# MATH 409 Lecture 12

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## Uniform Continuity

A function defined on a set is called **uniformly continuous** on if for every there exists such that and imply .

First introduced by Cauchy with all his work on 's and 's.

Recall that is continuous at a point if for every there exists such that and imply .

Therefore the uniform continuity of is a stronger property than the continuity of on .

### Examples

**Constant function.** is uniformly continuous

Indeed for any and .

**Identity function.**

Since , we choose and we have and .

**Sine function.**

It was shown in the previous lecture that for all , therefore whenever

### Lipschitz Functions

A function is called a **Lipschitz function** if there exists a constant such that for all .

**Theorem.** Any Lipschitz function is uniformly continuous.

*Proof.* Using notatino of the the definition, let for any , then implies

for all .

This is nice because we don't have to work with 's and 's to show that a function is uniformly continuous.

**Theorem.** The function is uniformly continuous on , but not Lipschitz.

*Proof.* For any , we have . It follows that is not Lipschitz.

Given , let In order to change the function by , we can only choose a difference in and smaller than .

Suppose , where . To estimate , we consider two cases:

- : we use the fact that is strictly increasing. Then . Since the function is strictly increasing, the farthest that and can be is at the extremes of an interval. Note that is not included in this interval, hence the strict "less than".
- : we have . Then .

Thus is uniformly continuous.

**Theorem.** is not uniformly continuous on .

Let and an arbitrary . Let be a natural number such that . Further, let and .

(The numbers are large, but the distance between them is small)

Then while .

We conclude that is not uniformly continuous.

**Theorem.** The function is Lipschitz (and hence uniformly continuous) on any bounded interval .

*Proof.* For any we obtain

**Theorem.** Any function continuous on a closed bounded interval is also uniformly continuous on .

*Proof by contraposition.* Assume that is not uniformly continuous on . We have to show that is not continuous on ,

By assumption, there exists a tsuch that for any , we can find two points satisfying and . In particular, for any , there exist points such that while .

By construction, is a bounded sequence. According to the Bolzano-Weierstrass theorem, there is a subsequence converging to a limit . Moreover, belongs to as .

Since for all , the subsequence also converges to . However, the inequalities imply that at least one of the sequences and is not converging to . It follows that the function is not continuous at .

**Theorem.** Suppose that a function is uniformly continuous on . Then it maps Cauchy sequences to Cauchy sequences, that is, for any Cauchy sequence , the sequence is also Cauchy.

*Proof.* Let be a Cauchy sequence. Since is uniformly continuous on , for every there exists such that and imply .

Since is a Cauchy sequence, there exists (note that depends on , so ultimately depends on ) such that for all . Then for all . We conclude that is a Cauchy sequence.

Cauchy sequences are nice because every Cauchy sequence is a convergent sequence (and vice versa), so the image of a convergent sequence is also a convergent sequence that converges to the limit of the function.

## Dense Subsets

Given a set and its subset , we say that is **dense** in if for any point and any the interval contains an element of .

### Examples

- An open bounded interval is dense in the closed interval .
- The set of rational numbers is dense in (previously shown without notion of density.

**Theorem.** A subset of the set is dense in if and only if for any there exists a sequence converging to .

*Proof.* Suppose that for any point there is a sequence converging to . Then any -neighborhood of contains an element of that sequence.

Conversely, suppose that is dense in . Then, given , for any there is a point . Clearly as .

### Continuous Extension

*Main function of the day.*

**Theorem.** Suppose that is dense in . Then any uniformly continuous function can be extended to a continuous function on . Moreover, the extension is unique and uniformly continuous.

*Proof.* First let us show that a continuous extension of the function to the set is unique (assuming it exists). Suppose are two continuous extensions of . Since the set is dense in , for any there is a sequence converging to . Since and are continuous at , we get and as (by sequential characterization of continuity). However for all . Hence . This shows the uniqueness of the extension.

Now to prove existence. given , let be a sequence of elements of converging to . The sequence is Cauchy. Since is uniformly continuous, it follows that the sequence is also Cauchy. Hence it converges to a limit . We claim that the limit depends only on and does not depend on the choice of the sequence .

Indeed, let be another sequence converging to . Then a sequence also converges to . Consequently, is convergent. The limit is since the subsequence converges to . Another subsequence is , hence it converges to as well. Now we set , which defines a function .

The continuity of the function implies that for , that is, is an extension of .

Finally, we need to show that the extension is uniformly continuous.

Given , let . Since is uniformly continuous, there is such that implies for all . For any we can find sequences and of elements of such that and as .

By construction of , we have and as .

If , then for all sufficiently large . Consequently, for all sufficiently large , which implies (by comparison theorem of taking limit of inequality). Thus and is uniformly continuous.

The implication of this theorem is the definition of the exponential function. We will define it for rational exponents and then extend it to the real line.

## Exponential Functions

**Theorem.** For any , there exists a unique function satisfying the following conditions:

- for all
- is continuous at .

This function is denoted and called **the exponential function with base** .

*Sketch of proof of uniqueness.*
We know how to prove this function is not negative: at any point . Let's take and . Thus

(nonzero)

If it's continuous at 0 and strictly positive, we can show it must be continuous at any point.

*Sketch of proof of existence.*
Let , , and for all . Further, let for all .

**Lemma 1.** a^{m+n} = a^m \, a^n and for all .

**Lemma 2.** If and satisfy , then .

For any , let , where and are chosen so that .

**Lemma 3.** and for all .

**Lemma 4.** The function , , is monotone.

**Lemma 5.** as .

**Lemma 6.** The function for is uniformly continuous on for any bounded interval .