Lecture 17: Uniform Continuity and the Definition of the Derivative

18.100A: Complete Lecture Notes Lecture 17: Uniform Continuity and the Definition of the Derivative Uniform Continuity Recall 1 Recall the definition of continuity: f : S → R is continuous on S if ∀c ∈ S and ∀ > 0, ∃δ = δ(, c) > 0 such that ∀x ∈ S, |x − c| < δ =⇒ |f (x) − f (c)| < . Here, δ(, c) denotes the fact that δ can depend on  and c. Example 2 Consider the function f (x) = x1 . f is continuous on (0, 1). Proof : Let  > 0. Choose δ = min 1 < 2c . Therefore, Thus, |x| n c c2 2, 2  o . Suppose |x − c| < δ. Then, |x − c| < c 2 =⇒ |x| > c − |x − c| > 2c . 1 1 |x − c| − = x c |xc| δ < |x||c| 2 < 2δ c 2 c2  ≤ 2 = . c 2  As shown in the previous example. δ depended on both  and c. Definition 3 (Uniformly Continuous) Let f : S → R. Then, f is uniformly continuous on S if ∀ > 0, ∃δ = δ() > 0 such that ∀x, c ∈ S, |x − c| < δ =⇒ |f (x) − f (c)| < . Remark 4. Thus, in the definition of uniform continuity, δ only depends on ! Example 5 The function f (x) = x2 is uniformly continuous on [0, 1]. Proof : Let  > 0. Choose δ = 2 . Then, if x, c ∈ [0, 1] then |x − c| < δ implies that |x2 − c2 | = |x + c||x − c| ≤ 2|x − c| < 2δ = . 1  However, there are of course continuous functions that are not uniformly continuous. For example, we will show that f (x) = x1 is not uniformly continuous on (0,1), but first we consider the negation of the definition. Negation 6 (Not Uniformly Continuous) Let f : S → R. Then, f is not uniformly continuous on S if ∃0 > 0, ∀δ > 0 such that ∃x, c ∈ S with |x − c| < δ and |f (x) − f (c)| ≥ 0 . Proof : Choose 0 = 2 (in fact, any 0 > 0 will show that x1 is not uniformly continuous on (0, 1)). Then, let δ > 0.  Choose c = min δ, 21 and x = 2c . Then, |x − c| = 2c ≤ 2δ < δ and 2 1 1 1 1 1 = = ≥ 1 = 2. − − x c c c c 2 Theorem 7 Let f : [a, b] → R. Then, f is continuous if and only if f is uniformly continuous. Proof : ( ⇐= ) This direction is left as an exercise to the reader. ( =⇒ ) Suppose f is continuous and assume for the sake of contradiction that f is not uniformly continuous. Then, ∃0 > 0 such that for all n ∈ N, ∃xn , cn ∈ [a, b] such that |xn − cn | < 1 and |f (xn ) − f (cn )| > 0 . n By Bolzano-Weierstrass, ∃ a subsequence {xnk } of {xn } and x ∈ [a, b] such that limk→∞ xnk = x. Similarly, by Bolzano-Weierstrass, ∃ a subsequence {cnk } of {cn } and c ∈ [a, b] such that limk→∞ cnk = c. Note that subsequence {xnkj } of {xnk } satisfies limj→∞ xnkj = x. Then, 1 − 0. |x − c| = lim |xnkj − cnkj | ≤ lim j→∞ j→∞ nkj Thus, x = c. But, since f is continuous at c, 0 = |f (c) − f (c)| = lim |f (xnkj ) − f (cnkj )| ≥ 0 . j→∞ This is a contradiction. Derivative Definition 8 Let I be an interval, let f : I → R, and let c ∈ I. We say that f is differentiable at c if the limit lim x→c f (x) − f (c) x−c exists. 2 Notation 9 If f is differentiable at c, we write f (x) − f (c) . x→c x−c f 0 (c) := lim Furthermore, if f is differentiable at every c ∈ I, we write f 0 or df dx for the function f 0 (x). Example 10 Consider the function f (x) = ax + b. Then, for all c ∈ R, f 0 (c) = a. Proof : This follows as lim x→c f (x) − f (c) ax + b − (ac + b) x−c = lim = a lim = lim a = a. x→c x→c x − c x→c x−c x−c  Example 11 (The Power Rule) For all n ∈ N, if f (x) = αxn , then for all c ∈ R, f 0 (c) = αncn−1 . Proof : We note that for all n ∈ N, (x − c) n−1 X xn−1−j cj = j=0 n−1 X xn−j cj − j=0 n−1 X xn−1−j cj+1 . j=0 Letting ` = j + 1, we obtain (x − c) n−1 X j=0 xn−1−j cj = n−1 X xn−j cj − j=0 =x xn−` c` `=1 n−0 0 c −x n n X n−n n c n =x −c . Therefore, n−1 n−1 X X αxn − αcn = α lim xn−1−j cj = α cn−1−j cj = αncn−1 . x→c x→c x−c j=0 j=0 lim  3