Understanding Uniform Continuity in Functions

1. Chapter 5 Limits and Continuity

2. Section5.4 Uniform Continuity

3. For a function f : Dto be continuous on D, it is required that for every x0  D and for every  > 0 there exists a  > 0 such that | f (x) – f (x0)|< whenever |x – x0| < and x  D. Note the order of the quantifiers:  may depend on both  and the point x0. If it happens that, given > 0, there is a > 0 that works for all x0 in D, then f is said to be uniformly continuous. Definition 5.4.1 Let f :D . We say that f is uniformly continuous on Dif for every  > 0 there exists a  > 0 such that | f (x) – f (y)|< whenever |x – y| < and x, y  D .

4. Example 5.4.2* We claim that the function f(x) = 3x is uniformly continuous on . Given any > 0, we want to make | f (x) – f (y)| <  by making x sufficiently close to y. We have | f (x) – f (y)| = |3x – 3y| = 3|x – y|. So we may take  =  /3. Then whenever |x – y| <  we have | f (x) – f (y)| = 3|x – y| < 3 = . We conclude that f is uniformly continuous on .

5. Example 5.4.4 The function f(x) = x2 is not uniformly continuous on . Let’s look at this graphically. Given a point x1 close to 0 and an -neighborhood about f(x1), the required  can be fairly large. f(x) f(x) = x2 But as x increases, the value of  must decrease. f(x2) This means the continuity is not uniform. -neighborhood f(x1) x1 x2 x

6. Example 5.4.4 The function f(x) = x2 is not uniformly continuous on . As a prelude to proving this, let’s write out the statement of uniform continuity and its negation. The function f is uniformly continuous on D if  > 0  > 0 such that  x, y  D, |x – y| <  implies | f (x) – f (y)| < . So the function f fails to be uniformly continuous on D if  > 0 such that  > 0,  x, y  D such that |x – y| <  and | f (x) – f (y)|. Suppose we take  = 1. (Any > 0 would work.) Now for the proof. We must show that given any > 0, there exist x, y in such that |x – y| <  and | f (x) – f (y)| = |x2 – y2| = |x + y|  | x – y|1. For any x, if we let y = x +  /2, then |x – y| =  /2 <  . Thus to make 1  |x + y|  | x – y| = |x + y|  ( /2), we need to have |x + y|  2/ . This prompts us to let x = 1/ . Here is the formal proof.

7. Example 5.4.4 The function f(x) = x2 is not uniformly continuous on . Proof: Then given any > 0, let x = 1/ and y = 1/+  /2. Let  = 1. Then |x – y| =  /2 <  , but | f (x) – f (y)| = |x2 – y2| = |x + y|  | x – y| Thus f is not uniformly continuous on . 

8. Example 5.4.5* The function f(x) = x2is uniformly continuous on D if D is a bounded set. For example, let D = [– 3, 3]. Then |x + y|  6. So given > 0, if  =  /6 and |x –y| <, we have | f (x) – f (y)| = |x2 – y2| = |x + y|  | x – y|  6| x – y | < 6 =  . This is a special case of the following theorem.

9. Theorem 5.4.6 Suppose f : D is continuous on a compact set D. Then f is uniformly continuous on D. Proof: Let > 0 be given. Since f is continuous on D, f is continuous at each x  D. Thus for each x  D, there exists a x > 0 such that | f (x) – f (y)| <  /2 whenever | x – y | < x and y  D. is an open cover of D. Now the family of neighborhoods F SinceD is compact, Fcontains a finite subcover. That is, there exist x1, …, xn in D such that Now let  = min If x, y D with |x – y| <  , then it can be shown (in the book) that there is some index i such that It follows that | f (x) – f (xi)| <  /2 and | f ( y) – f (xi)| <  /2, Open cover of D so that | f(x) – f(y)|  | f(x) – f(xi)| + | f(xi) – f(y)| <  /2 +  /2 = .  D

10. How does uniform continuity relate to sequences? Recall that the continuous image of a convergent sequence need not be convergent if the limit of the sequence is not in the domain of the function. For example, let f(x) = 1/x and xn = 1/n. Then (xn) converges to 0, but since f(xn) = n for all n, the sequence (f(xn)) diverges to +. But with uniform continuity we have the following: Theorem 5.4.8 Let f : D be uniformly continuous on D and suppose that (xn) is a Cauchy sequence in D. Then (f (xn)) is a Cauchy sequence. Proof: Given any > 0, since f is uniformly continuous on D there exists a > 0 such that | f(x) – f(y)|< whenever |x – y|< and x, y  D. Since (xn) is Cauchy, there exists N such that |xn – xm| <  whenever m, n  N. Thus for m, n N we have | f(xn) – f(xm)|< , so(f (xn)) is Cauchy. 

11. ~ ~ f f Using Theorem 5.4.8 we can derive a useful test to determine if a function is uniformly continuous on a bounded open interval. We say that a function : E is an extension of f : D  if DE and f(x) = (x) for all x  D. ~ ~ ~ ~ f f f f Theorem 5.4.9 A function f : (a, b) is uniformly continuous on (a, b) iff it can be extended to a function that is continuous on [a, b]. Proof: If f can be extended to a function that is continuous on the compact set [a, b], then is uniformly continuous on [a, b] by Theorem 5.4.6. It follows that (and hence f) is also uniformly continuous on the subset (a, b).

12. ~ ~ ~ ~ ~ ~ f f f f f f f(x) if a <x < b, p if x = a, q if x = b. (x) = Theorem 5.4.9 Proof: Conversely, suppose that f is uniformly continuous on (a, b). ~ f We claim that limxaf(x) and limxbf(x) both exist as real numbers. To see this, let (sn) be a sequence in (a, b) that converges to a. A function f : (a, b) is uniformly continuous on (a, b) iff it can be extended to a function that is continuous on [a, b]. Then (sn) is Cauchy, so Theorem 5.4.8 implies that (f(sn)) is also Cauchy. Theorem 4.3.12 then implies that (f(sn)) converges to some real number, say p. It follows (Theorem 5.1.10) that limxaf(x)=p. Similarly, we have limxbf(x)=q, for some real number q. Now define : [a,b]  by Then is an extension of f. Since f is continuous on (a, b), so is . But is also continuous at a and b, so is continuous on [a, b]. 