3.5: Uniform Continuity

We discuss here a stronger notion of continuity.

Definition $\PageIndex{1}$: Uniformly Continuous

Let $D$ be a nonempty subset of $\mathbb{R}$. A function $f: D \rightarrow \mathbb{R}$ is called uniformly continuous on $D$ if for any $\varepsilon > 0$, there exists $\delta > 0$ such that if $u,v \in D$ and $|u-v|<\delta$, then

$$|f(u)-f(v)|<\varepsilon .$$

Example $\PageIndex{1}$

Any constant function $f: D \rightarrow \mathbb{R}$, is uniformly continuous on its domain.

Solution

Indeed, given $\varepsilon > 0$, $|f(u)-f(v)|=0<\varepsilon$ for all $u,v \in D$ regardless of the choice of $\delta$.

The following result is straightforward from the definition.

Theorem $\PageIndex{1}$

If $f: D \rightarrow \mathbb{R}$ is uniformly continuous on $D$, then $f$ is continuous at every point $x_{0} \in D$.

Example $\PageIndex{2}$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x)=7 x-2$. We will show that $f$ is uniformly continuous on $\mathbb{R}$.

Solution

Let $\varepsilon > 0$ and choose $\delta=\varepsilon / 7$. Then, if $u,v \in \mathbb{R}$ and $|u-v|<\delta$, we have

$$|f(u)-f(v)|=|7 u-2-(7 v-2)|=|7(u-v)|=7|u-v|<7 \delta=\varepsilon \nonumber$$

Example $\PageIndex{3}$

Let $f:[-3,2] \rightarrow \mathbb{R}$ be given by $f(x)=x^{2}$. This function is uniformly continuous on $[-3,2]$.

Solution

Let $\varepsilon > 0$. First observe that for $u,v \in [-3,2]$ we have $|u+v| \leq|u|+|v| \leq 6$. Now set $\delta=\varepsilon / 6$. Then, for $u,v \in [-3,2]$ satisfying $|u-v|<\delta$, we have

$$|f(u)-f(v)|=\left|u^{2}-v^{2}\right|=|u-v||u+v| \leq 6|u-v|<6 \delta=\varepsilon. \nonumber$$

Example $\PageIndex{4}$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x)=\frac{x^{2}}{x^{2}+1}$. We will show that $f$ is uniformly continuous on $\mathbb{R}$.

Solution

Let $\varepsilon > 0$. We observe first that

$$\begin{align*} \mid \frac{u^{2}}{u^{2}+1} -\frac{v^{2}}{v^{2}+1} \mid &= | \frac{u^{2}\left(v^{2}+1\right)-v^{2}\left(u^{2}+1\right)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} | \\[4pt] &= \frac{|u-v||u+v|}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \leq \frac{|u-v|(|u|+|v|)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \\[4pt] &\leq \frac{|u-v|\left(\left(u^{2}+1\right)+\left(v^{2}+1\right)\right)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \\[4pt] &\leq|u-v|\left(\frac{1}{v^{2}+1}+\frac{1}{u^{2}+1}\right) \leq 2|u-v| , \end{align*}$$

(where we used that $|x| \leq x^{2}+1$ for all $x \in \mathbb{R}$, which can be easily seen by considering separately the cases $|x|<1$ and $|x| \geq 1)$.

Now set $\delta=\varepsilon / 2$. In view of the previous calculation, given $u,v \in \mathbb{R}$ satisfying $|u-v|<\delta$ we have

$$|f(u)-f(v)|=\left|\frac{u^{2}}{u^{2}+1}-\frac{v^{2}}{v^{2}+1}\right| \leq 2|u-v|<2 \delta=\varepsilon . \nonumber$$

Definition $\PageIndex{2}$: Hölder Continuity

Let $D$ be a nonempty subset of $\mathbb{R}$. A function $f: D \rightarrow \mathbb{R}$ is said to be Hölder continuous if there are constants $\ell \geq 0$ and $\alpha > 0$ such that

$$|f(u)-f(v)| \leq \ell|u-v|^{\alpha} \text { for every } u, v \in D .$$

The number $\alpha$ is called Hölder exponent of the function. If $\alpha = 1$, then the function $f$ is called Lipschitz continuous.

Theorem $\PageIndex{2}$

If a function $f: D \rightarrow \mathbb{R}$ is Hölder continuous, then it is uniformly continuous.

Proof

Since $f$ is Hölder conitnuous, there are constants $\ell \geq 0$ and $\alpha > 0$ such that

$$|f(u)-f(v)| \leq \ell|u-v|^{\alpha} \text { for every } u, v \in D . \nonumber$$

If $\ell = 0$, then $f$ is constant and, thus, uniformly continuous. Suppose next that $\ell > 0$. For any $\varepsilon > 0$, let $\delta=\left(\frac{\varepsilon}{\ell}\right)^{1 / \alpha}$. Then, whenever $u,v \in D$, with $|u-v|<\delta$ we have

$$|f(u)-f(v)| \leq \ell|u-v|^{\alpha}<\ell \delta^{\alpha}=\varepsilon . \nonumber$$

The proof is now complete. $\square$

Example $\PageIndex{5}$

1. Let $D=[a, \infty)$, where $a > 0$.(2) Let $D=[0, \infty)$.

Solution

1. Then the function $f(x)=\sqrt{x}$ is Lipschitz continuous on $D$ and, hence, uniformly continuous on this set. Indeed, for any $u,v \in D$, we have

$$|f(u)-f(v)|=|\sqrt{u}-\sqrt{v}|=\frac{|u-v|}{\sqrt{u}+\sqrt{v}} \leq \frac{1}{2 \sqrt{a}}|u-v| ,$$

which shows $f$ is Lipschitz with $\ell=1 /(2 \sqrt{a})$.

Figure $3.4$: The square root function.

2. Then $f$ is not Lipschitz continuous on $D$, but it is Hölder continuous on $D$ and, hence, $f$ is also uniformly continuous on this set.

Indeed, suppose by contradiction that $f$ is Lipschitz continuous on $D$. Then there exists a constant $\ell>0$ such htat

$$|f(u)-f(v)|=|\sqrt{u}-\sqrt{v}| \leq \ell|u-v| \text { for every } u, v \in D .$$

Thus, for every $n \in \mathbb{N}$, we have

$$\left|\frac{1}{\sqrt{n}}-0\right| \leq \ell\left|\frac{1}{n}-0\right| .$$

This implies

$$\sqrt{n} \leq \ell \text { or } n \leq \ell^{2} \text { for every } n \in \mathbb{N} .$$

This is a contradiction. Therefore, $f$ is not Lipschitz continuous on $D$.

Let us show that $f$ is Hölder continuous on $D$. We are going to prove that

$$|f(u)-f(v)| \leq|u-v|^{1 / 2} \text { for every } u, v \in D .$$

The inequality in (3.9) holds obviously for $u = v = 0$. For $u > 0$ or $v > 0$, we have

$$\begin{aligned} |f(u)-f(v)| &=|\sqrt{u}-\sqrt{v}| \\ &=\left|\frac{u-v}{\sqrt{u}+\sqrt{v}}\right| \\ &=\sqrt{|u-v|} \frac{\sqrt{|u-v|}}{\sqrt{u}+\sqrt{v}} \\ & \leq \frac{\sqrt{|u|+|v|}}{\sqrt{u}+\sqrt{v}} \sqrt{|u-v|} \\ &=\sqrt{|u-v|} \end{aligned} .$$

Note that one can justify that inequality

$$\frac{\sqrt{|u|+|v|}}{\sqrt{u}+\sqrt{v}} \leq 1$$

by squaring both sides since they are both positive. Thus, (3.9) is satisfied.

While every uniformly continuous function on a set $D$ is also continuous at each point of $D$, the converse is not true in general. The following example illustrates this point.

Example $\PageIndex{6}$

Let $f:(0,1) \rightarrow \mathbb{R}$ be given by

$$f(x)=\frac{1}{x} .$$

Figure $3.5$: Continuous but not uniformly continuous on $(0, \infty)$.

Solution

We already know that this function is continuous at every $\bar{x} \in(0,1)$. We will show that $f$ is not uniformly continuous on $(0,1)$. Let $\varepsilon = 2$ and $\delta > 0$. Set $\delta_{0}=\min \{\delta / 2,1 / 4\}$, $x=\delta_{0}$, and $y=2 \delta_{0}$. Then $x,y \in (0,1)$ and $|x-y|=\delta_{0}<\delta$, but

$$|f(x)-f(y)|=\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{y-x}{x y}\right|=\left|\frac{\delta_{0}}{2 \delta_{0}^{2}}\right|=\left|\frac{1}{2 \delta_{0}}\right| \geq 2=\varepsilon .$$

This shows $f$ is not uniformly continuous on $(0,1)$.

The following theorem offers a sequential characterization of uniform continuity analogous to that in Theorem 3.3.3.

Theorem $\PageIndex{3}$

Let $D$ be a nonempty subset of $\mathbb{R}$ and $f: D \rightarrow \mathbb{R}$. Then $f$ is uniformly continuous on $D$ if and only if the following condition holds

(C) for every two sequences $\left\{u_{n}\right\}$, $\left\{v_{n}\right\}$ in $D$ such that $\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0$, it follows that $\lim _{n \rightarrow \infty}\left(f\left(u_{n}\right)-f\left(v_{n}\right)\right)=0$.

Proof

Suppose first that $f$ is uniformly continuous and let $\left\{u_{n}\right\}$, $\left\{v_{n}\right\}$ be sequences in $D$ such that $\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0$. Let $\varepsilon > 0$. Choose $\delta > 0$ such that $|f(u)-f(v)|<\varepsilon$ whenever $u,v \in D$ and $|u-v|<\delta$. Let $N \in \mathbb{N}$ be such that $\left|u_{n}-v_{n}\right|<\delta$ for $n \geq N$. For such $n$, we have $\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right|<\varepsilon$. This shows $\lim _{n \rightarrow \infty}\left(f\left(u_{n}\right)-f\left(v_{n}\right)\right)=0$.

To prove the converse, assume condition (C) holds and suppose, by way of contradiction, that $f$ is not uniformly continuous. Then there exists $\varepsilon_{0}>0$ such that for any $\delta > 0$, there exists $u,v \in D$ with

$$|u-v|<\delta \text { and }|f(u)-f(v)| \geq \varepsilon_{0} .$$

Thus, for every $n \in \mathbb{N}$, there exist $u_{n}, v_{n} \in D$ with

$$\left|u_{n}-v_{n}\right| \leq 1 / n \text { and }\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right| \geq \varepsilon_{0} .$$

It follows that for such sequences, $\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0$, but $\left\{f\left(u_{n}\right)-f\left(v_{n}\right)\right\}$ does not converge to zero, which contradicts the assumption. $\square$

Example $\PageIndex{7}$

Using this theorem, we can give an easier proof that the function in Example 3.5.6 is not uniformly continuous.

Solution

Consider the two sequences $u_{n}=1 /(n+1)$ and $v_{n}=1 / n$ for all $n \geq 2$. Then clearly, $\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0$, but

$$\lim _{n \rightarrow \infty}\left(f\left(u_{n}\right)-f\left(v_{n}\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{1 /(n+1)}-\frac{1}{1 / n}\right)=\lim _{n \rightarrow \infty}(n+1-n)=1 \neq 0 .$$

The following theorem shows one important case in which continuity implies uniform continuity.

Theorem $\PageIndex{4}$

Let $f: D \rightarrow \mathbb{R}$ be a continuous function. Suppose $D$ is compact. Then $f$ is uniformly continuous on $D$.

Proof

Suppose by contradition that $f$ is not uniformly continuous on $D$. Then there exists $\varepsilon_{0}>0$ such that for any $\delta > 0$, there exists $u,v \in D$ with

$$|u-v|<\delta \text { and }|f(u)-f(v)| \geq \varepsilon_{0} .$$

Thus, for every $n \in \mathbb{N}$, there exists $u_{n}, v_{n} \in D$ with

$$\left|u_{n}-v_{n}\right| \leq 1 / n \text { and }\left|f\left(u_{n}\right)-f\left(v_{n}\right)\right| \geq \varepsilon_{0} .$$

Since $D$ is compact, there exist $u_{0} \in D$ and a subsequence $\left\{u_{n_{k}}\right\}$ of $\left\{u_{n}\right\}$ such that

$$u_{n_{k}} \rightarrow u_{0} \text { as } k \rightarrow \infty .$$

Then

$$\left|u_{n_{k}}-v_{n_{k}}\right| \leq \frac{1}{n_{k}} .$$

for all $k$ and, hence, we also have

$$v_{n_{k}} \rightarrow u_{0} \text { as } k \rightarrow \infty .$$

By the continuity of $f$,

$$f\left(u_{n_{k}}\right) \rightarrow f\left(u_{0}\right) \text { and } f\left(v_{n_{k}}\right) \rightarrow f\left(u_{0}\right) .$$

Therefore, $f$ converges to zero, which is a contradiction. The proof is now complete.

$\square$

We now prove a result that characterizes uniform continuity on open bounded intervals. We first make the observation that if $f: D \rightarrow \mathbb{R}$ is uniformly continuous on $D$ and $A \subset D$, then $f$ is uniformly continuous on $A$. More precisely, the restriction $f_{\mid A}: A \rightarrow \mathbb{R}$ is uniformly continuous on $A$ (see Section 1.2 for the notation). This follows by noting that if $|f(u)-f(v)|<\varepsilon$ whenever $u,v \in D$ with $|u-v|<\delta$, then we also have $|f(u)-f(v)|<\varepsilon$ when we restrict $u,v$ to be in $A$.

Theorem $\PageIndex{5}$

Let $a,b \in \mathbb{R}$ and $a < b$. A function $f:(a, b) \rightarrow \mathbb{R}$ is uniformly continuous if and only if $f$ can be extended to a continuous function $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ (that is, there is a continuous function $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ such that $f=\tilde{f}_{\mid(a, b)}$).

Proof

Suppose first that there exists a continuous function $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ such that $f=\tilde{f}_{\mid(a, b)}$. By Theorem 3.5.4, the function $\tilde{f}$ is uniformly continuous on $[a,b]$. Therefore, it follows from our early observation that $f$ is uniformly continuous on $(a,b)$.

For the converse, suppose $f:(a, b) \rightarrow \mathbb{R}$ is uniformly continuous. We will show first that $\lim _{x \rightarrow a^{+}} f(x)$ exists. Note that the one sided limit corresponds to the limit in Theorem 3.2.2. We will check that the $\varepsilon-\delta$ condition of Theorem 3.2.2 holds.

Let $\varepsilon > 0$. Choose $\delta_{0}>0$ so that $|f(u)-f(v)|<\varepsilon$ whenever $u,v \in (a,b)$ and $|u-v|<\delta_{0}$. Set $\delta=\delta_{0} / 2$. Then, if $u, v \in(a, b)$, $|u-a|<\delta$, and $|v-a|<\delta$ we have

$$|u-v| \leq|u-a|+|a-v|<\delta+\delta=\delta_{0}$$

and, hence, $|f(u)-f(v)|<\varepsilon$. We can now invoke Theorem 3.2.2 to conclude $\lim _{x \rightarrow a^{+}} f(x)$ exists. In a similar way we can show that $\lim _{x \rightarrow b^{-}} f(x)$ exists. Now define, $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ by

$$\tilde{f}(x)=\left\{\begin{array}{ll} f(x), & \text { if } x \in(a, b) \text {;} \\ \lim _{x \rightarrow a^{+}} f(x), & \text { if } x=a \text {;} \\ \lim _{x \rightarrow b^{-}} f(x), & \text { if } x=b \text {.} \end{array}\right .$$

By its definition $\tilde{f}_{\mid(a, b)}=f$ and, so, $tilde{f}$ is continuous at every $x \in (a,b)$. Moreover, $\lim _{x \rightarrow a^{+}} \tilde{f}(x)= \lim _{x \rightarrow a^{+}} f(x)=\tilde{f}(a)$ and $\lim _{x \rightarrow b^{-}} \tilde{f}(x)=\lim _{x \rightarrow b^{-}} f(x)=\tilde{f}(b)$, so $\tilde{f}$ is also continuous at $a$ and $b$ by Theorem 3.3.2. Thus $\tilde{f}$ is the desired continuous extension of $f$. $\square$

Exercise $\PageIndex{1}$

Prove that each of the following functions is uniformly continuous on the given domain:

1. $f(x)=a x+b, a, b \in \mathbb{R}$, on $\mathbb{R}$.
2. $f(x)=1 / x \text { on }[a, \infty)$, where $a > 0$.

Answer

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Exercise $\PageIndex{2}$

Prove that each of the following functions is not uniformly continuous on the given domain:

1. $f(x)=x^{2}$ on $\mathbb{R}$.
2. $f(x)=\sin \frac{1}{x}$ on $(0,1)$.
3. $f(x)=\ln (x)$ on $(0, \infty)$.

Answer

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Exercise $\PageIndex{3}$

Determine which of the following functions are uniformly continuous on the given domains.

1. $f(x)=x \sin \left(\frac{1}{x}\right)$ on $(0,1)$.
2. $f(x)=\frac{x}{x+1}$ on $[0, \infty)$.
3. $f(x)=\frac{1}{|x-1|}$ on $(0,1)$.
4. $f(x)=\frac{1}{|x-2|}$ on $(0,1)$.

Answer

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Exercise $\PageIndex{4}$

Let $D \subset \mathbb{R}$ and $k \in \mathbb{R}$. Prove that if $f, g: D \rightarrow \mathbb{R}$ are uniformly continuous on $D$, then $f+g$ and $kf$ are uniformly continuous on $D$.

Answer

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Exercise $\PageIndex{5}$

Give an example of a subset $D$ of $\mathbb{R}$ and uniformly continuous functions $f, g: D \rightarrow \mathbb{R}$ such that $fg$ is not uniformly continuous on $D$.

Answer

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Exercise $\PageIndex{6}$

Let $D$ be a nonempty subset of $\mathbb{R}$ and let $f: D \rightarrow \mathbb{R}$. Suppose that $f$ is uniformly continuous on $D$. Prove that if $\left\{x_{n}\right\}$ is a cauchy sequence with $x_{n} \in D$ for every $n \in \mathbb{N}$, then $\left\{f\left(x_{n}\right)\right\}$ is also a Caucy sequence.

Answer

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Exercise $\PageIndex{7}$

Let $a,b \in \mathbb{R}$ and let $f:(a, b) \rightarrow \mathbb{R}$.

1. Prove that if $f$ is uniformly continuous, then $f$ is bounded.
2. Prove that if $f$ is continuous, bounded, and monotone, then it is uniformly continuous.

Answer

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Exercise $\PageIndex{8}$

Let $f$ be a continuous function on $[a, \infty)$. Suppose

$$\lim _{x \rightarrow \infty} f(x)=c .$$

1. Prove that $f$ is bounded on $[a, \infty)$.
2. Prove that $f$ is uniformly continuous on $[a, \infty)$.
3. Suppose further that $c > f(a)$. Prove that there exists $x_{0} \in[a, \infty)$ such that

$$f\left(x_{0}\right)=\inf \{f(x): x \in[a, \infty)\}.$$

Answer

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