3.5: Uniform Continuity

Lafferriere, Lafferriere, and Nguyen
Portland State University via PDXOpen: Open Educational Resources

We discuss here a stronger notion of continuity.

Definition $\PageIndex$: Uniformly Continuous

Let $D$ be a nonempty subset of $\mathbb$. A function $f: D \rightarrow \mathbb$ 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|

Example $\PageIndex$

Any constant function $f: D \rightarrow \mathbb$, is uniformly continuous on its domain. Solution Indeed, given $\varepsilon > 0$, \(|f(u)-f(v)|=0

The following result is straightforward from the definition.

Theorem $\PageIndex$

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

Example $\PageIndex$

Let $f: \mathbb \rightarrow \mathbb$ be given by $f(x)=7 x-2$. We will show that $f$ is uniformly continuous on $\mathbb$. Solution Let $\varepsilon > 0$ and choose $\delta=\varepsilon / 7$. Then, if $u,v \in \mathbb$ and \(|u-v|

Example $\PageIndex$

Let $f:[-3,2] \rightarrow \mathbb$ be given by $f(x)=x^$. 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^-v^\right|=|u-v||u+v| \leq 6|u-v|

Example $\PageIndex$

Definition $\PageIndex$: Hölder Continuity

Let $D$ be a nonempty subset of $\mathbb$. A function $f: D \rightarrow \mathbb$ is said to be Hölder continuous if there are constants $\ell \geq 0$ and $\alpha > 0$ such that \[|f(u)-f(v)| \leq \ell|u-v|^ \text < for every >u, v \in D .\] The number $\alpha$ is called Hölder exponent of the function. If $\alpha = 1$, then the function $f$ is called Lipschitz continuous.

Theorem $\PageIndex$

If a function $f: D \rightarrow \mathbb$ 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 \[|f(u)-f(v)| \leq \ell|u-v|^ \text < for every >u, v \in D . \nonumber\] If $\ell = 0$, then $f$ is constant and, thus, uniformly continuous. Suppose next that $\ell > 0$. For any $\varepsilon > 0$, let $\delta=\left(\frac\right)^$. Then, whenever $u,v \in D$, with $|u-v|<\delta$ we have \[|f(u)-f(v)| \leq \ell|u-v|^<\ell \delta^=\varepsilon . \nonumber\] The proof is now complete. $\square$

Example $\PageIndex$

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

Solution

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

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

$Annotation 2020-08-30 161036.png$

Figure $3.4$: The square root function.

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

Thus, for every $n \in \mathbb$, we have

\[\sqrt \leq \ell \text < or >n \leq \ell^ \text < for every >n \in \mathbb .\]

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

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

Note that one can justify that inequality

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 $\PageIndex$

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

$Annotation 2020-08-30 162555.png$

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

Solution

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

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 $\PageIndex$

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

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

Proof

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

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

\[\left|u_-v_\right| \leq 1 / n \text < and >\left|f\left(u_\right)-f\left(v_\right)\right| \geq \varepsilon_ .\]

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

Example $\PageIndex$

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 $u_=1 /(n+1)$ and $v_=1 / n$ for all $n \geq 2$. Then clearly, $\lim _\left(u_-v_\right)=0$, but

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

Theorem $\PageIndex$

Let $f: D \rightarrow \mathbb$ 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 $\varepsilon_>0$ such that for any $\delta > 0$, there exists $u,v \in D$ with

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

\[\left|u_-v_\right| \leq 1 / n \text < and >\left|f\left(u_\right)-f\left(v_\right)\right| \geq \varepsilon_ .\]

Since $D$ is compact, there exist $u_ \in D$ and a subsequence $\left\>\right\>$ of $\left\\right\>$ such that

\[u_> \rightarrow u_ \text < as >k \rightarrow \infty .\]

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

\[v_> \rightarrow u_ \text < as >k \rightarrow \infty .\]

By the continuity of $f$,

\[f\left(u_>\right) \rightarrow f\left(u_\right) \text < and >f\left(v_>\right) \rightarrow f\left(u_\right) .\]

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

We now prove a result that characterizes uniform continuity on open bounded intervals. We first make the observation that if $f: D \rightarrow \mathbb$ is uniformly continuous on $D$ and $A \subset D$, then $f$ is uniformly continuous on $A$. More precisely, the restriction $f_<\mid A>: A \rightarrow \mathbb$ is uniformly continuous on $A$ (see Section 1.2 for the notation). This follows by noting that if \(|f(u)-f(v)|

Theorem $\PageIndex$

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

Proof

Suppose first that there exists a continuous function $\tilde:[a, b] \rightarrow \mathbb$ such that $f=\tilde_<\mid(a, b)>$. By Theorem 3.5.4, the function $\tilde$ 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 $f:(a, b) \rightarrow \mathbb$ is uniformly continuous. We will show first that $\lim _> f(x)$ 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.

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

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

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