McGill Uiversity Math 354: Hoors Aalysis 3 Fall 212 Assigmet 3 Solutios to selected problems Problem 1. Lipschitz uctios. Let M K be the set o all uctios cotiuous uctios o [, 1] satisyig a Lipschitz coditio with costat K >, i.e. such that For M K, deie the orm by Prove that (x) (y) K x, y [, 1]. (1) sup x [,1] (x) + sup x,y [,1] (x) (y). i) deies the orm o M K, i.e. c c ad that + g + g. ii) Coclude that d(, g) : g deies a distace o M K. iii) (extra credit) M K is closed, ad that it is the closure o the set o all dieretiable uctios o [, 1] satisyig (t) K. iv) The set M K M K is ot closed. v) (ot or credit). What do you thi is the closure o M? Solutio: For (i), we remar that liearity is obvious rom the deiitio o. The triagle iequality ollows rom the triagle iequality or the sup-orm, ad rom taig the supremum over x y i the ollowig iequality: (x) + g(x) (y) g(y) (x) (y) + g(x) g(y). The statemet o (ii) ollows by a stadard argumet o how a orm deies a distace. For (iii), assume that { (x)} is a sequece i Lip K, ad that (x) (x) as, i.e. that or ay ɛ > there exists N such that or all N, we have (x) (x) (y) + (y) sup (x) (x) + sup < ɛ. (2) x x y Cosider the expressio (x) (y) /. It ollows rom (2) that or N, we have (x) (y) ɛ + (x) (y) ɛ + K. Sice ɛ was arbitrary, we see that Lip K (ad is thereore automatically cotiuous), hece Lip K is closed. Next, i C 1 ([, 1]) with sup t (t) K, the Lip K by the itermediate value theorem, sice (y) (x) (y x) (θ), where θ [x, y]. Next, assume that (x) satisies (1) ad cosider the Berstei polyomial B (, x). It suices to show that
Lemma. I Lip K, the B (, x) Lip K. Ideed, B is clearly dieretiable or all, approximates uiormly as by a result proved i class. Also, the iequality B (, y) B (, x) K y x implies that B (, x) K (i we assume that B (, y) > K or some y [, 1], we shall get a cotradictio with Lipschitz iequality by choosig x close eough to y ad applyig the itermediate value theorem). Proo o the Lemma. (Tae rom the ote Lipschitz costats or the Berstei polyomials o a Lipschitz cotiuous uctio by B. Brow, D. Elliott ad D. Paget, Joural o Approximatio Theory 49, 196 199, 1987). Let x < y 1. The B (, y) j j j ( ) (1 y) j j ( ) (1 y) j j j ( ) j (x + (y x)) j ( j ) [ j!x (y x) j (1 y) j!(j )!( j)! ( ] j )x (y x) j ( ) j (3) Ater chagig the order o summatio ad writig + l j, (3) becomes B (, y) l We ext write a similar idetity or B (, x): B (, x) ( ) x ( ) x l!x (y x) l (1 y) l!(l)!( l)! ( ) ((y x) + (1 y)) ( ( ) + l ) [ ( ] )(y x) l (1 y) l l l!x (y x) l (1 y) l!l!( l)! ( ) (4) (5) Subtractig (5) rom (4) we id that B (, y) B (, x) By the Lipschitz coditio, l ( ) + l so it ollows rom the precedig iequality that B (, y) B (, x) K!x (y x) l (1 y) l!l!( l)! l ( ) ( ) l K, [ ( + l!x (y x) l (1 y) l!l!( l)! ) ( ) l. ( )]
The last expressio is equal to K K!(y x) l l!( l)! l ( l l ( l ) (y x) l ( l K B ((z) z, y x) K(y x), ) [ l ( ] l )x (1 y) l ) (x + 1 y) l where i the last lie we have used the idetity B ((z) z, x) x. That idetity ollows easily rom the secod combiatorial idetity give i the hadout about Berstei approximatio theorem. Summarizig, we have show that B (, y) B (, x) K(y x), which iishes the proo o the Lemma, as well as the part (iii) o Problem 1. For items (iv) ad (v), we ote that approximatio is uderstood i terms o the d (uiorm) distace. Thus, it ollows rom Berstei approximatio theorem that the closure o K Lip K is the whole C([, 1]). Ideed, or ay cotiuous uctio o [, 1], there exists N such that or ay > N, sup x (x) B (, x) ɛ. Now, we claim that B (, x) K Lip K. Ideed, (d/dx)b (, x) is cotiuous ad thus taes a maximum value, say K. As discussed beore, that shows that B (, x) Lip K. To see that ot every cotiuous uctio lies i K Lip K, cosider the uctio (x) x. The (d/dx)(x) 1/(2 x) goes to iiity as x. It is also easy to see that or ay K >, there exist < x < y < 1 such that y x 1 > K. y x y + x This happes i x ad y are small eough. Thus, x / Lip K or ay K, ad hece it is ot cotaied i their uio. Problem 2. Fredholm equatio. Use the ixed poit theorem to prove the existece ad uiqueess o the solutio to homogeeous Fredholm equatio (x) λ K(x, y)(y)dy. Here K(x, y) is a cotiuous uctio o [, 1] 2 satisyig K(x, y) M is called the erel o the equatio. Cosider the mappig o C([, 1]) ito itsel give by (A)(x) λ K(x, y)(y)dy. Let d d be the usual maximum distace betwee uctios. Prove that i) Prove that d(a, Ag) λm d(, g). ii) Coclude that A has a uique ixed poit i C([, 1]) or λ < 1/M, e.g. there exists a uique C([, 1]) such that A. iii) Prove that is a solutio o the Fredholm equatio.
Solutio: For (i), we id that A(x) Ag(x) λ 1 K(x, y)((xy) g(y))dy λk(x, y) (y) g(y) dy. The last expressio is λ M d(, g)dy λ Md(, g). For (ii), we remar that it ollows rom (i) that A is a cotractio mappig provided λ < 1/M. The existece o a uique ixed poit ollows rom stadard results. Item (iii) ollows rom the deiitio o A. Problem 3. Relative topology. Let X be a metric space, ad let Y be a subset o X (with the iduced distace). Prove that a set B is ope i Y i ad oly i B Y A, where A is ope i X. Solutio: Suppose B Y is ope i Y. So, or every y B there exists a positive umber r y such that U Y (y, r y ) : {z Y : d(y, z) < r y } B. Let A y B U X (y, r y ). Clearly, A is a ope subset o X (it is a uio o ope balls). Also, A Y y B (U X (y, r y ) Y ) y B U Y (y, r y ) B, sice U Y (y, r y ) B. For the coverse, let V be ope i X, ad let y V Y. The U X (y, t) V or some t >. But U Y (y, t) U X (y, t) Y V Y. Thus, V Y is ope i Y, QED. Problem 4. Let X X, where X is ope or all. Suppose that the restrictio X is cotiuous or all ; prove that is cotiuous o X. Solutio: Let : X Y be our map, ad let V Y be ope. To prove cotiuity, we have to show that 1 (V ) : {x X : (x) V } is ope. Now, 1 (V ) 1 (V ) A. But the latter set is the preimage o V uder the restrictio A, ad hece is ope sice A is cotiuous or all. Thus, 1 (V ) is ope as a uio o ope sets. A dieret proo uses sequeces. Let x y X. We wat to show that (x ) (y). Sice ope sets A m cover the whole space, we have y A m or some m. Sice A m is ope, we have B(y, r) A m or some r >. It ollows that x B(y, r) or large eough, so x A m. Sice Am is cotiuous, we have (x ) (y), QED. Problem 5. Cosider C([a, b]), the vector space o all cotiuous uctios o [a, b], equipped with the usual orm p, 1 p. Cosider a map Φ : C([a, b]) C([a, b]) deied by Φ() 2. For what values o p is this map cotiuous? Please justiy careully your aswer. Solutio: We have Φ( + h) Φ() 2h + h 2. First let p. Let C([a, b]), deote M, ad choose < ɛ < 1/3. Let h < mi(ɛ/(3m), ɛ). The 2h + h 2 2Mɛ 3M + ɛ2 ɛ(2/3 + ɛ) < ɛ, hece Φ is cotiuous at, hece Φ is cotiuous or p. Cosider ext 1 p <, ad assume WLOG that [a, b] [, 1]. Let (x), the Φ( + h) Φ() h 2. We would lie to choose h C([a, b]) such that h p is small, but h 2 p is large to prove that Φ is ot cotiuous at.
Let h(x) δ 1/p or x [, 1/], h(x) or x [2/, 1], ad let h(x) be liear or x [1/, 2/]. The it is easy to show that δ p (h(x)) p dx 2δ p, hece δ h p δ 2 1/p, ad the expressio goes to as δ. O the other had, it is also easy to show that δ 2p (h(x)) 2p dx 2δ 2p, hece δ 1/(2p) h 2 p δ (2) 1/(2p), ad the expressio diverges as, showig that Φ is ot cotiuous at. Problem 6. Let M be a bouded subset i C([, 1]). Prove that the set o uctios F (x) x (t)dt, M (6) has compact closure (i the space o cotiuous uctios with the uiorm distace d ). Solutio: Let F be a amily o uctios deied by (6). By Arzela-Ascoli Theorem, it suices to show that F is bouded ad equicotiuous. We irst remar, that sice M is bouded i C([, 1]) (with d, or uiorm, distace), there exists C > such that (t) < C or all t [, 1] ad or all M. It ollows that or ay x [, 1] ad or ay M, x x F (x) (t)dt (t) dt Cx C, so F is bouded i C([, 1]). To prove that F is equicotiuous, we ix ɛ >, ad let δ ɛ/c. Suppose < δ. Assume (without loss o geerality) that x < y. The or ay F F, y y F (y) F (x) (t)dt (t) dt < δ C ɛ, x so F is equicotiuous. This iishes the proo. Problem 7. Let X be a compact metric space with a coutable base, ad let A : X X be a map satisyig d(ax, Ay) < d(x, y) or all x, y X. Prove that A has a uique ixed poit i X. Solutio: Cosider the uctio (x) d(x, Ax). We irst show that is cotiuous. Ideed, i d(y, x) < ɛ, the d(ay, Ax) < ɛ as well sice A is cotractig, thereore d(x, Ax) d(y, Ay) < 2ɛ. Sice ɛ was arbitrary, cotiuity ollows. A cotiuous uctio o a compact set attais its miimum, say at a poit y X. Suppose that the miimum is positive, i.e. that Ay y. The d(a 2 y, Ay) < d(ay, y) sice A is cotractig, which cotradicts the miimality. Thereore, d(a, Ay) ad so y is a ixed poit. Uiqueess ollows rom the cotractig property o A i the usual way. Problem 8 (extra credit). Give a example o a o-compact but complete metric space X ad a map A : X X as i Problem 7 such that A does t have a ixed poit. Solutio: Cosider the example rom Problem 4, Assigmet 2: X N with d(m, ) 1+1/(m+ ). That distace deies discrete topology i X, so X is certaily complete (ay Cauchy sequece is evetually costat). Cosider ay icreasig map o N N, or example A() 2 + 1. The A decreases the distace. Ideed, 1 + 1/(m + ) > 1 + 1/(m 2 + 2 + 2), or all m >. It is also clear that A has o ixed poits. x
Problem 9 (extra credit). Let C([, 1]). Prove that or ay ɛ > ad N N there exists a uctio g C([, 1]) such that d 1 (, g) < ɛ ad g 2 > N. Solutio. The idea is based o a observatio that a uctio x α, 1/2 α < 1 is itegrable but ot square-itegrable o the iterval [, 1]. So, we ix 1/2 α < 1. We also let M : max x (x). Next, give ɛ > we ca choose < δ such that x α dx (2δ)1 α 1 α < ɛ/3, as well as (x) dx < ɛ/3. We also let M : max x (x). I additio, we require δ 1 α < ɛ/3 ad δ < /epsilo/3m We costruct g(x) as ollows: or 2δ x 1, we let (x) g(x). For delta x 2δ, we let x (1 + t)δ, t 1, ad deie g(x) (1 t)δ α + t(2δ) (i.e. we iterpolate liearly betwee ad g. O the iterval [, δ], we remar that as η, we have δ η x 2α dx, so we ca choose η > so that δ η x 2α dx > N 2. We ially let g(x) x α or x [η, δ], ad g(x) η α or x [, η]. We eed to veriy that g has the required properties. For the irst property we remar that (x) g(x) dx (x) g(x) dx (x) dx + g(x) dx The irst itegral i the right-had side is less tha ɛ/3 by the choice o δ. The secod itegral is less tha δ x α dx + δ max(δ α, M) ɛ 3 + max(δ1 α, δm) < 2ɛ 3, also by the choice o δ. Addig the two estimates, we id that (x) g(x) dx < ɛ. For the secod iequality, we id that g(x) 2 dx δ g(x) 2 dx δ η x 2α dx > N 2, so g 2 > N ad the secod requiremet is satisied. Problem 1. Tube Lemma. Let X be a metric space, ad let Y be a compact metric space. Cosider the product space X Y. I V is a ope set o X Y cotaiig the slice {x } Y o X Y, the V cotais some tube W Y about {x } Y, where W is a eighborhood o x i X. Give a example showig that the Tube Lemma does ot hold i Y is ot compact. Solutio. Let ρ be the distace o X ad σ the distace o Y. We deie the maximum distace d o X Y by d((x 1, y 1 ), (x 2, y 2 )) max(ρ(x 1, x 2 ), σ(y 1, y 2 )). (7) This deies the d distace i case R 2 R R. It easy to see that ope balls or the metric d have the orm U V, where U is a ope ball i X, ad V is a ope ball i Y (ad similarly or closed balls). It is also easy to see that the topology deied by the distace d max(ρ, sigma) is equivalet to topologies deied by d p : (ρ p + σ p ) 1/p, just lie or R 2, (i.e. ope ad closed sets coicide or all distaces), so we ca mae our calculatio usig the distace d without loss o geerality.
The poit (x, y) is a iterior poit o V or all y Y, hace there exist r r(y) > such that the ball U X (x, r(y)) U Y (y, r(y)) cetered at (x, y) is cotaied i V. Call the correspodig balls U X (y) ad U Y (y). The balls {U X (y) U Y (y)} y Y orm a ope cover o {x } Y. Sice {x } Y is isometric to Y, it is compact. Accordigly, there exist iitely may y Y, say y 1, y 2..., y such that j1 U X(y j ) U Y (y j ) cover {x } Y. Let r mi 1 j {r(y j )}. The we ca let W U(x, r) ad the coclusio will hold. For the couterexample i case o ocompact Y, let X Y R, x (so that {x } Y is the y-axis), ad cosider the ope set V be {(x, y) : y < 1/ x, or y.} Problem 11. Let B deote the set o all sequeces (x ) such that lim x. Cosider l 1 as a subset o l. Prove that the closure o l 1 i l is equal to B. Problem 12. a) Let A X be coected, ad let {A α } α I be a amily o coected subsets o X. Show that A A α or all α I, the A ( α I A α ) is coected. b) Let X ad Y be coected metric spaces. Show that X Y is coected. Solutio. a) Let B A ( α I A α ). Suppose or cotradictio that B is ot coected. The by Lemma we ca assume that B C D, where C ad D are disjoit ope subsets o X that have oempty itersectio with B. By a result proved i class, we ow that A A α is coected or all α. The A A α has to lie etirely i C or etirely i D, otherwise they will separate A A α. Thus A A α C (say). But this must the hold or all α, so B C ad D B. Cotradictio iishes the proo. (b) Give x X, cosider the map : Y X Y give by (y) (x, y). The map is cotiuous ad Y is coected, so {x } Y is coected or every Y. Next, ix y Y. We similarly id that X {y } is coected. I Problem 5, let A X {y }, ad let A x {x} Y, where the idex α is replaced by x X. Now, A A x (x, y ). It ollows rom Problem 5 that (X {y }) ( x X {x} Y ) is coected. But the above set is just X Y, so the proo is iished.