REAL ANALYSIS II: PROBLEM SET 1 - SOLUTIONS 18th Feb, 016 Defiitio (Lipschitz fuctio). A fuctio f : R R is said to be Lipschitz if there exists a positive real umber c such that for ay x, y i the domai of f, f(x) f(y) c x y. Exercise 1. Prove that every uiformly cotiuous real-valued fuctio is a cotiuous fuctio. Prove that every real-valued Lipschitz fuctio is a uiformly cotiuous fuctio. Part 1a. Let f : A R be a uiformly cotiuous fuctio. The f : A R is a cotiuous fuctio. Proof. We will show that f is cotiuous at a poit ad the ote that our choice of poit was arbitrary, which will give us our result of cotiuity o A. Let ε > 0 ad assume f : A R be a uiformly cotiuous fuctio. Choose ay a A. We will choose δ a = δ > 0 where δ > 0 is provided to us by the uiform cotiuity of f. Now, for ay x A such that x a < δ a, we ca coclude by the uiform cotiuity of f that f(x) f(a) < ε. Hece, f is cotiuous at a A. As our choice of a was arbitrary, we have that f is cotiuous o A. Part 1b. Let f : A R be a Lipschitz fuctio with Lipschitz costat 0 < c <. The f : A R is a uiformly cotiuous fuctio. Proof. Let ε > 0 ad assume f : A R be a Lipschitz fuctio with positive Lipschitz costat c. Choose δ < ε. Now, for ay x, y A such that x y < δ, we use this combied with c what we kow about Lipschitz fuctios to yield the followig: f(x) f(y) c x y < cδ < c ε c = ε. So, for ay ε > 0, we ca choose a δ > 0 such that for ay x y < δ, f(x) f(y) < ε. Therefore, f is uiformly cotiuous. Exercise. Costruct a example of a real-valued fuctio that is cotiuous, but ot uiformly cotiuous. Costruct a example of a real-valued fuctio that is uiformly cotiuous, but ot Lipschitz. Date: 18th Feb, 016.
Part a. Cosider the fuctio f : x R x R. This fuctio f is cotiuous but ot uiformly cotiuous o R. Proof. Before we begi the proof, we first observe that f is a cotiuous fuctio. Let ε > 0 ad δ > 0. Sice the real umbers are a Archimedia Field, there exists x > 0 such that ε < δ x. (1) Now, cosider x, ( x + 4) δ R. The differece betwee betwee these two poits, ( x + δ ) x 4 = δ 4 = δ 4 < δ. However, ow lookig at the image of these two poits uder the fuctio f, we discover that ( x + δ x 4) = x + xδ 4 + δ 16 x = xδ + δ 16. Because x ad δ are both positive umbers, we are able to drop the absolute value sig. Furthermore, sice xδ δ ad are both positive, their sum is strictly larger tha both umbers 16 idividually. xδ + δ 16 = xδ + δ 16 > xδ. By (1), this last term is strictly larger tha ε, that is, xδ > ε. Thus, for ay choice of positive ε ad δ, we ca always fid a pair of poits x ad x + δ 4 whose distace apart is less tha δ but whose image, f(x) ad f ( x + 4) δ, are further apart tha ε. Therefore, f is ot uiformly cotiuous but is cotiuous. Part b. Cosider the fuctio f : [0, [ R defied by f(x) = x. uiformly cotiuous, but is ot Lipschitz. This fuctio is Claim 1. The fuctio f : [0, [ R defied by f(x) = x is uiformly cotiuous. Proof. First ote that for 0 x y, the ( y x ) = y xy +x. Because multiplyig by positive umbers preserves order, the followig holds: x y x xy x xy x xy x xy x y xy + x y x ( y x ) y x ( y x ) y x.
Adaptig this yields that for ay x, y [0, [, y x y x. The, for ay ε > 0, choose δ = ε. For ay x, y [0, [, if y x < δ, the y x y x < δ = ε. Therefore f is uiformly cotiuous. Claim. The fuctio f : [0, [ R defied by f(x) = x is ot Lipschitz. Proof. Let 0 < c <. We will show that there exist choices of x, y [0, [ so that c x y < x y. Choose x = 0 ad y > 0 such that c y < 1 which is guarateed by the Archimedia property. We ca rewrite this iequality as cy < y. Now we ca see that f is ot Lipschitz. c y x = c y 0 = cy < y = y 0 = y x. Exercise 3 (L A TEX). Prove the Itermediate Value Property Property (Itermediate Value). Let f be a cotiuous real-valued fuctio o the iterval [a, b]. If f(a) < f(b) ad if c is a umber such that f(a) < c < f(b), the there exists a poit x (a, b) such that f(x) = c. Proof. Let f : [a, b] R be a cotiuous fuctio ad assume further that f(a) < f(b). Let c be ay umber satisfyig f(a) < c < f(b). Defie the followig sets A = f 1 [(, c)] = {x [a, b] : f(x) < c} [a, b], ad B = f 1 [(c, )] = {x [a, b] : c < f(x)} [a, b]. Because A ad B are the iverse images of ope sets, ad f is a cotiuous fuctio, A ad B are both oempty ope sets. Further, by costructio, A B =. The set A is bouded above by b, ad so has a least upper boud, call x = sup A. By costructio, c = sup f[a]. Sice A [a, b], f[a] f[a, b] ad by the Heie-Borel Theorem ad the fact that the image of a compact set is compact, c f[a, b]. We will show that f(x) = c. Let ε > 0 ad δ > 0 be give by the cotiuity of f at c. Furthermore, let x be ay sequece i A such that x x as. So, there exists some N N such that for all > N, x x < δ. I tur, this gives us that f(x ) f(x) < ε. To recap, for ay {x } A N with x x = sup A as, we have that f(x ) f(x) as. This gives us that f(x) is the least upper boud of f[a] ad so f(x) = c. Alterate Proof. As the image of coected sets is agai coected, f[a, b] must agai be a coected set. This gives us that [f(a), f(b)] f[a, b] ad so f takes o ever valued betwee f(a) ad f(b).
Exercise 4. Suppose f is a real-valued fuctio o R which satisfies (f(x + h) f(x h)) = 0 h 0 for every x R. Prove that f is cotiuous or provide a couter-example that satisfies this coditio ad is ot cotiuous. Part 4a. Cosider the Dirichlet fuctio f : R R defied by { 1, if x Q f(x) = 0, otherwise. This fuctio satisfies the property give but is ot cotiuous. Proof. To see that the fuctio is ot cotiuous at ay c R, choose 1 > ε > 0 ad let δ > 0. For ay x R with x c < δ, by the desity of the ratioal i the reals, we ca fid a x 1 betwee x ad c such that, if c is irratioal, x 1 is ratioal or if c is ratioal, x 1 is irratioal. Thus, x 1 c < δ ad f(x 1 ) f(c) = 1 > ε. Therefore f is ot cotiuous at ay poit o R. To see that f satisfies the ecessary property, let x R. The, for ay h, there exists N such that 1 < h. Let h be the umber such that h = 1 ad hh > 0. By costructio, x h ad x + h are both irratioal or ratioal at the same time. Hece f(x + h) = f(x h) ad so f satisfies the give property. Exercise 5. Let {a } =1 be a coverget sequece. Prove that a 1 + a + + a = a. Costruct a sequece {b } =1 such that b 1+b + +b coverges but b does ot coverge. Defiitio. Give a sequece {a } =1, we will say that the sequece is Cesàro Coverget if a 1 + a + + a exists ad is fiite. Part 5a. Let {a } =1 be a coverget sequece. The a 1 + a + + a = a.
Proof. Let a = a ad ε > 0. We will utilize the followig otatioal defiitio of τ = a 1 + a + + a. We ca create a ew sequece {b } =1 where we defie b = a a. Hece, b = (a a) = a a = a a = 0. From this quick calculatio, we kow that b 0 as ad so is a coverget sequece. We also kow that the set B = { b = a a } is bouded. Let B be a upper boud for the set B. Let ε > 0 ad N N such that for all > N, b < ε which is guarateed by our above discussio. Let M N be the atural umber, larger tha N, give by the Archimedia property of the Real umbers that guaratees BN < Mε. Later, we will use this rewritte, as BN < ε. M Now, give ay > M, τ a = a 1 + a + + a a = (a 1 a) + (a a) + + (a a) a 1 a + + a N a + a N+1 a + + a a where the last step follows from the triagle iequality. Rewritig this usig the sequece {b } =1, we yield the followig expressio: τ a b 1 + + b N + b N+1 + + b. We kow that for ay m > N, b m < ε ad so we ca use this for the terms after N. For boudig the terms up to N, we will boud these above by the upper boud B we earlier determied existed. This gives us b 1 + + b N + b N+1 + + b B + + B + ε + + ε = NB < NB + ε + N < ε + ε = ε. Thus, we have that for all ε > 0 there exists some M N such that for all > M, τ a ε. Therefore, we ca coclude that τ a as. Puttig everythig together, we have a 1 + a + + a = a = a. ε
{ Part 5b. Cosider the sequece b = ( 1+( 1) )} =1. The b 1 + b + + b coverges eve though the sequece {b } =1 does ot have a it. Proof. The sequece give, practically speakig, alterates betwee 0 ad 1. To see that this sequece is ot coverget, we will show that the sequece is ot eve Cauchy. Let ε = 1. The, for ay N, b b +1 = 1 > ε ad so the sequece is ot Cauchy ad therefore caot be coverget. Notice that, if is eve, b = 1, ad if is odd, b = 0. We the costruct a ew sequece {τ } =1 where τ = b 1 + b + + b. If is eve, the the sum b 1 + b + + b has half the terms 0 ad the other half 1. The b 1 + b + + b = ad τ = = 1. Otherwise, if is odd, the the sum b 1 + b + + b eds with a 0 ad has half of the remaiig 1 terms that are 1. Hece τ = 1 = 1 1. From ispectio of these two calculatios, we ca hypothesize that the it of {τ } =1 is 1. Let us ow demostrate this fact. 1 Let ε > 0. Choose N N such that for all > N, < ε which is guarateed by the Archimedia property of the Real umbers. Case ( is eve). If is eve, the τ 1 = 1 1 = 0 < ε. Case ( is odd). If is odd, the τ 1 = 1 1 1 = 1 1 < ε. Hece τ 1 < ε for all > N ad therefore τ = 1. Thus, this exhibited sequece is ot coverges ormally, but is Cesàro coverget. Exercise 6 (Dirichlet Series). Let σ > 0. Prove that if a σ is coverget, the Proof. Put ad =1 (a 1 + a + + a ) σ = 0. t = (a 1 + a + + a ) σ s = a 1 1 σ + + a σ
ad s = s. Does the expressio below coverge to zero? t σ ( + 1) σ (s s) = σ (s v s) (v σ (v + 1) σ ) + σ s. () v=1 Exercise 7. Let f : D R R be cotiuous o a dese subset of the domai of f. Prove f ca be exteded to a cotiuous o the etire domai of the fuctio. Proof. Let A be a dese subset of D such that f : A R is cotiuous. By desity, for ay x D \ A, there exists a sequece {x } A N such that x x as. Furthermore, by the cotiuity of f, we have that {f(x )} is a coverget sequece i R. Now, defie f : D R as follows: f(x) = { f(x) if x A, f(x ) if x D \ A ad ay {x } A N with x x. To fiish off the problem, we must show that f is cotiuous. Let ε > 0 be give ad {y } A N, y A, {x } (D \ A) N ad x D \ A. Case 1. Suppose y y as. By the cotiuity of f, f(y ) f(y) as ad thus f(y ) f(y). Case. Suppose y x as. From the defiitio of f, f(y ) f(x) as. Case 3. Suppose x y as. We will costruct a sequece of elemets i A i the followig maer; let y be a poit i A such that x y < 1. We are guarateed to always fid a elemet of A ear a elemet of x because A is dese i D \ A. As its are uique i R (Prove This), y y as. By the cotiuity of f, there exists a δ > 0 ad N 1 N such that for all > N 1, if y y < δ, the f(y ) f(y) < ε. Now, for the give δ, we have that there exists N N such that for > N, y x < δ. Fially, by the costructio of f, there exists some N3 N such that for > N 3, f(x ) f(y ) < ε. Let N = max {N 1, N, N 3 }. This gives us that for ay > N, f(x ) f(y ) < ε, f(y) f(y ) < ε, x y < δ, ad y y < δ. Hece, x y x y + y y < δ + δ = δ which gives us that f(x ) f(y) f(y) f(y ) + f(x ) f(y ) < ε + ε = ε. Therefore f(x ) f(y) as.
Case 4. Suppose x x as. Let y be a poit i A such that x y < 1. The the sequece y x as. By Case, where exists N 1 N ad δ > 0 such that for > N 1, y x < δ gives us that f(y ) f(x) < ε. For the give δ, we have that there exists N N such that for > N, y x < δ. Fially, by the costructio of f, there exists some N3 N such that for > N 3, f(x ) f(y ) < ε. Let N = max {N 1, N, N 3 }. This gives us that for ay > N, f(x ) f(y ) < ε, f(x) f(y ) < ε, x y < δ, ad x y < δ. Hece, x x x y + x y < δ + δ = δ which gives us that f(x ) f(x) f(x) f(y ) + f(x ) f(y ) < ε + ε = ε. Therefore f(x ) f(x) as. The, for ay coverget sequece {z } i D with it z there is a coverget subsequece that falls ito oe of Case 1 through 4 ad by liftig from a subsequece of a coverget sequece, the properties must be preserved ad so f(z ) f(z) as. Therefore f is cotiuous o D. Exercise 8. State ad Prove the Extreme Value Theorem. Theorem. Let f : A R R be cotiuous ad A be compact. The, there exist a, b A such that for all x A, f(b) f(x) f(a). Proof. Let f : A R R be cotiuous ad A be compact. As the image of a compact set is compact, f[a] is a compact set. By the Heie Borel theorem, f[a] is closed ad bouded. Because f[a] is bouded, f[a] has a supremum ad ifimum ad has sequece i f[a] that coverge to the supremum ad ifimum. Utilizig the closedess of f[a], f[a] cotais the its of all coverget sequeces of its poits. Hece the supremum ad ifimum of f[a] are elemets of f[a] ad so f attais its maximum ad miimum values.