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Uniformly continuous function


Any uniformly continuous function is actually a continuous function. However, as we will see below, the converse is not true. This important class of functions satisfies useful properties. In this article, we will discuss all these facts in detail.

The uniform continuity of a function

In some mathematical problems, continuity is not enough to make a decision. In fact, the function needs more regularities. In this section, we will see how a uniformly continuous function can help produce important properties for the function. Before that, let us first recall the definition of the norm continuity of a function.

The definition and first properties of the uniform continuity

Definition: Let $I$ be s subset of the real number set $\mathbb{R}$. We say that $f:I\to\mathbb{R}$ is a uniformly continuous function on $I,$ if for any $\varepsilon>0,$ there exists $\alpha>0$ such that for any $x,y\in I$, we have $$ |x-y|<\alpha \Rightarrow |f(x)-f(y)|<\varepsilon.$$

Uniform continuity mainly means that if two arbitrary points in the interval $I$ are very close to each other, then their images must also be very close to each other. This also implies that the graph of the function $f$ is located in a band of small width.

Proposition: Every uniformly continuous function is continuous.

Proof: Let the function $f:I\to \mathbb{R}$ be uniformly continuous, and let $a\in I$. By definition, for any $x\in I,$ closed to $a$, there exists $\alpha>0$ such that $$ |x-a|<\alpha \Rightarrow |f(x)-f(a)|<\varepsilon.$$ This maens that $f$ is continuous at $a$.

Examples: 1- The function $x\mapsto \sin(x)$ is uniformly continuous on $\mathbb{R}$. In fact, we know that the sinus function is differentiable on $\mathbb{R}$. Thus by the mean value theorem, for any $x,y\in \mathbb{R}$ there exists a real number $c$ between $x$ and $y$ such that $\sin(x)-\sin(y)=\cos(c)(x-y)$. Thus $|\sin(x)-\sin(y)|\le |x-y|$. Now take $\varepsilon>0,$ and select $\alpha=\varepsilon$. thus $|x-y|<\alpha$ implies that $|\sin(x)-\sin(y)|<\varepsilon$. This ends the proof.

2- Similarly, the function $x\mapsto \arctan(x)$ is uniformly continuous on $\mathbb{R}$. In fact, again the mean value theorem, for any $x,y$, we have $|\arctan(x)-\arctan(y)|\le |x-y|$. We then use the same technique as for the sinus function.

3- The square root function is uniformly continuous on $\mathbb{R}^+$. In fact, as a simple exercise one can see that for any $x,y\in \mathbb{R}^+,$ $$ |\sqrt{x}-\sqrt{y}|\le \sqrt{|x-y|}.$$ Let now $\varepsilon>0$ and choose $\alpha=\varepsilon^2$. We then have for any $x,y\in\mathbb{R}^+$, if $|x-y|<\alpha= \varepsilon^2,$ then $\sqrt{|x-y|}< \varepsilon$. This implies that $|\sqrt{x}-\sqrt{y}|< \varepsilon$. So that the function $x\mapsto \sqrt{x}$ is uniformly continuous on $\mathbb{R}^+$.

Example of a continuous not uniformly continuous function

We first state and prove a result that characterizes the norm continuity of functions.

Theorem: A function $f:I\to\mathbb{R}$ is uniformly continuous on $I$ if and only if, for any sequences $(u_n)_n,(v_n)-n\subset I$ sucn that $u_n-v_n$ to $0$ as $n\to\infty,$ we have $f(u_n)-f(v_n)\to 0$ as $n\to \infty$.

Proof: Let $\varepsilon>0$. We first prove the direct implication. On the one hand, there exists $\alpha>0$ such that for any $x,y\in I$, we have $$ |x-y|<\alpha \Rightarrow |f(x)-f(y)|<\varepsilon.$$ On the other hand, as $u_n-v_n$ to $0$ as $n\to\infty,$ then there exists $N\in\mathbb{N}$ such that for any $n\in\mathbb{N},$ $n>N$ implies that $|u_n-v_n|<\alpha$. This implies that $|f(u_n)-f(v_n)|<\varepsilon,$ whenever $n>N$. Thus $f(u_n)-f(v_n)\to 0$ as $n\to\infty$. Conversely, assume that $f$ is not uniformly continuous on $I$. This means that there exists $\varepsilon>0$, for all $\alpha>0$, there exist $x,y\in I$ such that $|x-y|<\alpha$ and $|f(x)-f(y)|\ge \varepsilon$. Take $\alpha=\frac{1}{n}$ for any $n\in\mathbb{N}^\ast$. There exists $x_n,y_n\in I$ for each $n$ such that $|x_n-y_n|<\alpha$ and $|f(x_n)-f(y_n)|\ge \varepsilon$. This implies that that $x_n-y_n\to 0$ and $|f(x_n)-f(x_n)|$ does not goes to zero as $n\to \infty$. This is a contradiction.

The example: The function $f(x)=\sin(x^2)$ is not uniformly continuous on $\mathbb{R}$. In fact, we will apply the above theorem. Let’s consider the sequences $$ u_n=\sqrt{\pi n},\quad v_n=\sqrt{\left( n+\frac{1}{2}\right)\pi}.$$ First, we have \begin{align*} u_n-v_n&= \frac{\pi n- ( n+\frac{1}{2}}{\sqrt{\pi n}+\sqrt{\left( n+\frac{1}{2}\right)\pi}}\cr& = -\frac{\pi}{2} \frac{1}{\sqrt{\pi n}+\sqrt{\left( n+\frac{1}{2}\right)\pi}}.\end{align*} Clearly $u_n-v_n\to 0$ as $n\to \infty$. On the other hand, \begin{align*}f(u_n)-f(v_n)=\sin(n\pi)-\sin(\frac{\pi}{2}+n\pi)=-\cos(n\pi)=(-1)^{n+1}.\end{align*} Thus the sequence $(|f(u_n)-f(v_n)|)_n$ does not goes to zero. This ends the proof.

The Continuous Extension Theorem

We recall that a $(u_n)_n$ is a Cauchy sequence if for any $\varepsilon>0,$ there exists positive integer $N\in \mathbb{N}$ such that for any $p,q\in \mathbb{N},$ $$ p>N,\;q>N \Rightarrow |u_p-u_q|<\varepsilon.$$

Theorem: A function $f:I\to\mathbb{R}$ is uniformly continuous on $I$ if and only if for any Cauchy sequence $(u_n)_n\subset I$, $(f(u_n))_n$ is a Cauchy sequence.

Proof: The proof is very similar to the proof of the previous theorem, we omit it.

Theorem: Let $a$ and $b$ be two real numbers with $a<b$ and let $f:(a,b)\to \mathbb{R}$ be a uniformly continuous function. The we can extend $f$ to a continuous function $\tilde{f}$ on $[a,b]$.

Proof: It suffices to prove that the limits of $f$ at the points $a$ and $b$ exist. Since $a$ and $b$ play the same role, we only focus on point $a$. Let $(u_n)_n\subset (a,b)$ such that $u_n\to a$ and let us prove that $(f(u_n))_n$ has a limit. As $(u_n)_n$ is a convergent sequence, then it is a Cauchy sequence. So, by the above theorem $(f(u_n))_n$ is a Cauchy sequence. As we work in $\mathbb{R},$ the sequence $(f(u_n))_n$ has a limit. This ends the proof.

The extension of $f$ is the function $\tilde{f}:[a,b]\to\mathbb{R}$ defined by \begin{align*}\tilde{f}(x)=\begin{cases} \ell_1,& x=a,\cr f(x),& x\in (a,b),\cr \ell_1,& x=b,\end{cases}\end{align*} where $$ \ell_1=\lim_{x\to a}f(x),\quad \ell_2=\lim_{x\to b}f(x).$$

This is a great theorem of continuous extension functions. Also, the result of this page can also be extended to vector-valued spaces where the normed vector space is complete, that is, in which every Cauchy sequence converges in the same space.

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