Math 28 Spring 2008 Exam 2

Math 28 Spring 2008: Exam 2 Instructions: Each problem is scored out of 10 points for a total of 40 points. You may not use any outside materials(eg. notes or books). You have 50 minutes to complete this exam. Problem 1. (a) Let f : A → R where A ⊂ R. State the definition for f to be uniformly continuous on A. (b) Which of the following functions are uniformly continuous on [0, ∞)? (i) f (x) = sin(x2 ) (ii) f (x) = 1 x+1 Proof. (a) A function f : A → R is uniformly continuous on A if for every ² > 0 there exists a δ > 0 such that for every x, y ∈ A, |x − y| < δ implies that |f (x) − f (y)| < ². (b) (i) No. Let the sequence of points (xn ) be defined by xn = 2nπ and the sequence of points (yn ) be defined by yn = 2nπ. Then we have r ¯ ¯ ¯ √ π ¯¯ ¯ |f (xn ) − f (yn )| = ¯f ( 2nπ) − f ( 2nπ + )¯ = |sin(2nπ) − sin(2nπ + π/2)| = 1. 2 Also notice that |xn − yn | → 0 as n → ∞. So for any ² ≤ 1 and for any δ > 0 there exists an N ∈ N such that |xn − yn | < δ, but we have that |f (xN ) − f (yN )| = 1 ≥ ². (ii) Yes. Let ² < 0 and let δ = ² Then for x, y ∈ A with |x − y| < δ we have ¯ ¯ ¯ ¯ ¯ 1 ¯ 1 ¯¯ ¯¯ y−x ¯ ¯ − = ¯ 1 + x y + 1 ¯ ¯ (1 + x)(1 + y) ¯ |x − y| = δ = ². ≤ 1 where the inequality comes from the fact that x, y ∈ [0, ∞) and so 1 1 1+x , 1+y ≥ 1. Problem 2. Let C be the Cantor set on [0, 1] obtained in the standard way by successively removing the middle third of each interval. Define g : [0, 1] → R by ( 1 x∈C g(x) = 0 x 6∈ C. (a) Show that g is discontinuous at every point in C. (b) Show that g is continuous at every point not in C. Proof. We will use the topological criterion for continuity. 1 (a) Let c ∈ C and let ² = 12 . Then for every δ > 0, the neighborhood Vδ (c) in not a subset of C (since we proved in class that C contains no intervals). Thus there exists a point x ∈ Vδ (c) with x 6∈ C and hence g(x) = 0 6∈ V² (g(c)). (b) Let c 6∈ C and let ² > 0. Since C we proved in class that is closed, its complement is open. Hence, there exists a δ > 0 such that Vδ (c) ⊆ C c . Take any x ∈ Vδ (c), then x ∈ C c and hence g(x) = 0. Hence we have x ∈ Vδ (c) implies g(x) ∈ V² (g(c)) and g is continuous at c. Problem 3. (a) State the Generalized Mean Value Theorem. (b) Let f : R → R be a differentiable function and suppose that f 0 is bounded. Show that f is uniformly continuous. Proof. (a) If f and g are continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c ∈ (a, b) where (f (b) − f (a))g 0 (c) = (g(b) − g(a))f 0 (c). If g 0 (x) is never 0 on (a, b), then the conclusion can be stated as f (b) − f (a) f 0 (c) = . 0 g (c) g(b) − g(a) (b) We are given that f 0 (x) is bounded. So let M > 0 be a bound for f 0 (x). Let ² > 0 and let δ = ² M. Consider x > y ∈ R such that |x − y| < δ. Then we have by the Mean Value Theorem that there exists a c ∈ (x, y) such that f (y) − f (x) = f 0 (c) y−x and hence |f (y) − f (x)| =M |y − x| so we have |f (y) − f (x)| = M |y − x| < M ² = ². M Problem 4. (a) State the definition for a function f : A → R to be differentiable on an interval A. 2 (b) Let f : [−1, 1] → R be the function defined by ( x2 sin x12 f (x) = 0 x 6= 0 x = 0. Show that f is differentiable, but that its derivative is unbounded. Proof. (a) Let f : A → R be a function defined on an interval A. Given c ∈ A, the derivative of f at c is defined by f (x) − f (c) f 0 (c) = lim x→c x−c provided the limit exists. If f 0 (c) exists for all be points c in A, then we say that f is differentiable on A (b) To show differentiable, we first note that x2 and sin x are continuous differentiable functions and hence their product is differentiable. Since sin x12 is defined everywhere but x = 0, we can use the standard differentiation rules for x 6= 0 and only need to consider the limit definition for the case x = 0. Consider f (x) − 0 1 lim = lim x sin 2 . x→0 x→0 x x This is bounded by 1 −x ≤ x sin 2 ≤ x x so by the Squeeze Theorem the limit approaches 0. Hence f 0 (0) also exists. The derivative is given by ( 2x sin x12 − f (x) = 0 0 2 x cos x12 x 6= 0 x = 0. where the case x 6= 0 is from standard differentiation rules (product rule and chain rule). To see that the derivative is unbounded, consider the sequence xn = √ 1 with (xn ) → 0 and (2n+1)π 1 1 1 1 lim − cos 2 = lim − p (−1) = lim p = ∞. n→∞ n→∞ n→ inf ty x x (2n + 1) (2n + 1) Hence we have values of f 0 (x) which can be arbitrarily large as x → 0. 3