Selected problems and solutions 1. Prove that |(0, 1)| = |R|. Solution. Define f : (0, 1) → R by 1 x−0.5 + 2, x < 0.5 1 f (x) = 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) → 2N by using binary expansions. Then define a surjective function g : 2N → (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 {an }∞ n=1 such that ∞ X x= an 2−n n=1
where an ∈ {0, 1} for all n ∈ N. (The expansion is typically written 0.a1 a2 a3 . . ..) It may be assumed that the an ’s do not have an “infinite trail” of 1’s; that is, that {an }∞ n=1 does not converge to 1; this ensures that the binary expansions are unique. Now define f : (0, 1) → 2N by f (x) = {n : an = 1} where {an }∞ 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 : 2N → (0, 1) as follows. If S ⊂ N and S 6= ∅, S 6= N, define g(S) = x where x has the binary expansion {an }∞ n=1 given by ( 1, n ∈ S an = 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
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 = 1 and note that either 1 ∈ / (a − , a + ) or −1 ∈ / (a − , a + ) (or both). Let N be an arbitrary natural number; we will find n ≥ N such that |an − a| ≥ . If 1 ∈ / (a − , a + ) then n = 2N suffices; if −1 ∈ / (a − , a + ) 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 = x − y. Then (x − , x + ) 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 {an }∞ n=1 is a sequence converging to a, and assume b is an accumulation point of {an : n ∈ N}. Prove that a = b. Solution. Assume that a 6= b, and let = |b − a|/2. As b is an accumulation point of {an : n ∈ N}, for any natural number N there exists n ≥ N such that an ∈ (b − , b + ). (This is because every neighborhood of b contains infinitely many elements of {an : n ∈ N}, so in particular must contain an ’s with n ≥ N .) So for any N , there exists n ≥ N such that |an − a| ≥ , which means that {an }∞ 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) 6= 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 > 0. Choose δ > 0 such that x ∈ [a, b] and 0 < |x − c| < δ ⇒ |f (x) − L| < (1) Choose u, v ∈ (a, b) such that c − δ < u < c < v < c + δ. Then L − < f (u) ≤ g(c) ≤ f (c) ≤ f (v) < L + We justify the inequalities in (2) as follows: 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
(2)
The fourth inequality comes from the fact that f is increasing. As > 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 S = {a ∈ [0, 1] : lim f (x) does not exist} x→a
is countably infinite. Solution: (i) Define f : R → R by ( m, f (x) = 0,
if x = else
n m
∈ Q is in lowest terms
and let I = (a, b) be a (bounded) interval with K = max{|a|, |b|}. Assume f is bounded by M . Then n n o |Q ∩ I| ≤ : |m| ≤ M, |n| ≤ M K ≤ (2M + 1)(2M K + 1) m a contradiction since |Q ∩ I| is infinite. (ii) Define f : [0, 1] → R by f (x) =
( 1/n, 0,
x∈
1 1 n+1 , n
i
,n∈N
x=0
Then S = { n1 : 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 1 1 1 1 |x − y| = − = < <δ n n + 1 n(n + 1) n 3
but |f (x) − f (y)| = | cos(nπ) − cos((n + 1)π)| = 2 ≥ 1
9. 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 limx→c f (x) = f (c), then limx→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 limx→c f (x) = f (c). We will consider two cases: g(c) = f (c) and g(c) 6= f (c). Assume g(c) = f (c). Let > 0 and choose δ > 0 such that |x − c| < δ implies1 |f (x) − f (c)| < . 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) + for c ≤ x < v. Thus x ∈ (u, v) and (∗) imply g(c) − = f (c) − < f (u) ≤ g(x) ≤ max{g(c), f (c) + } = g(c) + . It easily follows that limx→c g(x) = g(c). Now assume g(c) 6= f (c) and let = |f (c) − g(c)|. Choose δ such that |x − c| < δ implies |f (x) − f (c)| < . 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 limx→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 6= 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 0.
11. 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 > 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
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 < , 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 > 0 and pick any δ > 0. Then x, y ∈ A and |x − y| < δ imply |f (x) − f (y)| = |0 − 0| = 0 < . To see that f is not uniformly continuous on A ∪ B, let = 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 ≥ .
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) =
1 . 1 + x2
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 > 0. Observe that if f (y) ≥ , then y ∈ Q and x = n/m where m ≤ 1/ and 1 ≤ n ≤ m. Thus, S = {y ∈ (0, 1) : f (y) ≥ } is finite. Pick δ = min{|x − y| : y ∈ S, y 6= x}. Then y ∈ (0, 1) and 0 < |x − y| < δ implies y∈ / S, so that |f (y) − 0| = f (y) < . This proves that limy→x f (y) = 0. Since x is a limit point of (0, 1), f is continuous at x if and only if limy→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 6= 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
(ii) x < v < u < y
(iii) v < x < u < y
(iv) x < u < y < v 5
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 = x2 + 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 6= 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 x1 ∈ S, then x2 ∈ S \ {x1 }, then x3 ∈ S \ {x1 , x2 }, ... in this way we have (by induction) distinct points xn ∈ S for all n ∈ N. Let T = {xn : n ∈ N} and define f : T → T by f (xn ) = x2n . Since the xn ’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
18. Consider Q as a metric space with the usual distance function d(x, y) = |x − y|, and define S = {x ∈ Q : 2 < x2 < 3}. Show that S is closed and bounded in Q, but that S is not compact. √ √ Solution. Note that S c = {x ∈ Q : x2 < 2 or x2 > 3} since ± 2 ∈ / Q, ± 3 ∈ / Q. Let x ∈ S c . Then x2 > 3 or x2 < 2. If x2 > 3 and x > 0 choose2 δ > 0 such that (x − δ)2 > 3. If x2 > 3 and x ≤ 0 choose δ > 0 such that (x + δ)2 > 3. If x2 < 2 and x > 0 choose δ > 0 such that (x + δ)2 < 2. If x2 < 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 ⊂ √B1 (2), for example. √ To see that S is not compact, for each n ∈ N choose qn ∈ Q ∩ S such that 3 − 1/n < qn < 3. Then it is not hard to see that {qn : n ∈ N} is an infinite subset of S with no limit point in S.
19. Let X be a metric space. A collection {Uα } 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 ∈ Uα ⊂ V for some Uα . Prove that the collection S = {(q − , q + ) : q ∈ Q, > 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 − , x + ) of x such that (x − , x + ) ⊂ U . Choose q ∈ Q such that q ∈ (x − /3, x + /3), and choose δ such that /3 < δ < 2/3. Then x ∈ (q − δ, q + δ) ⊂ (x − , x + ) ⊂ 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 x1 ∈ X. Then pick x2 ∈ X \ Bδ (x1 ). Next pick x3 ∈ X \ (Bδ (x1 ) ∪ Bδ (x2 )). Continuing in this way, show that X \ (Bδ (x1 ) ∪ . . . ∪ Bδ (xm )) must eventually be empty for some m. Then consider neighborhoods of xi with δ = 1/n.) Solution. Assume every infinite subset of X has a limit point. Let δ > 0 and choose xj ’s as in the hint. If X \ (Bδ (x1 ) ∪ . . . ∪ Bδ (xm )) is nonempty for every m, then we obtain an infinite subset {x1 , x2 , . . .} of X. This subset has a limit point x. Consider the neighbhorhood Bδ/2 (x) of x. This neighborhood contains infinitely many xj ’s, so in particular it contains xi 6= xj . But xi , xj ∈ Bδ/2 (x) implies that xj ∈ Bδ (xi ), contrary to the construction of the xj ’s. We conclude that X ⊂ Bδ (x1 ) ∪ . . . ∪ Bδ (xm ) for some m. Let S be the set consisting of the all the xj ’s chosen in this way, for each δ = 1/n, n = 1, 2, . . .. Since for each δ = 1/n we have finitely many xj ’s, S is countable. To see that S is dense in X, let x ∈ X and let B (x) be a neighborhood of x. Choose n so that 1/n < , and pick an xj from S such that x ∈ B1/n (xj ). (This is possible since S includes the centers of a collection of 1/n-neighborhoods which cover X.) Then xj ∈ B (x) as desired. 2
See also HW1, problem 2.
7
21. The Cantor set consists of real numbers which admit a ternary (base 3) decimal expansion of the form .a1 a2 a3 ... where an ∈ {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 → 2N by f (.a1 a2 a3 . . .) = {n ∈ N : an = 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 = .a1 a2 a3 . . . be a point in R \ C, expressed as a ternary expansion. Let m = min{n : an ∈ / {0, 2}} and k = min{n > m : an 6= 2}. Note −k −k that (x − 3 , x + 3 ) contains no points of C. This shows the complement of C is open, so C is closed. Now let y = .b1 b2 b3 . . . be any point in C, written as a ternary expansion with bn ∈ {0, 2} for all n. Let yn = .c1 c2 c3 . . . where cj = bj for j 6= n and cn = 0 if bn = 2, and cn = 2 if bn = 0. Then yn ∈ C, yn 6= y for all n and |yn − y| = 2/3n . Any neighborhood (y − , y + ) contains yn for n sufficiently large, showing y is a limit point of C.
d 23. Prove that if Un is a dense open subset of Rd for n = 1, 2, . . ., then ∩∞ n=1 Un is dense in R .
Solution. Let x ∈ Rd and let N be a neighborhood of x. Since U1 is dense in Rd we may pick a point x1 ∈ U1 such that x1 ∈ N . Since U1 ∩ N is open we may choose a closed ball3 B1 around x1 such that B1 ⊂ U1 ∩ N . Suppose we have chosen closed balls B1 , . . . , Bn around x1 , . . . , xn , respectively, such that (∗) Bj+1 ⊂ Bj , Bj ⊂ Uj . Since Un+1 is dense in Rd and Bn contains an open ball B, we may pick a point xn+1 ∈ Un+1 such that xn+1 ∈ B ⊂ Bn . Now B ∩ Un+1 is open so we may pick a closed ball Bn+1 around xn+1 such that Bn+1 ⊂ B ∩ Un+1 . In particular Bn+1 ⊂ Bn and Bn+1 ⊂ Un+1 . By induction ∞ we have closed balls B1 , B2 , . . . satisfying (∗). So there is a point y ∈ ∩∞ n=1 Bn ⊂ ∩n=1 Un . Since ∞ d also y ∈ N , we see that ∩n=1 Un is dense in R as desired.
24. Let X be a metric space and {an } a sequence in X. (i) Suppose that the range of {an } is bounded and has exactly one limit point, a. Must it be true that {an } converges to a? Prove it, or provide a counterexample. (ii) Prove that {an } converges if and only if every subsequence of {an } converges. (iii) Prove that if two subsequences of {an } converge to different limits, then {an } does not converge. Solution. Consider first (i). Let X = R and an = 1 for n odd and an = 1/n for n even. Then 3
That is, a set of the form {y ∈ Rd : |x − y| ≤ } where x ∈ Rd and > 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
the range of {an } has exactly one limit point, namely 0, but {an } does not converge. Consider now (ii). Assume {an } converges, say to a. Let > 0. Choose N such that n ≥ N implies d(an , a) < . Let {ank } be a subsequence of {an }. Then k ≥ N implies nk ≥ k ≥ N and so d(ank , a) < . This shows that every subsequence of {an } converges to a. Notice that we have just proved (iii). Conversely, if every subsequence of {an } converges, then {an } 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 xn ∈ E such that |xn − 2| < 1/n. Then {xn } has no √subsequence which converges to a point of E. (To this, for a given y ∈ E, let = |y − 2|/2; for any n > 1/ we have √ see √ |xn − y| ≥ |y − 2| − | 2 − xn | > 2 − = .)
26. Define a0 = 2 and an+1 = φ(an ) for n ≥ 0, where φ(x) :=
x + x2 . 2
4 Prove that {an }∞ n=1 is decreasing and bounded below , hence convergent. Then let X = √ [√ 2, 2] ⊂ R and use our result on contraction mappings to show that {an }∞ n=1 converges to 2. What can be said about the rate of convergence?
Solution. By the geometric-arithmetic mean inequality s √ 2 φ(x) ≥ x = 2 x and so also
1 + x22 φ(x) = ≤ 1. x 2
√ In particular this shows {an }∞ bounded below by 2. Now we show {an }∞ n=1 is decreasing and √ 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 x−y y−x x−y φ(x) − φ(y) = + ∈ ,0 2 xy 4 which shows |φ(x)−φ(y)| ≤ k|x−y|. By solving x∗ = φ(x∗ ) and selecting the positive solution we √ conclude that {an }∞ converges to x = 2. It is not hard to see that the rate of convergence ∗ n=1 4
√
You may use without justification the geometric-arithmetic mean inequality, which states that (x + y)/2 ≥ xy for positive real numbers x, y.
9
satisfies the upper bound kn 2 |xn − x∗ | ≤ |x0 − x1 | = 1−k 3
n 1 . 4
27. 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 f
−1
( x, x ∈ [0, 1) (x) = 4 − x, x ∈ [1, 2]
is not continuous.
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 {an } be a sequence in D \ {a} converging to a. By Problem 4 on Midterm 2, it suffices to show that {f (an )} converges. Let > 0. Using uniform continuity, pick δ > 0 such that x, y ∈ D and |x − y| < δ imply |f (x) − f (y)| < . Since convergent sequences are Cauchy, we may pick N such that n, m ≥ N implies |an − am | < δ. Then n, m ≥ N implies |f (an ) − f (am )| < . So {f (an )} is Cauchy, hence convergent. ¯ → R by (ii) Define an extension f¯ : D ( f (x), x∈D f¯(x) = ¯ \D limy→x f (y), x ∈ D Then f¯ is well-defined by part (i), and it is continuous by construction. ¯ ⊂ R is closed by definition and bounded since D is bounded. So D ¯ ⊂R (iii) Note that D ¯ ¯ ¯ ¯ is compact. Now f is continuous so f (D) ⊂ R is compact, hence bounded. As f (D) = f (D) ⊂ ¯ we conclude that f (D) is bounded. f¯(D),
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). 10
(3)
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 > 0 and pick δ = /K. Then x, y ∈ X and d(x, y) < δ imply d(f (x), f (y)) ≤ K d(x, y) < Kδ = . √ 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 2 .
30. 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 > 0, (v − , v + ) is not a subset of f (U ): for example v + /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 0 (x) 6= 0 for all x ∈ (a, b), then f is injective. (iii) If f is differentiable and f 0 (a) < 0 < f 0 (b), then there is c ∈ (a, b) such that f 0 (c) = 0. (iv) If f is differentiable and f 0 (a) < d < f 0 (b), then there is c ∈ (a, b) such that f 0 (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 0 (c) = [f (y) − f (x)]/(y − x) = 0. 11
(iii) Assume f is differentiable on (a, b), and f 0 (x) 6= 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 6= y ∈ (a, b), [f (y) − f (x)]/(y − x) ≥ 0, which shows f 0 (x) ≥ 0. If f is decreasing, then for any x 6= y ∈ (a, b), [f (y) − f (x)]/(y − x) ≤ 0, which shows f 0 (x) ≤ 0. (iv) Suppose f is differentiable and f 0 (a) < d < f 0 (b). Define g(x) = f (x) − xd for x ∈ [a, b]. Then g is differentiable and g 0 (a) < 0 < g 0 (b), so by (iii), there exists c ∈ (a, b) such that g 0 (c) = 0. Thus f 0 (c) = g 0 (c) + d = d.
32. Give an example of a function f such that f 0 exists on [0, 1] but is not continuous on [0, 1]. Note that by Problem 31, such f 0 will have the intermediate value property, despite being discontinuous. Solution. Define f : [0, 1] → R by ( x2 sin(1/x), x ∈ (0, 1] f (x) = 0, x=0 Then f 0 (x) = 2x sin(1/x) − cos(1/x) for x > 0 and x2 sin(1/x) − 0 = lim x sin(1/x) = 0. x→0 x→0 x−0
f 0 (0) = lim
Thus f 0 exists on [0, 1] but is not continuous at 0, since f 0 does not have a limit at 0.
33. Suppose f is defined in a neighborhood of x and f 00 (x) exists. Prove that lim
h→0
f (x + h) + f (x − h) − 2f (x) = f 00 (x). h2
Give an example in which the limit above exists but f 00 does not exist. Solution. Note that the assumptions imply that f 0 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) = h2 .
Now p and q are differentiable on (0, δ/2), q 0 (h) 6= 0 for all h ∈ (0, δ/2), p(0) = q(0) = 0, and by the chain rule, f 0 (x + h) − f 0 (x − h) p0 (h) = lim h→0 q 0 (h) h→0 2h 0 f (x + h) − f 0 (x) f 0 (x) − f 0 (x − h) = lim + h→0 2h 2h 00 00 f (x) f (x) = + = f 00 (x). 2 2 lim
12
So by L’Hospital’s rule, lim
h→0
p(h) = f 00 (x), q(h)
as desired. For the example, define f : R → R by ( f (x) =
x2 2 , 2 − x2 ,
x>0 x≤0
Then f 0 (x) = |x|, which we have seen is not differentiable at 0. However, f (h) + f (−h) − 2f (0) h2 − (−h)2 = lim = 0. h→0 h→0 h2 h2 lim
34. Let f 0 be differentiable on [a − h, a + h] such that f 00 is continuous at a. If f 0 (a) = 0 and f 00 (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 00 is continuous at a necessary? Solution. Taylor’s theorem shows that f (x) − f (a) 1 = f 00 (ζ), x−a 2 where ζ is between x and a. Since f 00 (a) < 0 and f 00 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 0 (x) − f 0 (a) f 0 (x) = x−a x−a is negative for x in a neighborhood (a−δ, a+δ) of a. This shows that f 0 (x) > 0 for x ∈ (a−δ, a) and f 0 (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, dX ) and (Y, dY ) be metric spaces with X compact, let Z = {f : X → Y | f is continuous}, and define dZ : Z × Z → R by dZ (f, g) = sup dY (f (x), g(x)). x∈X
Prove that dZ is a metric. Solution. Fix f, g ∈ Z and define φ : X → R by φ(x) = dY (f (x), g(x)). We first claim that φ is continuous. Let > 0 and x ∈ X. Pick δ > 0 such that dX (y, x) < δ implies
13
dY (f (y), f (x)) < /2 and dY (g(y), g(x)) < /2. Then dX (y, x) < δ implies |φ(y) − φ(x)| = |dY (f (y), g(y)) − dY (f (x), g(x))| ≤ |dY (f (y), g(y)) − dY (f (x), g(y))| + |dY (f (x), g(y) − dY (f (x), g(x))| ≤ dY (f (y), f (x)) + dY (g(y), g(x)) < with the last line coming from the triangle inequality in Y . Since X is compact it follows that dZ (f, g) = supx∈X φ(x) is finite. Note that dZ (f, g) ≥ 0 and for any h ∈ Z, dZ (f, g) = sup dY (f (x), g(x)) = sup dY (g(x), f (x)) = dZ (g, f ), x∈X
x∈X
dZ (f, g) = 0 ⇐⇒ sup dY (f (x), g(x)) = 0 x∈X
⇐⇒ dY (f (x), g(x)) = 0 ∀ x ∈ X ⇐⇒ f (x) = g(x) ∀ x ∈ X, dZ (f, g) = sup dY (f (x), g(x)) ≤ sup [dY (f (x), h(x)) + dY (h(x), f (x))] x∈X
x∈X
≤ sup dY (f (x), h(x)) + sup dY (h(x), g(x)) = dZ (f, h) + dZ (h, g). x∈X
x∈X
These statements follow from the fact that dY is a metric.
36. Let Z, X and Y be as in Problem 35, and consider continuous functions fn : X → Y . Prove that fn converges uniformly if and only if {fn } converges in (Z, dZ ). Solution. Suppose {fn } converges uniformly to f . Then since each fn is continuous, f is continuous, so f ∈ Z. Let > 0 and pick N such that n ≥ N implies dY (fn (x), f (x)) < for all x ∈ X. Then n ≥ N implies dZ (fn , f ) = sup dY (fn (x), f (x)) ≤ , x∈X
which shows that {fn } converges to f in (Z, dZ ). Conversely assume that {fn } converges to f in (Z, dZ ). Pick N such that n ≥ N implies dZ (fn , f ) = sup dY (fn (x), f (x)) < . x∈X
Then n ≥ N implies dY (fn (x), f (x)) < for all x ∈ X, which shows uniform convergence.
37. Give an example of sequences {fn }, {gn } of uniformly converging functions such that {fn gn } does not converge uniformly. Solution. Let fn (x) = x + 1/n = gn (x) be defined on [0, ∞). Given > 0, pick N > 1/. Then n ≥ N implies |fn (x) − x| = 1/n < for all x ∈ [0, ∞), which shows uniform convergence to x. Observe that (fn gn )(x) = x2 + 2x/n + 1/n2 converges pointwise to x2 . For = 1, observe that for any N , 2x 1 |(fN gN )(x) − x2 | = + 2 >1= N N 14
when x ≥ N/2. Hence, the convergence is not uniform.
38. Define fn (x) =
x . 1 + nx2
Prove that {fn } converges uniformly to a function f. Show that f 0 (x) = limn→∞ fn0 (x) except when x = 0. Solution. Since each fn is odd, it suffices to consider convergence on [0, ∞). On this interval we have 1 − nx2 fn0 (x) = = 0 ⇐⇒ x = n−1/2 . (1 + nx2 )2 and fn (n−1/2 ) = We claim that 0 ≤ fn (x) ≤ Mn :=
1 . 2n1/2
1 , 2n1/2
x ∈ [0, ∞).
(4)
If not, there is y ∈ [0, ∞) such that fn (y) > Mn . Suppose y < n−1/2 . Since f (0) = 0, IVT implies there is u ∈ (0, y) such that f (u) = Mn , and consequently Rolle’s theorem implies there is v ∈ (u, n−1/2 ) such that fn0 (v) = 0, contradiction. Suppose then that y > n−1/2 . Since limx→∞ fn (x) = 0, IVT implies there is u ∈ (y, ∞) such that f (u) = Mn , and so Rolle’s theorem implies there is v ∈ (n−1/2 , u) such that f 0 (v) = 0, contradiction. Note that (4) implies uniform convergence, since Mn does not depend on x and limn→∞ Mn = 0. The last statement is straightforward to check.
39. Give an example of a sequence of equicontinuous functions {fn } that converges pointwise but not uniformly. Solution. Define fn : R → R by fn (x) = x − (n − 1) for x ∈ [n − 1, n], fn (x) = (n + 1) − x for x ∈ [n, n + 1] and fn (x) = 0 otherwise. To establish equicontinuity, given > 0 let δ = ; then |fn (x) − fn (y)| < whenever |x − y| < δ and n ∈ N. To see pointwise convergence, given > 0 and x ∈ R, pick N > x + 1. Then n ≥ N implies fn (x) = 0, showing that {fn } converges to 0 pointwise. To see that the convergence is not uniform, let = 1/2, take any N , and note that |fN (N ) − 0| = |1 − 0| = 1 > .
40. Let fn : [a, b] → R be monotone for each n. Suppose {fn } converges pointwise to a continuous function. Show that it converges uniformly. Solution. Let f be the pointwise limit of {fn }, which is uniformly continuous since [a, b] is compact. Let > 0 and pick δ > 0 such that |x − y| < δ implies |f (x) − f (y)| < /2. As
15
{(x − δ/2, x + δ/2)}x∈[a,b] is an open cover of [a, b], there is a finite subcover, δ δ δ δ [a, b] ⊂ x1 − , x1 + ∪ . . . ∪ xk − , xk + . 2 2 2 2 where WLOG x1 < . . . < xk . Observe that then xi − xi−1 < δ for i = 2, . . . , k. For i = 1, . . . , k, pick Ni such that n ≥ Ni implies |fn (xi ) − f (xi )| < /2. Let N = max{N1 , . . . , Nk } and let n ≥ N . Assume WLOG that fn is increasing. Fix any x ∈ [a, b]; we have x ∈ [xi−1 , xi ] for some i = 2, . . . , k. Then fn (xi−1 ) − f (x) ≤ fn (x) − f (x) ≤ fn (xi ) − f (x) and so |fn (x) − f (x)| ≤ max{|fn (xi−1 ) − f (x)|, |fn (xi ) − f (x)|}, while for j = i − 1 or j = i, |fn (xj ) − f (x)| ≤ |fn (xj ) − f (xj )| + |f (xj ) − f (x)| <
+ = . 2 2
Combining the last two expressions gives |fn (x) − f (x)| < .
41. Let X be compact, fn : X → R, and {fn } equicontinuous. Suppose {fn } converges pointwise. Prove that it converges uniformly. Solution. Let > 0. Pick δ > 0 such that d(x, y) < δ implies |fn (x) − fn (y)| < /3 for all n. Since {Bδ (x)}x∈X is an open cover of X, there is a finite subcover X ⊂ Bδ (x1 ) ∪ . . . ∪ Bδ (xk ). For i = 1, . . . , k, pick Ni such that m, n ≥ Ni implies |fn (xi ) − fm (xi )| < /3, and let N = max{N1 , . . . , Nk }. Let m, n ≥ N and let x ∈ X be arbitrary. Then x ∈ Bδ (xi ) for some i and |fn (x) − fm (x)| ≤ |fn (x) − fn (xi )| + |fn (xi ) − fm (xi )| + |fm (xi ) − fm (x)| <
+ + = . 3 3 3
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 0 = 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 Z x Ψ(f )(x) = c + φ(t, f (t)) dt, x ∈ [0, 1] 0
16
is a contraction mapping on Z. Solution. Observe that Ψ(Z) ⊂ Z and dZ (Ψ(f ), Ψ(g)) = sup |Ψ(f )(x) − Ψ(g)(x)| x∈[0,1]
Z x = sup [φ(t, f (t)) − φ(t, g(t))] dt x∈[0,1] 0 Z x ≤ sup |φ(t, f (t)) − φ(t, g(t))| dt x∈[0,1] 0
≤ sup |φ(x, f (x)) − φ(x, g(x))| x∈[0,1]
≤ sup K|f (x) − g(x)| x∈[0,1]
= KdZ (f, g), proving that Ψ is a contraction mapping. Note also that Z is complete: if {fn } 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 x
Z |Ψ(f )(x) − c| = |
φ(t, f (t)) dt| ≤ M 0
and
Z |Ψ(f )(x) − Ψ(f )(y)| =
x
y
φ(t, f (t)) dt ≤ M |x − y|.
Thus, Ψ(E) ⊂ E. Note that E is uniformly bounded (by M + c) and equicontinuous: given > 0, pick δ = /M ; then |x − y| < δ implies |f (x) − f (y)| ≤ M |x − y| = 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 theorem5 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
44. Let fn , f : X → R be such that f is continuous at x and fn → f uniformly. Show that xn → x in X implies fn (xn ) → f (x). Solution. Assume xn → x and let > 0. Pick Nf such that n ≥ N implies |fn (y) − f (y)| < /2 for all y ∈ X, pick δ > 0 such that d(y, x) < δ implies |f (y) − f (x)| < /2, and pick Nx such that n ≥ Nx implies d(xn , x) < δ. Let N = max{Nf , Nx }. Then n ≥ N implies |fn (xn ) − f (x)| ≤ |fn (xn ) − f (xn )| + |f (xn ) − f (x)| <
+ = . 2 2
45. Let fn , f : X → R be such that X is compact, f is continuous, and fn (xn ) → f (x) whenever xn → x in X. Show that {fn } converges uniformly to f . Solution. Suppose that {fn } does not converge uniformly to f . Then there is > 0 such that for all N , there exists n ≥ N and y ∈ X such that |fn (y) − f (y)| ≥ . This allows us to build inductively a sequence {ynk } such that |fnk (ynk ) − f (ynk )| ≥ for all k. Since X is compact, {ynk } has a subsequence {xn } converging to some point x ∈ X. Thus, fn (xn ) → f (x) but |fn (xn ) − f (xn )| ≥
for all n.
Pick δ > 0 such that d(y, x) < δ implies d(f (y), f (x)) < /2, and pick N such that n ≥ N implies d(xn , x) < δ. Then n ≥ N implies |fn (xn ) − f (x)| ≥ |fn (xn ) − f (xn )| − |f (xn ) − f (x)| > −
= , 2 2
so that {fn (xn )} does not converge to f (x), contradiction.
46. Let f : [0, ∞) → R and for all n define fn : [0, ∞) → R by fn (x) = f (xn ). Under what conditions on f is {fn } equicontinuous? Solution. If f is a constant function, then of course {fn } is equicontinuous. So suppose f is nonconstant, say f (s) = a and f (t) = b for some s 6= t ∈ [0, ∞). Let = |b − a|/2. If f is equicontinuous, there is δ > 0 such that |x − y| < δ implies |fn (x) − fn (y)| < for all n. Let √ √ n n sn = s and tn = t. Note that |fn (sn ) − fn (tn )| = |f (s) − f (t)| = |b − a| > for all n, but we may choose n large enough so that |sn − tn | < δ. Thus, {fn } cannot be equicontinuous. We have shown that {fn } is equicontinuous if and only if f is constant.
47. Let P∞fn : [a, b] → (0, ∞) be continuous, such that f (x) = that n=1 fn (x) converges uniformly on [a, b].
P∞
n=1 fn (x)
is continuous. Show
P Solution. Let gn (x) = f (x) − nk=1 fk (x) for x ∈ [a, b]. By our assumptions gn is nonnegative and continuous, gn (x) ≤ gn+1 (x) for all x ∈ [a, b], and gn → 0 pointwise. Let > 0 and 18
define Un = {x ∈ [a, b] : gn (x) < }. Then U1 ⊂ U2 ⊂ . . . since gn ≤ gn+1 , pointwise convergence implies that ∪∞ n=1 Un = [a, b], and continuity of gn shows that Un is open in [a, b]. Compactness of [a, b] yields a finite subcover of the Un ’s, say [a, b] = Un1 ∪ . . . ⊂ Unk = UN where N = max{n1 , . . . , nk }. Moreover, n ≥ N implies [a, b] = UN ⊂ Un and so gn (x) < for all x ∈ [a, b]. (This is a special case of Dini’s theorem.)
48. Let {fn } be a sequence of continuous functions [a, b] → R. Suppose fn → 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 {fn } are all continuous and converge pointwise to 0, but the convergence is not uniform since fn (1/n) = 1 for all n (see Problem 44).
49. Let fn : X → R be a uniformly convergent sequence of continuous functions on X. Give an example to show that {fn } may not be equicontinuous. Solution. Let fn (x) = sin(1/x)/n be defined on (0, 1). Since |fn (x)| ≤ 1/n for all x ∈ (0, 1), {fn } converges uniformly to 0. However, {fn } is not equicontinuous because (each) fn is not uniformly continuous.
50. Let fn : X → R be such that X is compact and {fn } is pointwise bounded and equicontinuous. Let φ(x) = supn∈N |fn (x)|. Show φ is continuous.
Solution. Note that φ is well-defined by pointwise boundedness. Let > 0 and pick δ > 0 such that d(x, y) < δ implies |fn (x) − fn (y)| < for every n. Fix x, y ∈ X be such that d(x, y) < δ. Then for every n, |fn (x)| < |fn (y)| + , |fn (y)| < |fn (x)| + , and so φ(x) ≤ φ(y) + ,
φ(y) ≤ φ(x) + .
Thus, |φ(x) − φ(y)| ≤ .
51. Suppose fn : [a, b] → R is such that
P
fn converges uniformly. Show that fn → 0 uniformly.
Solution. Let sn be the nth partial sum, let > 0 and pick N such that n ≥ N implies ∞ ∞ X X s (x) − f (x) = f (x) n−1 < . n k 2 n=1
k=n
19
for all x ∈ [a, b]. Then n ≥ N implies ∞ ∞ X X |fn (x)| = fk (x) − fk (x) < + = 2 2 k=n+1
k=n
for all x ∈ [a, b].
52. Let f : [0, 1] → R be infinitely differentiable, such that f (n) (0) = 0 for n = 0, 1, 2, . . . but f P (n) is not identically zero. Suppose that an f converges uniformly. Show that lim n!an = 0. Solution. Pick y ∈ [0, 1] such that f (y) 6= 0. By Taylor’s theorem, |f (y)| =
|f (n) (ζn )|y n , n!
n = 1, 2, . . . ,
where ζn ∈ (0, y). Thus, |f (y)n!an | ≤
|f (y)n!an | = |an f (n) (ζn )|. yn
(5)
By Problem 51, {an f (n) } converges uniformly to 0. So given > 0, we may choose n such that n ≥ N implies |an f (n) (x)| < |f (y)| for all x ∈ [0, 1]. Then from (5), n ≥ N implies |n!an | < .
53. Suppose fn : Rk → R is such that {fn } 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, {fn } converges uniformly on B δ (x) and so f is continuous on B δ (x), and in particular at x.
54. Let f : R → Rn (n ≥ 2) be differentiable with f 0 (t) 6= 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 = t0 , where f (t0 ) = q. Thus, 0 = φ0 (t0 ) = 2(f (t0 ) − p) · f 0 (t0 ).
55. Suppose f : Rn → R is such that f (tx) = tf (x) for all t ∈ R and x ∈ Rn , 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
Solution. We prove the contrapositive. Suppose that f differentiable. Then f (tv) − f (0) − f 0 (0)(tv) = f (v) − f 0 (0)v, t→0 t
0 = lim
showing f (v) = f 0 (0)v, that is, f is linear. An example is f (x, y) = (x1/3 + y 1/3 )3 .
56. Define f : R2 → 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)| |xy| x2 + y 2 =p ≤p = |(x, y)| → 0 as (x, y) → 0. |(x, y)| x2 + y 2 x2 + y 2
57. Let E ⊂ Rn be open and f : E → R differentiable. Prove that if f has a local maximum at a ∈ E, then f 0 (a) = 0. Solution. Suppose f has a local maximum at a. Then for i = 1, . . . , n and sufficiently small t, f (a + tei ) − f (a) f (a + tei ) − f (a) ≤ 0, if t > 0, ≥ 0, if t < 0. t t This shows that Di f (a) = 0 for i = 1, . . . , n and so f 0 (a) = (D1 f (a), . . . , Dn f (a)) = 0.
58. For E ⊂ R2 , suppose f : E → R is differentiable with D1 f (x) = 0 for all x ∈ E. Under what condition on E can we say f depends on x2 only? Solution. This will be true if for each x2 ∈ R, the set Ex2 := {x1 ∈ R : (x1 , x2 ) ∈ E} is connected (the empty set being considered connected vacuously). Fix x2 ∈ R and define g(x1 ) = f (x1 , x2 ) on Ex2 . Suppose g(x1 ) 6= g(x01 ). Our condition implies that Ex2 contains the interval between x1 and x01 , so by the single variable MVT there is t between x1 and x01 such that g 0 (t) 6= 0. But g 0 (t) = D1 f (t, x2 ), contradiction.
59. Let f : U → Rm be of class C 1 , with U ⊂ Rn an open set containing the line segment L from a to a + h. Suppose T : Rn → Rm is linear with matrix A. Show that |f (a + h) − f (a) − T (h)| ≤ |h| max ||f 0 (x) − A||. x∈L
Solution. Define g(x) = f (x) − T (x). The multivariate MVT implies |g(a + h) − g(a)| ≤ |h| max ||g 0 (x)||. x∈L
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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 ||f 0 (x) − A||. x∈L
60. Let f : Rm → Rm be of class C 1 on the unit ball B1 (0). Suppose that f (0) = 0, f 0 (0) = I and ||f 0 (x) − I|| < for all x ∈ B1 (0). Use Problem 59 to show that f (B1 (0)) ⊂ B1+ (0). Solution. With 1 the identity map, x, y ∈ B1 (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 ||f 0 (z) − I|| < |x − y|, z∈L
where the first inequality above comes from Problem 59, and the second from assumption6 . Thus, |f (x) − f (y)| < (1 + )|x − y|. Taking y = 0 and using f (0) = 0, we get |f (x)| < (1 + )|x| ≤ 1 + .
61. Let f : Rn → Rm be of class C 1 at a and suppose dfa : Rn → Rm is injective. Use Problem 59 to show that f is injective in a neighborhood of a. Solution. First notice that x |dfa (x)| = |x| dfa ≥ |x| min |dfa (y)| = c|x| |x| |y|=1 where c > 0 since dfa is injective. Next, note that since f is class C 1 at a, there is δ > 0 such that ||f 0 (z) − f 0 (a)|| < < c for all z ∈ Bδ (a). So with x 6= y ∈ Bδ (a) and L the line segment from x to y, we have ||f (x) − f (y)| − |dfa (x − y)|| ≤ |f (x) − f (y) − dfa (x − y)| ≤ |x − y| max ||f 0 (z) − f 0 (a)|| < |x − y| z∈L
by Problem 59. Thus, |f (x) − f (y)| > |dfa (x − y)| − |x − y| ≥ c|x − y| − |x − y| = (c − )|x − y| > 0.
62. 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, Kn d(f (x), x). d(f n (x), x∗ ) ≤ 1−K 6
Note that convexity of B1 (0) has also been used.
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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−1 + . . . + K n d(f (x), x) Kn ≤ d(f (x), x). 1−K
(6)
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).
63. Let G : R2 → R be class C 1 , such that G(a, b) = 0 and D2 G(a, b) 6= 0. Prove7 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 fn+1 (x) = fn (x) −
G(x,fn (x)) D2 G(a,b) .
Solution. Using continuity of D2 G, choose δ > 0 such that 1 − D2 G(x, z) ≤ 1 D2 (a, b) 2 for (x, z) ∈ (a − δ, a + δ) × (b − δ, b + δ). Using continuity of G, choose ∈ (0, δ) such that G(x, b) δ D2 G(a, b) < 2 whenever x ∈ (a − , a + ). For x ∈ (a − , a + ), define φx : (b − δ, b + δ) → R by φx (z) = z − Note that |φ0x (z)|
G(x, z) . D2 G(a, b)
D2 G(x, z) 1 = 1 − ≤ . D2 (a, b) 2
Thus, φx is Lipschitz continuous with Lipschitz constant 1/2. Also, G(x, b) 1 < δ. |φx (z) − b| ≤ |φx (z) − φx (b)| + |φx (b) − b| ≤ |z − b| + 2 D2 G(a, b) 7
Don’t assume the implicit function theorem here!
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This shows that φx maps into (b−δ, b+δ). Thus, φx 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 φa (b) = b and the fixed point is unique. Convergence to the fixed point satisfies (see Problem 62) |φnx (z) − f (x)| ≤ δ21−n . (7) Consider the function sequence {fn (x)} from the hint, defined for x ∈ (a − , a + ), with f0 (x) ≡ b. Since f0 and G are continuous, a simple induction argument shows that fn is continuous for each n. Note that fn (x) = φnx (b), so (7) shows that {fn } converges uniformly to f . Thus, f is continuous.
64. Let A be a symmetric n × n matrix and define q : Rn → R by q(x) = xt 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 : Rn → R by g(x) = |x|2 − 1 and let M = {x ∈ Rn : g(x) = 0}. Since g 0 (x) = 2x 6= 0 for all x ∈ M , we may use the method of Lagrange multipliers: q 0 (v) = λg 0 (v),
some λ ∈ R.
(8)
Note that, with Ak the kth row of A, n n n n X X X X Dk q(x) = Dk (xt Ax) = Dk xi Aij xj = Akj xj + xi Aik = 2 Akj xj = 2Ak x, i,j=1
j=1
i=1
j=1
where the last equality uses the fact that A is symmetric. Thus, q 0 (x) = 2Ax, and so from (8) we get Av = λv.
65. Define f : (R+ )n → R by f (x) = n−1 (x1 + . . . + xn ). Find the minimum value of f on the surface S = {x ∈ Rn : g(x) = 0}, where g(x) = x1 . . . xn − c (c > 0 constant). Use this to deduce the geometric-arithmetic mean inequality: for positive reals a1 , . . . , an , (a1 . . . an )1/n ≤ n−1 (a1 + . . . + an ).
Solution. Let h = f |S be the restriction of f to S. We claim it suffices to minimize h in the cube C := [0, nc1/n ]n . Since C is compact, h|C attains a minimum value at a point a ∈ C. Suppose a ∈ int C. Then since g 0 (a) 6= 0, we can use Lagrange multipliers to get f 0 (a) = λg 0 (a), i.e., n−1 = λa1 . . . abi . . . an ,
i = 1, . . . , n.
It follows that a1 = . . . = an , so since a1 . . . an = c, we have a1 = . . . = an = c1/n and f (a) = c1/n . If a ∈ / int C, then ai ≥ nc1/n for some i and so f (a) ≥ c1/n . Thus, h cannot attain a smaller value than c1/n . The geometric-arithmetic mean inequality follows. 24
66. A set P ⊂ Rn 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 ⊂ Rn−1 such that P = {x ∈ Rn : πi (x) ∈ U and xi = h(πi (x))}, where πi (x) := (x1 , . . . , x ˆi , . . . , xn ). A set M ⊂ Rn is called an (n − 1)-manifold if for each x ∈ M there is an open set W ⊂ Rn containing x such that W ∩ M is a (n − 1)-patch. Suppose g : Rn → R is class C 1 and let M = {x ∈ Rn : g(x) = 0}. Suppose g 0 (x) 6= 0 for all x ∈ M . Show that then M is an (n − 1)-manifold in Rn . Solution. Let c ∈ M . Then g 0 (c) 6= 0 so Di g(c) 6= 0, some i = 1, . . . , n. WLOG suppose Dn g(c) 6= 0, and write c = (a, b) with a ∈ Rn−1 , b ∈ R. By the implicit function theorem there is a neighborhood U of a (in Rn−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) ∈ Rn : x ∈ U, y = h(x)}. Note that this is an (n − 1)-patch, so we are done.
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