McGill University Math 354: Honors Analysis 3 Fall 2012 Solutions to selected problems

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1 McGill Uiversity Math 354: Hoors Aalysis 3 Fall 212 Assigmet 3 Solutios to selected problems Problem 1. Lipschitz fuctios. Let Lip K be the set of all fuctios cotiuous fuctios o [, 1] satisfyig a Lipschitz coditio with costat K >, i.e. such that For f Lip K, defie the orm f by Prove that f(x) f(y) K x, y [, 1]. (1) f = sup x [,1] f(x) + sup x,y [,1] x y f(x) f(y). i) f defies the orm o Lip K, i.e. c f = c f ad that f + g f + g. ii) Coclude that d(f, g) := f g defies a distace o Lip K. iii) (extra credit) Lip K is closed, ad that it is the closure of the set of all differetiable fuctios o [, 1] satisfyig f (t) K. iv) The set M = K Lip K is ot closed. v) (ot for credit). What do you thik is the closure of M? Solutio: For (i), we remark that liearity is obvious from the defiitio of f. The triagle iequality follows from the triagle iequality for the sup-orm, ad from takig the supremum over x y i the followig iequality: f(x) + g(x) f(y) g(y) f(x) f(y) + g(x) g(y). The statemet of (ii) follows by a stadard argumet o how a orm defies a distace. For (iii), it may be helpful to view Lip K as a subset of Lip K for K < K or as a subset of C([, 1]). We wat to show that Lip K is closed i either of those two sets with respect to the usual d metric. Assume {f (x)} is a sequece i Lip K, ad that f (x) f(x) as, i.e. that for ay ɛ > there exists N such that for all N, we have f(x) f (x) f(y) + f (y) sup f (x) f(x) + sup < ɛ. (2) x x y Cosider the expressio f(x) f(y) /. It follows from (2) that for N, we have f(x) f(y) ɛ + f (x) f (y) ɛ + K. Sice ɛ was arbitrary, we see that f Lip K (ad is therefore automatically cotiuous), hece Lip K is closed.

2 Next, if f C 1 ([, 1]) with sup t f (t) K, the f Lip K by the itermediate value theorem, sice f(y) f(x) = (y x) f (θ), where θ [x, y]. Next, assume that f(x) satisfies (1) ad cosider the Berstei polyomial B (f, x). It suffices to show that Lemma. If f Lip K, the B (f, x) Lip K. Ideed, B is clearly differetiable for all, approximates f uiformly as by a result proved i class. Also, the iequality B (f, y) B (f, x) K y x implies that B (f, x) K (if we assume that B (f, y) > K for some y [, 1], we shall get a cotradictio with Lipschitz iequality by choosig x close eough to y ad applyig the itermediate value theorem). Proof of the Lemma. (Take from the ote Lipschitz costats for the Berstei polyomials of a Lipschitz cotiuous fuctio by B. Brow, D. Elliott ad D. Paget, Joural of Approximatio Theory 49, , 1987). Let x < y 1. The B (f, y) = = = j= j= j= k= ( ) (1 y) j f j ( ) (1 y) j f j j ( ) j (x + (y x)) j ( j ) [ j k=!x k (y x) j k (1 y) j f k!(j k)!( j)! ( ] j )x k (y x) j k k ( ) j After chagig the order of summatio ad writig k + l = j, (3) becomes k!x k (y x) l (1 y) k l ( ) k + l B (f, y) = f k!(l)!( k l)! k= k= l= We ext write a similar idetity for B (f, x): ( ) ( ) k B (f, x) = x k f ((y x) + (1 y)) k k = = k= ( ) x k f k k k= l= Subtractig (5) from (4) we fid that k B (f, y) B (f, x) = By the Lipschitz coditio, ( k ) [ k ( ] k )(y x) l (1 y) k l l l=!x k (y x) l (1 y) k l f k!l!( k l)! k= l= ( ) k + l f f so it follows from the precedig iequality that B (f, y) B (f, x) K!x k (y x) l (1 y) k l k!l!( k l)! k k= l= ( ) k ( ) ( ) k l K, [ f ( k + l!x k (y x) l (1 y) k l k!l!( k l)! ) f ( ) l. ( )] k (3) (4) (5)

3 The last expressio is equal to K K!(y x) l l!( l)! l= ( l l= ( l ) (y x) l ( l K B (f(z) = z, y x) = K(y x), ) [ l ( ] l )x k (1 y) k l = k k= ) (x + 1 y) l = where i the last lie we have used the idetity B (f(z) = z, x) = x. That idetity follows easily from the secod combiatorial idetity give i the hadout about Berstei approximatio theorem. Summarizig, we have show that B (f, y) B (f, x) K(y x), which fiishes the proof of the Lemma, as well as the part (iii). For items (iv) ad (v), we ote that approximatio is uderstood i terms of the d (uiform) distace. Thus, it follows from Berstei approximatio theorem that the closure of K Lip K is the whole C([, 1]). Ideed, for ay cotiuous fuctio f o [, 1], there exists N such that for ay > N, sup x f(x) B (f, x) ɛ. Now, we claim that B (f, x) K Lip K. Ideed, (d/dx)b (f, x) is cotiuous ad thus takes a maximum value, say K. As discussed before, that shows that B (f, x) Lip K. To see that ot every cotiuous fuctio lies i K Lip K, cosider the fuctio f(x) = x. The (d/dx)f(x) = 1/(2 x) goes to ifiity as x. It is also easy to see that for ay K >, there exist < x < y < 1 such that y x 1 = > K. y x y + x This happes if x ad y are small eough. Thus, x / Lip K for ay K, ad hece it is ot cotaied i their uio. Problem 2. Fredholm equatio. Use the fixed poit theorem to prove the existece ad uiqueess of the solutio to homogeeous Fredholm equatio f(x) = λ K(x, y)f(y)dy. Here K(x, y) is a cotiuous fuctio o [, 1] 2 satisfyig K(x, y) M is called the kerel of the equatio. Cosider the mappig of C([, 1]) ito itself give by (Af)(x) = λ K(x, y)f(y)dy. Let d = d be the usual maximum distace betwee fuctios. Prove that i) Prove that d(af, Ag) λm d(f, g). ii) Coclude that A has a uique fixed poit i C([, 1]) for λ < 1/M, e.g. there exists a uique f C([, 1]) such that Af = f.

4 iii) Prove that f is a solutio of the Fredholm equatio. Solutio. For (i), we fid that Af(x) Ag(x) = λ 1 K(x, y)(f(xy) g(y))dy λk(x, y) f(y) g(y) dy. The last expressio is λ M d(f, g)dy = λ Md(f, g). For (ii), we remark that it follows from (i) that A is a cotractio mappig provided λ < 1/M. The existece of a uique fixed poit follows from stadard results (we should also metio here that the fixed poit theorem depeds o the fact that C([, 1]) is complete with respect to the d metric). Item (iii) follows from the defiitio of A. Problem 3. Relative topology. Let X be a metric space, ad let Y be a subset of X (with the iduced distace). Prove that a set B is ope i Y if ad oly if B = Y A, where A is ope i X. Solutio: Suppose B Y is ope i Y. So, for 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 of X (it is a uio of 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 for 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 be a topological space, X = X, where X is ope for all. Suppose that the restrictio f X is cotiuous for all ; prove that f is cotiuous o X. Solutio. Let f : X Y be our map, ad let V Y be ope. To prove cotiuity, we have to show that f 1 (V ) := {x X : f(x) V } is ope. Now, f 1 (V ) = f 1 (V ) ( X ) = f 1 (V ) X. But the latter set is the preimage of V uder the restrictio f X, ad hece is ope sice f X is cotiuous for all. Thus, f 1 (V ) is ope as a uio of ope sets. A differet proof uses sequeces i the case where X is a metric space. Let x y X. We wat to show that f(x ) f(y). Sice ope sets X m cover the whole space, we have y X m for some m. Sice X m is ope, we have B(y, r) X m for some r >. It follows that x B(y, r) for large eough, so x X m. Sice f Xm is cotiuous, we have f(x ) f(y), QED. Problem 5. Cosider C([a, b]), the vector space of all cotiuous fuctios o [a, b], equipped with the usual orm f p, 1 p. Cosider a map Φ : C([a, b]) C([a, b]) defied by Φ(f) = f 2. For what values of p is this map cotiuous? Please justify carefully your aswer.

5 Solutio: We have Φ(f + h) Φ(f) = 2fh + h 2. First let p =. Let f C([a, b]), deote f = M, ad choose < ɛ < 1/3. Let h < mi(ɛ/(3m), ɛ). The 2fh + h 2 2Mɛ 3M + ɛ2 = ɛ(2/3 + ɛ) < ɛ, hece Φ is cotiuous at f, hece Φ is cotiuous for p =. Cosider ext 1 p <, ad assume WLOG that [a, b] = [, 1]. Let f(x), the Φ(f + h) Φ(f) = h 2. We would like 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 f. Let h(x) = δ 1/p for x [, 1/], h(x) = for x [2/, 1], ad let h(x) be liear for 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 2 1/p h 2 p δ 2 (2) 1/p, ad the expressio diverges as, showig that Φ is ot cotiuous at f. Problem 6. Let M be a bouded subset i C([, 1]). Prove that the set of fuctios F (x) = x f(t)dt, f M (6) has compact closure (i the space of cotiuous fuctios with the uiform distace d ). Solutio: Let F be a family of fuctios defied by (6). By Arzela-Ascoli Theorem, it suffices to show that F is bouded ad equicotiuous. We first remark, that sice M is bouded i C([, 1]) (with d, or uiform, distace), there exists C > such that f(t) < C for all t [, 1] ad for all f M. It follows that for ay x [, 1] ad for ay f M, x x F (x) = f(t)dt f(t) dt Cx C, so F is bouded i C([, 1]). To prove that F is equicotiuous, we fix ɛ >, ad let δ = ɛ/c. Suppose < δ. Assume (without loss of geerality) that x < y. The for ay F F, y y F (y) F (x) = f(t)dt f(t) dt < δ C = ɛ, so F is equicotiuous. This fiishes the proof. x x Problem 7. Let X be a compact metric space with a coutable base, ad let A : X X be a map satisfyig d(ax, Ay) < d(x, y) for all x, y X. Prove that A has a uique fixed poit i X.

6 Solutio. Cosider the fuctio f(x) = d(x, Ax). We first show that f is cotiuous. Ideed, if d(y, x) < ɛ, the d(ay, Ax) < ɛ as well sice A is cotractig, therefore as a result of applyig the triagle iequality we get d(x, Ax) d(y, Ay) < 2ɛ. Sice ɛ was arbitrary, cotiuity follows. A cotiuous fuctio 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. Therefore, d(a, Ay) = ad so y is a fixed poit. Uiqueess follows from the cotractig property of A i the usual way. Problem 8 (extra credit). Give a example of 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 fixed poit. Solutio. Cosider X = N with d(m, ) = 1 + 1/(m + ). That distace defies discrete topology i X, so X is certaily complete (ay Cauchy sequece is evetually costat). Cosider ay icreasig map of N N, for example A() = The A decreases the distace. Ideed, 1 + 1/(m + ) > 1 + 1/(m ), for all m >. It is also clear that A has o fixed poits. Problem 9 (extra credit). Let f C([, 1]). Prove that for ay ɛ > ad N N there exists a fuctio g C([, 1]) such that d 1 (f, g) < ɛ ad g 2 > N. Solutio. The idea is based o a observatio that a fuctio x α, 1/2 α < 1 is itegrable but ot square-itegrable o the iterval [, 1]. So, we fix 1/2 α < 1. We also let M := f = max x f(x). Next, give ɛ > we ca choose < δ such that 2δ x α dx = (2δ)1 α 1 α < ɛ/3, as well as 2δ f(x) dx < ɛ/3. We also let M := f = max x f(x). I additio, we require δ 1 α < ɛ/3 ad δ < ɛ/3m We costruct g(x) as follows: for 2δ x 1, we let f(x) = g(x). For δ x 2δ, we let x = (1 + t)δ, t 1, ad defie g(x) = (1 t)δ α + tf(2δ) (i.e. we iterpolate liearly betwee f ad g. O the iterval [, δ], we remark 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 fially let g(x) = x α for x [η, δ], ad g(x) = η α for x [, η]. We eed to verify that g has the required properties. For the first property we remark that f(x) g(x) dx = 2δ f(x) g(x) dx 2δ f(x) dx + 2δ g(x) dx The first itegral i the right-had side is less tha ɛ/3 by the choice of 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 of δ. Addig the two estimates, we fid that f(x) g(x) dx < ɛ. For the secod iequality, we fid that g(x) 2 dx δ g(x) 2 dx δ η x 2α dx > N 2,

7 so g 2 > N ad the secod requiremet is satisfied. Problem 1. Tube Lemma. Let X be a metric space, ad let Y be a compact metric space. Cosider the product space X Y. If V is a ope set of X Y cotaiig the slice {x } Y of X Y, the V cotais some tube W Y about {x } Y, where W is a eighborhood of x i X. Give a example showig that the Tube Lemma does ot hold if Y is ot compact. Solutio. Let ρ be the distace o X ad σ the distace o Y. We defie 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 defies the d distace i case R 2 = R R. It easy to see that ope balls for the metric d have the form U V, where U is a ope ball i X, ad V is a ope ball i Y (ad similarly for closed balls). It is also easy to see that the topology defied by the distace d = max(ρ, σ) is equivalet to topologies defied by d p := (ρ p + σ p ) 1/p, just like for R 2, (i.e. ope ad closed sets coicide for all distaces), so we ca make our calculatio usig the distace d without loss of geerality. The poit (x, y) is a iterior poit of V for 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 form a ope cover of {x } Y. Sice {x } Y is isometric to Y, it is compact. Accordigly, there exist fiitely may y Y, say y 1, y 2..., y k such that k j=1 U X(y j ) U Y (y j ) cover {x } Y. Let r = mi 1 j k {r(y j )}. The we ca let W = U(x, r) ad the coclusio will hold. For the couterexample i case of ocompact Y, let X = Y = R, x = (so that {x } Y is the y-axis), ad cosider the ope set V = {(x, y) : xy < 1}. Problem 11. Let B deote the set of all sequeces (x ) such that lim x =. Cosider l 1 as a subset of l. Prove that the closure of l 1 i l is equal to B. Solutio. Suppose that x = (x 1, x 2,...) B. Cosider the sequece (x ) =1 of elemets i l 1 where x = (x 1, x 2,..., x,,,...). The because lim x = we have sup x i x i = d(x, x) as. So B cl(l 1 ). To show the reverse iclusio, let x cl(l 1 ) ad choose a sequece (x ) =1 of elemets i l 1 such that sup x i x i as. Let ɛ > be give. Choose N N such i N that N sup i N x i x i ɛ 2. For the elemet xn, choose M N such that x N i ɛ 2 wheever i M. The i M, we have (usig reverse triagle iequality) x i x N i ɛ 2 x i ɛ. So x B. i N Problem 12. (a) Let A X be coected, ad let {A α } α I be a family of coected subsets of X. Show that A A α for 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.

8 Solutio. (a) Let B = A ( α I A α ). Suppose for cotradictio that B is ot coected. The by Lemma we ca assume that B C D, where C ad D are disjoit ope subsets of X that have oempty itersectio with B. By a result proved i class, we kow that A A α is coected for 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 for all α, so B C ad D B =. Cotradictio fiishes the proof. (b) Give x X, cosider the map f : Y X Y give by f(y) = (x, y). The map f is cotiuous ad Y is coected, so {x } Y is coected for every Y. Next, fix y Y. We similarly fid that X {y } is coected. Referrig to part (a), 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 follows from (a) that (X {y }) ( x X {x} Y ) is coected. But the above set is just X Y, so the proof is fiished.