Math 316, Intro to Analysis: Uniform continuity Today we talk about uniform continuity. Before we begin let’s see a non-example. In order to illustrate what uniform continuity means start with a f : U → R. Saying that f is continuous on U means that

∀p ∈ U, f is continuous at p Expand out the definition of “f is continuous at p.”

∀p ∈ U On what variables can δ depend?

1 • Below is the proof that f(x) = is continuous on (0, ∞). Fill in the choice of δ which x completes the proof.

Proof. Consider any a ∈ (0, ∞) and any  > 0 Let δ = δ(a, ) = . Assume that |x − a| < δ Then, using that a > 0

1 1 |a − x| (1) − = < x a |xa| If we had a lower bound on |x| depending on a and δ then we could try to set the thing on the right for  and get our choice for δ. Let’s get that lower bound. Using the :

|a| = |x+(a−x)| ≤

so that |x| ≥ , and we have that lower bound. Replacing the |x| in (1) will result in something difficult to solve for δ. Instead let’s be sloppy. Our choice of δ doesn’t need to be optimal. What is a choice for δ which will guarantee that this lower bound is positive.

δ = produces the bound |x| ≥ . Thus we get that:

1 1 |a − x| − = < x a |xa| Thus, for all  > 0 there is a δ > 0 (Namely δ = ) such that if

1 1 |x − a| < δ then − < . This function is continuous. x a  Our proof above has lots of the scratchwork still contained. Take a moment and write down the proof below 1 Claim: The function f : (0, ∞) → R given by f(x) = x is continuous Consider any a ∈ (0, ∞) and any  > 0. Let δ = . Suppose that x ∈ (0, ∞) and that |x − a| < δ. 1 2

Then |x| =

So that |f(x) − f(a)| =

So that for all a ∈ (0, ∞ and any  > 0 there is a δ = such that if |x − a| < δ then |f(x) − f(a)| < . We conclude that f is continuous. The proof above gives δ as a function depending on a and on . If you fix an  then what can you say about lim δ(a, )? a→0

This function is the prototypical nonexample of uniform continuity. Definition 1. Let f : U → R be a function. f is called uniformly continuous on U if for all  > 0 there is a δ > 0 such that for all a, x ∈ U, if |x − a| < δ then |f(x) − f(a)| < . The idea is that for a function to be uniformly continuous δ cannot depend on a. In the below write down with all the formal quantifiers the definitions of continuity on a set U and of uniform continuity on U

• f is continuous on U means:

• f is uniformly continuous on U means:

What is the difference between the two definitions? On what can δ depend? Examples: • We claimed that f(x) = 1/x is not uniformly continuous on (0, ∞). Our argument above does not prove this. Indeed some critic might point out that there may be a better choice of δ which did not depend on a and were were not clever enough to find it. Let us vindicate ourselves. 1 Claim f(x) = x is not uniformly continuous on (0, ∞). By negating the definition of uniform continuity, discover what we need to prove:

Let  = (we get to choose ) consider any δ > 0. We must find x, a ∈ (0, ∞) such that |x − a| < δ and |f(x) − f(a)| ≥  = 1. You’ll need to fill in the choice of . I will give you x and a. 1 1 Consider any n ∈ and let x = and a = . Then N n n + 1 n n

|xn − an| = . 3

Since this goes to 0 as n goes to ∞ there is an n such that |xn − an| < δ. Now we compare |f(xn) − f(an)| with  to contradict the definition of uniform continuity.

|f(xn) − f(an)| = |n + 1 − n| = ≥ . Thus f is not uniformly continuous. • For you: Prove that f(x) = x2 is not uniformly continuous on (0, ∞). Do so by letting  = 1, considering any δ, letting xn = n and yn = n + δ/2 and then seeing 2 2 what happens to |xn − yn| as n → ∞.

1 • We want an example of uniform continuity. Let f : [1, ∞) → be given by f(x) = . R x Show that f is uniformly continuous on [1, ∞). 1 In order to motivate the proof notice that when we proved was continuous we x used δ0 = . I am saying δ0 since this δ is not good enough for uniform continuity: It depends on a. Can you think of a positive quantity independent of a which is less than δ for all a ∈ [1, ∞)? Use this result for the δ in the following proof. 1 Prove that f(x) = is continuous on [1, ∞). x Consider any  > 0. Let δ = δ() = . Consider any x, a ∈ [1, ∞] Assume that |x − a| < δ.

Then using that x ≥ 1 and a ≥ 1 we see that |f(x) − f(a)| =

Thus, for all  > 0 there exists a δ = such that for all x, a ∈ [1, ∞), if |x − a| < δ then |f(x) − f(a)| < . f is uniformly continuous on [0, ∞).

So that |f(x) − f(a)| <  and f is uniformly continuous. 4

In fact we can find lots of uniformly continuous functions Theorem 2 (Continuity on a compact set =⇒ uniform continuity). Let f : U → R be a on a compact set. Then f is uniformly continuous. Proof. The proof will be motivated by the argument that x2 is not uniformly continuous on (0, ∞). If f is not uniformly continuous then perhaps we can find a pair of sequences which sees the failure of uniform continuity. Suppose that f is not uniformly continuous. Then there exists a  > 0 such that for all δ > 0 there is a pair x, y such that and . We want to get sequences in play. Take the  from the line above. We have a conclusion 1 for all δ. Let δ = . Then there exist x , y ∈ U such that |x − y | < δ = n n n n n n n and |f(x) − f(y)| ≥ . Regard (xn) and (yn) as sequences. Since U is compact we see that

.

Let x = lim xkn Since |xkn − ykn | < → and xkn → x, ykn → .

But then by continuity f(xkn ) → and f(ykn ) → . Since  > 0, this means that there is an N such that for all n > N

|f(xkn ) − f(x)| <

|f(ykn ) − f(x)| < So that

|f(xkn ) − f(ykn )| =

But this contradicts the assumption that |f(xn) − f(yn)| >  for all n. f must be uniformly continuous!  Why should we bother studying uniformly continuous functions? How are they better than continuous functions? What good does uniform continuity hypothesis do us? For one thing, uniformly continuous functions preserve the property of boundedness. Theorem 3. If U is a bounded set and f : U → R is uniformly continuous then f(U) is bounded. Proof. Let U be a bounded set, and f : U → R be uniformly continuous. Suppose that f(U) is not bounded. Thean in particular for any natural number n ∈ N, n is not an upper bound and

Since U is bounded xn has a convergent subsequence xnk . Since U is not assumed to be closed, we can’t do what we might like to to and try to evaluate f at the limit point, since it might not be in the domain.

Instead we simply remark that xnk is Cauchy. Since f is uniformly continuous, for any  > 0, there exists a δ() > 0 such that for all x, y ∈ U, if then . 5

Take  = 1 and appeal to the fact that xnk is Cauchy: Since δ(1) > 0 there exists an N such that if k, ` > N then , so that . In particular, let ` = N + 1, then for all k > N we see that

But then f(xnk ) = ynk is a bounded sequence, but ynk → ∞, a contradiction. Thus it must be that f(U) is bounded,  Nice properties of uniform continuity. With the rest of today we will make whatever progress we can on the following facts. It is entirely possible that we will not make it to these at all. (1) If f : U → R is uniformly continuous and un is a of real numbers, then f(un) is Cauchy. (We even proved this while we were proving Theorem 3.) (2) If p is a limit point of U and f : U → is uniformly continuous then lim f(x) exists. R x→p (3) Continuous extension For any set U ⊆ R let U be the union of U with its set of limit points. We call this set the Closure of U. If f : U → R is uniformly continuous then there is a continuous function F : U → R such that F (u) = f(u) for all u ∈ U. A topological question to meditate on: If U is bounded then is U compact? If so then we have a strong correspondence between uniform continuity and compactness for bounded sets.