Examples of Uniformly Continuous Functions

From General Criteria to Concrete Functions

The previous tutorial, Lipschitz Implies Uniform Continuity, showed how one linear estimate on the change in a function’s values gives a single radius that works throughout the domain. This is a powerful way to produce examples: establish a Lipschitz bound, then apply that theorem. But uniform continuity can also follow from a weaker estimate, in which the output difference is bounded by a positive power of the input difference.

The examples here illustrate both approaches. Some functions have a linear bound on all of $\mathbb{R}$; others have such a bound only on a restricted domain. The cube-root function provides a useful contrast: it is uniformly continuous on an unbounded domain, even though its changes near zero do not obey any fixed linear bound.

Theorem (Positive-Power Estimate Implies Uniform Continuity): Let $E\subseteq\mathbb{R}$, and suppose $f:E\to\mathbb{R}$ satisfies $$ |f(x)-f(y)|\leq C|x-y|^\alpha $$ for all $x,y\in E$, where $C\geq0$ and $\alpha>0$. Then $f$ is uniformly continuous on $E$. If $C>0$, one may take $\delta=(\varepsilon/C)^{1/\alpha}$ for a given $\varepsilon>0$; if $C=0$, any $\delta>0$ works.

Proof. First suppose $C>0$. Fix $\varepsilon>0$ and set $\delta=(\varepsilon/C)^{1/\alpha}$, which is positive because $\varepsilon$, $C$, and $\alpha$ are positive. If $x,y\in E$ and $|x-y|<\delta$, then

|f(x)-f(y)| \leq C|x-y|^\alpha <C\delta^\alpha =C\left(\frac{\varepsilon}{C}\right) =\varepsilon.

This radius depends on $\varepsilon$, $C$, and $\alpha$, but not on the points $x$ and $y$, so $f$ is uniformly continuous. If $C=0$, the assumed estimate gives $|f(x)-f(y)|=0$ for all $x,y\in E$. Thus any positive $\delta$ works for every $\varepsilon>0$. This proves the theorem. $\square$

The Lipschitz case is the special case $\alpha=1$. When $0<\alpha<1$, the estimate allows function values to change more rapidly than a linear bound would allow, particularly for very small input distances. Nevertheless, because $|x-y|^\alpha$ tends to zero with $|x-y|$, it still gives a uniform continuity radius.

Examples with Linear Bounds

A direct way to establish uniform continuity on all of $\mathbb{R}$ is to find a Lipschitz constant that works for every pair of real inputs. The sine function is one example. Its values oscillate, but the trigonometric difference identity controls how much they can change.

Theorem (Sine Is Lipschitz on $\mathbb{R}$): For all $x,y\in\mathbb{R}$, $$ |\sin x-\sin y|\leq |x-y|. $$ Consequently, sine is uniformly continuous on $\mathbb{R}$.

Proof. The difference identity for sine gives

|\sin x-\sin y| =2\left|\sin\left(\frac{x-y}{2}\right) \cos\left(\frac{x+y}{2}\right)\right|.

Use $|\cos t|\leq1$ for every real $t$, and the elementary inequality $|\sin t|\leq|t|$. Then

|\sin x-\sin y| \leq 2\left|\sin\left(\frac{x-y}{2}\right)\right| \leq 2\left|\frac{x-y}{2}\right| =|x-y|.

Thus sine is Lipschitz with constant $1$. By the theorem Lipschitz Implies Uniform Continuity, it is uniformly continuous on $\mathbb{R}$; explicitly, $\delta=\varepsilon$ works for every $\varepsilon>0$. $\square$

Worked Example: A Rational Function on the Real Line

Consider $r(x)=1/(1+x^2)$ for $x\in\mathbb{R}$. The denominator is always positive, and for any $x,y\in\mathbb{R}$, subtraction gives

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

Since $|x+y|\leq |x|+|y|$, we can bound the final factor as follows:

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

For the last inequality, $2|t|\leq1+t^2$ for every real $t$, so $|t|/(1+t^2)\leq1/2$. We have proved that $|r(x)-r(y)|\leq|x-y|$ for every pair of real inputs. Therefore $r$ is Lipschitz with constant $1$, and $\delta=\varepsilon$ proves uniform continuity on all of $\mathbb{R}$. This example shows that an unbounded domain need not prevent a global estimate: the denominator controls the factor involving the inputs.

Worked Example: A Polynomial on a Bounded Interval

Let $p(x)=x^2$ on $[-3,2]$. For any $x,y\in[-3,2]$, factor the difference and use the bounds on the inputs:

|p(x)-p(y)| =|x^2-y^2| =|x-y||x+y| \leq 6|x-y|,

because $|x|\leq3$ and $|y|\leq3$, so $|x+y|\leq|x|+|y|\leq6$. Thus $p$ is Lipschitz on this interval with constant $6$. Given $\varepsilon>0$, choose $\delta=\varepsilon/6$. If $x,y\in[-3,2]$ and $|x-y|<\delta$, then

|p(x)-p(y)|\leq6|x-y|<6\delta=\varepsilon.

The domain restriction matters to this particular argument: the bound on $|x+y|$ came from the bounded interval. On the whole real line, $|x+y|$ has no fixed upper bound, so this calculation does not give a global Lipschitz constant for the square function.

A Uniformly Continuous Function Without a Lipschitz Bound

A positive-power estimate can succeed even when no linear estimate is available. Consider the cube-root function on the nonnegative real numbers. The key inequality comes from comparing two nonnegative cube roots.

Worked Example: The Cube Root on an Unbounded Domain

Define $c(x)=\sqrt[3]{x}$ for $x\in[0,\infty)$. Suppose first that $x\geq y\geq0$, and put $a=\sqrt[3]{y}$ and $b=\sqrt[3]{x}-\sqrt[3]{y}$. Then $a,b\geq0$, $\sqrt[3]{x}=a+b$, and

x-y=(a+b)^3-a^3=3a^2b+3ab^2+b^3\geq b^3.

Because $b=|\sqrt[3]{x}-\sqrt[3]{y}|$, this proves

|\sqrt[3]{x}-\sqrt[3]{y}|^3\leq x-y.

If $y\geq x\geq0$, interchange $x$ and $y$ in the same argument. In either case, taking nonnegative cube roots gives the estimate

|\sqrt[3]{x}-\sqrt[3]{y}|\leq |x-y|^{1/3}.

Apply the positive-power estimate theorem with $C=1$ and $\alpha=1/3$. For a given $\varepsilon>0$, $\delta=\varepsilon^3$ works: whenever $x,y\in[0,\infty)$ and $|x-y|<\delta$,

|c(x)-c(y)| \leq |x-y|^{1/3} <\delta^{1/3} =\varepsilon.

Thus the cube-root function is uniformly continuous on an unbounded domain. It is not Lipschitz there. For $t>0$, compare the inputs $t^3$ and $0$; the ratio of output distance to input distance is

\frac{|c(t^3)-c(0)|}{|t^3-0|} =\frac{t}{t^3} =\frac{1}{t^2}.

These ratios are unbounded as $t$ approaches zero. Hence no finite Lipschitz constant can satisfy the required inequality for all pairs, despite the uniform continuity estimate just proved.

Choosing an Estimate That Fits the Function

The examples suggest a practical approach. First identify the domain, since a useful estimate may depend on its restrictions. Next, subtract two function values and look for a bound in terms of $|x-y|$. If the result is a constant multiple of $|x-y|$, the function is Lipschitz and the previous tutorial’s theorem applies. If the bound instead involves $|x-y|^\alpha$ for some $\alpha>0$, the positive-power estimate theorem still supplies a uniform radius.

A common pitfall is to conclude that a function is not uniformly continuous just because it is not Lipschitz. The cube-root example rules out that inference: it has no global linear bound, but its fractional-power estimate tends to zero with the input distance and works uniformly across the whole domain. A second pitfall is to use a bound that depends on the particular inputs without finding a fixed estimate valid throughout the domain. Uniform continuity requires one radius for all pairs; estimates that vary with where the pair is located do not establish that requirement by themselves.

These examples also distinguish the role of the domain from the role of the formula. A polynomial can have a simple Lipschitz estimate on a bounded interval, while the same calculation may provide no global constant on an unbounded domain. Conversely, the rational function and sine examples have estimates that work everywhere. The useful question is not simply whether a formula looks complicated or whether its domain is bounded, but whether the change in its output can be controlled uniformly as the inputs become close.

Check Your Understanding

Use the estimates and arguments in this tutorial to answer the following questions.

How does the choice of $\delta$ in the positive-power estimate depend on $C$, $\alpha$, and $\varepsilon$?
Which trigonometric identities and inequalities give a Lipschitz bound for sine?
Why does $1/(1+x^2)$ have a Lipschitz bound that works on all of $\mathbb{R}$?
Where does the boundedness of $[-3,2]$ enter the estimate for the square function?
How does the cube-root estimate prove uniform continuity without proving Lipschitz continuity?

Examples of Uniformly Continuous Functions

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