Chapter 5 Limits and Continuity

Section 5.4 Uniform Continuity

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 D if for every  > 0 there exists a  > 0 such that | f (x) – f ( y)| <  whenever | x – y | <  and x, y  D .

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 .

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

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)|  . Now for the proof. Suppose we take  = 1. (Any  > 0 would work.) 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.

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

Example 5.4.5* The function f (x) = x2 is 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.

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. Now the family of neighborhoods F is an open cover of D. Since D is compact, F contains a finite subcover. That is, there exist x1, …, xn in D such that If x, y  D with | x – y | <  , then it can be shown (in the book) that there is some index i such that Now let  = min Open cover of D It follows that | f (x) – f (xi)| <  /2 and | f ( y) – f (xi)| <  /2, so that | f (x) – f ( y)|  | f (x) – f (xi)| + | f (xi) – f ( y)| <  /2 +  /2 = .  D

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. 

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 ~ 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]. f ~ 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. f ~ It follows that (and hence f ) is also uniformly continuous on the subset (a, b). 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]. f ~ Proof: Conversely, suppose that f is uniformly continuous on (a, b). We claim that lim x  a f (x) and lim x  b f (x) both exist as real numbers. To see this, let (sn) be a sequence in (a, b) that converges to a. 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 lim x  a f (x) = p. Similarly, we have lim x  b f (x) = q, for some real number q. Now define : [a, b]  by f ~ Then is an extension of f . f ~ (x) = f (x) if a < x < b, p if x = a, q if x = b. f ~ Since f is continuous on (a, b), so is . f ~ But is also continuous at a and b, so is continuous on [a, b].  f ~