Problem List MATH 5143 Fall, 2013

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1 Problem List MATH 5143 Fall, 2013 On any problem you may use the result of any previous problem (even if you were not able to do it) and any information given in class up to the moment the problem was assigned. 1. Show that for k N, k is rational only if k is a perfect square (that is, is the square of an element of N, so that in fact k N). Or, stated differently, k is irrational unless k is a perfect square. (The converse is trivial: if k = m 2 then k = m N, which is rational.) Begin your proof as follows, and use the known fact of number theory that every element of N can be expressed uniquely as an ordered product of primes (e.g., 10 = , 12 = , 360 = ): Let k, m, n N be such that ( m n )2 = k, and assume without loss of generality that m and n have no common factors. Let R be the least element of N so that if p 1, p 2,..., p R denote the first R prime numbers, then m = p m1 1 p m R n = p n1 1 pn R R, R, and k = p k1 1 pk R R, where at least one of m R, n R, and k R is nonzero. [Due Wednesday, August 28] 2. Show that if r is rational and s is irrational then both r+s and r s are irrational. [Due Wednesday, August 28] 3. Suppose x and y are any two real numbers, x < y. Show that there exists an irrational number between them. [Due Wednesday, August 28] 4. Ch 1, Ex 32. [Due Wednesday, August 28] 5. Complete the proof of Theorem 1.3.4(i) as follows. Fix an upper bound M of S once and for all (hypothesis 2). For any n N define b n by b n = M k 2 for the smallest k N {0} such that b n n is not an upper bound of S. a. Explain why k = 0 is impossible. b. Explain why k must exist (you must use a hypothesis and a property of R). By construction b n+1 = b n or b n+1 = b n + 1 2, hence b n+1 1 b 2 b 3 M. Also by construction for any n N, b n is an upper bound of S. n By Completeness Property there exists b R such that lim n b n = b. We will show that b = sup(s). c. Prove Claim 1: b n b for all n N. Hint. Proof by contradiction: what if b < b K for some K N? d. Prove Claim 2: b is an upper bound of S. Hint. Proof by contradiction. e. Explain how it follows from the claims that b n < b for all n N. f. Use Proposition to show that b = sup(s). Hint. b n b, b n < b, b n is not an upper bound, b is. [Due Wednesday, September 4]

2 2 6. Prove Theorem directly (not using Theorem 1.4.3) as follows. Let (x n ) be a Cauchy sequence in R. a. Use the Cauchy Property with ɛ = 1 to prove that (x n ) is bounded, i.e., that the set {x n : n N} is bounded. For each n N let s n = sup{x j : j n}. b. Show that s n actually exists and that (s n ) is a monotonic non-increasing sequence. If B is such that B x n B for all n N (B exists by part (a)), then B inf{x j : j n} sup{x j : j n} = s n, hence by Property the sequence (s n ) has a limit x R. We will show that (x n ) converges to x. Hence let ɛ > 0 be given. c. Show that there exists N 1 N such that if m, n N 1 then x m x n < ɛ 3. (1) d. Show that there exists N 2 N, N 2 N 1 such that e. Show that there exists K N, K N 2 such that x s N2 < ɛ 3. (2) x K s N2 < ɛ 3. (3) f. Use these estimates to show that for sufficiently large n (specify how large) x x n < ɛ, as required. [Due Wednesday, September 4] 7. In a metric space (M, d) (actually, in any topological space) prove that a set A is open if and only if A = Int(A). You may assume that Int(A) is defined either by Definition or by Int(A) = {U : U is open and U A}. [Due Wednesday, September 4] 8. Ch 2, Ex 8. [Due Wednesday, September 11] 9. Ch 2, Ex 20a. The set D(A, ɛ) is called the ɛ-neighborhood of the set A. When A is just a single point x, then it is what we have denoted D(x, ɛ). [Due Wednesday, September 11] 10. Let (x n ) be a sequence in R. a. Prove that (x n ) converges only if it has a unique cluster point. (There are really two assertions being made here.) Work from the definitions of the concepts involved, not theorems. b. Either prove the converse, or produce a counterexample. (To produce a counterexample means two things: (i) exhibit the sequence and (ii) prove that it has the properties desired.) [Due Wednesday, September 11] 11. Ch 3, Ex 1, only compactness (not connectedness). You may quote any theorem in this chapter. [Due Wednesday, September 18]

3 3 12. Ch 3, Ex 4. [Due Wednesday, September 18] 13. Ch 3, Ex 5. [Due Wednesday, September 18] 14. Ch 2, Ex 20a, second attempt. Hint. In order to use the triangle inequality you will need to work with points of M, since there is no triangle inequality for distances from points to sets (unless you formulate one and prove it). Use the analogue of Proposition for infima to get your hands a point you need, but note that the point is in M but the infimum is a number in R, not M. 15. Ch 4, Ex 1a. 16. Let (M, d) and (N, ρ) be metric spaces, let A be a subset of M, let f be a mapping from A into N, and let x 0 be a point in A. Consider the following two statements, the first of which is our definition of continuity of the mapping f at the point x 0 : and ( ɛ > 0) ( δ > 0) (x A & d(x, x 0 ) < δ) ρ(f(x), f(x 0 )) < ɛ (1) ( ɛ > 0) ( δ > 0) (x A & d(x, x 0 ) δ) ρ(f(x), f(x 0 )) ɛ. (2) Show that statements (1) and (2) are equivalent: if statement (1) is true, then statement (2) is true, and if statement (2) is true, then statement (1) is true. Remark. Your exposition will probably be much easier if the ɛ and δ in display (1) are subscripted with a 1 and those in display (2) are subscripted with a Let (M, µ), (N, η), and (P, ρ) be metric spaces. Let f : A M N and g : B N P be mappings, and let x 0 be a point in dom(g f). Show that if f is continuous at x 0 and g is continuous at f(x 0 ) then g f is continuous at x 0. Use the definition of continuity of a mapping at a point that was given in class, display (1) in the previous problem. 18. Let A, B, and P be sets and f and g mappings from A to B and from B to P, respectively. Prove that for any subset U of P, (g f) 1 (U) = f 1 (g 1 (U)). 19. Suppose M is a metric space and f : A M R is continuous at x 0 A. Show that if f(x 0 ) 0 then there is a neighborhood U of x 0 such that for all x in A U, f(x) 0. Without loss of generality you may assume that a := f(x 0 ) > Prove that every continuous self-map of an interval has a fixed point. That is, prove that if f : [a, b] [a, b] is continuous then there exists c [a, b] such that f(c) = c. Note that it is not assumed that f maps onto [a, b]; that is, it is not assumed that every y [a, b] is the image of some x in [a, b]. 21. Let (M, d) and (N, ρ) be metric spaces and f : A M N. Then f is

4 4 i. Lispchitz continuous at x 0 A if there exist δ, K R + such that d(x, x 0 ) < δ implies ρ(f(x), f(x 0 )) K d(x, x 0 ); ii. locally Lipschitz on A if it is Lipschitz continuous at every x 0 A; and iii. Lipschitz continuous on A (or just Lipschitz on A) if there exists K R + such that for all x 1, x 2 A, ρ(f(x 1 ), f(x 2 )) K d(x 1, x 2 ). Note that in the third case there is one uniform Lipschitz constant K, and it works for every pair of points in the whole set A regardless of their distance apart. a. Show that f(x) = 3 x is not Lipschitz continuous at x 0 = 0. b. Show that for every a > 0, f(x) = 3 x is Lipschitz on [a, ). You may use the Mean Value Theorem if you like, but I would prefer a direct proof. 22. Ch 4, Exercise Ch 4, Exercise 21. (For each function you must either prove uniform continuity or prove that the condition for uniform continuity fails.) 24. Ch4, Exercise 24a. The context of the problem is that (M, d) and (N, ρ) are metric spaces. Avoid a proof by contradiction. [Due Wednesday, October 9] 25. Ch 4, Exercise 24b. Additionally, construct a concrete example of a continuous map f : A M N and a Cauchy sequence (x k ) in A such that (f(x k )) is not a Cauchy sequence in N. [Due Wednesday, October 9] 26. Ch 4, Exercise 25. Give only the direct proof; do not apply Exercise 24c. You need only consider x = 0, since precisely the same ideas are used at x = 1. What this exercise is saying is that under the hypotheses, f can be continuously extended to [0, 1]. Hint. To say that f is bounded means that there exists a number M such that f (x) M for all x (0, 1). One idea for the proof is to sample f(x) near 0 so as to create a Cauchy sequence that converges to what the limit L ought to be, then prove that the limit exists and is L. [Due Wednesday, October 9] 27. Prove the following proposition: a bounded function f : [a, b] R is integrable if and only if for every ɛ > 0 there exists a partition P of [a, b] such that U(f, P ) L(f, P ) < ɛ. (One direction is sketched out for you in the first paragraph of the proof of on page 221 of the text, but you must supply the details.) [Due Wednesday, October 16] 28. Prove the following proposition: if a bounded function f : [a, b] R is integrable then the function f is integrable. Hint. Use Problem 27. Employ the following notation: for a fixed partition P = {a = x 0, x 1,..., x N 1, x N = b} of [a, b], for j {0,..., N 1}: M j = sup{f(x) : x [x j, x j+1 ]},

5 5 M j = sup{ f(x) : x [x j, x j+1 ]}, m j = inf{f(x) : x [x j, x j+1 ]}, and m j = inf{ f(x) : x [x j, x j+1 ]}. One idea is to compare upper and lower sums for f and f for the same partition P ; it could help to divide the set of indices {0,..., N 1} into three subsets: D = {j : M j m j < 0}; E = {j : m j 0}; and N = {j : M j 0}. Bonus. Can you find a counterexample to the reverse implication? [Due Wednesday, October 16] 29. Define f : R R by f(x) = 1 if x = 1 n for some n N, f(x) = 0 otherwise. Note that f is discontinuous at infinitely many points in [0, 1] and is neither increasing nor decreasing on [0, 1]. Show that nevertheless 1 f exists and find 0 its value. [Due Wednesday, October 16] 30. Suppose f is continuous on [a, b]. Prove that there must exist a number ξ [a, b] such that b [Due Wednesday, October 16] a f(x) dx = f(ξ)(b a). 31. The version of the Fundamental Theorem of Calculus given in class was: If f is continuous on [a, b] then for every number c [a, b] the function F c (x) defined by F c (x) def = x f(u) du exists and is continuous on [a, b], differentiable c on (a, b), and satisfies F c(x) = f(x) for all x (a, b). a. Use this theorem to show that the function L : (0, ) R defined by the integral L(x) = x 1/u du is differentiable on (0, ) and to compute its derivative. 1 b. For any positive real number a and any rational number r the number a r can be defined purely in terms of algebra, without recourse to limits. Prove that for all such a and r, L(a r ) = rl(a). Hint. Compare the derivatives of the functions L(x r ) and rl(x). [Due Wednesday, October 23] 32. Assume the truth of the statements in the previous problem. a. Show that L(4) > 1 by choosing a partition P of [1, 4] such that the lower sum based on P satisfies L( 1 u, P ) > 1 and using the definition of integrability. b. Use part (a) of this exercise and part (b) of the previous exercise to show that for any positive real number M there exist numbers x ± M (0, ) such that L(x M ) < M and L(x+ M ) > M. c. Prove that L maps (0, ) onto R. [Due Wednesday, October 23]

6 6 33. Assume the truth of the statements in the previous two problems. a. Use the result of Problem 31(a) and Theorem (which is true if a is or if b is ) to show that L has an inverse function E. Explicitly give the domain and range of E. b. Explain why, for a any positive real number and r any irrational number, the number E(rL(a)) exists and is a suitable definition for the symbol a r. [Due Wednesday, October 23] 34. Ch 5, Exercise 2abc. What is meant is the following: first determine whether or not the sequence (f k ) of functions has a pointwise limit on the set indicated, call it A. If there is no pointwise limit on A, prove it. If there is a pointwise limit f then prove either that the convergence to f is uniform on A or that it is not uniform on A. State whether f is continuous on A or not (but there is no need to prove your assertion on this point). [Due Wednesday, October 23] 35. Ch 5, Exercise 4. [Due Wednesday, October 30] 36. Ch 5, Exercise 8. That is, suppose that K is a compact subset of a metric space M with metric d, that for each n N the function f n : K R is continuous, and that there is a continuous function f : K R such that (f n ) converges to f pointwise on K. Either prove that (f n ) converges to f uniformly on K, or construct a counterexample. [Due Wednesday, October 30] 37. Ch 5, Exercise 18 [Due Wednesday, October 30] 38. a. What hypothesis in Theorem was omitted when it was stated in class? b. Indicate the exact point in the proof presented in class where the omitted hypothesis is needed, and was used implicitly. [Due Monday, November 4] 39. Ch 6, 3, Exercise 1 (on page 338). Use Definition to prove differentiability (which implies continuity, which you therefore need not consider), and do not quote any general theorems in your discussion. [Due Wednesday, November 6] 40. Ch 6, Exercise 1. [The authors give a hint in the back of the book. You may use Theorem ] [Due Wednesday, November 6] 41. Ch 6, Exercise 5abcdeg. [You may assume that the derivative exists, making this as simple as it appears: just compute the matrix of first partial derivatives of the appropriate dimension.] [Due Wednesday, November 13] 42. Suppose f : A R n R m is differentiable at x 0 IntA. Prove that for every v R n, lim t 0 1 t [f(x 0 + tv) f(x 0 )] exists and is Df x0 v.

7 Hint. First establish the result for v = 0. Then use the linearity of Df x0 to establish that lim 1 t 0 t [f(x 0 + tv) f(x 0 )] Df x0 v exists and is zero based on existence of Df x0, the alternative display f(x 0 + h) f(x 0 ) Df x0 h lim h 0 h for Definition 6.1.1, and the fact that h can be chosen to be tv. (Additional explanation and justification will be needed at several stages in this argument.) [Due Wednesday, November 13] 43. Ch 6, Exercise 31. Explicitly identify the derivative in your solution. Remarks. The norm on C(A, R) is f = sup{ f(x) : x A} (see the first few paragraphs of 5.5). The definition of the derivative is Definition on page 328, which makes sense because the only properties of R m and R n used in the definition are that they are normed vector spaces, which is true of C(A, R) and R (absolute value is a norm). [Due Wednesday, November 13] 44. Ch 6, Exercise 32. [Due Wednesday, November 13] 45. For f, g : R R, if f (x) = g (x) for all x then there exists c R such that f(x) = g(x) + c for all x. Prove or provide a counterexample for the analogous statement for f, g : R n R m. 46. Consider Taylor s Theorem, Theorem 6.8.5, in the one-dimensional case with the roles of x and y reversed (this is just a notational convenience for the next problem), y = 0, and A an open interval that contains 0. The theorem then says that if f has k + 1 continuous derivatives on A then for any x A f(x) = f(0) + f (0)x + 1 2! f (0)x k! f (k) (0)x k + 1 (k+1)! f (k+1) (ξ)x k+1 where ξ is a number between 0 and x. Suppose there exists a number R > 0 and a function f on an open interval U that contains [ R, R] such that (i) f is C 2 on U (which means that both f and f exist and are continuous on U) and (ii) f(0) = f (0) = 0. Use the theorem to show that there exists a constant C > 0 such that the estimate f(x) Cx 2 holds on [ R, R]. 47. Ch 6, Exercise 41. Prove that your example truly is an example and explain how you found it. (There are examples that are quite simple, once you find them.) Hint. Problem Ch 4, Exercise 45. There is a long hint at the back of the text which you may use. 49. Write out the terms through order three of the Taylor series of f(x, y) = sin e x+y in two ways: a. by a formal manipulation using the series for the sine and exponential functions; 7

8 8 b. using Taylor s Theorem, and in this case including a remainder term. 50. Ch 7, Exercise 3. [Due Wednesday, December 4] 51. Ch 7, Exercise 12. That is, prove the Inverse Function Theorem using the Implicit Function Theorem. The proof appears in many books, but you should not to look for it. Instead come up with the ideas on your own. Consideration of the one-dimensional case should give you the clue you need. Incidentally, the proof is pretty short, because all the hard work was done in proving the Inverse Function Theorem originally. Once you prove one of these two theorems, then you can get the other one readily, but one or the other has to be proved independently in order to avoid circularity. [Due Wednesday, December 4]