$$$$

19  $★$ Uniform Continuity

Continuity is a local property: to verify that a function is continuous, we look near individual points. But what if we want a global notion — one that controls the behavior of a function across its entire domain at once?

This brings us to uniform continuity, a stronger condition that ensures a single δ works for all points in the domain. While every uniformly continuous function is continuous, the reverse is not always true.

In this chapter, we will: - Define uniform continuity and explore how it differs from ordinary continuity. - Test whether standard operations like addition, multiplication, and inversion preserve uniform continuity. - Prove two deep results that make uniform continuity especially useful: that it ensures extendability and automatically holds for continuous functions on closed intervals.

Definition 19.1 (Uniform Continuity: $\epsilon -\delta$) A function $f$ is uniformly continuous on a domain $D\subset \mathbb{R}$ if for every $\epsilon$ there exists a $\delta$ such that for any $x,y\in D$ with $|x-y|<\delta$, it follows that $|f(x)-f(y)|<\epsilon$.

Here’s an example showing how to use the definition, proving $x^2$ is uniformly continuous on an interval.

Example 19.1 $f(x)=x^2$ is uniformly continuous on the interval $[1,3]$.

Proof (Scratch). Here’s some scratch work: let $\epsilon >0$. Then at any $a$ we see that $|f(x)-f(a)|=|x^2-a^2|=|x+a||x-a|$. If $|x-a|<\delta$ and we want $|f(x)-f(a)|<\epsilon$, this tells us that we want $$|x+a|\delta <\epsilon$$ We don’t know what $x$ and $a$ are, but we do know they are points in the interval $[1,3]$! So, the smallest $x+a$ could be is $1+1=2$, and the biggest is $3+3=6$. This means that $$|x+a|\delta \le 6\delta$$ So, if we can make $6\delta <\epsilon$, we are good! This is totally possible: just set $\delta =\epsilon /6$. Below is the rigorous proof.

Proof (Rigorous). Let $\epsilon >0$, and set $\delta =\epsilon /6$. Note that for any $a\in [1,3]$ and any $x$ within $\delta$ of $a$, we know $a\le 3$ and $x\le 3$ so $x+a\le 6$. But this implies that $$|x^2-a^2|=|x+a||x-a|\le 6|x-a|<6\delta <6\frac{\epsilon }{6}=\epsilon$$ And so $f$ is uniformly continuous, as this single choice of $\delta$ works for every point $a\in [1,3]$.

For normal continuity, we had a way to test using sequences. This proved quite useful since we are so good at working with sequences these days. There is an analog for uniform continuity as well

Exercise 19.1 (Sequences and Uniform Continuity) A function $f$ is uniformly continuous if and only if for every pair of sequences $u_n,v_n$ in the domain with $lim{u}_n-v_n=0$, then $limf(u_n)-f(v_n)=0$.

Uniform continuity is stricter than regular continuity: there are functions which are continuous but are not uniformly continuous. Here we see $x^2$ is such an example, using the sequence criterion

Example 19.2 The function $f(x)=1/x$ is continuous, but not uniformly continuous on $(0,1)$. Looking at the sequence $1/n$ we see $f(1/n)=1/(1/n)=n$. So, consider the two sequences $s_n=1/(n+1)$ and $t_n=1/n$. These have $s_n-t_n\to 0$ by the limit theorems (as each individually goes to zero) yet $f(s_n)-f(t_n)=(n+1)-n=1$ is a constant sequence not converging to zero.

The sequence $1/n$ used in this example provides a hint of one way to detect uniformly continuous functions: $1/n$ is Cauchy but $f(1/n)$ was not, and we were able to use this to show $f$ was not uniformly continuous.

These examples show that uniform continuity is genuinely stronger than ordinary continuity. In particular, the failure of uniform continuity for f(x) = 1/x on (0,1) highlights that local control is not enough — behavior near the edges matters, and if the function “does something crazy” (here a vertical asymptote) we won’t be able to find a uniform $\delta$.

Theorem 19.1 (Uniformly Continuity Preserves Cauchy Sequences) If $f$ is uniformly continuous and $x_n$ is cauchy, then $f(x_n)$ is cauchy.

Proof. Let $x_n$ be an arbitrary Cauchy sequence in the domain of $f$, and choose arbitrary $\epsilon >0$. Then by uniform continuity there is a $\delta$ such that for $|x-y|<\delta$ we know $|f(x)-f(y)|<\epsilon$. Since $x_n$ is Cauchy, given this $\delta$ we can find an $N$ such that $n,m>N$ implies $|x_n-x_m|<\delta$, and hence $|f(x_n)-f(x_m)|<\epsilon$. But this is precisely the definition of $\{f(x_n)\}$ being a Cauchy sequence, so we are done.

Great way to check if a function is not uniformly continuous: can you find a cauchy seq taken to a non-cauchy sequence?

Example (PICTURE) functions like $\sin (1/x)$ are also not uniformly continuous on $(0,1)$ even though it is bounded.

WARNING: does not work in reverse: the function $x\mapsto x^2$ takes Cauchy sequences to cauchy seqs but is not uniformly continuous.

Definition 19.2 (Cauchy Continuous Functions) A real valued function $f$ on a domain $D\subset \mathbb{R}$ is Cauchy Continuous if for every cauchy sequence $\{d_n\}$ in $D$, the sequence $f(d_n)$ is also Cauchy.

19.1 Properties of Uniformly Continuous Functions

In a previous chapter, we showed that continuous functions behave well under addition, multiplication, and composition. Now we ask: do these same operations preserve uniform continuity? The answers are a little more nuanced. Let’s go through them carefully.

Proposition 19.1 (Constant Multiples of Uniformly Continuous Functions) Let $f$ be uniformly continuous, and $k\in \mathbb{R}$. Then $kf$ is uniformly continuous.

Proof. If $k=0$ then $kf$ is the constant zero function, so we ignore that case. For $k\ne 0$, let $\epsilon \ge 0$ consider $\epsilon /|k|$ and take the corresponding uniform $\delta$ for $f$. For $|x-y|<\delta$ we see $|f(x)-f(y)|<\epsilon /|k|$, and so $$|kf(y)-kf(x)|<|k||f(y)-f(x)|\le |k|\frac{\epsilon }{|k|}=\epsilon$$

Exercise 19.2 (Sums of Uniformly Continuous Functions) Let $f$ and $g$ be uniformly continuous. Then $f+g$ is uniformly continuous.

From these it follows that $f-g$ is uniformly continuous (as its equal to $f+(-1)g$) and $af+bg$ are for any $a,b\in \mathbb{R}$ are uniformly continuous as well. It might be tempting to believe, after seeing the above proofs that all of the limit laws should have analogs for uniform continuity, just as they did for continuity. But this is not true!

Example 19.3 (Reciprocals need not be Uniformly Continuous) The function $y=x$ is uniformly continuous and nonzero on $(0,1)$ but its reciprocal $f(x)=1/x$ is not.

Proof. Fx any $\delta >0$, and note that given any $1/n<\delta$ we have $$\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}<\frac{1}{n}<\delta$$ but applying $f$, $$f\left(\frac{1}{n+1}\right)-f\left(\frac{1}{n}\right)=n+1-n=1$$ Thus, fixing any $\epsilon <1$ there can’t be a uniform $\delta$, as its always possible to find points separated by less than $\delta$ mapped to points separated by a distance of 1.

This generalizes directly to reciprocals: if $f$ is uniformly continuous then $1/f$ need not be

Exercise 19.3 Let $f$ be uniformly continuous and bounded away from zero: $f(x)\ge b>0$ for all $x$ in the domain. Prove that $1/f$ is uniformly continuous.

What about products? Again we need a boundedness assumption:

Exercise 19.4 (Uniform Continuity and Products) Let $f$ and $g$ be uniformly continuous bounded functions with the same domain. Then $f(x)g(x)$ is uniformly continuous.

Proof. Since $f,g$ are bounded we chan choose an $M>0$ with $|f(x)|<M$ and $|g(x)|<M$ for all $x$ in the domain. Let $\epsilon >0$ be arbitrary, and using uniform continuity for $f,g$ choose ${\delta }_f$ such that $|x-y|<{\delta }_f$ implies $|f(x)-f(y)|<\epsilon /2M$ and an analogous ${\delta }_g$ for $g$. Set $\delta =min\{{\delta }_f,{\delta }_g\}$ and for any $x,y$ with $|x-y|<\delta$ we compute $$|f(y)g(y)-f(x)g(x)|=|f(y)g(y)-f(y)g(x)+f(y)g(x)-f(x)g(x)|$$ $$\le |f(y)g(y)-f(y)g(x)|+|f(y)g(x)-f(x)g(x)|=|f(y)||g(y)-g(x)|+|g(x)||f(y)-f(x)|$$ As both $|f|$ and $|g|$ are bounded by $M$, this is less than or equal to $M(|g(y)-g(x)|+|f(y)-f(x)|)$, and each of these terms is less than $\epsilon /2M$ by hypothesis, so $$|f(y)g(y)-f(x)g(x)|\le M(\frac{\epsilon }{2M}+\frac{\epsilon }{2M})=\epsilon$$ as required.

Exercise 19.5 Show that this boundedness assumption is necessary by giving an example of two uniformly continuous functions whose product is not uniformly continuous.

Proposition 19.2 (Composition of Uniformly Continuous) Let $f$ and $g$ be uniformly continuous functions. Then the composition $f\circ g(x)=f(g(x))$ is uniformly continuous.

Proof. Choose $\epsilon >0$ and let ${\delta }_f$ be a uniform delta for $f(x)$. Use this to select a uniform ${\delta }_g$ for $g$, such that whenever $|x-y|<{\delta }_g$, we have $|g(x)-g(y)|<{\delta }_f$. This turns out to be the right uniform value for the composition $f\circ g$, as $|g(x)-g(y)|<{\delta }_f⟹|f(g(x))-f(g(y))|<\epsilon$.

Like reciprocals, inverses pose a problem:

Exercise 19.6 (Inverses and Uniform Continuity) Give an example of a uniformly continuous function whose inverse is not uniformly continuous.

Exercise 19.7 Prove that $f(x)=x^2$ is not uniformly continuous on the entire real line, using either the $\epsilon -\delta$ definition or the sequence definition.

19.2 Continuous Extension

So far uniform continuity seems to be a slightly more restrictive definition (requiring one to prove their choice of $\delta$ works everywhere) with consequently weaker theorems (inverses and products need not be uniformly continuous, even though they are continuous, for one). So why would one care about this harder to verify and harder to work with notion of continuity? The main reason is that the stricter definition of uniform continuity allows us to prove some (very useful) things which are just not true about the standard version! The chief among these is perhaps the continuous extension theorem.

Roughly speaking, if a function is uniformly continuous on an open interval, then we can define its values at the endpoints in a way that makes it continuous on the closed interval. This is not always possible for merely continuous functions, as we saw with f(x) = 1/x on (0,1). Uniform continuity makes all the difference.

Theorem 19.2 (Extending Uniform Continuity to Endpoints) If $f:(a,b)\to \mathbb{R}$ is uniformly continuous, then there exists a continuous extension of $\tilde{f}$ of $f$ to $[a,b]$.

We could stop to prove this here, but in fact the same technique proves a more general extension theorem of which this is a special case:

Theorem 19.3 (The Continuous Extension Theorem)  

Proof. Proof sketch: $D$ dense in $X$, define $f(x)$ by $limf(d_n)$ for $d_n\to x$. Need to check (1) this defines a value, (2) its well defined, independent of sequence.

For (1): if $d_n\to x$ then $d_n$ is convergent, hence Cauchy. $f$ is uniformly continuous so it takes Cauchy sequences to cauchy sequences. Thus $f(x_n)$ is Cauchy, hence convergent.

Next for (2): if $c_n,d_n\to x$ are two such sequences, make the interleaved sequence $c_1,d_1,c_2,d_2,c_3,d_3\cdots$. This converges to $x$ as well so is Cauchy. Thus applying $f$ yields a cauchy (hence convergent) sequence, and all subsequences have the same limit. Since $c_n$ and $d_n$ are subsequences, we see $f(c_n)$ and $f(d_n)$ converge to the same value.

Note in the proof above we only used one property of uniformly continuous functions: that they take cauchy sequences to cauchy sequences. So this actually applies more generally, to Cauchy Continuous functions.

Corollary 19.1 (Continuous Extension of Cauchy Continuous Functions) If $f$ is Cauchy continuous on a set $D$ which is dense in $X$, then there exists a unique continuous extension $\tilde{f}$ of $f$ to $X$.

19.3 Continuous on a Closed Interval

The continuous extension theorems provide a first (of several) motivations for being interested in this stronger notion of continuity. Hence its useful to develop some results for telling when a function which known a priori only to be continuous is in fact uniformly continuous. The most useful of these provides a surprisingly simple condition: so long as the domain is a closed interval, continuity and uniform continuity are equivalent!

Theorem 19.4 (Continuous $+$ Closed $⟹$ Uniform) Let $f$ be a continuous function defined on the closed interval $[a,b]$. Then $f$ is in fact uniformly continuous on this interval.

Proof. Assume for the sake of contradiction that $f$ is not uniformly continuous, and fix $\epsilon >0$. Then there is no fixed $\delta$ that works, so for any proposed $\delta$, there must be some $a$ where it fails.

We can use this to produce a sequence: for $\delta =1/n$ let $a_n\in I$ be a point where this $\delta$ fails: there is some $x_n$ within $1/n$ of $a_n$ but $|f(x_n)-f(a_n)|>\epsilon$.

Thus, in fact we have two sequences $x_n$ and $a_n$! We know very little about either except that they are in a closed interval $I$, so we can apply Bolzano Weierstrass to get convergent subsequences (we have to be a bit careful here, see the exercise below).

We will call the subsequences $X_n$ and $A_n$ (with capital letters). Now that we know they both converge, we can see that they also have the same limit: (as, by construction $|X_n-A_n|<\frac{1}{n}$). Call that limit $L$.

Then since $f$ is continuous at $L$, we know that $$limf(X_n)=f(lim{X}_n)=f(L)=f(lim{A}_n)=limf(A_n)$$ Thus, $limf(X_n)-f(A_n)=0$. However this is impossible, since for all values of $n$ we know $|f(X_n)-f(A_n)|>\epsilon$! This is a contradiction, and thus there must have been some uniform $\delta$ that worked all along.

In proof, use that we can simultaneously apply bolzano weierstrass to two sequences: this appears as an Exercise 9.14 back in the chapter on subsequences. If you didn’t do it then, you should prove this for yourself now.

Exercise 19.8 (Periodic Continuous Functions are Uniformly Continuous) Let $f$ be a periodic continuous function on $\mathbb{R}$. Then $f$ is uniformly continuous.