Analysis III Math 414 Spring 27 Professor Ben Richert Exam 1 Solutions Problem 1 Let X be the set of all continuous real valued functions on [, 1], and let ρ : X X R be the function ρ(f, g) = sup f g (1) Prove that (X, ρ) is a metric space Solution We are required to show that: (a) for all f, g X, ρ(f, g) = ρ(g, f), (b) ρ(f, g) = if and only if f = g, and (c) for all f, g, h X, ρ(f, g) ρ(f, h) + ρ(g, h) The first fact is obvious since ρ(f, g) = sup f g = sup g f = ρ(g, f) The second fact is similarly immediate since ρ(f, g) = t [, 1], that is, if f = g, and obviously f = g implies that ρ(f, g) = sup observe that ρ(f, g) = sup f g = sup f h + h g sup where the last inequality holds since sup f h + h g f h + h g sup f g = can only occur if f = g for all f g = sup = Finally we ( f h + h g ) for each t [, 1] by the usual triangle inequality We finish by observing that ( ) f h + h g sup f h + sup h g = ρ(f, h) + ρ(h, g) = ρ(f, h) + ρ(g, h) as required (2) Prove that the collection P of polynomials with domains taken to be [, 1] is dense in X Solution We need to show that the closure of P in X is all of X Since the closure of X is the union of P with all its accumulation points, it is enough to show that any f X is an accumulation point of P That is, it is enough to show that for f X and ɛ >, there is p P such that ρ(p, f) < ɛ By the Weierstrass approximation theorem, there does exist a polynomial p such that f(x) p(x) < ɛ for all x [, 1] It follows immediately that ρ(p, f) = sup f p < ɛ, and we are finished (3) Suppose that f j = j ( 1) k x 2k+1 (2k + 1)! for j = 1, 2, Show that f j sin x as elements of X k= Solution Let ɛ > be given We need to demonstrate that there is N > such that n > N implies ρ(sin x, f n ) = sin x f n (x) < ɛ sup
We know that sin x is infinitely differentiable on ( 2, 2), and hence by Taylor s approximation theorem, (with f(x) = sin x for the sake of notation) j sin x = f (k) () xk k! + f (j+1) (x t)j dt j! Of course, The pattern is that dk dx k sin x x= = and thus for each j N, = sin x = j ( 1) k x 2k+1 (2k + 1)! + k= This implies that sin x f j (x) = k= d dx sin x x= = cos() = 1 d 2 dx 2 sin x x= = sin() = d 3 dx 3 sin x x= = cos() = 1 d 4 dx 4 sin x x= = sin() = d 5 dx 5 sin x x= = cos() = 1 2j+1 k= 1 if k = 4l + 1 1 if k = 4l + 3 if k is even f (k) () xk k! + dt = f j (x) + dt f (2j+2) Since f (2j+2) = ± sin t or ± cos t, it follows that f (2j+2) 1 and hence f (2j+2) (x t) 2j+1 dt (x t) 2j+1 dt But (x t) 2j+1 is positive on [, x], so (x t) 2j+1 dt = (x t) 2j+1 (x t)2j+2 dt = (2j + 2)! since x [, 1] Now let N be such that (2N + 2)! > 1 It follows that for all n > N, ɛ x dt (x t) 2j+1 dt dt = ()2j+2 (2j + 2)! + (x)2j+2 (2j + 2)! 1 (2j + 2)! sin x f n (x) 1 (2n + 2)! 1 (2N + 2)! < ɛ, and hence ρ(sin x, f n ) < ɛ for all n > N This completes the proof (4) Suppose that F is a closed, infinite, equicontinuous, equibounded family of functions in X Prove that F is compact in X
Solution We must demonstrate that any infinite {f n } F has a subsequence which converges in F Since F is closed, it contains its limit points, so it is simply enough to identify a convergent subsequence Of course, {f n } is an equicontinuous, equibounded family of functions, and hence, by the Arzela-Ascoli theorem, has an accumulation point, call it f Choose n 1 N such that ρ(f n1, f) < 1, an integer n 2 > n 1 such that ρ(f n2, f) < 1/2, and in general, n j > n j 1 such that ρ(f nj, f) < 1/j (in each case we can do this because f is an accumulation point, and hence there are infinitely many sequence elements of index larger than any given N in B(f, ɛ)) It follows that the subsequence {f nj } converges to f (given ɛ > let N = 1 ɛ, whence ρ(f n, f) < 1 n < 1 ɛ for all n > N as required) We N conclude that any sequence of F has a subsequence which converges in F, which completes the proof (5) Suppose that F is a closed family of functions of X with f(s) f < s t for all f F, s, t [, 1] Suppose further that there is a function g : [, 1] R such that F is compact 1 g(x) dx exists and g(x) f(x) for all f F Prove that Solution Since g is Riemann integrable on [, 1] it is bounded there, and since f(x) g(x) for all x [, 1], f F, it follows that F is equibounded Note also that F is equicontinuous To wit, for ɛ > given, let δ = ɛ 2 Then for all s, t [, 1] such that s t < δ, we have f(s) f < s t < δ = ɛ 2 = ɛ for all f F It follows that F is a closed, equicontinuous, equibounded family, so by the previous problem, it is compact Problem 2 In class we proved that countable unions of measurable sets are measurable Repeat this construction, only this time from the point of view of intersections Note: I expect you to redevelop the theory from the definitions, focusing on intersections where in class we considered unions Solution We want to show that if E j is measurable for all j N, then j=1e j is measurable Defintion 1 Given E R let m (E) = inf I j where the infimum is taken over coverings of E by collections {I j } E ji j of open intervals and I j is the usual length Lemma 1 If A B R, the m (A) m (B) Proof This is clear since A B I j for each open covering of B Lemma 2 For sets A, B R, m (A B) m (A) + m (B) Proof This is clear If A j I j and B i J i, then A B j I j i J i j Defintion 2 A set E R is said to be measurable if for all A R, m (A) = m (A E) + m (A E c ) Lemma 3 A set E R is measurable if and only if E c is measurable Proof This follows because the equation m (A) = m (A E) + m (A E c ) is symmetric in E and E c Lemma 4 To show that E R is measurable, it is enough to show that m (A) m (A E) + m (A E c ) for an arbitrary A R Proof Since A = (A E) (A E c ), it follows by lemma (2) that m (A) m (A E) + m (A E c ) holds automatically Lemma 5 Suppose that E 1 and E 2 are measurable sets Then E 1 E 2 is measurable Proof Let A R, and note that A (E 1 E 2 ) c = ((A E 1 ) E 2 ) (A E c 1) Then m (A (E 1 E 2 )) + m (A (E 1 E 2 ) c ) m ((A E 1 ) E 2 ) + m ((A E 1 ) E c 2) + m (A E c 1) = m (A E 1 ) + m (A E1) c = m (A) The first inequality holds by lemma (2), the second holds because E 2 is measurable, and the last holds because E 1 is measurable This is enough by lemma (4) Lemma 6 Suppose that E 1 and E 2 are measurable sets Then E 1 E 2 is measurable Proof This follows immediately from lemma (5) and lemma (3) Lemma 7 Suppose that E 1,, E n is a collection of measurable sets Then n j=1e j is measurable
Proof We do induction In fact, if n = 1 this is a tautology If n = 2, then the result follows from lemma (5) Suppose that n > 2 By induction, we can conclude that E := E 1 E n 1 is measurable, and again by lemma (5) E E n is measurable Since E E n = E 1 E n, we are finished Lemma 8 Suppose that {E i } is a collection of measurable sets Then there are measurable sets {F i } such that F i F j = R for i j E j = F j Proof Let F 1 = E 1 F 2 = E 2 E1 c F 3 = E 3 (E 2 E 1 ) c F j = E j (E j 1 E 1 ) c By lemmas (3,6,7), F i is measurable for all i If j > i, then ( F j F i = E j (E j 1 E 1 ) c) (E i (E i 1 E 1 ) c) = E j E c j 1 E c i E c 1 E i E c i 1 E c 1 = R since both E i and Ei c appear in the union Furthermore, note that since E i F i, it follows that E i F i If x F i, then by definition x E i (E i 1 E 1 ) c Of course, it is now enough to show that x E i for all i, that is, it is enough to show that the set A = {t x E t } is empty Note that 1 A by definition If A, then there is a smallest element, 1 < a A Thus, x E a 1 E 1, but x E a But x F a = E a (E a 1 E 1 ) c, which must then imply that x (E a 1 E 1 ) c, and this is a contradiction Lemma 9 Suppose that F 1,, F n are measurable sets and F i F j = R for i j Then F c n (F 1 F n 1 ) c = Proof This is clear since (F c n (F 1 F n 1 ) c ) c = (F n (F 1 F n 1 )) = (F n F 1 ) (F n F 1 ) = R R = R Lemma 1 Suppose that E 1 and E 2 are measurable sets and E 1 E 2 = The m (E 1 ) + m (E 2 ) = m (E 1 E 2 ) Proof Since E 1 is measurable, we have But E 1 E 2 =, and E 2 E c 1 = E 2, so the result is proved m (E 2 E 1 ) + m (E 2 E c 1) = m (E 2 ) Lemma 11 Suppose that F 1,, F n are measurable sets F i F j = R for i j Then m (A (F 1 F n ) c ) = Proof We do induction on n If n = 1 this is a tautology If n > 1, then write m (A (F 1 F n ) c ) = m (A ((F 1 F n 1 ) c F c n)) = m ((A F c n) (A ((F 1 F n 1 ) c )) = m (A Fn) c + m (A (F 1 F n 1 ) c ) n 1 n = m (A Fn) c + m (A Fi c ) = m (A Fi c ) n m (A Fi c ) where the third equality holds because Fn c and (F 1 F n 1 ) c are measurable and disjoint (see lemmas (9,1)), and the fourth inequality holds by the induction hypothesis Lemma 12 Suppose that {E j } is a countable collection of measurable sets Then m (A Ej c ) m (A ( E j ) c )
Proof Let ɛ > be given and suppose that m (A Ej c ) = e j Then (by the definition of outer measure) there is {I (j) i } such that i I (j) i (A Ej c ) and I (j) i < e j + ɛ/2 j Of course i so m (A ( j E j ) c ) j completes the proof i I (j) i = j j i I (j) i j (A E c j ) = A ( j E j ) c, e j + ɛ/2 j = ɛ + j e j = ɛ + j m (A E c j ) Since this is true for each ɛ >, this Lemma 13 Suppose that {E j } is a countable collection of measurable sets Then j E j is measurable Proof First, we find {F j } such that F i F j = R for i j and E j = F j (as guaranteed to exist by lemma (8)) Let G = F i = E i and G n = F 1 F n Then G n is measurable for all n N by lemma (7), and hence m (A) = m (A G n ) + m (A G c n) for all A R Since G n = F 1 F n F i = G, it follows that m (A G n ) m (A G) (by lemma (1)), and thus Now by lemma (11), we can write and this holds for all n We conclude that But by lemma (12), m (A) = m (A G n ) + m (A G c n) m (A G) + m (A G c n) m (A) m (A G) + m (A G c n) = m (A G) + m (A) m (A G) + n m (A Fi c ) m (A Fi c ) m (A Fi c ) m (A ( Fi c )) = m (A ( F i ) c ), and this certainly completes the proof, since now we have that m (A) m (A G) + m (A Fi c ) m (A G) + m (A ( F i ) c ) = m (A G) + m (A G c ), which is enough by lemma (4) Problem 3 Let {f j } be a sequence of continuous, real valued functions on R with the property that {f j (x)} is unbounded for all x Q Use the Baire Category theorem to prove that it cannot then be true that whenever t is irrational then the sequence {f j } is bounded Note: this is a somewhat challenging problem You need to start by deciding what complete metric space the Baire Category theorem should give you information about Solution We will show that if the sequence {f j } is bounded at each irrational t, then we can write R as a countable union of nowhere dense sets, which contradicts the Baire Category theorem (since R is a complete metric space) For j N, let { } A j = t R Q { f j } is bounded (not strictly) by j Let {q j } be an enumeration of Q, and let Q j = {q j } By hypothesis, every t R Q is in some A j, and of course j=1q j = Q, so clearly ( j=1q j ) ( i j=nftya j ) = R Also, this is a countable union, so to obtain the contradiction it is enough to show that Q j and A j for all j N are nowhere dense That Q j is nowhere dense is immediate, since Q j = {q j } consists of a single point For A j, we first show that it is closed Suppose that r is an accumulation point of A j We need to demonstrate that f N (r) < j for all N N, that is, we need to show that r A j So fix N N Then given ɛ >, there is δ > (since f N is
continuous) such that x r < δ implies that f N (x) f N (r) < ɛ But r is an accumulation point of A j, so there is t A j such that t r < δ and hence f N (r) < f N + ɛ j + ɛ Since ɛ was arbitrary, we conclude that f N (r) j But N was also arbitrary Hence the conclusion that {f j (r)} is bounded above by j and thus r A j as required Next we show that A j contains no open balls (this is enough since A j is its own closure) In fact, this is easy because (by definition) Q A j =, but there are rationals arbitrarily close to any irrational To be precise, suppose that t A j and ɛ > is given Then the ball B(t, ɛ) = {x R x t < ɛ} Aj because (by the density of the rationals in the reals) there is q Q such that q t < ɛ, but q A j We conclude that A j is nowhere dense By the Baire Category theorem, this completes the proof