The Weierstrass Approximation Theorem

Transcription

1 The Weierstrss Approximtio Theorem Jmes K. Peterso Deprtmet of Biologicl Scieces d Deprtmet of Mthemticl Scieces Clemso Uiversity Februry 26, 2018 Outlie The Wierstrss Approximtio Theorem MtLb Implemettio Compositios of Riem Itegrble Fuctios

2 The ext result is idispesble i moder lysis. Fudmetlly, it sttes tht cotiuous rel-vlued fuctio defied o compct set c be uiformly pproximted by smooth fuctio. This is used throughout lysis to prove results bout vrious fuctios. We c ofte verify property of cotiuous fuctio, f, by provig logous property of smooth fuctio tht is uiformly close to f. We will oly prove the result for closed fiite itervl i R. The geerl result for compct subset of more geerl set clled Topologicl Spce is modifictio of this proof which is ctully ot tht more difficult, but tht is other story. Theorem Let f be cotiuous rel-vlued fuctio defied o [0, 1]. For y ɛ > 0, there is polyomil, p, such tht f (t) p(t) < ɛ for ll t [0, 1], tht is p f < ɛ We first derive some equlities. We will deote the itervl [0, 1] by I. By the biomil theorem, for y x I, we hve x (1 x) = (x + 1 x) = 1. (α)

3 Differetitig both sides of Equtio α, we get ( ) ( ) 0 = x 1 (1 x) x ( )(1 x) 1 ( ( ) = )x 1 (1 x) 1 (1 x) x( ) ( ( ) = )x 1 (1 x) 1 x Now, multiply through by x(1 x), to fid 0 = Differetitig gi, we obti x (1 x) ( x). 0 = ( ) ( ) d x (1 x) ( x). dx This leds to series of simplifictios. It is pretty messy d my texts do ot show the detils, but we thi it is istructive. 0 = = [ x (1 x) +( x) (( )x (1 x) 1 + x 1 (1 x) )] [ x (1 x) +( x)(1 x) 1 x 1( )] ( )x + (1 x)

4 or 0 = ( x (1 x) + ( x) 2 (1 x) 1 x 1) = x (1 x) + ( x) 2 x 1 (1 x) 1 Thus, sice the first sum is 1, we hve = ( x) 2 x 1 (1 x) 1 d multiplyig through by x(1 x), we hve x(1 x) = ( x) 2 x (1 x) or x(1 x) = ( ) ( x )2 x (1 x) This lst equlity the leds to the ( ) ( x )2 x (1 x) x(1 x) = (β) We ow defie the th order Berstei Polyomil ssocited with f by ( ) B(x) = x (1 x) f. Note f (x) B(x) = ( ) [ ( )] x (1 x) f (x) f.

5 Also ote tht f (0) B(0) = f (1) B(1) = 0, so f d B mtch t the edpoits. It follows tht f (x) B(x) x (1 x) f ). (x) f (γ) Now, f is uiformly cotiuous o I sice it is cotiuous. So, give ɛ > 0, there is δ > 0 such tht x < δ f (x) f ( ) < ɛ 2. Cosider x to be fixed i [0, 1]. The sum i Equtio γ hs oly + 1 terms, so we c split this sum up s follows. Let {K1, K2} be prtitio of the idex set {0, 1,..., } such tht K1 x < δ d K2 x δ. The f (x) B(x) x (1 x) f ) (x) f K1 + x (1 x) f ). (x) f K2 which implies f (x) B(x) ɛ x (1 x) 2 K1 + x (1 x) f ) (x) f K2 = ɛ 2 + x (1 x) f ). (x) f K2

6 Now, f is bouded o I, so there is rel umber M > 0 such tht f (x) M for ll x I. Hece x (1 x) f ) (x) f 2M x (1 x). K2 K2 Sice K2 x δ, usig Equtio β, we hve δ 2 K2 x (1 x) ( K2 ) ( x )2 x (1 x) x(1 x). This implies tht K2 x (1 x) x(1 x) δ 2. d so combiig iequlities 2M x (1 x) K2 2Mx(1 x) δ 2 We coclude the tht x (1 x) f ) 2Mx(1 x) (x) f δ 2. K2 Now, the mximum vlue of x(1 x) o I is 1 4, so x (1 x) f ) M (x) f 2δ 2. K2

7 Filly, choose so tht > M δ 2 ɛ. The M δ < ɛ implies M 2 2δ < ɛ 2 2. So, Equtio γ becomes f (x) B(x) ɛ 2 + ɛ 2 = ɛ. Note tht the polyomil B does ot deped o x I, sice oly depeds o M, δ, d ɛ, ll of which, i tur, re idepedet of x I. So, B is the desired polyomil, s it is uiformly withi ɛ of f. Commet A chge of vrible trsltes this result to y closed itervl [, b]. Let s write some code to implemet Berstei polyomils I Octve/ MtLb. We use the fuctio Berstei (f,,b,) lie this: X = l i s p c e ( 0, 1 0, ) ; f x ) e. ˆ (. 3 x ). c o s ( 2 x ) ; B3 = B e r s t e i ( f, 0, 1 0, 3 ) ; p l o t (X, f (X),X, B3 (X) ) ; 5 B10 = B e r s t e i ( f, 0, 1 0, 1 0 ) ; p l o t (X, f (X),X, B3 (X),X, B10 (X) ) ; B25 = B e r s t e i ( f, 0, 1 0, 2 5 ) ; p l o t (X, f (X),X, B3 (X),X, B10 (X),X, B25 (X) ) ; B45 = B e r s t e i ( f, 0, 1 0, 4 5 ) ; 10 p l o t (X, f (X),X, B3 (X),X, B10 (X),X, B25 (X),X, B45 (X) ) ; B150 = B e r s t e i ( f, 0, 1 0, ) ; p l o t (X, f (X),X, B3 (X),X, B10 (X),X, B25 (X),X, B45 (X),X, B150 (X) ) ; l e g e d ( f, B3, B10, B25, B45, B140 ) ; x l b e l ( x ) ; y l b e l ( y ) ; 15 t i t l e ( f (x) = e^{ -3x } cos (2x + 0.3) o [0,10] d Berstei Polyomils ) ;

8 f u c t i o p = B e r s t e i ( f,, b, ) % compute B e r s t e i p o l y o m i l p p r o x i m t i o % o f o r d e r o t he i t e r v l [, b ] to f % f i s the fuctio 5 % i s the order of the Berstei polyomil p x ) 0 ; f o r i = 1 : +1 = i 1; % c o v e r t t h e i t e r v l [, b ] to [ 0, 1 ] h e r e 10 y x ) ( x ) /( b ) ; % c o v e r t t h e p o i t s we e v l u t e f t to be i [, b ] % i s t e d o f [ 0, 1 ] z = + ( b ) / ; q x ) choose (, ) ( y ( x ). ˆ ). ((1 y ( x ) ). ˆ ( ) ) f ( z ) ; 15 p x ) ( p ( x ) + q ( x ) ) ; ed ed We delibertely tried to pproximte oscilltig fuctio with the Berstei polyomils d we expected this to be difficult becuse we chose the itervl [0, 10] to wor o. The Berstei polyomils require us to clculte the biomil coefficiets. Whe is lrge. such s i our exmple with = 150, ruig this code geertes errors bout the possible loss of precisio i the use of choose which is the biomil coefficiet fuctio. You c see how we did with the pproximtios i the ext figure.

9 Exmple If b f (s)s ds = 0 for ll with f cotiuous, the f = 0 o [, b]. Solutio Let B (f ) be the Berstei polyomil of order ssocited with f. From the ssumptio, we see b f (s)b(f )(s)ds = 0 for ll lso. Now cosider b b f 2 b (s)ds f (s)b (f )(s)ds = f (s)(f (s) B (f )(s))ds b f B (f ) f (s) ds The give ɛ > 0, there is N so tht > N implies f B (f ) < ɛ/(( f + 1)(b )). The we hve > N implies

10 Solutio b b f 2 (s)ds f (s)b(f )(s)ds < ɛ b f (s) ds ( f + 1)(b ) ɛ < f (b ) < ɛ ( f + 1)(b ) This tells us b f (s)b(f )(s)ds b f 2 (s)ds. Sice b f (s)b(f )(s)ds = 0 for ll, we see b f 2 (s)ds = 0. It the follows tht f = 0 o [, b] usig this rgumet. Assume f 2 is ot zero t some poit c i [, b]. We c ssume c is iterior poit s the rgumet t edpoit is similr. Sice f 2 (c) > 0 d f is cotiuous, there is r with (r c, r + c) [, b] d f 2 (x) > f 2 (c)/2 o tht itervl. Hece Solutio b c r c+r b 0 = f 2 (s)ds = f 2 ds + f 2 (s)ds + f 2 (s)ds c r c+r > (2r)f 2 (c)/2 > 0 This is cotrdictio d so f 2 = 0 o [, b] implyig f = 0 o [, b].

11 We lredy ow tht cotiuous fuctios d mootoe fuctios re clsses of fuctios which re Riem Itegrble o the itervl [, b]. Hece, sice f (x) = x is cotiuous o [0, M] for y positive M, we ow f is Riem itegrble o this itervl. Wht bout the compositio g where g is just ow to be o egtive d Riem itegrble o [, b]? If g were cotiuous, sice compositios of cotiuous fuctios re lso cotiuous, we would hve immeditely tht g is Riem Itegrble. However, it is ot so esy to hdle the cse where we oly ow g is Riem itegrble. Let s try this pproch. Usig the Weierstrss Approximtio Theorem, we ow give fiite itervl [c, d], there is sequece of polyomils {p(x)} which coverge i to x o [c, d]. Of course, the polyomils i this sequece will chge if we chge the itervl [c, d], but you get the ide. To pply this here, ote tht sice g is Riem Itegrble o [, b], g must be bouded. Sice we ssume g is o egtive, we ow tht there is positive umber M so tht g(x) is i [0, M] for ll x i [, b]. Thus, there is sequece of polyomils {p} which coverge i to o [0, M]. Next, we ow polyomil i g is lso Riem itegrble o [, b] (f 2 = f f so it is itegrble d so o). Hece, p(g) is Riem itegrble o [, b]. The give ɛ > 0, we ow there is positive N so tht p(u) u < ɛ, if > N d u [0, M].

12 Thus, i prticulr, sice g(x) [0, M], we hve p(g(x)) g(x) < ɛ, if > N d x [, b]. We hve therefore proved tht p g coverges to g o [0, M]. Wht we eed ext is ew theorem: If f coverges to f i o [, b] with ll f Riem itegrble, the the limit fuctio f is lso Riem itegrble d b f(s)ds b f (s)ds. We c ideed prove such theorem. Hece, we ow g is Riem itegrble whe g is Riem Itegrble. Note the id of rgumets we use here. The ides of fuctios covergig i the sup orm d the questios wht properties re retied i the limit re of gret importce. Note this id of proof mes us thi tht compositios of cotiuous fuctio ( lie ) with Riem itegrble fuctio will give us ew Riem itegrble fuctios s the rgumet we used bove would trslte esily to the ew cotext. We will explore this more lter. I geerl, the compositio of Riem Itegrble fuctios is ot Riem itegrble. Here is the stdrd couterexmple. Defie f o [0, 1] by f (y) = { 1 if y = 0 0 if 0 < y 1 d g o [0, 1] by 1 if x = 0 g(x) = 1/q if x = p/q, (p, q) = 1, x (0, 1] d x is rtiol 0 if x (0, 1] d x is irrtiol Let s show g is RI o [0,1]. First defie g to be 0 t x = 1. If we show this modifictio of g is RI, g will be too. Let q be prime umber bigger th 2. The form the uiform prtitio πq = {0, 1/q,..., (q 1)/q, 1}. O ech subitervl of this prtitio, we see mj = 0. Iside the subitervl [(j 1)/q, j/q], the mximum vlue of g is 1/q. Hece, we hve Mj = 1/q.

13 This gives U(g, πq) L(g, πq) = πq (1/q) xi = 1/q Give ɛ > 0, there is N so tht 1/N < ɛ. Thus if q0 is prime with q0 > N, we hve U(g, πq0) L(g, πq0) < ɛ. So if π is y refiemet of, we hve U(g, π) L(g, π) < ɛ lso, πq0 Thus g stisfies the Riem Criterio d so g is RI o [0, 1]. It is lso esy to see 1 g(s)ds = 0. Now f g becomes 0 f (1) if x = 0 f (g(x)) = f (1/q) if x = p/q, (p, q) = 1, x (0, 1] d x rtiol f (0) if 0 < x < 1 d x irrtiol 1 if x = 0 = 0 if if x rtiol (0, 1) 1 if if x irrtiol (0, 1) The fuctio f g bove is ot Riem itegrble s U(f g) = 1 d L(f g) = 0. Thus, we hve foud two Riem itegrble fuctios whose compositio is ot Riem itegrble! Homewor For f (x) = si 2 (3x), grph the Berstei polyomils B5(f ), B10(f ) d B15(f ) log with f simulteously o the itervl [ 2, 4] o the sme grph i MtLb. This is word doc report so write your code, documet it d prit out the grph s prt of your report For f (x) = si(x) + x, grph the Berstei polyomils B5(f ), B10(f ) d B15(f ) log with f simulteously o the itervl [0, 6] o the sme grph i MtLb. This is word doc report so write your code, documet it d prit out the grph s prt of your report.

14 Homewor Fix x 0 i R. Let the sequece () be defied by = cos(3 x). Prove this sequece does ot lwys hve limit but it does hve t lest oe subsequetil limit. Of course, becuse of periodicity, it is eough to loo t poits x [0, 2π]. For exmple, there is limit if x = π/2 d other specil poits. Let s loo t x = 1 crefully. Prove this sequece does ot hve limit but it does hve t lest oe subsequetil limit. Hit The Bolzo - Weierstrss Theorem tells us there is subsequece (cos(3 ) which coverges to umber α i [ 1, 1]. There re idetities which show us cos(3θ) = 4 cos 3 (θ) 4 cos(θ). Hece, we ow cos(3 (3 )) 4α 3 4α Homewor Cotiued Hit If α 0, use the bove fct to show the subsequeces cos(3 +1 ) d cos(3 ) coverge to differet vlues. This implies the limit c ot exist. If α = 0, the rgumet is hrder. This tells us si(3 ) 1. Now show si(3θ) = si(θ) (4 cos 2 (θ) 1) d lso show tht the si(3θ) idetity implies si(3 ) c ot hve limit. Filly, sice si(θ) = ± 1 cos2 (θ), if we ssume the limit of cos(3 ) exists d equls β, the by looig t ll the qudrt choices tht the vlue of β implies, we fid the limit of si(3 ) must exist. This is cotrdictio d so the limit of cos(3 ) does ot exist.

15 Homewor Cotiued Hit This sequece is hrd to lyze s usig the 2π periodicty of cos we c recst the sequece usig remiders. 3 = 2π + r, r (0, 1) L cos(3 ) = 2π + r ) = cos(2πr ) cos(l We hve o ide how the remider terms re distributed i (0, 1) r d tht is why this lyis is very otrivil.