Chapter Limits and Continuity Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide Section 5.4 Uniform Continuity Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide For a function f : D → to be continuous on D, it is required that for every x0 ∈ D and for every ε > there exists a δ > such that | f (x) – f (x0)| < ε whenever | x – x0 | < δ and x ∈ D Note the order of the quantifiers: δ may depend on both ε and the point x0 If it happens that, given ε > 0, there is a δ > 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 ε > there exists a δ > such that | f (x) – f ( y)| < ε whenever | x – y | < δ and x, y ∈ D Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide Example 5.4.2* We claim that the function f (x) = 3x is uniformly continuous on Given any ε > 0, we want to make | f (x) – f ( y)| < ε by making x sufficiently close to y We have So we may take δ = ε /3 | f (x) – f ( y)| = | 3x – 3y| = 3| x – y| Then whenever | x – y | < δ we have | f (x) – f ( y)| = 3| x – y| < 3δ = ε We conclude that f is uniformly continuous on Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide Example 5.4.4 The function f (x) = x is not uniformly continuous on Let’s look at this graphically Given a point x1 close to and an ε -neighborhood about f (x1), f (x) f (x) = x the required δ can be fairly large But as x increases, the value f (x2) of δ must decrease This means the continuity is ε -neighborhood not uniform f (x1) x1 Copyright © 2013, 2005, 2001 Pearson Education, Inc x2 x Section 5.4, Slide Example 5.4.4 The function f (x) = x 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 ∀ ε > ∃ δ > such that ∀ x, y ∈ D, | x – y | < δ implies | f (x) – f ( y)| < ε So the function f fails to be uniformly continuous on D if ∃ ε > such that ∀ δ > 0, (Any ε > would work.) Suppose we take ε = Now for the proof | x – y | < δ and | f (x) – f ( y)| ≥ ε ∃ x, y ∈ D such that We must show that given any δ > 0, there exist x, y in such that | x – y | < δ and 2 | f (x) – f ( y)| = | x – y | = | x + y | ⋅ | x – y | ≥ For any x, if we let y = x + δ /2, then | x – y | = δ /2 < δ ≤ | x + y| ⋅ | x – y| we need to have | x + y | ≥ 2/δ = Thus to make | x + y | ⋅ (δ /2), This prompts us to let x = 1/δ Here is the formal proof Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide Example 5.4.4 The function f (x) = x is not uniformly continuous on Proof: Let ε = Then given any δ > 0, let x = 1/δ and y = 1/δ + δ /2 Then | x – y | = δ /2 < δ , but 2 | f (x) – f ( y)| = | x – y | = | x + y | ⋅ | x – y | = Thus f is not uniformly continuous on Copyright © 2013, 2005, 2001 Pearson Education, Inc 1 δ δ δ + + > ÷ ÷ = δ δ ÷ δ ♦ Section 5.4, Slide Example 5.4.5* The function f (x) = x is uniformly continuous on D if D is a bounded set For example, let D = [– 3, 3] Then | x + y | ≤ So given ε > 0, if δ = ε /6 and | x – y | < δ , we have 2 | f (x) – f ( y)| = | x – y | = | x + y | ⋅ | x – y | ≤ 6| x – y | < 6δ = ε This is a special case of the following theorem Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide Theorem 5.4.6 Suppose f : D → is continuous on a compact set D Then f is uniformly continuous on D Proof: Let ε > be given Since f is continuous on D, f is continuous at each x ∈ D Thus for each x ∈ D, there exists a δ > such that x | f (x) – f ( y)| < ε /2 whenever | x – y | < δ Now the family of neighborhoods F Since D is compact, F contains a finite subcover That is, there exist x1, …, xn in D such that { δ x / 2, K , δ x / 2} Now let δ = n and y ∈ D δ = N x ; x ÷: x ∈ D δ x1 D ⊆ N x1 ; is an open cover of D δ xn ÷∪ L ∪ N xn ; ÷ If x, y ∈ D with | x – y | < δ , then it can be shown (in the book) that there is some index i such that Open cover of D x | x − xi | < δ xi and | y − xi | < δ xi It follows that | f (x) – f (x )| < ε /2 and | f ( y) – f (x )| < ε /2, i i so that | f (x) – f ( y)| ≤ | f (x) – f (x )| + | f (x ) – f ( y)| i i < ε /2 + ε /2 = ε ♦ D Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide 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 ε > 0, since f is uniformly continuous on D there exists a δ > such that | f (x) – f ( y)| < ε whenever | x – y | < δ and x, y ∈ D Since (xn) is Cauchy, there exists N ∈ such that | xn – xm | < δ whenever m, n ≥ N Thus for m, n ≥ N we have | f (xn) – f ( xm)| < ε , so ( f (xn)) is Cauchy ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide 10 Using Theorem 5.4.8 we can derive a useful test to determine if a function is uniformly continuous on a bounded open interval ~ f We say that a function ~ f f (x) = :E→ is an extension of f : D → if D ⊆ E and (x) for all x ∈ D Theorem 5.4.9 A function f : (a, b) → ~ f to a function is uniformly continuous on (a, b) iff it can be extended that is continuous on [a, b] Proof: ~ If f can be extended to a function f ~ f It follows that then that is continuous on the compact set [a, b], is uniformly continuous on [a, b] by Theorem 5.4.6 ~ f f ) is also uniformly continuous on the subset (a, b) (and hence Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide 11 Theorem 5.4.9 A function f : (a, b) → ~ f to a function is uniformly continuous on (a, b) iff it can be extended that is continuous on [a, b] Proof: Conversely, suppose that f is uniformly continuous on (a, b) We claim that lim → f (x) and lim → f (x) both exist as real numbers x a x b 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 → f (x) = p.Similarly, we have x a Now define lim → f (x) = q, for some real number q x b ~ f (x) = f (x) if a < x < b, p if x = a, q But by ~ Then is an f extension of f Since f is continuous on (a, b), so is if x = b ~ is falso continuous at a and b, so ~ : [a, b] → f ~ f ~ is continuous on [a, b] ♦ f Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 5.4, Slide 12 ... Given a point x1 close to and an ε -neighborhood about f (x1), f (x) f (x) = x the required δ can be fairly large But as x increases, the value f (x2) of δ must decrease This means the continuity... 5.4.9 A function f : (a, b) → ~ f to a function is uniformly continuous on (a, b) iff it can be extended that is continuous on [a, b] Proof: ~ If f can be extended to a function f ~ f It follows... < ε So the function f fails to be uniformly continuous on D if ∃ ε > such that ∀ δ > 0, (Any ε > would work.) Suppose we take ε = Now for the proof | x – y | < δ and | f (x) – f ( y)| ≥ ε ∃ x,