ large enough that $1/n<\delta$, and take

x=n,\qquad y=n+\frac{1}{n}.

Their input distance satisfies

|y-x|=\frac{1}{n}<\delta.

But their output difference is

|y^2-x^2| =\left(n+\frac{1}{n}\right)^2-n^2 =2+\frac{1}{n^2} >1=\varepsilon_0.

Thus every proposed positive radius admits a pair of inputs closer than that radius whose output difference is not less than $\varepsilon_0$. No uniform radius exists for this tolerance, so the function is not uniformly continuous on $\mathbb{R}$. $\square$

There is no contradiction with continuity at each $a$. The continuity proof controls inputs near one fixed center, and its radius can account for the size of that center. The pairs in the proof above move farther out as $n$ grows; their input separation shrinks, but the square function changes substantially there. Uniform continuity must control pairs throughout the domain, including those far from any one fixed point.

Worked Example: The Reciprocal Function on the Positive Half-Line

Let $g(x)=1/x$ on $E=(0,\infty)$. First fix $a>0$ and $\varepsilon>0$. Choose

\delta=\min\left(\frac{a}{2},\frac{\varepsilon a^2}{2}\right).

If $x>0$ and $|x-a|<\delta$, then $x>a/2$. Therefore

\left|\frac{1}{x}-\frac{1}{a}\right| =\frac{|x-a|}{ax} <\frac{2|x-a|}{a^2} <\frac{2\delta}{a^2} \leq\varepsilon.

The reciprocal function is thus continuous at every point of its domain. The radius depends on $a$, and in particular becomes small as $a$ approaches zero. To test uniform continuity, fix $\varepsilon_0=1/2$. For each positive integer $n$, set $x_n=1/n$ and $y_n=1/(n+1)$. Direct calculation gives

|x_n-y_n|=\frac{1}{n(n+1)}\longrightarrow0, \qquad \left|\frac{1}{x_n}-\frac{1}{y_n}\right| =|(n)-(n+1)|=1.

For any proposed $\delta>0$, choose $n$ so large that $1/(n(n+1))<\delta$. The inputs are then less than $\delta$ apart, but their outputs differ by $1>1/2$. Hence $g$ is not uniformly continuous on $(0,\infty)$. The failure occurs near the omitted endpoint zero, even though every point of the domain has its own continuity radius.

When One Estimate Supplies a Shared Radius

The contrast is not that uniform continuity always fails on an unbounded domain. It may hold when a single estimate controls output differences for every pair. In that case, the estimate gives a radius independent of the location of the inputs.

Worked Example: A Uniformly Continuous Function on the Real Line

Define $h(x)=1/(1+x^2)$ for $x\in\mathbb{R}$. For arbitrary $x,y\in\mathbb{R}$, a common denominator gives

|h(x)-h(y)| =\frac{|x^2-y^2|}{(1+x^2)(1+y^2)} =|x-y|\frac{|x+y|}{(1+x^2)(1+y^2)}.

Since $|x+y|\leq|x|+|y|$, and $t/(1+t^2)\leq1/2$ for every $t\geq0$, we have

\frac{|x+y|}{(1+x^2)(1+y^2)} \leq \frac{|x|}{1+x^2}\frac{1}{1+y^2} + \frac{|y|}{1+y^2}\frac{1}{1+x^2} \leq 1.

The bound $t/(1+t^2)\leq1/2$ follows from $(t-1)^2\geq0$, which implies $2t\leq1+t^2$; also $1/(1+t^2)\leq1$. Thus the estimate above yields

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

Given any $\varepsilon>0$, choose $\delta=\varepsilon$. Whenever $|x-y|<\delta$, the estimate gives $|h(x)-h(y)|\leq|x-y|<\varepsilon$, for every pair of real inputs. Therefore $h$ is uniformly continuous on $\mathbb{R}$. Unlike a pointwise continuity argument, this estimate does not need to know where either input lies.

Reading a Continuity Proof Carefully

When examining an epsilon-delta proof, ask what has been fixed before the radius is chosen. If a point $a$ has already been specified and the radius may depend on it, the proof establishes continuity at that point. To establish uniform continuity, the radius must be selected using $\varepsilon$ without depending on a center or an input pair. A bound involving $|a|$, for instance, may be entirely suitable for pointwise continuity and still be inadequate for uniform continuity on an unbounded set.

A common error is to prove continuity at an arbitrary point and then conclude that the function is uniformly continuous because the point was arbitrary. “Arbitrary” means the argument works separately for each point; it does not show that the radii can be chosen equally for all points. The common-center characterization explains the missing step: one must establish a common radius for every center at each fixed output tolerance.

The examples also show why the domain is part of the question. The square and reciprocal examples fail uniform continuity because of pairs in particular regions of their domains, while the function $h$ has a global estimate that controls all pairs. Keep the quantifiers visible, and test a proposed uniform radius against pairs that can move throughout the domain.

Check Your Understanding

Use the definitions, theorem, and examples in this tutorial to answer the following questions.

In continuity at a fixed point $a$, which quantities may the radius depend on? Which dependence is forbidden in uniform continuity?
Explain why a common radius for all centered implications is sufficient for uniform continuity.
In the proof that $x^2$ is not uniformly continuous, why does the pair $n,n+1/n$ contradict the definition for $\varepsilon_0=1$?
For the reciprocal function, what happens to the input distance and output distance for the pairs $1/n,1/(n+1)$?
How does the estimate for $h(x)=1/(1+x^2)$ produce a radius independent of the input points?