Uniform Continuity

Continuity is defined on each point c \in D, we want to define a notion of continuity that is not pointwise, and is instead defined on the whole domain D.

Motivation

A uniformly continuous function has a single \delta that works on the whole domain, whereas a continuous function has a separate \delta for each point c \in D.

In other words, continuity is a local condition on a function. It is defined at each point c \in D, and cares only about an infinitesimally small neighborhood about said point. Uniform continuity is a stronger notion, defined over the entirety of the domain D of a function.

Roughly speaking, a function is uniformly continuous if there exists no rapid changes in the function. Verticle asymptotes on a function are rapid changes, where as linear changes aren’t. Another piece of intuition is that a function is uniformly continuous if an infinitesimally small change in x does not create an appreciable change in f(x).¹

We use the following definition to formalize our intuition on rapid changes.

Definition

A function f \colon D \to \mathbb{R} is uniformly continuous on D if for all \varepsilon > 0, there exists \delta > 0 such that for all x,y \in D, |x-y| < \delta \implies |f(x) - f(y)| < \varepsilon. Symbolically

\forall \varepsilon > 0, \exists \delta > 0 | \forall x,y \in D |x-y| < \delta \implies |f(x) - f(y)| < \varepsilon

Example (A uniformly continuous function)

Claim: The function f(x) = 2x is uniformly continuous on \mathbb{R}.

Given \varepsilon > 0, we want to find some \delta > 0 such that

\forall x,y \in D, |x-y| < \delta \implies |f(x) - f(y)| = |2x - 2y| = 2|x - y| < \varepsilon.

Seeing this its clear we should choose \delta = \varepsilon/2, as that gives us the relation

\forall x,y \in D, |x-y| < \frac{\varepsilon}{2} \implies 2|x-y| < 2\frac{\varepsilon}{2} = \varepsilon.

Thus we have f(x) = 2x is uniformly continuous.

Example (A non uniformly continuous function)

Claim: The function f(x) = x^2 is not uniformly continuous on \mathbb{R}.

To prove a function is not uniformly continuous is equivalent to saying there exists some \varepsilon > 0 such that for all \delta > 0, there exist some pair x,y \in D such that |x-y| < \delta \implies |f(x) - f(y)| \geq \varepsilon.

We first notice that

|f(x) - f(y)| = |x^2 - y^2| = |x + y| \cdot |x - y|.

Let \varepsilon = 1. We need to show that given any \delta > 0, there exists some pair x,y \in D such that

|x-y| < \delta \implies |x + y| \cdot |x - y| \geq 1.

Because we only need to find one pair x,y for each \delta, it makes sense to have our pair depend on \delta. For any x, let y = x + \delta/2. This gives us |x - y| = \delta/2, and |x + y| \cdot |x - y| = |x + y| \cdot \delta/2.

In order for |x + y| \cdot \delta/2 to be greater than 1, we need |x + y| \geq 2/\delta. We choose x = 1/\delta, as this gives y = 1/\delta + \delta/2. We now have enough information to construct our proof.

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Let \varepsilon = 1. Given any \delta > 0, let x = 1/\delta, y = 1/\delta + \delta/2. Then |x-y| = \delta/2 < \delta, and |f(x) - f(y)| = |x+y| \cdot |x-y| = |2/\delta + \delta/2| \cdot |\delta/2| \geq 1. Thus our function f is not continuous on \mathbb{R}.

Those two examples provide the usual process we use to prove a function is or is not continuous. However, a function being uniformly continuous depends on the domain used.

Example (A bounded uniformly continuous function)

Claim: The function f(x) = x^2 is uniformly continuous on D = [-5,5].

Note that for any x,y \in D, we have |x + y| \leq 10. For any \varepsilon > 0, let \delta = \varepsilon/10. Then

|x-y| < \delta \implies |x + y|\cdot|x - y| < 10 \cdot \delta = 10\frac{\varepsilon}{10} = \varepsilon

By bounding our domain, we were able to show the function is uniformly continuous on said domain. This leads us to the following theorem.

Theorem 2.4.1

If D is compact, and f \colon D \to \mathbb{R} is continuous, then f is uniformly continuous.

Proof

Let \varepsilon > 0. Since f is continuous on D, it is continuous at each x \in D. For each x \in D, there exists some \delta_x > 0 such that

|x-y| < \delta_x \implies |f(x) - f(y)| < \frac{\varepsilon}{2}.

The collection

F = \left\{ N\left(x,\frac{\delta_x}{2}\right) \mid x \in D \right\}

is an open cover of D. Since D is compact, F contains a finite subcover. Thus there exists x_1,\dots,x_n \in D such that D \subset N(x_1,\delta_{x_1}/2) \cup \cdots \cup N(x_n,\delta_{x_n}/2).

Let \delta = \min\left\{\frac{\delta_{x_1}}{2},\cdots,\frac{\delta_{x_n}}{2}\right\}. We want to show that this \delta fufills the conditions for uniform continuity. Suppose x,y \in D and |x - y| < \delta. Since \bigcup_{i=1}^nN(x_i,\delta_{x_i}/2) covers D, there exists some i such that x \in N(x_i,\delta_{x_i}/2).

Since |x-y| < \delta \leq \delta_{x_i}/2, we have

|y - x_i| = |y-x + x - x_i| \leq |y-x| + |x - x_i| < \frac{\delta_{x_i}}{2} + \frac{\delta_{x_i}}{2} = \delta_{x_i}.

Thus |f(y) - f(x_i)| < \varepsilon/2. We also have |x - x_i| < \delta_{x_i}, so |f(x) - f(x_i)| < \varepsilon/2. Through the triangle inequality

|f(x) - f(y)| \leq |f(x) - f(x_i)|+|f(x_i) - f(y)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.

Therefore our function is uniformly continuous.

Historically, Cauchy believed the following theorem was true of any continuous function, which was later found to be inaccurate.

Theorem 2.4.2

Let f \colon D \to \mathbb{R} be uniformly continuous on D and suppose (x_n) is a Cauchy sequence in D. Then (f(x_n)) is Cauchy.

Proof

Given any \varepsilon > 0, since f is uniformly continuous on D, there exists some \delta > 0 such that for all x,y \in D

|x-y| < \delta \implies |f(x) - f(y)| < \varepsilon.

Since (x_n) is Cauchy, there exists an integer N such that, when n,m \geq N

|x_n - x_m| < \delta.

Therefore, for m,n \geq N, we also have |f(x_n) - f(x_m)| < \varepsilon, so (f(x_n)) is a Cauchy sequence.

Properties

Definition

We say that a function \tilde{f} \colon E \to \mathbb{R} is an extension of a function f \colon D \to \mathbb{R} if D \subseteq E and f(x) = \tilde{f}(x) for all x \in D.

This definition allows us to take a function f, and create a new function \tilde{f} such that f(x) = \tilde{f}(x). If we extend f to a domain, and prove \tilde{f} is uniformly continuous on said domain, then clearly f is uniformly continuous on it’s sub-domain.

Theorem 2.4.3

A function f \colon (a,b) \to \mathbb{R} is uniformly continuous on (a,b) if and only if it can be extended to a function \tilde{f} that is continuous on [a,b].

Proof

We start by proving the reverse direction. By Theorem 2.4.1 if \tilde{f} is continuous on [a,b], it is uniformly continuous. It follws then that \tilde{f}, and hence f, is also uniformly continuous on the subset (a,b).

Conversly, suppose that f is uniformly continuous on (a,b). Let (s_n) be a sequence in (a,b) such that \lim s_n = a (We know this sequence exists, because a is an accumulation point). By Theorem 1.3.5, (s_n) is Cauchy. Then by Theorem 2.4.2, (f(s_n)) is Cauchy, and thus converges. By Theorem 2.1.4, \lim_{x\to a}f(x) = p for some p \in \mathbb{R}. Similarly, \lim_{x \to b}f(x) = q for some q \in \mathbb{R}.

We define \tilde{f} \colon [a,b] \to \mathbb{R} by

\tilde{f}(x) = \begin{cases} f(x), & a < x < b\\ p, & x = a\\ q, & x = b \end{cases}

Then \tilde{f} is an extension of f. Since f is continuous on (a.b), so is \tilde{f}. Since \tilde{f} is continuous on a,b it follws that \tilde{f} is continuous on [a,b].

Unrelated.

Theorem 2.4.4

Let f\colon D \to \mathbb{R} be a uniformly continuous function on the bounded set D. Prove that f is bounded on D.

Proof

Suppose that f(D) is not bounded. Then, for any M \in \mathbb{R} there exists a sequence (x_n) \in D such that for some natural number n, |f(x_n)| \geq M.

Since D is bounded, there exists a convergent sub-sequence (x^\prime_n) of (x_n) by Theorem 1.4.2, which is additionally a Cauchy sequence by Lemma 1.3.3. Since f is uniformly continuous, we would have that (f(x^\prime_n)) is a Cauchy sequence. But by Lemma 1.3.4, every cauchy sequence must be bounded. Thus our assumption that f(D) is unbounded is false, and f(D) must be bounded.

Practice

Problem

A function f \colon \mathbb{R} \to \mathbb{R} is said to be periodic if there exists a number k > 0 such that f(x + k) = f(x) for all x \in \mathbb{R}. Suppose that f \colon \mathbb{R} \to \mathbb{R} is continuous and periodic. Prove that f is bounded and uniformly continuous on \mathbb{R}.

Solution

Let f be a continuous function with period k. Take f \colon [0,k] \to \mathbb{R}. Since [0,k] is compact, f([0,k]) is compact, and therefore bounded (since it obtains a minimum and maximum). For every x \in \mathbb{R}, we have some y \in [0,k] such that x-y = nk for some integer n, thus f(x) = f(y + nk) = f(y), and f(\mathbb{R}) = f([0,k]). Since f([0,k]) is bounded, so is f(\mathbb{R}).

Since [0,2k] is compact, and f is continuous, f([0,2k]) is uniformly continuous. That is, for every \varepsilon > 0, there exists some \delta_k > 0 such that

|x_k - y_k| < \delta \implies |f(x_k) - f(y_k)| < \varepsilon, \forall x_k,y_k \in [0,2k].

Chose some \varepsilon > 0. Let \delta = \delta_k. For all x,y \in \mathbb{R}, if |x-y| < \delta there exists x_k,y_k, \in [0,2k] and such that x = x_k + nk and y = y_k + nk for some integer n. Then for all x,y\in \mathbb{R},

|x - y| = |(x_k - n) - (y_k - n)| = |x_k - y_k| < \delta \implies\\ |f(x) - f(y)| = |f(x_k + nk) - f(y_k + nk)| = |f(x_k) - f(y_k)| < \varepsilon

Thus f \colon \mathbb{R} \to \mathbb{R} is uniformly continuous. Note we choose the interval [0,2k] instead of [0,k], so there arise no complications when x < nk and y > nk.

Recap

In this section we proved.

Theorem 2.4.1

If D is compact, and f \colon D \to \mathbb{R} is continuous, then f is uniformly continuous.

Theorem 2.4.2

Let f \colon D \to \mathbb{R} be uniformly continuous on D and suppose (x_n) is a Cauchy sequence in D. Then (f(x_n)) is Cauchy.

Theorem 2.4.3

A function f \colon (a,b) \to \mathbb{R} is uniformly continuous on (a,b) if and only if it can be extended to a function \tilde{f} that is continuous on [a,b].

Theorem 2.4.4

Let f\colon D \to \mathbb{R} be a uniformly continuous function on the bounded set D. Prove that f is bounded on D.

Footnotes

1. The following Mathexchange thread provides very useful intuition.