Selected problems and solutions

Transcription

1 Selected problems and solutions 1. Prove that (0, 1) = R. Solution. Define f : (0, 1) R by 1 x , x < 0.5 f(x) = 1 x 0.5 2, x > 0.5 0, x = 0.5 Note that the image of (0, 0.5) under f is (, 0), the image of (0.5, 1) under f is (0, ), and f(0) = 0. This shows that f is surjective. Furthermore, it is easy to check that f is injective when restricted to (0, 0.5) or (0.5, 1). As the images of these intervals are disjoint from each other and 0, it follows that f is injective on (0, 1). So f is a bijection as desired. 2. Find an injective function f : (0, 1) 2 N by using binary expansions. Then define a surjective function g : 2 N (0, 1) using a similar idea. Though g is not one-to-one, it is two-to-one in a certain sense; why? Solution. Every x (0, 1) has a binary expansion in the sense that there exists a sequence {a n } n=1 such that x = a n 2 n n=1 where a n {0, 1} for all n N. (The expansion is typically written 0.a 1 a 2 a 3....) It may be assumed that the a n s do not have an infinite trail of 1 s; that is, that {a n } n=1 does not converge to 1; this ensures that the binary expansions are unique. Now define f : (0, 1) 2 N by f(x) = {n : a n = 1} where {a n } n=1 is the binary expansion of x (unique in the above sense). It is easy to see that f is injective. Now define g : 2 N (0, 1) as follows. If S N and S, S N, define g(s) = x where x has the binary expansion {a n } n=1 given by { 1, n S a n = 0, n / S Also define g( ) = 1/3 and g(n) = 2/3. It is easy to see that g is surjective. Furthermore g is two-to-one in the sense that {S : g(s) = x} 2 The latter follows from the following fact: every real number has at most two distinct binary expansions, and if a real number has two distinct binary expansions then it has a terminating binary expansion. (This was the reason for choosing 1/3 and 2/3 above; they have no terminating binary expansion.) 1

2 3. Prove that {( 1) n } n=1 is not convergent. Solution. We will show the sequence does not converge to a, where a is an arbitrary real number. Let ɛ = 1 and note that either 1 / (a ɛ, a + ɛ) or 1 / (a ɛ, a + ɛ) (or both). Let N be an arbitrary natural number; we will find n N such that a n a ɛ. If 1 / (a ɛ, a + ɛ) then n = 2N suffices; if 1 / (a ɛ, a + ɛ) then n = 2N + 1 suffices. This completes the proof. 4. Assume S R and that x is an accumulation point of S. Show that if x is also an upper bound of S, then x = sup S. Solution. We assume x is an upper bound of S, so it only must be shown that there is no smaller upper bound. Assume y is an upper bound of S such that y < x, and let ɛ = x y. Then (x ɛ, x + ɛ) is a neighborhood of x which contains no point of S, which contradicts the assumption that x is an accumulation point of S. 5. Suppose that {a n } n=1 is a sequence converging to a, and assume b is an accumulation point of {a n : n N}. Prove that a = b. Solution. Assume that a b, and let ɛ = b a /2. As b is an accumulation point of {a n : n N}, for any natural number N there exists n N such that a n (b ɛ, b + ɛ). (This is because every neighborhood of b contains infinitely many elements of {a n : n N}, so in particular must contain a n s with n N.) So for any N, there exists n N such that a n a ɛ, which means that {a n } n=1 does not converge to a, a contradiction. 6. Assume f : [a, b] R is increasing and define g : (a, b) R by g(x) = sup{f(z) : z < x}. Prove that g(c) f(c) for all c (a, b). Also prove that if g(c) f(c), then f does not have a limit at c. Solution. Let c (a, b). As f is increasing, f(c) is an upper bound of {f(z) : z < c}. Thus, g(c) = sup{f(z) : z < c} f(c). To prove the second statement in Problem 5 we show the contrapositive (that is, if f has a limit at c, then g(c) = f(c)). Assume f has the limit L at c, and let ɛ > 0. Choose δ > 0 such that x [a, b] and 0 < x c < δ f(x) L < ɛ (1) Choose u, v (a, b) such that c δ < u < c < v < c + δ. Then We justify the inequalities in (2) as follows: L ɛ < f(u) g(c) f(c) f(v) < L + ɛ (2) The first and fifth inequalities follow from the limit statement (1); The second inequality comes from the fact that f(u) {f(z) : z < c}; The third inequality comes from the first part of this problem; 2

3 The fourth inequality comes from the fact that f is increasing. As ɛ > 0 was arbitrary, (2) implies L = g(c) = f(c). 7. Give examples of: (i) A function f : R R which is unbounded in every open interval. (ii) A function f : [0, 1] R such that is countably infinite. S = {a [0, 1] : lim x a f(x) does not exist} Solution: (i) Define f : R R by { m, if x = n f(x) = m Q is in lowest terms 0, else and let I = (a, b) be a (bounded) interval with K = max{ a, b }. Assume f is bounded by M. Then n } Q I { m : m M, n MK (2M + 1)(2MK + 1) a contradiction since Q I is infinite. (ii) Define f : [0, 1] R by f(x) = { ( ] 1/n, x 1 n+1, 1 n, n N 0, x = 0 Then S = { 1 n : n N} is countably infinite. 8. Give an example of a bounded function which is continuous but not uniformly continuous. (If it is clear the function is bounded and continuous, just say so, but you should justify the fact that it is not uniformly continuous.) Solution. Define f : (0, 1) R by f(x) = cos(π/x). We will take for granted the fact that f is continuous. Assume (for contradiction) that f is uniformly continuous. Then there is δ > 0 such that x, y (0, 1) and x y < δ imply f(x) f(y) < 1. Let x = 1/n and y = 1/(n + 1) where n > 1/δ. Then x y = 1 n 1 n + 1 = 1 n(n + 1) < 1 n < δ 3

4 but f(x) f(y) = cos(nπ) cos((n + 1)π) = Let f : [a, b] R be a bounded function, define g : (a, b) R by g(x) = sup{f(y) : y < x}, and let c (a, b). Prove that if lim x c f(x) = f(c), then lim x c g(x) = g(c). Solution. We recall the following simple fact: ( ) If x, y (a, b) with x < y, then f(x) g(y). Assume lim x c f(x) = f(c). We will consider two cases: g(c) = f(c) and g(c) f(c). Assume g(c) = f(c). Let ɛ > 0 and choose δ > 0 such that x c < δ implies 1 f(x) f(c) < ɛ. Pick u, v such that c δ < u < c < v < c + δ. Note that f(x) g(c) for x < c and f(x) < f(c) + ɛ for c x < v. Thus x (u, v) and ( ) imply g(c) ɛ = f(c) ɛ < f(u) g(x) max{g(c), f(c) + ɛ} = g(c) + ɛ. It easily follows that lim x c g(x) = g(c). Now assume g(c) f(c) and let ɛ = f(c) g(c). Choose δ such that x c < δ implies f(x) f(c) < ɛ. Then either f(x) > g(c) for all x (c δ, c + δ), or f(x) < g(c) for all x (c δ, c + δ). The former is impossible by ( ), so the latter must hold, which implies g is constant on (c δ, c + δ). It easily follows that lim x c g(x) = g(c). 10. Let f and g be defined as in Problem 9, and let c (a, b). If f has a limit at c, must g have a limit at c? Either prove it, or provide a counterexample. Solution. No. Define f : [ 1, 1] R by f(x) = 0 for x 0 and f(0) = 1. Then g : ( 1, 1) R has the formula g(x) = 0 for x 0 and g(x) = 1 for x > 0. So f has a limit at 0, but g does not have a limit at Let A and B be disjoint subsets of R, and f : A B R a continuous function. Assume f is uniformly continuous on A and on B. Must it be true that f is uniformly continuous on A B? Prove it or provide a counterexample. Solution. No. Let A = (0, 1) and B = (1, 2), and define f : A B R by { 0, x A f(x) = 1, x B We first show f is continuous. Let x A B and ɛ > 0. If x A, pick δ > 0 such that (x δ, x + δ) A; if x B pick δ > 0 such that (x δ, x + δ) B. Let y A B be such 1 Since c (a, b), δ > 0 can be chosen small enough so that x c < δ also implies x [a, b]. 4

5 that x y < δ. Then either x and y are both in A, or they are both in B. In either case f(x) f(y) = 0 < ɛ, proving f is continuous. Now we show f is uniformly continuous on A (the proof that f is uniformly continuous on B is analogous). Let ɛ > 0 and pick any δ > 0. Then x, y A and x y < δ imply f(x) f(y) = 0 0 = 0 < ɛ. To see that f is not uniformly continuous on A B, let ɛ = 1 and take any δ > 0. Let x = 1 min{1/2, δ/4} and y = 1 + min{1/2, δ/4}. Then x, y A B and x y < δ but f(x) f(y) = 0 1 = 1 ɛ. 12. Let f : R R be continuous and A R compact. Must it be true that f 1 (A) is compact? Prove it or provide a counterexample. Solution. No. Define f : R R by f(x) = x 2. and observe that 0 f(x) 1 for all x R. Now [0, 1] R is compact but f 1 ([0, 1]) = R is not compact. 13. Define f : (0, 1) R by f(x) = 0 if x is irrational, and f(x) = 1/m if x = n/m where n, m N have no common prime divisors. Let x (0, 1). Prove that f is continuous at x if and only if x is irrational. Solution. Let ɛ > 0. Observe that if f(y) ɛ, then y Q and x = n/m where m 1/ɛ and 1 n m. Thus, S = {y (0, 1) : f(y) ɛ} is finite. Pick δ = min{ x y : y S, y x}. Then y (0, 1) and 0 < x y < δ implies y / S, so that f(y) 0 = f(y) < ɛ. This proves that lim y x f(y) = 0. Since x is a limit point of (0, 1), f is continuous at x if and only if lim y x f(y) = f(x). As f(x) = 0 if and only if x / Q, the result follows. 14. Let f : R R and suppose that: ( ) for each c R, the equation f(x) = c has exactly two solutions. Prove that f is not continuous. Solution. Let c, d R with c d. By ( ) we may pick p < q such that f(p) = f(q) = c and r < s such that f(r) = f(s) = d. One of the following must hold: (i) x < u < v < y (iii) v < x < u < y (ii) x < v < u < y (iv) x < u < y < v 5

6 In all of cases (i)-(iv) we have f(x) = f(y) = p, f(u) = f(v) = q. We will consider only cases (i) and (iii), as the proofs in cases (ii) and (iv) are analogous. Suppose (i) holds. Pick any c (u, v). Then c [x, y] so we must have p f(c) q. Combining this with ( ), we see that actually p < f(c) < q. Now by IVT, there exists a (x, u) and b (v, y) such that f(a) = f(b) = f(c). This contradicts ( ). Now suppose (iii) holds. Pick any r such that p < r < q. Then by IVT there exists a (v, x), b (x, u), and c (y, v) such that f(a) = f(b) = f(c) = r. 15. Prove that for each positive real number x, there is a real number y such that y 2 = x. (Hint: let y = sup{z R : z 2 < x}, and show that y 2 = x.) Solution. Let x be a positive real number. First we show that y is well-defined. Note that S {z R : z 2 < x} is nonempty since it contains 0. Let z S. If z > x + 1 then z 2 > (x + 1) 2 = x 2 + 2x + 1 > x, contradiction. So z x + 1 which shows x + 1 is an upper bound of S. So y = sup S exists. Note that x/2 is positive and (x/2) 2 {z R : z 2 < x}, which shows y is positive. Suppose that y 2 x. Assume first y 2 > x. Choose 0 < δ < (y 2 2)/(2y) such that also δ < y. Then (y δ) 2 > x. Let z S. If z > y δ then z 2 > (y δ) 2 > x, contradiction. So z y δ, which shows y δ is an upper bound of S, contrary to y being the least upper bound of S. Now assume y 2 < x. Choose 0 < δ < (x y 2 )/(2y + 1) such that also δ < 1. Then (y + δ) 2 < x, so y + δ S, contrary to y being an upper bound of S. 16. Let S be a set. Prove that S is infinite if and only if A = S for some proper subset A of S. Solution. Assume A = S where A is a proper subset of S. If S is finite, say S = {1,..., n}, then A = {1,..., n k} < {1,..., n} = S, where k is the number of elements in S \ A, contradiction. So S is infinite as desired. Conversely assume S is infinite. Choose x 1 S, then x 2 S \ {x 1 }, then x 3 S \ {x 1, x 2 },... in this way we have (by induction) distinct points x n S for all n N. Let T = {x n : n N} and define f : T T by f(x n ) = x 2n. Since the x n s are distinct f is injective, so f : T f(t ) is bijective. Now define g : S S by g(x) = f(x) if x T, and g(x) = x if x S \ T. Since f is injective so is g, so g : S g(s) is bijective. Now A g(s) has the same cardinality as S, yet since f(t ) is a proper subset of T, A = (S \ T ) f(t ) is a proper subset of S, as desired. 17. Give an example of a bounded countably infinite set of real numbers with a countably infinite set of limit points. Solution. Consider S = {1/n+1/m : n, m N}. Then S is bounded (S [0, 2]) and countably infinite, and the set of limit points of S is {1/n : n N} {0}, which is also countably infinite. 6

7 18. Consider Q as a metric space with the usual distance function d(x, y) = x y, and define S = {x Q : 2 < x 2 < 3}. Show that S is closed and bounded in Q, but that S is not compact. Solution. Note that S c = {x Q : x 2 < 2 or x 2 > 3} since ± 2 / Q, ± 3 / Q. Let x S c. Then x 2 > 3 or x 2 < 2. If x 2 > 3 and x > 0 choose 2 δ > 0 such that (x δ) 2 > 3. If x 2 > 3 and x 0 choose δ > 0 such that (x + δ) 2 > 3. If x 2 < 2 and x > 0 choose δ > 0 such that (x + δ) 2 < 2. If x 2 < 2 and x 0 choose δ > 0 such that (x δ) 2 < 2. Then (x δ, x + δ) S c, showing S c is open. We conclude S is closed. Now S is bounded since S B 1 (2), for example. To see that S is not compact, for each n N choose q n Q S such that 3 1/n < q n < 3. Then it is not hard to see that {q n : n N} is an infinite subset of S with no limit point in S. 19. Let X be a metric space. A collection {U α } of open subsets of X is called a base for X if for every x X and every open set V in X containing x, we have x U α V for some U α. Prove that the collection S = {(q ɛ, q + ɛ) : q Q, ɛ > 0 Q} of neighborhoods in R with rational centers and rational widths is a base for R. Solution. Let x R and let U be an open set containing x. Then there is a neighborhood (x ɛ, x + ɛ) of x such that (x ɛ, x + ɛ) U. Choose q Q such that q (x ɛ/3, x + ɛ/3), and choose δ such that ɛ/3 < δ < 2ɛ/3. Then x (q δ, q + δ) (x ɛ, x + ɛ) U. 20. A metric space is called separable if it has a countable dense subset. Let X be a metric space. Prove that X is separable if every infinite subset of X has a limit point. (Hint: Let x 1 X. Then pick x 2 X \ B δ (x 1 ). Next pick x 3 X \ (B δ (x 1 ) B δ (x 2 )). Continuing in this way, show that X \ (B δ (x 1 )... B δ (x m )) must eventually be empty for some m. Then consider neighborhoods of x i with δ = 1/n.) Solution. Assume every infinite subset of X has a limit point. Let δ > 0 and choose x j s as in the hint. If X \ (B δ (x 1 )... B δ (x m )) is nonempty for every m, then we obtain an infinite subset {x 1, x 2,...} of X. This subset has a limit point x. Consider the neighbhorhood B δ/2 (x) of x. This neighborhood contains infinitely many x j s, so in particular it contains x i x j. But x i, x j B δ/2 (x) implies that x j B δ (x i ), contrary to the construction of the x j s. We conclude that X B δ (x 1 )... B δ (x m ) for some m. Let S be the set consisting of the all the x j s chosen in this way, for each δ = 1/n, n = 1, 2,.... Since for each δ = 1/n we have finitely many x j s, S is countable. To see that S is dense in X, let x X and let B ɛ (x) be a neighborhood of x. Choose n so that 1/n < ɛ, and pick an x j from S such that x B 1/n (x j ). (This is possible since S includes the centers of a collection of 1/n-neighborhoods which cover X.) Then x j B ɛ (x) as desired. 2 See also HW1, problem 2. 7

8 21. The Cantor set consists of real numbers which admit a ternary (base 3) decimal expansion of the form.a 1 a 2 a 3... where a n {0, 2} for all n. Use this to prove that the Cantor set is uncountable. Solution. Let C be the Cantor set and write each element of C in the form described above. Define f : C 2 N by f(.a 1 a 2 a 3...) = {n N : a n = 2}. It is easy to check that f is bijective. 22. Use decimal expansions as in Problem 1 to prove that the Cantor set is perfect. Solution. Let C be the Cantor set. Let x =.a 1 a 2 a 3... be a point in R \ C, expressed as a ternary expansion. Let m = min{n : a n / {0, 2}} and k = min{n > m : a n 2}. Note that (x 3 k, x + 3 k ) contains no points of C. This shows the complement of C is open, so C is closed. Now let y =.b 1 b 2 b 3... be any point in C, written as a ternary expansion with b n {0, 2} for all n. Let y n =.c 1 c 2 c 3... where c j = b j for j n and c n = 0 if b n = 2, and c n = 2 if b n = 0. Then y n C, y n y for all n and y n y = 2/3 n. Any neighborhood (y ɛ, y + ɛ) contains y n for n sufficiently large, showing y is a limit point of C. 23. Prove that if U n is a dense open subset of R d for n = 1, 2,..., then n=1 U n is dense in R d. Solution. Let x R d and let N be a neighborhood of x. Since U 1 is dense in R d we may pick a point x 1 U 1 such that x 1 N. Since U 1 N is open we may choose a closed ball 3 B 1 around x 1 such that B 1 U 1 N. Suppose we have chosen closed balls B 1,..., B n around x 1,..., x n, respectively, such that ( ) B j+1 B j, B j U j. Since U n+1 is dense in R d and B n contains an open ball B, we may pick a point x n+1 U n+1 such that x n+1 B B n. Now B U n+1 is open so we may pick a closed ball B n+1 around x n+1 such that B n+1 B U n+1. In particular B n+1 B n and B n+1 U n+1. By induction we have closed balls B 1, B 2,... satisfying ( ). So there is a point y n=1 B n n=1 U n. Since also y N, we see that n=1 U n is dense in R d as desired. 24. Let X be a metric space and {a n } a sequence in X. (i) Suppose that the range of {a n } is bounded and has exactly one limit point, a. Must it be true that {a n } converges to a? Prove it, or provide a counterexample. (ii) Prove that {a n } converges if and only if every subsequence of {a n } converges. (iii) Prove that if two subsequences of {a n } converge to different limits, then {a n } does not converge. Solution. Consider first (i). Let X = R and a n = 1 for n odd and a n = 1/n for n even. Then 3 That is, a set of the form {y R d : x y ɛ} where x R d and ɛ > 0. We may choose an open ball (neighborhood) with the same property; to get a closed ball with this property simply divide the radius in half. The closed balls are bounded by definition, hence compact. 8

9 the range of {a n } has exactly one limit point, namely 0, but {a n } does not converge. Consider now (ii). Assume {a n } converges, say to a. Let ɛ > 0. Choose N such that n N implies d(a n, a) < ɛ. Let {a nk } be a subsequence of {a n }. Then k N implies n k k N and so d(a nk, a) < ɛ. This shows that every subsequence of {a n } converges to a. Notice that we have just proved (iii). Conversely, if every subsequence of {a n } converges, then {a n } converges (every sequence is a subsequence of itself). This finishes the proof of (ii). 25. Let X be a metric space and E a closed and bounded subset of X. Must it be true that every sequence in E has a subsequence which converges to a point in E? Either prove it or provide a counterexample. Solution. No. Let X = Q with d(x, y) = x y, let E = {x X : 2 < x < 3}, and pick x n E such that x n 2 < 1/n. Then {x n } has no subsequence which converges to a point of E. (To see this, for a given y E, let ɛ = y 2 /2; for any n > 1/ɛ we have x n y y 2 2 x n > 2ɛ ɛ = ɛ.) 26. Define a 0 = 2 and a n+1 = φ(a n ) for n 0, where φ(x) := x + 2 x. 2 Prove that {a n } n=1 is decreasing and bounded below4, hence convergent. Then let X = [ 2, 2] R and use our result on contraction mappings to show that {a n } n=1 converges to 2. What can be said about the rate of convergence? Solution. By the geometric-arithmetic mean inequality ( ) 2 φ(x) x = 2 x and so also φ(x) x = x In particular this shows {a n } n=1 is decreasing and bounded below by 2. Now we show {a n } n=1 satisfies the conditions of Problem 3 with X = [ 2, 2] and k = 1/4. The two displays above imply that φ(x) X. Let x, y X with x < y. Then φ(x) φ(y) = x y 2 + y x xy [ x y 4 which shows φ(x) φ(y) k x y. By solving x = φ(x ) and selecting the positive solution we conclude that {a n } n=1 converges to x = 2. It is not hard to see that the rate of convergence 4 You may use without justification the geometric-arithmetic mean inequality, which states that (x + y)/2 xy for positive real numbers x, y. ], 0 9

10 satisfies the upper bound x n x kn 1 k x 0 x 1 = 2 3 ( ) 1 n Give an example of a continuous bijective function f : X Y between metric spaces X and Y, such that f 1 is not continuous. Solution. Define f : [0, 1) [2, 3] [0, 2] by { x, x [0, 1) f(x) = 4 x, x [2, 3] Then f is continuous and bijective, but is not continuous. f 1 (x) = { x, x [0, 1) 4 x, x [1, 2] 28. Let D R be bounded and f : D R a uniformly continuous function. (i) Let a be a limit point of D. Prove that f has a finite limit at a. (ii) Use (i) to show that f can be extended to a continuous function on the closure of D. (iii) Conclude that f(d) is bounded. Solution. (i) Let {a n } be a sequence in D \ {a} converging to a. By Problem 4 on Midterm 2, it suffices to show that {f(a n )} converges. Let ɛ > 0. Using uniform continuity, pick δ > 0 such that x, y D and x y < δ imply f(x) f(y) < ɛ. Since convergent sequences are Cauchy, we may pick N such that n, m N implies a n a m < δ. Then n, m N implies f(a n ) f(a m ) < ɛ. So {f(a n )} is Cauchy, hence convergent. (ii) Define an extension f : D R by { f(x), x D f(x) = lim y x f(y), x D \ D Then f is well-defined by part (i), and it is continuous by construction. (iii) Note that D R is closed by definition and bounded since D is bounded. So D R is compact. Now f is continuous so f( D) R is compact, hence bounded. As f(d) = f(d) f( D), we conclude that f(d) is bounded. 29. Let X and Y be metric spaces. A function f : X Y is said to be Lipschitz continuous if there is K > 0 such that for all x, y X, d(f(x), f(y)) K d(x, y). (3) 10

11 Prove that Lipschitz continuous functions are uniformly continuous. Give an example to show that the converse is false. Solution. Let f : X Y satisfy equation (3) for all x, y X. Let ɛ > 0 and pick δ = ɛ/k. Then x, y X and d(x, y) < δ imply d(f(x), f(y)) K d(x, y) < Kδ = ɛ. To see that the converse is false, consider f : [0, 1] R, f(x) = x. We have seen that f is continuous. Since [0, 1] is compact, f is uniformly continuous. However, equation (3) will not hold when x = 0 and 0 y < 1/K Let f : R R be continuous and suppose that f(u) is open for each open set U R. Prove that f is monotonic. Solution. Suppose f is not monotonic. Then there exists x < y < z such that either f(x) < f(y) and f(y) > f(z), or f(x) > f(y) and f(y) < f(z). We will consider only the former case, as the latter is analogous. Since f is continuous, it attains a maximum value on [x, z], say f(u) = v = sup{f(z) : z [x, z]} for some u [x, z]. Since y (x, z) and f(y) > f(x), f(y) > f(z), we must actually have u (x, z). Now let U = (x, z) and note that v f(u). But for any ɛ > 0, (v ɛ, v + ɛ) is not a subset of f(u): for example v + ɛ/2 / f(u). So f(u) is not open, contradiction. 31. Let f : [a, b] R. Prove the following statements: (i) If f is continuous and injective, then f is monotone. (ii) If f is differentiable and f (x) 0 for all x (a, b), then f is injective. (iii) If f is differentiable and f (a) < 0 < f (b), then there is c (a, b) such that f (c) = 0. (iv) If f is differentiable and f (a) < d < f (b), then there is c (a, b) such that f (c) = d. Solution. (i) Suppose f is continuous and injective, yet f is not monotone. Then there exist x, y, z [a, b], with x < y < z, such that either (a) or (b) below holds: (a) f(x) < f(y) and f(y) > f(z) (b) f(x) > f(y) and f(y) < f(z) We will consider only case (a), as (b) is similar. If f(z) < f(x), then by IVT there exists w (y, z) such that f(w) = f(x), contradiction to injectivity of f. If f(z) > f(x), then there exists w (x, y) such that f(w) = f(z), again contrary to injectivity of f. (ii) Assume f is differentiable on (a, b). Suppose f is not injective. Then there is x < y [a, b] such that f(x) = f(y). By MVT, there exists c (x, y) (a, b) such that f (c) = [f(y) f(x)]/(y x) = 0. 11

12 (iii) Assume f is differentiable on (a, b), and f (x) 0 for all x (a, b). Then by (ii), f is injective, so by (i), f is monotone. If f is increasing, then for any x y (a, b), [f(y) f(x)]/(y x) 0, which shows f (x) 0. If f is decreasing, then for any x y (a, b), [f(y) f(x)]/(y x) 0, which shows f (x) 0. (iv) Suppose f is differentiable and f (a) < d < f (b). Define g(x) = f(x) xd for x [a, b]. Then g is differentiable and g (a) < 0 < g (b), so by (iii), there exists c (a, b) such that g (c) = 0. Thus f (c) = g (c) + d = d. 32. Give an example of a function f such that f exists on [0, 1] but is not continuous on [0, 1]. Note that by Problem 31, such f will have the intermediate value property, despite being discontinuous. Solution. Define f : [0, 1] R by f(x) = { x 2 sin(1/x), x (0, 1] 0, x = 0 Then f (x) = 2x sin(1/x) cos(1/x) for x > 0 and f x 2 sin(1/x) 0 (0) = lim = lim x sin(1/x) = 0. x 0 x 0 x 0 Thus f exists on [0, 1] but is not continuous at 0, since f does not have a limit at Suppose f is defined in a neighborhood of x and f (x) exists. Prove that f(x + h) + f(x h) 2f(x) lim h 0 h 2 = f (x). Give an example in which the limit above exists but f does not exist. Solution. Note that the assumptions imply that f is defined in a neighborhood (x δ, x + δ) of x. For h [0, δ/2], define p(h) = f(x + h) + f(x h) 2f(x), q(h) = h 2. Now p and q are differentiable on (0, δ/2), q (h) 0 for all h (0, δ/2), p(0) = q(0) = 0, and by the chain rule, p (h) lim h 0 q (h) = lim h 0 = lim h 0 = f (x) 2 f (x + h) f (x h) 2h ( f (x + h) f (x) 2h + f (x) = f (x) f (x) f ) (x h) 2h

13 So by L Hospital s rule, as desired. For the example, define f : R R by p(h) lim h 0 q(h) = f (x), f(x) = { x 2 2, x > 0 x2 2, x 0 Then f (x) = x, which we have seen is not differentiable at 0. However, f(h) + f( h) 2f(0) h 2 ( h) 2 lim h 0 h 2 = lim h 0 h 2 = Let f be differentiable on [a h, a + h] such that f is continuous at a. If f (a) = 0 and f (a) < 0, use Taylor s theorem to show that f has a strict local maximum at a, that is, f(x) < f(a) for x in a neighborhood of a. Is the assumption that f is continuous at a necessary? Solution. Taylor s theorem shows that f(x) f(a) x a = 1 2 f (ζ), where ζ is between x and a. Since f (a) < 0 and f is continuous at a, the RHS above is negative for x in a neighborhood of a. However, continuity at a is not needed. By definition of (second) derivative, f (x) f (a) = f (x) x a x a is negative for x in a neighborhood (a δ, a+δ) of a. This shows that f (x) > 0 for x (a δ, a) and f (x) < 0 for x (a δ, a). Finally MVT shows that f is strictly increasing on [a δ, a] and strictly decreasing on [a, a + δ], which allows us to conclude. 35. Let (X, d X ) and (Y, d Y ) be metric spaces with X compact, let Z = {f : X Y f is continuous}, and define d Z : Z Z R by Prove that d Z is a metric. d Z (f, g) = sup d Y (f(x), g(x)). x X Solution. Fix f, g Z and define φ : X R by φ(x) = d Y (f(x), g(x)). We first claim that φ is continuous. Let ɛ > 0 and x X. Pick δ > 0 such that d X (y, x) < δ implies 13

14 d Y (f(y), f(x)) < ɛ/2 and d Y (g(y), g(x)) < ɛ/2. Then d X (y, x) < δ implies φ(y) φ(x) = d Y (f(y), g(y)) d Y (f(x), g(x)) d Y (f(y), g(y)) d Y (f(x), g(y)) + d Y (f(x), g(y) d Y (f(x), g(x)) d Y (f(y), f(x)) + d Y (g(y), g(x)) < ɛ with the last line coming from the triangle inequality in Y. Since X is compact it follows that d Z (f, g) = sup x X φ(x) is finite. Note that d Z (f, g) 0 and for any h Z, d Z (f, g) = sup d Y (f(x), g(x)) = sup d Y (g(x), f(x)) = d Z (g, f), x X x X d Z (f, g) = 0 sup d Y (f(x), g(x)) = 0 x X d Y (f(x), g(x)) = 0 x X f(x) = g(x) x X, d Z (f, g) = sup x X d Y (f(x), g(x)) sup[d Y (f(x), h(x)) + d Y (h(x), f(x))] x X sup d Y (f(x), h(x)) + sup d Y (h(x), g(x)) = d Z (f, h) + d Z (h, g). x X x X These statements follow from the fact that d Y is a metric. 36. Let Z, X and Y be as in Problem 35, and consider continuous functions f n : X Y. Prove that f n converges uniformly if and only if {f n } converges in (Z, d Z ). Solution. Suppose {f n } converges uniformly to f. Then since each f n is continuous, f is continuous, so f Z. Let ɛ > 0 and pick N such that n N implies d Y (f n (x), f(x)) < ɛ for all x X. Then n N implies d Z (f n, f) = sup d Y (f n (x), f(x)) ɛ, x X which shows that {f n } converges to f in (Z, d Z ). Conversely assume that {f n } converges to f in (Z, d Z ). Pick N such that n N implies d Z (f n, f) = sup d Y (f n (x), f(x)) < ɛ. x X Then n N implies d Y (f n (x), f(x)) < ɛ for all x X, which shows uniform convergence. 37. Give an example of sequences {f n }, {g n } of uniformly converging functions such that {f n g n } does not converge uniformly. Solution. Let f n (x) = x + 1/n = g n (x) be defined on [0, ). Given ɛ > 0, pick N > 1/ɛ. Then n N implies f n (x) x = 1/n < ɛ for all x [0, ), which shows uniform convergence to x. Observe that (f n g n )(x) = x 2 + 2x/n + 1/n 2 converges pointwise to x 2. For ɛ = 1, observe that for any N, (f N g N )(x) x 2 = 2x N + 1 N 2 > 1 = ɛ 14

15 when x N/2. Hence, the convergence is not uniform. 38. Define f n (x) = x 1 + nx 2. Prove that {f n } converges uniformly to a function f. Show that f (x) = lim n f n(x) except when x = 0. Solution. Since each f n is odd, it suffices to consider convergence on [0, ). On this interval we have f n(x) = 1 nx2 (1 + nx 2 ) 2 = 0 x = n 1/2. and We claim that f n (n 1/2 ) = 1 2n 1/2. 0 f n (x) M n := 1, x [0, ). (4) 2n1/2 If not, there is y [0, ) such that f n (y) > M n. Suppose y < n 1/2. Since f(0) = 0, IVT implies there is u (0, y) such that f(u) = M n, and consequently Rolle s theorem implies there is v (u, n 1/2 ) such that f n(v) = 0, contradiction. Suppose then that y > n 1/2. Since lim x f n (x) = 0, IVT implies there is u (y, ) such that f(u) = M n, and so Rolle s theorem implies there is v (n 1/2, u) such that f (v) = 0, contradiction. Note that (4) implies uniform convergence, since M n does not depend on x and lim n M n = 0. The last statement is straightforward to check. 39. Give an example of a sequence of equicontinuous functions {f n } that converges pointwise but not uniformly. Solution. Define f n : R R by f n (x) = x (n 1) for x [n 1, n], f n (x) = (n + 1) x for x [n, n + 1] and f n (x) = 0 otherwise. To establish equicontinuity, given ɛ > 0 let δ = ɛ; then f n (x) f n (y) < ɛ whenever x y < δ and n N. To see pointwise convergence, given ɛ > 0 and x R, pick N > x + 1. Then n N implies f n (x) = 0, showing that {f n } converges to 0 pointwise. To see that the convergence is not uniform, let ɛ = 1/2, take any N, and note that f N (N) 0 = 1 0 = 1 > ɛ. 40. Let f n : [a, b] R be monotone for each n. Suppose {f n } converges pointwise to a continuous function. Show that it converges uniformly. Solution. Let f be the pointwise limit of {f n }, which is uniformly continuous since [a, b] is compact. Let ɛ > 0 and pick δ > 0 such that x y < δ implies f(x) f(y) < ɛ/2. As 15

16 {(x δ/2, x + δ/2)} x [a,b] is an open cover of [a, b], there is a finite subcover, [a, b] ( x 1 δ 2, x 1 + δ ) (... x k δ 2 2, x k + δ ). 2 where WLOG x 1 <... < x k. Observe that then x i x i 1 < δ for i = 2,..., k. For i = 1,..., k, pick N i such that n N i implies f n (x i ) f(x i ) < ɛ/2. Let N = max{n 1,..., N k } and let n N. Assume WLOG that f n is increasing. Fix any x [a, b]; we have x [x i 1, x i ] for some i = 2,..., k. Then f n (x i 1 ) f(x) f n (x) f(x) f n (x i ) f(x) and so while for j = i 1 or j = i, f n (x) f(x) max{ f n (x i 1 ) f(x), f n (x i ) f(x) }, f n (x j ) f(x) f n (x j ) f(x j ) + f(x j ) f(x) < ɛ 2 + ɛ 2 = ɛ. Combining the last two expressions gives f n (x) f(x) < ɛ. 41. Let X be compact, f n : X R, and {f n } equicontinuous. Suppose {f n } converges pointwise. Prove that it converges uniformly. Solution. Let ɛ > 0. Pick δ > 0 such that d(x, y) < δ implies f n (x) f n (y) < ɛ/3 for all n. Since {B δ (x)} x X is an open cover of X, there is a finite subcover X B δ (x 1 )... B δ (x k ). For i = 1,..., k, pick N i such that m, n N i implies f n (x i ) f m (x i ) < ɛ/3, and let N = max{n 1,..., N k }. Let m, n N and let x X be arbitrary. Then x B δ (x i ) for some i and f n (x) f m (x) f n (x) f n (x i ) + f n (x i ) f m (x i ) + f m (x i ) f m (x) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. 42. Let φ : [0, 1] R R be continuous. Suppose there is 0 < M < 1 such that φ(r, s) φ(r, t) M s t for all r [0, 1] and s, t R. Prove there is a solution to y = φ(x, y), y(0) = c as follows: Let Z be the set of continuous functions [0, 1] R with the sup metric (see HW5, Problem 1), and show that Ψ(f)(x) = c + x 0 φ(t, f(t)) dt, x [0, 1] 16

17 is a contraction mapping on Z. Solution. Observe that Ψ(Z) Z and d Z (Ψ(f), Ψ(g)) = sup x [0,1] = sup x [0,1] 0 x sup x [0,1] Ψ(f)(x) Ψ(g)(x) x 0 [φ(t, f(t)) φ(t, g(t))] dt φ(t, f(t)) φ(t, g(t)) dt sup φ(x, f(x)) φ(x, g(x)) x [0,1] sup K f(x) g(x) x [0,1] = Kd Z (f, g), proving that Ψ is a contraction mapping. Note also that Z is complete: if {f n } is a Cauchy sequence in Z then it is uniformly convergent (see also Problem 36), so its limit must be a continuous function, hence an element of Z. This allows us to conclude. Observe that M < 1 is not needed; we can divide the differential equation by any nonzero constant and it still holds. 43. Let φ : [0, 1] R R be continuous. Suppose there is M > 1 such that φ(r, s) M for all r [0, 1] and s R. Let Z and Ψ be as in Problem 42 and define E = {f Z : f(x) c M and f(x) f(y) M x y for all x, y [0, 1]}. Show that Ψ(E) E and use the Arzela-Ascoli theorem to show that E is compact. Solution. Let f E. Then and Ψ(f)(x) c = Ψ(f)(x) Ψ(f)(y) = x 0 x y φ(t, f(t)) dt M φ(t, f(t)) dt M x y. Thus, Ψ(E) E. Note that E is uniformly bounded (by M + c) and equicontinuous: given ɛ > 0, pick δ = ɛ/m; then x y < δ implies f(x) f(y) M x y = ɛ for all f E. It is easy to see that E is also closed. So the Arzela-Ascoli theorem shows that E is compact. One can also check that E is convex, and then the Brouwer fixed point theorem 5 shows that Ψ E has a fixed point. (This gives an alternate proof of existence of a solution to the ODE in Problem 42.) 5 We did not discuss this theorem in class, you don t need to know it 17

18 44. Let f n, f : X R be such that f is continuous at x and f n f uniformly. Show that x n x in X implies f n (x n ) f(x). Solution. Assume x n x and let ɛ > 0. Pick N f such that n N implies f n (y) f(y) < ɛ/2 for all y X, pick δ > 0 such that d(y, x) < δ implies f(y) f(x) < ɛ/2, and pick N x such that n N x implies d(x n, x) < δ. Let N = max{n f, N x }. Then n N implies f n (x n ) f(x) f n (x n ) f(x n ) + f(x n ) f(x) < ɛ 2 + ɛ 2 = ɛ. 45. Let f n, f : X R be such that X is compact, f is continuous, and f n (x n ) f(x) whenever x n x in X. Show that {f n } converges uniformly to f. Solution. Suppose that {f n } does not converge uniformly to f. Then there is ɛ > 0 such that for all N, there exists n N and y X such that f n (y) f(y) ɛ. This allows us to build inductively a sequence {y nk } such that f nk (y nk ) f(y nk ) ɛ for all k. Since X is compact, {y nk } has a subsequence {x n } converging to some point x X. Thus, f n (x n ) f(x) but f n (x n ) f(x n ) ɛ for all n. Pick δ > 0 such that d(y, x) < δ implies d(f(y), f(x)) < ɛ/2, and pick N such that n N implies d(x n, x) < δ. Then n N implies f n (x n ) f(x) f n (x n ) f(x n ) f(x n ) f(x) > ɛ ɛ 2 = ɛ 2, so that {f n (x n )} does not converge to f(x), contradiction. 46. Let f : [0, ) R and for all n define f n : [0, ) R by f n (x) = f(x n ). Under what conditions on f is {f n } equicontinuous? Solution. If f is a constant function, then of course {f n } is equicontinuous. So suppose f is nonconstant, say f(s) = a and f(t) = b for some s t [0, ). Let ɛ = b a /2. If f is equicontinuous, there is δ > 0 such that x y < δ implies f n (x) f n (y) < ɛ for all n. Let s n = n s and t n = n t. Note that f n (s n ) f n (t n ) = f(s) f(t) = b a > ɛ for all n, but we may choose n large enough so that s n t n < δ. Thus, {f n } cannot be equicontinuous. We have shown that {f n } is equicontinuous if and only if f is constant. 47. Let f n : [a, b] (0, ) be continuous, such that f(x) = n=1 f n(x) is continuous. Show that n=1 f n(x) converges uniformly on [a, b]. Solution. Let g n (x) = f(x) n k=1 f k(x) for x [a, b]. By our assumptions g n is nonnegative and continuous, g n (x) g n+1 (x) for all x [a, b], and g n 0 pointwise. Let ɛ > 0 and 18

19 define U n = {x [a, b] : g n (x) < ɛ}. Then U 1 U 2... since g n g n+1, pointwise convergence implies that n=1 U n = [a, b], and continuity of g n shows that U n is open in [a, b]. Compactness of [a, b] yields a finite subcover of the U n s, say [a, b] = U n1... U nk = U N where N = max{n 1,..., n k }. Moreover, n N implies [a, b] = U N U n and so g n (x) < ɛ for all x [a, b]. (This is a special case of Dini s theorem.) 48. Let {f n } be a sequence of continuous functions [a, b] R. Suppose f n f uniformly and f is continuous. Must the convergence be uniform? Solution. No, Let [a, b] = [0, 1] and let f(x) = 0 for x [2/n, 1], f(x) = 2/n x/n for x [1/n, 2/n], and f(x) = x/n for x [0, 1/n]. Then {f n } are all continuous and converge pointwise to 0, but the convergence is not uniform since f n (1/n) = 1 for all n (see Problem 44). 49. Let f n : X R be a uniformly convergent sequence of continuous functions on X. Give an example to show that {f n } may not be equicontinuous. Solution. Let f n (x) = sin(1/x)/n be defined on (0, 1). Since f n (x) 1/n for all x (0, 1), {f n } converges uniformly to 0. However, {f n } is not equicontinuous because (each) f n is not uniformly continuous. 50. Let f n : X R be such that X is compact and {f n } is pointwise bounded and equicontinuous. Let φ(x) = sup n N f n (x). Show φ is continuous. Solution. Note that φ is well-defined by pointwise boundedness. Let ɛ > 0 and pick δ > 0 such that d(x, y) < δ implies f n (x) f n (y) < ɛ for every n. Fix x, y X be such that d(x, y) < δ. Then for every n, f n (x) < f n (y) + ɛ, f n (y) < f n (x) + ɛ, and so Thus, φ(x) φ(y) ɛ. φ(x) φ(y) + ɛ, φ(y) φ(x) + ɛ. 51. Suppose f n : [a, b] R is such that f n converges uniformly. Show that f n 0 uniformly. Solution. Let s n be the nth partial sum, let ɛ > 0 and pick N such that n N implies s n 1(x) f n (x) = f k (x) < ɛ 2. n=1 k=n 19

20 for all x [a, b]. Then n N implies f n (x) = for all x [a, b]. k=n+1 f k (x) f k (x) < ɛ 2 + ɛ 2 = ɛ k=n 52. Let f : [0, 1] R be infinitely differentiable, such that f (n) (0) = 0 for n = 0, 1, 2,... but f is not identically zero. Suppose that a n f (n) converges uniformly. Show that lim n!a n = 0. Solution. Pick y [0, 1] such that f(y) 0. By Taylor s theorem, where ζ n (0, y). Thus, f(y) = f (n) (ζ n ) y n, n = 1, 2,..., n! f(y)n!a n f(y)n!a n y n = a n f (n) (ζ n ). (5) By Problem 51, {a n f (n) } converges uniformly to 0. So given ɛ > 0, we may choose n such that n N implies a n f (n) (x) < ɛ f(y) for all x [0, 1]. Then from (5), n N implies n!a n < ɛ. 53. Suppose f n : R k R is such that {f n } is equicontinuous and converges pointwise to f. Show that f is continuous. Solution. Fix x X and δ > 0. Observe that B δ (x) is compact. Recall that a sequence of continuous functions on a compact space converges uniformly if and only if it is equicontinuous and converges pointwise. Hence, {f n } converges uniformly on B δ (x) and so f is continuous on B δ (x), and in particular at x. 54. Let f : R R n (n 2) be differentiable with f (t) 0 for all t. Fix p / f(r) and let q be a point on f(r) with minimal distance to p (assumed to exist). Show that p q is orthogonal to f(r) at q. Solution. Define φ : R R by φ(t) = f(t) p 2. By assumption φ has a minimum at t = t 0, where f(t 0 ) = q. Thus, 0 = φ (t 0 ) = 2(f(t 0 ) p) f (t 0 ). 55. Suppose f : R n R is such that f(tx) = tf(x) for all t R and x R n, but f is not linear. Show that f has directional derivatives at the origin, but is not differentiable there. Give an example of such a function. 20

21 Solution. We prove the contrapositive. Suppose that f differentiable. Then f(tv) f(0) f (0)(tv) 0 = lim = f(v) f (0)v, t 0 t showing f(v) = f (0)v, that is, f is linear. An example is f(x, y) = (x 1/3 + y 1/3 ) Define f : R 2 R by f(x, y) = xy. Use the definition of derivative to show that f is differentiable everywhere, with df (a,b) (x, y) = bx + ay. Solution. Define L(x, y) = bx + ay. Then f(a + x, b + y) f(a, b) L(x, y) (x, y) = xy x 2 + y x2 + y 2 = (x, y) 0 as (x, y) 0. 2 x 2 + y2 57. Let E R n be open and f : E R differentiable. Prove that if f has a local maximum at a E, then f (a) = 0. Solution. Suppose f has a local maximum at a. Then for i = 1,..., n and sufficiently small t, f(a + te i ) f(a) t 0, if t > 0, f(a + te i ) f(a) t 0, if t < 0. This shows that D i f(a) = 0 for i = 1,..., n and so f (a) = (D 1 f(a),..., D n f(a)) = For E R 2, suppose f : E R is differentiable with D 1 f(x) = 0 for all x E. Under what condition on E can we say f depends on x 2 only? Solution. This will be true if for each x 2 R, the set E x2 := {x 1 R : (x 1, x 2 ) E} is connected (the empty set being considered connected vacuously). Fix x 2 R and define g(x 1 ) = f(x 1, x 2 ) on E x2. Suppose g(x 1 ) g(x 1 ). Our condition implies that E x 2 contains the interval between x 1 and x 1, so by the single variable MVT there is t between x 1 and x 1 such that g (t) 0. But g (t) = D 1 f(t, x 2 ), contradiction. 59. Let f : U R m be of class C 1, with U R n an open set containing the line segment L from a to a + h. Suppose T : R n R m is linear with matrix A. Show that f(a + h) f(a) T (h) h max x L f (x) A. Solution. Define g(x) = f(x) T (x). The multivariate MVT implies g(a + h) g(a) h max x L g (x). 21

22 By linearity of T and the fact that its derivative matrix at any point is A, this can be rewritten f(a + h) f(a) T (h) h max x L f (x) A. 60. Let f : R m R m be of class C 1 on the unit ball B 1 (0). Suppose that f(0) = 0, f (0) = I and f (x) I < ɛ for all x B 1 (0). Use Problem 59 to show that f(b 1 (0)) B 1+ɛ (0). Solution. With 1 the identity map, x, y B 1 (0), and L the line segment from x to y, f(x) f(y) (x y) = f(x) f(y) 1(x y) x y max z L f (z) I < ɛ x y, where the first inequality above comes from Problem 59, and the second from assumption 6. Thus, Taking y = 0 and using f(0) = 0, we get f(x) f(y) < (1 + ɛ) x y. f(x) < (1 + ɛ) x 1 + ɛ. 61. Let f : R n R m be of class C 1 at a and suppose df a : R n R m is injective. Use Problem 59 to show that f is injective in a neighborhood of a. Solution. First notice that df a (x) = x df a ( ) x x min x df a(y) = c x y =1 where c > 0 since df a is injective. Next, note that since f is class C 1 at a, there is δ > 0 such that f (z) f (a) < ɛ < c for all z B δ (a). So with x y B δ (a) and L the line segment from x to y, we have f(x) f(y) df a (x y) f(x) f(y) df a (x y) x y max z L f (z) f (a) < ɛ x y by Problem 59. Thus, f(x) f(y) > df a (x y) ɛ x y c x y ɛ x y = (c ɛ) x y > Let f : X X, with X a complete metric space and f Lipschitz continuous with constant 0 < K < 1. Prove that there is a unique fixed point x X satisfying f(x ) = x, and for each x X, d(f n (x), x ) Kn d(f(x), x). 1 K 6 Note that convexity of B 1(0) has also been used. 22

23 Solution. Fix x X. For m > n, d(f m (x), f n (x)) d(f m (x), f m 1 (x)) d(f n+1 (x), f n (x)) ( K m K n) d(f(x), x) Kn d(f(x), x). 1 K So {f n (x)} is Cauchy. Since X is complete, {f n (x)} converges, say to x. By continuity of f, ( ) f(x ) = f lim f n (x) = lim f n+1 (x) = x. n n If y satisfies f(y ) = y then d(x, y ) = d(f(x ), f(y )) Kd(x, y ), which is impossible unless x = y. The estimate on the rate of convergence to x comes from letting m in (6). (6) 63. Let G : R 2 R be class C 1, such that G(a, b) = 0 and D 2 G(a, b) 0. Prove 7 that there exists a continuous real-valued function f, defined on an open interval around a, such that f(a) = b and G(x, f(x)) 0. Hint: Consider the sequence f n+1 (x) = f n (x) G(x,fn(x)) D 2 G(a,b). Solution. Using continuity of D 2 G, choose δ > 0 such that 1 D 2G(x, z) D 2 (a, b) 1 2 for (x, z) (a δ, a + δ) (b δ, b + δ). Using continuity of G, choose ɛ (0, δ) such that G(x, b) D 2 G(a, b) < δ 2 whenever x (a ɛ, a + ɛ). For x (a ɛ, a + ɛ), define φ x : (b δ, b + δ) R by Note that φ x (z) = z G(x, z) D 2 G(a, b). φ x(z) = 1 D 2G(x, z) D 2 (a, b) 1 2. Thus, φ x is Lipschitz continuous with Lipschitz constant 1/2. Also, φ x (z) b φ x (z) φ x (b) + φ x (b) b 1 2 z b + G(x, b) D 2 G(a, b) < δ. 7 Don t assume the implicit function theorem here! 23

24 This shows that φ x maps into (b δ, b+δ). Thus, φ x is a contraction mapping (see Problem 62). Let f(x) be its unique fixed point, and note that φ(f(x)) = f(x) implies G(x, f(x)) = 0. Also, b = f(a) since φ a (b) = b and the fixed point is unique. Convergence to the fixed point satisfies (see Problem 62) φ n x(z) f(x) δ2 1 n. (7) Consider the function sequence {f n (x)} from the hint, defined for x (a ɛ, a + ɛ), with f 0 (x) b. Since f 0 and G are continuous, a simple induction argument shows that f n is continuous for each n. Note that f n (x) = φ n x(b), so (7) shows that {f n } converges uniformly to f. Thus, f is continuous. 64. Let A be a symmetric n n matrix and define q : R n R by q(x) = x t Ax. Suppose the restriction of q to the unit sphere {x : x = 1} attains a maximum or minimum at the point v. Show that Av = λv for some λ R. Solution. Define g : R n R by g(x) = x 2 1 and let M = {x R n : g(x) = 0}. Since g (x) = 2x 0 for all x M, we may use the method of Lagrange multipliers: Note that, with A k the kth row of A, n D k q(x) = D k (x t Ax) = D k x i A ij x j = q (v) = λg (v), some λ R. (8) i,j=1 n A kj x j + j=1 n x i A ik = 2 i=1 n A kj x j = 2A k x, where the last equality uses the fact that A is symmetric. Thus, q (x) = 2Ax, and so from (8) we get Av = λv. j=1 65. Define f : (R + ) n R by f(x) = n 1 (x x n ). Find the minimum value of f on the surface S = {x R n : g(x) = 0}, where g(x) = x 1... x n c (c > 0 constant). Use this to deduce the geometric-arithmetic mean inequality: for positive reals a 1,..., a n, (a 1... a n ) 1/n n 1 (a a n ). Solution. Let h = f S be the restriction of f to S. We claim it suffices to minimize h in the cube C := [0, nc 1/n ] n. Since C is compact, h C attains a minimum value at a point a C. Suppose a int C. Then since g (a) 0, we can use Lagrange multipliers to get f (a) = λg (a), i.e., n 1 = λa 1... â i... a n, i = 1,..., n. It follows that a 1 =... = a n, so since a 1... a n = c, we have a 1 =... = a n = c 1/n and f(a) = c 1/n. If a / int C, then a i nc 1/n for some i and so f(a) c 1/n. Thus, h cannot attain a smaller value than c 1/n. The geometric-arithmetic mean inequality follows. 24

25 66. A set P R n is called an (n 1)-patch if for some i {1,..., n}, there exists a real-valued differentiable function h defined on an open set U R n 1 such that P = {x R n : π i (x) U and x i = h(π i (x))}, where π i (x) := (x 1,..., ˆx i,..., x n ). A set M R n is called an (n 1)-manifold if for each x M there is an open set W R n containing x such that W M is a (n 1)-patch. Suppose g : R n R is class C 1 and let M = {x R n : g(x) = 0}. Suppose g (x) 0 for all x M. Show that then M is an (n 1)-manifold in R n. Solution. Let c M. Then g (c) 0 so D i g(c) 0, some i = 1,..., n. WLOG suppose D n g(c) 0, and write c = (a, b) with a R n 1, b R. By the implicit function theorem there is a neighborhood U of a (in R n 1 ) and a unique C 1 function h : U R such that h(a) = b and g(x, h(x)) = 0 for x U. Moreover, from the proof of the implicit function there is a neighborhood W of c such that W M is the graph of h, that is, W M = {(x, y) R n : x U, y = h(x)}. Note that this is an (n 1)-patch, so we are done. 25