We will begin by supplying the proof to (a).

Transcription

1 (The solutios of problem re mostly from Jeffrey Mudrock s HWs) Problem 1. There re three sttemet from Exmple 5.4 i the textbook for which we will supply proofs. The sttemets re the followig: () The spce C k ([, b]) of k-times cotiuously differetible fuctios o [, b] is ot Bch spce with respect to the sup-orm for k 1, sice the uiform limit of cotiuously differetible fuctios eed ot be differetible. (b) C k ([, b]) is Bch spce with respect to the C k -orm. (c) Covergece with respect to the C k -orm implies uiform covergece of fuctios d their first k derivtives. We will begi by supplyig the proof to (). Let (f ) =1 be the sequece of fuctios i C 1 ([ 1, 1]) give by f (x) = x (Note: We kow from 2 clculus tht ll the fuctios i this sequece re cotiuously differetible). Now, suppose f : [ 1, 1] R is give by f(x) = x. Now, for y N we clculte tht: f f = sup x 1 x x = sup x 1 x x 2 1 = sup + x 2 x x 1 x x sup 2 1 x 1 = 1. So, we hve tht for ech N, f f 1. The, by the squeeze theorem, we coclude tht lim f f =. Thus, (f ) =1 coverges uiformly to f, d we kow from clculus tht f is ot differetible t. We coclude tht the spce C k ([, b]) of k-times cotiuously differetible fuctios o [, b] is ot Bch spce with respect to the sup-orm for k 1, sice the uiform limit of cotiuously differetible fuctios eed ot be differetible. 1 2 Now, we tur our ttetio to sttemet (b). We will prove tht C k ([, b]) is Bch spce with respect to the C k -orm for k N by iductio o k. For the bse cse, we ote tht the sttemet is true for k = 1 by Exercise 2.5 (problem 5) from homework ssigmet umber 3. Now, for the iductive step, suppose tht k N d k 2. We ssume tht C l ([, b]) is Bch spce with respect to the C l -orm for ll turl umbers 1

2 l which stisfy 1 l k 1. To complete the iductive step we must show tht this implies tht C k ([, b]) is Bch spce with respect to the C k -orm. Suppose tht (f ) =1 is rbitrry Cuchy sequece i C k ([, b]) with respect to the C k -orm. This mes tht for rbitrry positive rel umber ɛ, there is N N such tht f f m C k = i= f (i) m < ɛ wheever, m N (Note: f (i) is the i th derivtive of f). This mes tht if, m N, the k 1 f f m C k 1 = i= f (i) m i= f (i) m < ɛ. So, we hve tht (f ) =1 is lso Cuchy sequece i C k 1 ([, b]) with respect to the C k 1 -orm. Sice we kow tht C k 1 ([, b]) is Bch spce with respect to the C k 1 -orm, we kow tht (f ) =1 coverges to some fuctio f C k 1 ([, b]) with respect to the C k 1 -orm. Similrly, if, m N, the f (k) f (k) m i= f (i) m < ɛ. So, we hve tht (f (k) ) =1 is Cuchy sequece i C([, b]) with respect to the sup-orm. By Theorem 2.4, we kow tht this mes tht (f (k) ) =1 coverges to some fuctio g C([, b]) with respect to the sup-orm. Now, sice (f ) =1 coverges to f C k 1 ([, b]) with respect to the C k 1 -orm, we kow tht if ɛ 1 is rbitrry positive rel umber, there is N 1 N such tht f f C k 1 < ɛ 1 wheever N 1. This mes tht if N 1, the f (k 1) f (k 1) f f C k 1 < ɛ 1. So, we hve tht (f (k 1) ) =1coverges to f (k 1) with respect to the sup-orm. Suppose tht x is rbitrry rel umber i [, b]. We will ow show tht lim f (k) (t)dt = g(t)dt. We first ote tht if x =, the the bove equtio is true sice both sides equl zero i this cse. Now, suppose tht x >. Sice (f (k) (x)) =1 coverges to g C([, b]) with respect to the sup-orm, we kow tht for y positive rel umber ɛ 2, there is N 2 N such tht f (k) g = sup t b f (k) (t) g(t) < 2

3 ɛ 2 x for ll N 2. So, if N 2, we hve by bsic properties of itegrls tht x f (k) (t)dt g(t)dt = (f (k) (t) g(t))dt sup f (k) (t) g(t) (x ) So, we hve tht for y x [, b], lim f (k) (t)dt = g(t)dt. t b < ɛ 2 x (x ) = ɛ 2. We kow from clss d the text tht sice (f (k 1) (x)) =1 coverges to f (k 1) C([, b]) with respect to the sup-orm, the sequece (f (k 1) (x)) =1 lso coverges to f (k 1) poitwise. So, we hve tht for y x [, b], lim f (k 1) (x) = f (k 1) (x). We ote tht by the Fu- Now, suppose x is rbitrry elemet i [, b]. dmetl Theorem of Clculus f (k 1) (x) f (k 1) () = f (k) (t)dt for every N. If we tke the limit of both sides of this equtio s goes to ifiity we obti (by the fcts we hve show): f (k 1) (x) f (k 1) () = g(t)dt. This equtio clerly holds for ll x [, b] sice x ws tke rbitrrily. The, by the Fudmetl Theorem of Clculus, we hve tht f (k) (x) = g(x) for ech x [, b]. Sice g C([, b]), we kow tht f C k ([, b]). We filly clim tht (f ) =1 coverges to f with respect to the C k -orm. To see why this is so, suppose tht ɛ 3 is rbitrry positive rel umber. Sice (f ) =1 coverges to f with respect to the C k 1 -orm, we kow tht there is N 3 N such tht k 1 f f C k 1 = f (i) f (i) < ɛ 3 2 i= wheever N 3. Similrly sice (f (k) ) =1 coverges to g with respect to the sup-orm d f (k) (x) = g(x) for ech x [, b], we kow there is N 4 N such tht f (k) f (k) < ɛ 3 2 wheever N 4. So, if mx{n 3, N 4 }, we hve tht f f C k = i= k 1 f (i) f (i) = f (i) f (i) + f (k) f (k) < ɛ 3 2 +ɛ 3 2 = ɛ 3. i= 3

4 Thus, (f ) =1 coverges to f with respect to the C k -orm. Sice (f ) =1 ws rbitrry Cuchy sequece i C k ([, b]), we coclude tht C k ([, b]) is Bch spce with respect to the C k -orm. This completes the iductive step, d we re fiished provig (b). Now, we tur our ttetio to provig (c). Suppose tht (f ) =1 is sequece of fuctios i C k ([, b]) which coverges to f C k ([, b]) with respect to the C k -orm (where k 1). Now, suppose tht is iteger such tht k, d suppose tht ɛ is rbitrry positive rel umber. Sice (f ) =1 coverges to f with respect to the C k -orm, we kow tht there is N N such tht f f C k = i= wheever N. This mes tht if N, the f () f () i= f (i) < ɛ f (i) < ɛ. So, we hve tht (f () ) =1 coverges uiformly to f (). This mes tht covergece with respect to the C k -orm implies uiform covergece of fuctios d their first k derivtives. Problem 2. We will rewrite Exmple 5.11, mkig correctios whe eeded. A Schuder bsis (f ) =1 of C([, 1]) my be costructed from tet fuctios. For =, 1, we defie f (x) = 1 d f 1 (x) = x. For 2 k 1 < 2 k, where k 1, we defie f (x) = 2 k [x (2 k (2 2) 1)] if x I, 1 2 k [x (2 k (2 1) 1)] if x J, otherwise, where I = [2 k (2 2) 1, 2 k (2 1) 1) d J = [2 k (2 1) 1, 2 k (2) 1). The grphs of these fuctios form sequece of tets of height oe d width 2 k+1 (where k 1) tht sweep cross the itervl [, 1]. If f C([, 1]), the we my compute the coefficiets c i the expsio f(x) = c f (x) = by equtig the vlues of f d the series t the poits x = 2 k m for k N d m =, 1,..., 2 k. The uiform cotiuity of f implies tht the resultig series coverges uiformly to f. 4

5 Problem 3. We begi this problem by recllig two fcts. I prticulr, i the first homework we proved tht i y ormed lier spce the orm of the zero vector is. Also, i clss we lered tht lier opertor T : X Y betwee lier spces X d Y mps the zero vector i X to the zero vector i Y. Now, we tur our ttetio to the problem. Usig the ottio of the book, we must show tht if{m : M for ll x X}, sup x, sup 1, d sup =1 re equl (where T : X Y is bouded lier mp betwee the ormed lier spces X d Y, d the X d Y orms re both deoted s ). We let = if{m : M for ll x X}, b = sup x, c = sup 1, d d = sup =1. Suppose tht x is rbitrry ozero vector i X. orm, we kow tht x = 1 = 1. By the properties of So, we hve tht d T x. By the properties of orms d lier mps, we kow tht ) T = ( x T ( 1 ) x = 1. So, we hve tht ( ) d x T = d 1 = d. Moreover, if x is the zero vector, we hve tht d, sice i this cse both sides of the iequlity would equl. So, by wht we hve show, we kow tht d {M : M for ll x X} which immeditely implies tht d. Now, suppose tht x is rbitrry vector i X such tht = 1. Sice c = sup 1, we hve tht c. So, c is upper boud of the set { : x X, = 1}. This immeditely implies tht d c. Now, suppose tht x is rbitrry ozero vector i X such tht 1. Sice b = sup x, we hve tht b = b. Moreover, whe x is the zero vector we hve tht b sice i this cse both sides of the iequlity equl. So, we hve tht b for y 5

6 x X tht stisfies 1. This mes tht if z is rbitrry vector i X such tht z 1, we hve T z b z b. We coclude tht b is upper boud of the set { : x X, 1}. This immeditely implies tht c b. At this poit we hve show tht d c b. We will ow show tht b. To see why this is so, suppose for the ske of cotrdictio tht b >. This mes tht there is rel umber, α, i {M : M for ll x X} such tht α < b (This is becuse if o such α existed, we would hve tht b is lower boud of {M : M for ll x X} which would cotrdict the fct tht is the gretest lower boud of {M : M for ll x X}). Now, for y ozero vector x i X, we kow tht α = α. So, we hve tht α is upper boud of { } : x X, x which implies tht b α. This however is cotrdictio, d we coclude tht b. So, we hve tht d c b which implies tht = b = c = d. Thus, we hve tht the expressios i (5.2) d (5.3) for the orm of bouded lier opertor re equivlet. Problem 4. Suppose tht f is rbitrry fuctio i C([, 1]). We ote (by property of supremum) tht δ(f) = f() sup f(x) = f. x 1 So, we hve tht δ(f) f for ll f C([, 1]), d we kow tht δ is bouded by defiitio Now, suppose tht g is rbitrry fuctio i C([, 1]) such tht g = 1 By wht we hve lredy show we kow tht δ(g) g = 1. This implies tht 1 is upper boud of { δ(f) : f C([, 1]), f = 1} which mes tht δ 1 (sice δ = sup{ δ(f) : f C([, 1]), f = 1}). Now, cosider the fuctio h : [, 1] R give by h(x) = 1. Clerly h C([, 1]) d h = 1. So, we hve tht δ δ(h) = h() = 1. 6

7 This mes tht 1 δ 1, d we hve tht the orm of δ equls 1. Now, we will show tht if C([, 1]) is equipped with the oe-orm, the δ is ubouded. For the ske of cotrdictio suppose tht δ is bouded whe C([, 1]) is equipped with the oe-orm. This mes tht there is costt M such tht δ(f) M f 1 for ll f C([, 1]). Now, for ech N let f : [, 1] R be give by f (x) = e x. We kow from clculus tht f C([, 1]) for ech N. Now, let m be the turl umber give by m = M + 1. We ote tht M f m 1 = M f m (x) dx = M me mx = M(1 e m ) M < m = f m () = δ(f m ). So, f m C([, 1]) d we hve tht δ(f m ) > M f m 1. This is cotrdictio d we coclude tht if C([, 1]) is equipped with the oe-orm, the δ is ubouded. Problem 5. Before provig the desired sttemets we mke some observtios d itroduce some ottio. First, we ote tht sice [, 1] [, 1] is closed d bouded subset of R 2, we kow it is compct. Similrly sice [, 1] is closed d bouded subset of R it is compct. We kow from the text tht sice k is cotiuous, it is bouded which mes there is positive rel umber M such tht k(x, y) M for ll (x, y) i [, 1] [, 1]. We lso kow from the text d clculus tht k(x, y) is uiformly cotiuous, d by the properties of itegrls, we kow tht r : [, 1] R give by r(x) = k(x, y) dy is cotiuous fuctio. Sice [, 1] is compct, we kow tht there is x [, 1] such tht r(x ) = sup x [,1] r(x). The, we kow tht d we let { } r(x ) = mx k(x, y) dy, x 1 { } L = r(x ) = mx k(x, y) dy. x 1 Now, we tur our ttetio to showig tht K is bouded. Suppose tht f is rbitrry fuctio i C([, 1]). We ote by the properties of itegrls d by wht we hve show: 7

8 Kf(x) = sup x 1 k(x, y)f(y)dy sup x 1 sup x 1 f sup x 1 k(x, y) f(y) dy k(x, y) f dy So, we hve tht K is bouded by defiitio Moreover, sice k(x, y) dy L f. K = if{m : Kf(x) M f for ll f C([, 1])}, we lso hve tht K L by the bove. We will ow prove tht K = L. Let g : [, 1] R be the fuctio give by g(y) = k(x, y). We immeditely hve tht g is cotiuous sice k is cotiuous. We ow let A = g 1 ((, )) d B = g 1 ((, )). Sice (, ) d (, ) re both ope subsets of R, we kow tht A d B re ope subsets of R (sice g is cotiuous). This immeditely implies tht both A d B re Lebesgue mesurble. Now, let X A : [, 1] R d X B : [, 1] R be chrcteristic fuctios of A d B respectively give by: { 1 if x A, X A (x) = otherwise, d X B (x) = { 1 if x B, otherwise. From rel lysis, we hve tht both X A d X B re Lebesgue mesurble fuctios sice A d B re Lebesgue mesurble. Now, cosider the fuctio f : [, 1] R give by f(x) = X A (x) X B (x). We hve tht f is Lebesgue mesurble sice the sum of two Lebesgue mesurble fuctios is Lebesgue mesurble. Moreover, sice A d B re disjoit, we hve tht f(x) dx 1 < (where the itegrl is tke i the Lebesgue sese). We lso ote tht by costructio k(x, y)f(y)dy = k(x, y) dy = L. From rel lysis we kow tht C C ([, 1]) is dese i L 1 ([, 1]) (this follows from Lusi s theorem). This mes tht there is sequece of fuctios, (f ) =1 i C C ([, 1]) which coverges to f with respect to the L 1 -orm (the L 1 -orm is itroduced o pge 92 of the book). Furthermore sice f is bouded bove by 1 d below by -1, we my ssume tht f 1 for ll N gi from Lusi s theorem (oe c fid this i Folld s lysis text for exmple). Suppose m is rbitrry turl umber. We kow there is N m N such tht f f 1 = f (x) f(x) dx < 1 Mm 8

9 wheever N m. Now, sice K = sup f 1 Kf(x), we hve tht if N m (by the properties of itegrls): K Kf (x) = L k(x, y)f (y)dy k(x, y)f(y)dy + L M k(x, y) f (y) f(y) dy k(x, y)(f (y) f(y))dy f (y) f(y) dy > L M 1 Mm = L 1 m. Thus, we hve tht K > L 1 m for every m N (sice m ws rbitrry). If we tke the limit of both sides of this iequlity s m pproches ifiity, we obti K L. So, we hve ow show tht L K L which mes tht K = L s desired. Problem 6. A bsic trigoometric idetity tells us tht for y x, y [, 1], si(π(x y)) = si(πx πy) = si(πx) cos(πy) cos(πx) si(πy). So, by the properties of itegrls, we hve tht for y f C([, 1]) Kf(x) = si(π(x y))f(y)dy = si(πx) cos(πy)f(y)dy cos(πx) Now, suppose tht c 1 d c 2 re rel umber costts such tht c 1 si(πx) c 2 cos(πx) = for ll x [, 1] (clerly such costts exist sice we could tke c 1 d c 2 to both be d the equtio would be stisfied). We clim tht it must be the cse tht c 1 = c 2 =. To see why this is so, ote tht if we let x = 1 2, c 1 si(πx) c 2 cos(πx) = becomes c 1 =. Similrly, if we let x = 1, c 1 si(πx) c 2 cos(πx) = becomes c 2 =. So, we hve tht c 1 = c 2 =. Now, by defiitio, we kow tht the kerel of K cosists of ll fuctios f C([, 1]) such tht Kf(x) = si(πx) cos(πy)f(y)dy cos(πx) si(πy)f(y)dy = for ll x [, 1]. By wht we hve show thus fr, we hve tht f ker K if d oly if We coclude tht { ker K = f C([, 1]) : cos(πy)f(y)dy = d cos(πy)f(y)dy = d 9 si(πy)f(y)dy =. } si(πy)f(y)dy =. si(πy)f(y)dy.

10 Now, we clim tht the rge of K is A = { si(πx) + b cos(πx) : x [, 1],, b R}. To see why this is so, suppose tht g C([, 1]) is rbitrry elemet of r K. This mes tht there is fuctio f C([, 1]) such tht Kf(x) = g(x) for ll x [, 1]. By wht we hve show this mes tht g(x) = si(πx) cos(πy)f(y)dy cos(πx) for ech x [, 1]. By the properties of itegrls, we hve tht cos(πy)f(y)dy R d si(πy)f(y)dy si(πy)f(y)dy R. Thus, we hve tht g A which immeditely implies tht r K A. Now, suppose tht h is rbitrry elemet of A. This mes tht h C([, 1]) d there re rel umber costts, d b, such tht h(x) = si(πx) + b cos(πx) for ech x [, 1]. We let f : [, 1] R be the fuctio give by f(x) = 2 cos(πx) 2b si(πx). We kow from clculus tht f C([, 1]). Also, bsed upo wht we hve show thus fr d wht we kow from clculus, we kow tht for ech x [, 1], Kf(x) = si(πx) = si(πx) ( 2 cos 2 (πy) 2b cos(πy) si(πy) ) ( dy cos(πx) 2 cos(πy) si(πy) 2b si 2 (πy) ) dy [ y + 2π si(2πy) b ] 1 π si2 (πy) y= = si(πx)[] cos(πx)[ b] = si(πx) + b cos(πx) = h(x). [ cos(πx) π si2 (πy) The bove computtio immeditely implies tht h rk. This mes tht A rk. We coclude tht A = rk, d we re fiished with the problem. ( by b )] 1 2π si(2πy y= 1