Why Uniform Continuity Is Different

One Local Scale or One Scale for the Whole Domain?

In the previous tutorial, we planned continuity proofs by fixing a point $a$ and then choosing an input radius that controls the output error near that point. The choice can depend on $a$. Uniform continuity asks for something stronger: once an output tolerance is fixed, a single input radius must work for every pair of points in the domain. That change in the order of the quantifiers is small to write down, but it can change whether a function has the property.

At a point $a$, continuity says that for every $\varepsilon>0$, there is a $\delta>0$ such that whenever $x$ is in the domain and $|x-a|<\delta$, the output difference $|f(x)-f(a)|$ is less than $\varepsilon$. Here the radius may depend on both $\varepsilon$ and $a$. Uniform continuity replaces the fixed point $a$ by two arbitrary domain points and requires the radius to be chosen without knowing where those points are.

Definition: 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 is a $\delta>0$ such that for all $x,y\in E$, $$ |x-y|<\delta \quad\Longrightarrow\quad |f(x)-f(y)|<\varepsilon. $$ The chosen $\delta$ may depend on $\varepsilon$, on $f$, and on $E$, but not on the particular points $x$ and $y$.

The difference is not that one condition measures small output changes and the other does not. Both do. The difference is where the points may be located when the input distance is required to be small. Pointwise continuity controls behavior near one specified point at a time. Uniform continuity controls the same output tolerance at every location in the domain, using the same input distance bound.

Uniform Continuity Gives Continuity Everywhere

The stronger condition immediately gives the weaker one. To check continuity at $a\in E$, use the uniform radius for pairs $x,y\in E$, and then set $y=a$. This observation is useful, but its converse is not automatic: continuity at every point allows the radius to vary from point to point.

Theorem (Uniform Continuity Implies Continuity): If $f:E\to\mathbb{R}$ is uniformly continuous on $E$, then $f$ is continuous at every $a\in E$.

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

This is the continuity condition at $a$. Since $a$ was arbitrary, $f$ is continuous at every point of $E$. $\square$

The proof uses the same radius that uniform continuity supplies; it does not need a new choice for each point. This is exactly why uniform continuity is stronger. The theorem says every uniformly continuous function is continuous, but it does not say that every continuous function is uniformly continuous.

Worked Example: A Cubic Is Continuous at Each Point

Consider $f(x)=x^3$ on $\mathbb{R}$. Earlier in the course, continuity of polynomial functions was established. The following estimate also shows how a pointwise radius can depend on the point. Fix $a\in\mathbb{R}$ and suppose $|x-a|<1$. Then $|x|<|a|+1$, and factoring the difference of cubes gives

|x^3-a^3| =|x-a|\,|x^2+ax+a^2| \leq |x-a|\bigl(|x|^2+|a||x|+|a|^2\bigr) \leq 3(|a|+1)^2|x-a|.

For the last inequality, each of $|x|^2$, $|a||x|$, and $|a|^2$ is at most $(|a|+1)^2$. Given $\varepsilon>0$, choose

\delta=\min\left\{1,\frac{\varepsilon}{3(|a|+1)^2}\right\}.

If $|x-a|<\delta$, then $|x-a|<1$, so the estimate applies, and

|x^3-a^3| \leq 3(|a|+1)^2|x-a| <3(|a|+1)^2\delta \leq\varepsilon.

This proves continuity at $a$. Notice that the displayed choice can shrink as $|a|$ grows. A pointwise proof permits that dependence; uniform continuity would require a single choice that works for all $a$.

When the Required Scale Breaks Down

A useful way to expose the difference is to look for pairs of inputs that get arbitrarily close while their outputs do not get close. If such pairs exist, no single radius can guarantee a chosen output tolerance throughout the domain. The next theorem makes this test precise.

Theorem (Sequential Criterion for Uniform Continuity): 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 two sequences $(x_n)$ and $(y_n)$ in $E$, $$ |x_n-y_n|\longrightarrow 0 \quad\Longrightarrow\quad |f(x_n)-f(y_n)|\longrightarrow 0. $$

Proof. First suppose that $f$ is uniformly continuous, and let $(x_n)$ and $(y_n)$ be sequences in $E$ such that $|x_n-y_n|\to0$. Let $\varepsilon>0$. Choose $\delta>0$ from uniform continuity. Since $|x_n-y_n|\to0$, there is an index $N$ such that $n\geq N$ implies $|x_n-y_n|<\delta$. Uniform continuity then gives $|f(x_n)-f(y_n)|<\varepsilon$ for every $n\geq N$. Thus the output differences tend to zero.

Conversely, suppose the stated property holds, but $f$ is not uniformly continuous. Negating the definition, there is an $\varepsilon_0>0$ such that for every $\delta>0$, there are $x,y\in E$ satisfying

|x-y|<\delta \quad\text{and}\quad |f(x)-f(y)|\geq\varepsilon_0.

For each positive integer $n$, use $\delta=1/n$ to choose $x_n,y_n\in E$ with $|x_n-y_n|<1/n$ and $|f(x_n)-f(y_n)|\geq\varepsilon_0$. The input differences tend to zero, but the output differences cannot tend to zero because they are all at least $\varepsilon_0$. This contradicts the assumed property. Therefore $f$ is uniformly continuous. $\square$

This criterion is especially effective for proving failure of uniform continuity: it is enough to construct one pair of sequences. The two sequences need not converge to a point in the domain. They may move farther and farther through the domain, or approach a boundary point that is not included in it. That freedom is one reason the criterion detects behavior that pointwise continuity does not.

Worked Example: Squaring on the Whole Real Line

Let $f(x)=x^2$ on $\mathbb{R}$. This polynomial is continuous at every real number, but it is not uniformly continuous on $\mathbb{R}$. To use the sequential criterion, set $x_n=n$ and $y_n=n+1/n$. Both sequences lie in $\mathbb{R}$, and

|x_n-y_n|=\left|n-\left(n+\frac1n\right)\right|=\frac1n\longrightarrow0.

However, substituting the same values into the output difference gives

|f(x_n)-f(y_n)| =\left|n^2-\left(n+\frac1n\right)^2\right| =\left|n^2-\left(n^2+2+\frac1{n^2}\right)\right| =2+\frac1{n^2}.

This quantity tends to $2$, not to zero. The sequential criterion therefore shows that squaring is not uniformly continuous on $\mathbb{R}$. The failure is compatible with continuity at every point: as the inputs move farther from the origin, small changes in input can produce output changes that do not stay small.

Worked Example: Reciprocation Near an Excluded Endpoint

Consider $g(x)=1/x$ on $E=(0,1)$. Earlier in the course, continuity of rational functions on their natural domains was established, so $g$ is continuous at every point of $E$. For integers $n\geq2$, take $x_n=1/n$ and $y_n=1/(n+1)$. Both points belong to $(0,1)$, and

|x_n-y_n| =\frac1n-\frac1{n+1} =\frac{1}{n(n+1)} \longrightarrow0.

Their outputs, however, are $g(x_n)=n$ and $g(y_n)=n+1$, so

$$ |g(x_n)-g(y_n)|=|n-(n+1)|=1 $$

for every $n$. The outputs do not approach one another, and the sequential criterion proves that $g$ is not uniformly continuous on $(0,1)$. The point $0$ is not in the domain, so continuity at each domain point only controls neighborhoods centered at positive points; it does not provide one scale that works all the way toward zero.

The Domain Can Change the Answer

Whether a function is uniformly continuous is a property of the function together with its domain. A formula that fails the test on a large domain may pass it after restriction. On a bounded interval, some algebraic factors that grow without limit on the whole real line acquire a fixed bound. The following estimate verifies this directly, without relying on a general theorem about compact sets.

Worked Example: A Cubic on a Bounded Interval

Let $h(x)=x^3$ on $E=[-2,2]$. For any $x,y\in E$, factor the difference and use $|x|\leq2$ and $|y|\leq2$:

|h(x)-h(y)| =|x-y|\,|x^2+xy+y^2| \leq |x-y|\bigl(|x|^2+|x||y|+|y|^2\bigr) \leq 12|x-y|.

Given $\varepsilon>0$, choose $\delta=\varepsilon/12$. For every $x,y\in[-2,2]$ with $|x-y|<\delta$, the estimate yields

|h(x)-h(y)|\leq12|x-y|<12\delta=\varepsilon.

The same input radius works for all pairs in the interval, so $h$ is uniformly continuous on $[-2,2]$. The calculation contrasts with the pointwise estimate for the cubic on $\mathbb{R}$: on $[-2,2]$, the factor $x^2+xy+y^2$ is bounded by $12$, independently of the pair.

What the Difference Does—and Does Not—Say

Uniform continuity is not a claim that a function changes at a constant rate. It says that for any specified output tolerance, some input tolerance works everywhere on the domain. The radius may become smaller when the requested output tolerance becomes smaller, but it cannot be adjusted separately at different locations. In particular, one should not mistake “continuous at every point” for “uniformly continuous on the domain.”

The examples show two distinct sources of failure. For $x^2$ on $\mathbb{R}$, the problematic pairs move out toward arbitrarily large inputs. For $1/x$ on $(0,1)$, they move toward an endpoint excluded from the domain. In both cases, continuity remains valid at every point that actually belongs to the domain. Uniform continuity asks for control across the entire domain, including regions where pointwise continuity alone supplies no common scale.

A reliable test is to write down the quantifiers before choosing a radius. If the proof fixes a point first, the radius may depend on it, and the conclusion is pointwise continuity. If the proof must handle arbitrary $x,y\in E$ with one radius, it is addressing uniform continuity. For a negative result, the sequential criterion often gives a clean certificate: find input pairs whose distances tend to zero while the output distances stay bounded away from zero.

Check Your Understanding

Use the definitions, examples, and sequential criterion to answer the following questions.

In the definition of continuity at $a$, which points may the choice of $\delta$ depend on, and how does that change for uniform continuity?
Why does uniform continuity imply continuity at every point of the domain?
For $f(x)=x^2$ on $\mathbb{R}$, what do the sequences $x_n=n$ and $y_n=n+1/n$ show about the output differences?
Why does the reciprocal example use points approaching zero even though zero is not in the domain?
What bound on $|x^2+xy+y^2|$ makes the cubic uniformly continuous on $[-2,2]$?