Lecture 26 - Uniform Continuity

The idea of uniform continuity is like a stronger, built on definition of Continuity. Because we think of continuous functions as "preserving the real line", this is a much stronger condition.

The question we'll first answer today, is given some $f : A \to R$ where $f$ is continuous on $A$ , then what sort of properties are preserved? For example, if $A$ is closed, open, compact, or connected, is the same true for $f (A) = {f (x) : x \in A}$ .

We could also ask this for things like Cauchy Sequences; namely if $(x_{n})$ is Cauchy, then is $f (x_{n})$ also Cauchy?

Continuity itself won't answer these questions. For example, consider $A = R$ (then $A$ is both open and closed). Then if we define $f : A \to R$ where $f (x) = \frac{1}{1 + x^{2}}$ :

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f(x) = 1/(1+x^2)

What is the range $f (A)$ ? Here it is $f (A) = (0, 1]$ which is neither open nor closed.

This gives us the following theorem:

Theorem

The continuous image of a compact set is also compact.

Proof

Let $f : K \to R$ be such that $f$ is continuous for all $c \in K$ , and $K$ is compact.

To show that $f (K)$ is compact, we'll take an arbitrary sequence in the set and show that it converges to a point in the set (see Closed iff Convergent Cauchy Sequences). Let $(y_{n}) \subseteq f (K)$ be such an arbitrary sequence. Since $y_{n} \in f (K)$ then $\exists x_{n} \in K$ where $f (x_{n}) = y_{n}$ . Since $K$ is compact, then $\exists (x_{n_{k}}) \subseteq K$ (a subsequence) such that $x_{n_{k}} \to x$ where $x \in K$ . Note that $f (x) \in f (K)$ automatically

But then since $f$ is continuous at $x \in K$ , then any sequence converging to $x$ like $x_{n_{k}}$ (like it does here) implies that $f (x_{n_{k}}) \to f (x)$ (so then we get the right immediately). But since $y_{n} = f (x_{n_{k}}) = y_{n_{k}}$ then $y_{n_{k}} \to f (x) \in f (K)$ as desired.

It is bounded trivially. Thus $f (K)$ must be compact.

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See Extreme Value Theorem for an immediate use of the theorem.

See Compactness Is Preserved By Continuity which this corollary comes from.

Extreme Value Theorem

If $f$ is continuous on a closed interval then $f$ attains an absolute maximum and an absolute minimum in that interval.

Proof

Let $a < b \in R$ . Take $f : [a, b] \to R$ to be continuous. By the Heine-Borel Theorem (Compact Equivalences) then our $[a, b]$ is compact. Therefore, by Compactness Is Preserved By Continuity then $f ([a, b])$ is compact. Compact gives that $f ([a, b])$ must be bounded, so then we can define $m := inf f ([a, b])$ and $M := sup f ([a, b])$ both exist.

$f ([a, b])$ are both closed (by being compact), so then our $m, M \in f ([a, b])$ . So then $\exists x_{1}, x_{2} \in [a, b]$ where $f (x_{1}) = m$ and $f (x_{2}) = M$ as desired (since these points $x_{1}, x_{2}$ ) are our minimum/maximum points.

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Back to Uniform Continuity

Before jumping into the definition of uniform continuity, let's look at some examples.

First let $A = (0, 1)$ . Let $f, g : A \to R$ where $f (x) = x^{2}$ and $g (x) = \frac{1}{x}$ . Both functions $f$ and $g$ , by the Algebraic Continuity Theorem, are continuous for each $c \in A = (0, 1)$ .

Now let's consider $ε - δ$ proofs of this (even though we've shown they are continuous). Let $c \in A = (0, 1)$ . To show continuity we'll want to let $ε > 0$ . What $δ$ do we need to pick such that the $ε$ challenge can be satisfied? In these cases:

For $f$ we'd need:

\begin{aligned} | f (x) - f (c) | & = | x^{2} - c^{2} | \\ = | x + c | | x - c | \\ < 2 | x - c | & (| x + c | < | 1 + c | = 2) \end{aligned}

Then to use this we'd need $2 | x - c | < 2 δ = ε$ implies $δ = \frac{ε}{2}$ .

For $g$ we'd need:

\begin{aligned} | g (x) - g (c) | & = | \frac{1}{x} - \frac{1}{c} | \\ = | \frac{x - c}{x c} | \\ = \frac{| x - c |}{x c} \end{aligned}

We'd like to get to $< \frac{| x - c |}{k}$ where $k$ doesn't depend on $x$ . Since $x$ approaches $c$ , it suggests that $x$ is within $\frac{c}{2}$ of $c$ . Then if we suppose $| x - c | < δ < \frac{c}{2}$ then $x > \frac{c}{2}$ . So then $\frac{1}{x} < \frac{2}{c}$ thus we have:

\begin{aligned} < \frac{| x - c |}{\frac{c^{2}}{2}} \\ = ε \end{aligned}

Thus we want $δ < \frac{c}{2}$ and $δ < \frac{c^{2}}{2} ε$ so then we want $δ = min {\frac{c}{2}, \frac{c^{2}}{2} ε}$ .

What's important here is that $δ$ for $f$ was independent of $c$ , meanwhile $δ$ for $g$ was dependent on $c$ (for all cases). The idea is having the independence is some stronger form of continuity. Uniform continuity.

Thus we can get into the definition:

uniform continuity

Let $f : A \to R$ is uniformly continuous on $A$ if $\forall ε > 0$ $\exists δ > 0$ such that $x, y \in A$ has:
$| x - y | < δ \Rightarrow | f (x) - f (y) | < ε$

In contrast, Continuity itself is continuous at a point $y$ specifically. Having to consider the $y$ may imply that $δ$ depends on that $y$ . Here, we don't care about specific points it is continuous at.

Theorem

A continuous function on a compact set is uniformly continuous.

The idea here is that in our example in Lecture 26 - Uniform Continuity#Back to Uniform Continuity the function $f$ would be still continuous if $A = [0, 1]$ instead (same definition of $f$ , just include the endpoints on the interval). In contrast, $g$ couldn't get that same extension.

Proof

Let $f : K \to R$ be continuous, with $K$ being compact.

To show $f$ is uniformly continuous, follow the definition. Let $ε > 0$ . Since $f$ is continuous, then $δ > 0$ where $\forall x \in K$ then picking $\forall x^{'} \in K$ has $0 < | x - x^{'} | < δ \to | f (x) - f (x^{'}) | < \frac{ε}{2}$ .

The problem here is that $δ$ may change with each $x$ , so first notice that if I consider the following collection (we denote the $δ$ we choose for each $x$ as $δ (x)$ ):
${V_{\frac{δ (x)}{2}} (x) | x \in K}$
This is an Open Cover of $K$ . By the Heine-Borel Theorem (Compact Equivalences) then there is a finite subcover $\exists x_{1}, x_{2}, \dots, x_{n}$ such that:
$K \subseteq V_{\frac{δ (x)}{2}} (x_{1}) \cup \dots \cup V_{\frac{δ (x_{n})}{2}} (x_{n})$
Now choose $δ = min {\frac{δ (x_{1})}{2}, \dots, \frac{δ (x_{n})}{2}}$ . Let $x, y \in K$ where $| x - y | < δ$ . Consider $x \in K$ then by our construction then $\exists x_{k}$ such that $x \in V_{\frac{δ (x_{k})}{2}} (x_{k})$ . Therefore, using the given for being continuous then:
$\Rightarrow | x - x_{k} | < \frac{δ (x_{k})}{2} < δ (x_{k}) \Rightarrow | f (x) - f (x_{k}) | < \frac{ε}{2}$

Now notice:
$\begin{aligned} | y - x_{k} | & \leq | y - x | + | x - x_{k} | \\ < δ + \frac{δ (x_{k})}{2} \\ \leq \frac{δ (x_{k})}{2} + \frac{δ (x_{k})}{2} \\ = δ (x_{k}) \\ \Rightarrow | f (y) - f (x_{k}) | < \frac{ε}{2} & Def’n Continuity \end{aligned}$
Now using that we got $| f (x) - f (x_{k}) | < \frac{ε}{2}$ then:
$\begin{aligned} | f (x) - f (y) | & \leq | f (x) - f (x_{k}) | + | f (x_{k}) - f (y) | \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}$

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