Bernstein Polynomials

Transcription

1 7 Bestei Polyomials 7.1 Itoductio This chapte is coceed with sequeces of polyomials amed afte thei ceato S. N. Bestei. Give a fuctio f o [0, 1, we defie the Bestei polyomial B (f; x = ( f =0 ( x (1 x (7.1 fo each positive itege. Thus thee is a sequece of Bestei polyomials coespodig to each fuctio f. As we will see late i this chapte, if f is cotiuous o [0, 1, its sequece of Bestei polyomials coveges uifomly to f o [0, 1, thus givig a costuctive poof of Weiestass s Theoem 2.4.1, which we stated i Chapte 2. Thee ae seveal poofs of this fudametal theoem, begiig with that give by K. Weiestass [55 i (See the Notes i E. W. Cheey s text [7. This cotais a lage umbe of histoical efeeces i appoximatio theoy. Bestei s poof [3 was published i Oe might wode why Bestei ceated ew polyomials fo this pupose, istead of usig polyomials that wee aleady kow to mathematics. Taylo polyomials ae ot appopiate; fo eve settig aside questios of covegece, they ae applicable oly to fuctios that ae ifiitely diffeetiable, ad ot to all cotiuous fuctios. We ca also dismiss aothe obvious cadidate, the itepolatig polyomials fo f costucted at equally spaced poits. Fo the latte sequece of polyomials does ot covege uifomly to f fo all f C[0, 1, ad the same is tue of itepolatio o ay othe fixed sequece of abscis-

2 Bestei Polyomials sas. Howeve, L. Fejé [19 used a method based o Hemite itepolatio i a poof published i 1930, which we will discuss i the ext sectio. Late i this sectio we will coside how Bestei discoveed his polyomials, fo this is ot immediately obvious. We will also see that although the covegece of the Bestei polyomials is slow, they have compesatig shape-pesevig popeties. Fo example, the Bestei polyomial of a covex fuctio is itself covex. It is clea fom (7.1 that fo all 1, B (f;0=f(0 ad B (f;1=f(1, (7.2 so that a Bestei polyomial fo f itepolates f at both edpoits of the iteval [0, 1. Example It follows fom the biomial expasio that B (1; x = =0 ( x (1 x =(x +(1 x =1, (7.3 so that the Bestei polyomial fo the costat fuctio 1 is also 1. Sice ( ( 1 = 1 fo 1, the Bestei polyomial fo the fuctio x is B (x; x = =0 ( x (1 x = x =1 ( 1 1 x 1 (1 x. Note that the tem coespodig to = 0 i the fist of the above two sums is zeo. O puttig s = 1 i the secod summatio, we obtai 1 ( 1 B (x; x =x s x s (1 x 1 s = x, (7.4 the last step followig fom (7.3 with eplaced by 1. Thus the Bestei polyomial fo the fuctio x is also x. We call B the Bestei opeato; it maps a fuctio f, defied o [0, 1, to B f, whee the fuctio B f evaluated at x is deoted by B (f; x. The Bestei opeato is obviously liea, sice it follows fom (7.1 that B (λf + µg =λb f + µb g, (7.5 fo all fuctios f ad g defied o [0, 1, ad all eal λ ad µ. Weow equie the followig defiitio.

3 7.1 Itoductio 249 Defiitio Let L deote a liea opeato that maps a fuctio f defied o [a, b to a fuctio Lf defied o [c, d. The L is said to be a mootoe opeato o, equivaletly, a positive opeato if f(x g(x, x [a, b (Lf(x (Lg(x, x [c, d, (7.6 whee we have witte (Lf(x to deote the value of the fuctio Lf at the poit x [a, b. We ca see fom (7.1 that B is a mootoe opeato. It the follows fom the mootoicity of B ad (7.3 that m f(x M, x [0, 1 m B (f; x M, x [0, 1. (7.7 I paticula, if we choose m = 0 i (7.7, we obtai f(x 0, x [0, 1 B (f; x 0, x [0, 1. (7.8 It follows fom (7.3, (7.4, ad the liea popety (7.5 that B (ax + b; x =ax + b, (7.9 fo all eal a ad b. We theefoe say that the Bestei opeato epoduces liea polyomials. We ca deduce fom the followig esult that the Bestei opeato does ot epoduce ay polyomial of degee geate tha oe. Theoem The Bestei polyomial may be expessed i the fom B (f; x = =0 ( f(0 x, (7.10 whee is the fowad diffeece opeato, defied i (1.67, with step size h =1/. Poof. Begiig with (7.1, ad expadig the tem (1 x,wehave B (f; x = ( ( f =0 Let us put t = + s. We may wite = =0 x t=0 =0 ( 1 s ( s x s. t, (7.11 sice both double summatios i (7.11 ae ove all lattice poits (, s lyig i the tiagle show i Figue 7.1. We also have ( ( ( ( t =, s t

4 Bestei Polyomials s 0 FIGURE 7.1. A tiagula aay of 1 ( + 1( + 2 lattice poits. 2 ad so we may wite the double summatio as ( t ( ( x t ( 1 t t f = t t=0 =0 t=0 ( t t f(0 x t, o usig the expasio fo a highe-ode fowad diffeece, as i Poblem This completes the poof. I (1.80 we saw how diffeeces ae elated to deivatives, showig that m f(x 0 h m = f (m (ξ, (7.12 whee ξ (x 0,x m ad x m = x 0 + mh. Let us put h =1/, x 0 = 0, ad f(x =x k, whee k. The we have f(0 = 0 fo >k ad k k f(0 = f (k (ξ =k!. (7.13 Thus we see fom (7.10 with f(x =x k ad k that B (x k ; x =a 0 x k + a 1 x k a k 1 x + a k, say, whee a 0 = 1 fo k = 0 ad k = 1, ad ( ( k! a 0 = k k = 1 1 ( 1 2 ( 1 k 1 fo k 2. Sice a 0 1 whe k 2, this justifies ou above statemet that the Bestei opeato does ot epoduce ay polyomial of degee geate tha oe.

5 Example With f(x =x 2,wehave ( 1 f(0=0, f(0 = f f(0 = 1 2, 7.1 Itoductio 251 ad we see fom (7.13 that 2 2 f(0 = 2! fo 2. Thus it follows fom (7.10 that ( ( B (x 2 x 2x 2 ; x = = x ( x 2, which may be witte i the fom B (x 2 ; x =x x(1 x. (7.14 Thus the Bestei polyomials fo x 2 covege uifomly to x 2 like 1/, vey slowly. We will see fom Vooovskaya s Theoem that this ate of covegece holds fo all fuctios that ae twice diffeetiable. We have aleady see i (7.7 that if f(x is positive o [0, 1, so is B (f; x. We ow show that if f(x is mootoically iceasig, so is B (f; x. Theoem The deivative of the Bestei polyomial B +1 (f; x may be expessed i the fom B +1(f; x =( +1 f =0 ( ( +1 x (1 x (7.15 fo 0, whee is applied with step size h =1/( + 1. Futhemoe, if f is mootoically iceasig o mootoically deceasig o [0, 1, so ae all its Bestei polyomials. Poof. The veificatio of (7.15 is omitted because it is a special case of (7.16, coceig highe-ode deivatives of the Bestei polyomials, which we pove i the ext theoem. To justify the above emak o mootoicity, we ote that if f is mootoically iceasig, its fowad diffeeces ae oegative. It the follows fom (7.15 that B +1(f; x is oegative o [0, 1, ad so B +1 (f; x is mootoically iceasig. Similaly, we see that if f is mootoically deceasig, so is B +1 (f; x. Theoem Fo ay itege k 0, the kth deivative of B +k (f; x may be expessed i tems of kth diffeeces of f as B (k + k! +k (f; x =(! k f =0 ( + k ( x (1 x (7.16 fo all 0, whee is applied with step size h =1/( + k.

6 Bestei Polyomials Poof. We wite +k B +k (f; x = f =0 ad diffeetiate k times, givig +k +k (f; x = f B (k d s dx s x = ( ( + k + k =0 ( ( + k + k x (1 x +k p(x, (7.17 whee p(x = dk dx k x (1 x +k. We ow use the Leibiz ule (1.83 to diffeetiate the poduct of x ad (1 x +k. Fist we fid that ad d k s dx k s (1 x+k =! ( s! x s, s 0, 0, s<0, k s ( + k! ( 1 ( + s! (1 x+s, s, 0, s>. Thus the kth deivative of x (1 x +k is p(x = ( ( 1 k s k! ( + k! s ( s! ( + s! x s (1 x +s, (7.18 s whee the latte summatio is ove all s fom 0 to k, subject to the costaits 0 s. We make the substitutio t = s, so that +k = =0 s t=0 k. (7.19 A diagam may be helpful hee. The double summatios i (7.19 ae ove all lattice poits (, s lyig i the paallelogam depicted i Figue 7.2. The paallelogam is bouded by the lies s =0,s = k, t = 0, ad t =, whee t = s. We also ote that ( + k! ( + k! ( + k! = ( s! ( + s!! ( s. (7.20

7 7.1 Itoductio 253 s k 0 + k FIGURE 7.2. A paallelogam of ( + 1(k + 1 lattice poits. It the follows fom (7.17, (7.18, (7.19, ad (7.20 that the kth deivative of B +k (f; x is ( + k!! t=0 k ( ( 1 k s k s ( ( t + s f + k t x t (1 x t. Fially, we ote fom Poblem that k ( ( ( ( 1 k s k t + s t f = k f, s + k + k whee the opeato is applied with step size h =1/(+k. This completes the poof. By usig the coectio betwee diffeeces ad deivatives, we ca deduce the followig valuable esult fom Theoem Theoem If f C k [0, 1, fo some k 0, the m f (k (x M, x [0, 1 c k m B (k (f; x c k M, x [0, 1, fo all k, whee c 0 = c 1 = 1 ad ( c k = 1 1 ( 1 2 ( 1 k 1, 2 k. Poof. We have aleady see i (7.7 that this esult holds fo k =0.Fo k 1 we begi with (7.16 ad eplace by k. The, usig (7.12 with h =1/, we wite ( k f whee /<ξ < ( + k/. Thus = f (k (ξ k, (7.21 k B (k (f; x = c k f (k (ξ x (1 x k, =0

8 Bestei Polyomials ad the theoem follows easily fom the latte equatio. Oe cosequece of this esult is that if f (k (x is of fixed sig o [0, 1, the B (k (f; x also has this sig o [0, 1. Fo example, if f (x exists ad is oegative o [0, 1, so that f is covex, the B (f; x is also oegative ad B (f; x is covex. Bestei s discovey of his polyomials was based o a igeious pobabilistic agumet. Suppose we have a evet that ca be epeated ad has oly two possible outcomes, A ad B. Oe of the simplest examples is the tossig of a ubiased coi, whee the two possible outcomes, heads ad tails, both occu with pobability 0.5. Moe geeally, coside a evet whee the outcome A happes with pobability x [0, 1, ad thus the outcome B happes with pobability 1 x. The the pobability of A happeig pecisely times followed by B happeig times is x (1 x. ( Sice thee ae ways of choosig the ode of outcomes out of, the pobability of obtaiig outcomes A ad outcomes B i ay ode is give by ( p, (x = x (1 x. Thus we have ( ( +1 x p, (x = p, 1 (x, 1 x ad it follows that p, (x >p, 1 (x if ad oly if <( +1x. We deduce that p, (x, egaded as a fuctio of, with x ad fixed, has a peak whe = x x, fo lage, ad is mootoically iceasig fo < x ad mootoically deceasig fo > x. We aleady kow that p, (x =B (1; x =1, ad i the sum =0 ( p, (xf = B (f; x, =0 whee f C[0, 1 ad is lage, the cotibutios to the sum fom values of sufficietly emote fom x will be egligible, ad the sigificat pat of the sum will come fom values of close to x. Thus, fo lage, ( x B (f; x f f(x, ad so B (f; x f(x as. While this is by o meas a igoous agumet, ad is thus ot a poof, it gives some isight ito how Bestei was motivated i his seach fo a poof of the Weiestass theoem.

9 7.1 Itoductio 255 Example To illustate Bestei s agumet coceig the polyomials p,, let us evaluate these polyomials whe = 8 ad x =0.4. The esultig values of p, (x ae give i the followig table: p, (x I this case, the lagest value of p, is attaied fo = x = 3, cosistet with ou above aalysis, which shows that x x =3.2. Theoem Give a fuctio f C[0, 1 ad ay ɛ>0, thee exists a itege N such that fo all N. f(x B (f; x <ɛ, 0 x 1, Poof. I othe wods, the above statemet says that the Bestei polyomials fo a fuctio f that is cotiuous o [0, 1 covege uifomly to f o [0, 1. The followig poof is motivated by the plausible agumet that we gave above. We begi with the idetity multiply each tem by give ( ( 2 x =0 ( x 2 = ( ( 2 2 ( x + x 2, x (1 x, ad sum fom =0to, to x (1 x = B (x 2 ; x 2xB (x; x+x 2 B (1; x. It the follows fom (7.3, (7.4, ad (7.14 that ( ( 2 x =0 x (1 x = 1 x(1 x. (7.22 Fo ay fixed x [0, 1, let us estimate the sum of the polyomials p, (x ove all values of fo which / is ot close to x. To make this otio pecise, we choose a umbe δ>0ad let S δ deote the set of all values of satisfyig x δ. We ow coside the sum of the polyomials p, (x ove all S δ. Note that x δ implies that 1 δ 2 ( x 2 1. (7.23

10 Bestei Polyomials The, usig (7.23, we have S δ ( x (1 x 1 δ 2 S δ ( x 2 ( x (1 x. The latte sum is ot geate that the sum of the same expessio ove all, ad usig (7.22, we have 1 δ 2 ( ( 2 x =0 S δ x (1 x x(1 x = δ 2. Sice 0 x(1 x 1 4 o [0, 1, we have ( x (1 x 1 4δ 2. (7.24 Let us wite = =0 S δ + / S δ, whee the latte sum is theefoe ove all such that x <δ. Havig split the summatio ito these two pats, which deped o a choice of δ that we still have to make, we ae ow eady to estimate the diffeece betwee f(x ad its Bestei polyomial. Usig (7.3, we have ad hece f(x B (f; x = =0 f(x B (f; x = S δ + / S δ ( ( ( f(x f ( ( ( f(x f ( ( ( f(x f We thus obtai the iequality f(x B (f; x ( ( f(x f S δ + / S δ ( ( f(x f x (1 x, x (1 x x (1 x. x (1 x x (1 x.

11 7.1 Itoductio 257 Sice f C[0, 1, it is bouded o [0, 1, ad we have f(x M, fo some M>0. We ca theefoe wite ( f(x f 2M fo all ad all x [0, 1, ad so ( ( f(x f S δ O usig (7.24 we obtai ( ( f(x f S δ x (1 x 2M S δ ( x (1 x x (1 x. M 2δ 2. (7.25 Sice f is cotiuous, it is also uifomly cotiuous, o [0, 1. Thus, coespodig to ay choice of ɛ>0 thee is a umbe δ>0, depedig o ɛ ad f, such that / S δ x x <δ f(x f(x < ɛ 2, fo all x, x [0, 1. Thus, fo the sum ove / S δ,wehave ( ( f(x f x (1 x < ɛ ( x (1 x 2 / S δ < ɛ 2 / S δ =0 ( x (1 x, ad hece, agai usig (7.3, we fid that ( ( f(x f x (1 x < ɛ 2. (7.26 O combiig (7.25 ad (7.26, we obtai f(x B (f; x < M 2δ 2 + ɛ 2. It follows fom the lie above that if we choose N>M/(ɛδ 2, the f(x B (f; x <ɛ fo all N, ad this completes the poof. Usig the methods employed i the above poof, we ca show, with a little geate geeality, that if f is meely bouded o [0, 1, the sequece (B (f; x =1 coveges to f(x ataypoitx whee f is cotiuous. We will ow discuss some futhe popeties of the Bestei polyomials.

12 Bestei Polyomials Theoem If f C k [0, 1, fo some itege k 0, the B (k (f; x coveges uifomly to f (k (x o[0, 1. Poof. We kow fom Theoem that the above esult holds fo k =0. Fo k 1 we begi with the expessio fo B (k +k (f; x give i (7.16, ad wite ( k f = f (k (ξ + k ( + k k, whee /( + k <ξ < ( + k/( + k, as we did similaly i (7.21. We the appoximate f (k (ξ, witig We thus obtai say, whee ad S 2 (x = ( f (k (ξ =f (k ( + f (k (ξ f (k (.!( + k k B (k +k ( + k! (f; x =S 1(x+S 2 (x, (7.27 S 1 (x = ( =0 =0 ( f (k ( f (k (ξ f (k ( x (1 x ( x (1 x. I S 2 (x, we ca make ξ <δfo all, fo ay choice of δ>0, by takig sufficietly lage. Also, give ay ɛ>0, we ca choose a positive value of δ such that ( f (k (ξ f (k <ɛ, fo all, because of the uifom cotiuity of f (k.thuss 2 (x 0 uifomly o [0, 1 as. We ca easily veify that!( + k k ( + k! 1 as, ad we see fom Theoem with f (k i place of f that S 1 (x coveges uifomly to f (k (x o[0, 1. This completes the poof. As we have just see, ot oly does the Bestei polyomial fo f covege to f, but deivatives covege to deivatives. This is a most emakable popety. I cotast, coside agai the sequece of itepolatig polyomials (p fo e x that appea i Example Although this sequece of polyomials coveges uifomly to e x o [ 1, 1, this does ot

13 7.1 Itoductio 259 hold fo thei deivatives, because of the oscillatoy atue of the eo of itepolatio. O compaig the complexity of the poofs of Theoems ad 7.1.6, it may seem supisig that the additioal wok equied to complete the poof of Theoem fo k 1 is so little compaed to that eeded to pove Theoem y D A C B x 1 x 3 x 2 x FIGURE 7.3. A ad B ae the poits o the chod CD ad o the gaph of the covex fuctio y = f(x, espectively, with abscissa x 3 = λx 1 +(1 λx 2. We ow state esults coceig the Bestei polyomials fo a covex fuctio f. Fist we defie covexity ad show its coectio with secodode divided diffeeces. Defiitio A fuctio f is said to be covex o [a, b if fo ay x 1,x 2 [a, b, λf(x 1 +(1 λf(x 2 f(λx 1 +(1 λx 2 (7.28 fo ay λ [0, 1. Geometically, this is just sayig that a chod coectig ay two poits o the covex cuve y = f(x is eve below the cuve. This is illustated i Figue 7.3, whee CD is such a chod, ad the poits A ad B have y-coodiates λf(x 1 +(1 λf(x 2 ad f(λx 1 +(1 λx 2, espectively. If f is twice diffeetiable, f beig covex is equivalet to f beig oegative. Of couse, fuctios ca be covex without beig diffeetiable. Fo example, we ca have a covex polygoal ac. Theoem A fuctio f is covex o [a, b if ad oly if all secodode divided diffeeces of f ae oegative. Poof. Sice a divided diffeece is uchaged if we alte the ode of its agumets, as we see fom the symmetic fom (1.21, it suffices to coside the divided diffeece f[x 0,x 1,x 2 whee a x 0 <x 1 <x 2 b. The we obtai fom the ecuece elatio (1.22 that f[x 0,x 1,x 2 0 f[x 1,x 2 f[x 0,x 1. (7.29

14 Bestei Polyomials O multiplyig the last iequality thoughout by (x 2 x 1 (x 1 x 0, which is positive, we fid that both iequalities i (7.29 ae equivalet to (x 1 x 0 (f(x 2 f(x 1 (x 2 x 1 (f(x 1 f(x 0, which is equivalet to (x 1 x 0 f(x 2 +(x 2 x 1 f(x 0 (x 2 x 0 f(x 1. (7.30 If we ow divide thoughout by x 2 x 0 ad wite λ =(x 2 x 1 /(x 2 x 0, we see that x 1 = λx 0 +(1 λx 2, ad it follows fom (7.30 that λf(x 0 +(1 λf(x 2 f(λx 0 +(1 λx 2, thus completig the poof. The poofs of the followig two theoems ae held ove util Sectio 7.3, whee we will state ad pove geealizatios of both esults. Theoem If f(x iscovexo[0, 1, the fo all 1. Theoem If f(x is covex o[0, 1, B (f; x f(x, 0 x 1, (7.31 B 1 (f; x B (f; x, 0 x 1, (7.32 fo all 2. The Bestei polyomials ae equal at x = 0 ad x =1, sice they itepolate f at these poits. If f C[0, 1, the iequality i (7.32 is stict fo 0 <x<1, fo a give value of, uless f is liea i each of the itevals B 1 (f; x =B (f; x. [ 1 1, 1, fo 1 1, whe we have simply Note that we have fom Theoem with k = 2 that if f (x 0, ad thus f is covex o [0, 1, the B (f; x is also covex o [0, 1. I Sectio 7.3 we will establish the stoge esult that B (f; x iscovexo[0, 1, povided that f is covex o [0, 1. We coclude this sectio by statig two theoems coceed with estimatig the eo f(x B (f; x. The fist of these is the theoem due to Elizaveta V. Vooovskaya ( , which gives a asymptotic eo tem fo the Bestei polyomials fo fuctios that ae twice diffeetiable. Theoem Let f(x be bouded o [0, 1. The, fo ay x [0, 1 at which f (x exists, lim (B (f; x f(x = 1 2 x(1 xf (x. (7.33

15 See Davis [10 fo a poof of Vooovskaya s theoem. 7.1 Itoductio 261 Fially, thee is the followig esult that gives a uppe boud fo the eo f(x B (f; x i tems of the modulus of cotiuity, which we defied i Sectio 2.6. Theoem If f is bouded o [0, 1, the f B f 3 ( 1 2 ω, (7.34 whee deotes the maximum om o [0, 1. See Rivli [48 fo a poof of this theoem. Example Coside the Bestei polyomial fo f(x = x 1 2, B (f; x = 1 ( 2 =0 x (1 x. The diffeece betwee B (f; x ad f(x atx = 1 2 is ( 2 =0 = e, say. Let us ow choose to be eve. We ote that ( 1 2 ( ( = 1 ( 2 fo all, ad that the quatities o each side of the above equatio ae zeo whe = /2. It follows that 2 e = 1 ( 2 =0 /2 =2 =0 ( 1 2 (. (7.35 y x FIGURE 7.4. The fuctio f(x = x 1 o [0, 1. 2

16 Bestei Polyomials Let us split the last summatio ito two. We obtai /2 ( ( ( = 1 ( + = 1 ( 2 /2 2 /2 =0 =0 ad sice ( ( 1 =, 1, 1 we fid that /2 2 =0 ( /2 =2 =1 It the follows fom (7.35 that e = ( 1 1 ( /2 = =1 ( π +2 1, =2 1. fo lage. The last step follows o usig Stilig s fomula fo estimatig! (see Poblem This shows that f B f 0 at least as slowly as 1/ fo the fuctio f(x = x 1 2, whee deotes the maximum om o [0, 1. Poblem Show that fo all 3. B (x 3 ; x =x x(1 x(1+(3 2x, 2 Poblem Show that B (e αx ; x =(xe α/ +(1 x fo all iteges 1 ad all eal α. Poblem Deduce fom Defiitio 7.1.2, usig iductio o, that a fuctio f is covex o [a, b if ad oly if ( λ f(x f λ x, =0 fo all 0, fo all x [a, b, ad fo all λ 0 such that λ 0 + λ λ =1. Poblem Fid a fuctio f ad a eal umbe λ such that f is a polyomial of degee two ad B f = λf. Also fid a fuctio f ad a eal umbe λ such that f is a polyomial of degee thee ad B f = λf. Poblem Veify Vooovskaya s Theoem diectly fo the two fuctios x 2 ad x 3. =0

17 7.2 The Mootoe Opeato Theoem The Mootoe Opeato Theoem I the 1950s, H. Bohma [5 ad P. P. Koovki [31 obtaied a amazig geealizatio of Bestei s Theoem They foud that as fa as covegece is coceed, the cucial popeties of the Bestei opeato B ae that B f f uifomly o [0, 1 fo f =1,x, ad x 2, ad that B is a mootoe liea opeato. (See Defiitio We ow state the Bohma Koovki theoem, followed by a poof based o that give by Cheey [7. Theoem Let (L deote a sequece of mootoe liea opeatos that map a fuctio f C[a, b to a fuctio L f C[a, b, ad let L f f uifomly o [a, b fo f =1,x, ad x 2. The L f f uifomly o [a, b fo all f C[a, b. Poof. Let us defie φ t (x =(t x 2, ad coside (L φ t (t. Thus we apply the liea opeato L to φ t, egaded as a fuctio of x, ad the evaluate L φ t (which is also a fuctio of x atx = t. Sice L is liea, we obtai (L φ t (t =t 2 (L g 0 (t 2t(L g 1 (t+(l g 2 (t, whee Hece g 0 (x =1, g 1 (x =x, g 2 (x =x 2. (L φ t (t =t 2 [(L g 0 (t 1 2t[(L g 1 (t t+[(l g 2 (t t 2. O witig to deote the maximum om o [a, b, we deduce that L φ t M 2 L g 0 g 0 +2M L g 1 g 1 + L g 2 g 2, whee M = max( a, b. Sice fo i = 0, 1, ad 2, each tem L g i g i may be made as small as we please, by takig sufficietly lage, it follows that (L φ t (t 0 as, (7.36 uifomly i t. Now let f be ay fuctio i C[a, b. Give ay ɛ>0, it follows fom the uifom cotiuity of f that thee exists a δ > 0 such that fo all t, x [a, b, t x <δ f(t f(x <ɛ. (7.37 O the othe had, if t x δ, wehave f(t f(x 2 f 2 f (t x2 δ 2 = αφ t (x, (7.38

18 Bestei Polyomials say, whee α =2 f /δ 2 > 0. Note that φ t (x 0. The, fom (7.37 ad (7.38, we see that fo all t, x [a, b, f(t f(x ɛ + αφ t (x, ad so ɛ αφ t (x f(t f(x ɛ + αφ t (x. (7.39 At this stage we make use of the mootoicity of the liea opeato L.We apply L to each tem i (7.39, egaded as a fuctio of x, ad evaluate each esultig fuctio of x at the poit t, to give ɛ(l g 0 (t α(l φ t (t f(t(l g 0 (t (L f(t ɛ(l g 0 (t+α(l φ t (t. Obseve that (L φ t (t 0, sice L is mootoe ad φ t (x 0. Thus we obtai the iequality f(t(l g 0 (t (L f(t ɛ L g 0 + α(l φ t (t. (7.40 If we ow wite L g 0 =1+(L g 0 g 0, we obtai ad so deive the iequality L g 0 1+ L g 0 g 0, f(t(l g 0 (t (L f(t ɛ(1 + L g 0 g 0 +α(l φ t (t. (7.41 We ow wite f(t (L f(t =[f(t(l g 0 (t (L f(t+[f(t f(t(l g 0 (t, ad hece obtai the iequality f(t (L f(t f(t(l g 0 (t (L f(t + f(t f(t(l g 0 (t. (7.42 I (7.41 we have aleady obtaied a uppe boud fo the fist tem o the ight of (7.42, ad the secod tem satisfies the iequality f(t f(t(l g 0 (t f L g 0 g 0. (7.43 The, o substitutig the two iequalities (7.41 ad (7.43 ito (7.42, we fid that f(t (L f(t ɛ +( f + ɛ L g 0 g 0 + α(l φ t (t. (7.44 O the ight side of the above iequality we have ɛ plus two oegative quatities, each of which ca be made less tha ɛ fo all geate tha some sufficietly lage umbe N, ad so f(t (L f(t < 3ɛ,

19 7.2 The Mootoe Opeato Theoem 265 uifomly i t, fo all >N. This completes the poof. Remak If we go though the above poof agai, we ca see that the followig is a valid alteative vesio of the statemet of Theoem We will fid this helpful whe we discuss the Hemite Fejé opeato. Let (L deote a sequece of mootoe liea opeatos that map fuctios f C[a, b to fuctios L f C[a, b. The if L g 0 g 0 uifomly o [0, 1, ad (L φ t (t 0 uifomly i t o [0, 1, whee g 0 ad φ t ae defied i the poof of Theoem 7.2.1, it follows that L f f uifomly o [0, 1 fo all f C[a, b. Example We see fom Examples ad that B (1; x =1, B (x; x =x, ad B (x 2 ; x =x x(1 x. Thus B (f; x coveges uifomly to f(x o[0, 1 fo f(x =1,x, ad x 2, ad sice the Bestei opeato B is also liea ad mootoe, it follows fom the Bohma Koovki Theoem that B (f; x coveges uifomly to f(x o[0, 1 fo all f C[0, 1, as we aleady foud i Bestei s Theoem We ow ecall the Hemite itepolatig polyomial p 2+1, defied by (1.38. If we wite q 2+1 (x = [a i u i (x+b i v i (x, (7.45 i=0 whee u i ad v i ae defied i (1.39 ad (1.40, the If we ow choose q 2+1 (x i =a i, q 2+1(x i =b i, 0 i. (7.46 a i = f(x i, b i =0, 0 i, (7.47 whee the x i ae the zeos of the Chebyshev polyomial T +1 ad f is a give fuctio defied o [ 1, 1, it follows that q 2+1 (x = f(x i u i (x =(L f(x, (7.48 i=0 say, whee u i is give by (2.103, ad so (L f(x = ( 2 T+1 (x f(x i 1 x ix +1 (x x i=0 i 2. (7.49

20 Bestei Polyomials We call L the Hemite Fejé opeato. It is clea that L is a liea opeato, ad sice 0 1 x i x 2 fo 1 x 1, (7.50 fo all i, we see that L is mootoe. We also ote that L epoduces the fuctio 1, sice the deivative of 1 is zeo ad (L 1(x itepolates 1 o the Chebyshev zeos. It is also obvious that L does ot epoduce the fuctios x ad x 2, sice thei fist deivatives ae ot zeo o all the Chebyshev zeos. Let us apply L to the fuctio φ t (x =(t x 2.We obtai ( 2 T+1 (t (L φ t (t = (1 x i t, +1 i=0 ad it follows fom (7.50 that (L φ t (t 2 +1, so that (L φ t (t 0 uifomly i t o [ 1, 1. Thus, i view of the alteative statemet of Theoem 7.2.1, give i the emak followig the poof of the theoem, we deduce the followig esult as a special case of the Bohma Koovki Theoem Theoem Let (L deote the sequece of Hemite Fejé opeatos, defied by (7.49. The L f f uifomly fo all f C[ 1, 1. Theoem 7.2.2, like Bestei s Theoem 7.1.5, gives a costuctive poof of the Weiestass theoem. A diect poof of Theoem 7.2.2, which does ot explicitly use the Bohma Koovki theoem, is give i Davis [10. We will give aothe applicatio of the Bohma Koovki theoem i the ext sectio. We ca show (see Poblem that the oly liea mootoe opeato that epoduces 1, x, ad x 2, ad thus all quadatic polyomials, is the idetity opeato. This puts ito pespective the behaviou of the Bestei opeato, which epoduces liea polyomials, but does ot epoduce x 2, ad the Hemite Fejé opeato, which does ot epoduce x o x 2. Poblem Let L deote a liea mootoe opeato actig o fuctios f C[a, b that epoduces 1, x, ad x 2. Show that (L φ t (t =0, whee φ t (x =(t x 2. By wokig though the poof of Theoem show that fo a give f C[a, b, (7.40 yields f(t (Lf(t ɛ fo all t [a, b ad ay give ɛ>0. Deduce that Lf = f fo all f C[a, b, ad thus L is the idetity opeato.

21 Poblem Deduce fom Theoem that ( 2 T+1 (x 1 x i x lim +1 (x x i 2 =1, whee the x i deote the zeos of T +1. i=0 7.3 O the q-iteges O the q-iteges I view of the may iteestig popeties of the Bestei polyomials, it is ot supisig that seveal geealizatios have bee poposed. I this sectio we discuss a geealizatio based o the q-iteges, which ae defied i Sectio 1.5. Let us wite [ 1 B (f; x = f x (1 q s x (7.51 =0 fo each positive itege, whee f deotes the value of the[ fuctio f at x =[/[, the quotiet of the q-iteges [ ad [, ad deotes a q-biomial coefficiet, defied i ( Note that a empty poduct i (7.51 deotes 1. Whe we put q = 1 i (7.51, we obtai the classical Bestei polyomial, defied by (7.1, ad i this sectio we cosistetly wite B (f; x to mea the geealized Bestei polyomial, defied by (7.51. I Sectio 7.5, wheeve we eed to emphasize the depedece of the geealized Bestei polyomial o the paamete q, we will wite B(f; q x i place of B (f; x. We see immediately fom (7.51 that B (f;0=f(0 ad B (f;1=f(1, (7.52 givig itepolatio at the edpoits, as we have fo the classical Bestei polyomials. It is show i Sectio 8.2 that evey q-biomial coefficiet is a polyomial i q (called a Gaussia polyomial with coefficiets that ae all positive iteges. It is thus clea that B, defied by (7.51, is a liea opeato, ad with 0 <q<1, it is a mootoe opeato that maps fuctios defied o [0, 1 to P. The followig theoem ivolves q-diffeeces, which ae defied i Sectio 1.5. This esult yields Theoem whe q =1. Theoem The geealized Bestei polyomial may be expessed i the fom [ B (f; x = =0 whee qf j = 1 q f j+1 q 1 1 q f j, 1, with 0 qf j = f j = f([j/[. qf 0 x, (7.53

22 Bestei Polyomials Poof. Hee we equie the idetity, 1 [ (1 q s x= ( 1 s q s(s 1/2 s x s, (7.54 which is equivalet to (8.12 ad educes to a biomial expasio whe we put q = 1. Begiig with (7.51, ad expadig the tem cosistig of the poduct of the factos (1 q s x, we obtai B (f; x = [ [ f x ( 1 s q s(s 1/2 s =0 x s. Let us put t = + s. The, sice [ [ s = [ t [ t, we may wite the latte double sum as [ t [ x t ( 1 t q (t (t 1/2 t t t=0 =0 f = t=0 [ t t qf 0 x t, o usig the expasio fo a highe-ode q-diffeece, as i ( This completes the poof. We see fom (1.33 ad (1.113 that k q f(x 0 q k(k 1/2 [k! = f[x 0,x 1,...,x k = f (k (ξ, k! whee x j =[j ad ξ (x 0,x k. Thus q-diffeeces of the moomial x k of ode geate tha k ae zeo, ad we see fom Theoem that fo all k, B (x k ; x is a polyomial of degee k. We deduce fom Theoem that B (1; x =1. (7.55 Fo f(x =x we have 0 qf 0 = f 0 = 0 ad 1 qf 0 = f 1 f 0 =1/[, ad it follows fom Theoem that B (x; x =x. (7.56 Fo f(x =x 2 we have 0 qf 0 = f 0 =0, 1 qf 0 = f 1 f 0 =1/[ 2, ad 2 qf 0 = f 2 (1 + qf 1 + qf 0 = ( 2 ( 2 [2 [1 (1 + q. [ [

23 7.3 O the q-iteges 269 We the fid fom Theoem that B (x 2 ; x =x 2 + x(1 x. (7.57 [ The above expessios fo B (1; x, B (x; x, ad B (x 2 ; x geealize thei coutepats give ealie fo the case q = 1 ad, with the help of Theoem 7.2.1, lead us to the followig theoem o the covegece of the geealized Bestei polyomials. Theoem Let (q deote a sequece such that 0 <q < 1 ad q 1as. The, fo ay f C[0, 1, B (f; x coveges uifomly to f(x o[0, 1, whee B (f; x is defied by (7.51 with q = q. Poof. We saw above fom (7.55 ad (7.56 that B (f; x =f(x fo f(x =1adf(x =x, ad sice q 1as, we see fom (7.57 that B (f; x coveges uifomly to f(x fo f(x =x 2. Also, sice 0 <q < 1, it follows that B is a mootoe opeato, ad the poof is completed by applyig the Bohma Koovki Theoem We ow state ad pove the followig geealizatios of Theoems ad Theoem If f(x iscovexo[0, 1, the fo all 1 ad fo 0 <q 1. Poof. Fo each x [0, 1, let us defie B (f; x f(x, 0 x 1, (7.58 x = [ [ ad λ = [ +1 x (1 q s x, 0. We see that λ 0 whe 0 <q 1 ad x [0, 1, ad ote fom (7.55 ad (7.56, espectively, that ad λ 0 + λ λ =1 λ 0 x 0 + λ 1 x λ x = x. The we obtai fom the esult i Poblem that if f is covex o [0, 1, ( B (f; x = λ f(x f λ x = f(x, =0 ad this completes the poof. =0

24 Bestei Polyomials Theoem If f(x is covex o[0, 1, B 1 (f; x B (f; x, 0 x 1, (7.59 fo all 2, whee B 1 (f; x ad B (f; x ae evaluated usig the same value of the paamete q. The Bestei polyomials ae equal at x = 0 ad x = 1, sice they itepolate f at these poits. If f C[0, 1, the iequality i (7.59 is stict fo 0 <x<1 uless, fo a give value of, the fuctio [ [ 1 [ 1, f is liea i each of the itevals we have simply B 1 (f; x =B (f; x. [ [ 1, fo 1 1, whe Poof. I the poof give by Davis [10 fo the special case of this theoem whe q = 1, the diffeece betwee two cosecutive Bestei polyomials is expessed i tems of powes of x/(1 x. This is ot appopiate fo q 1, ad ou poof follows that give by Ouç ad Phillips [40. Fo 0 <q<1, let us wite 1 (B 1 (f; x B (f; x (1 q s x 1 1 = f =0 f =0 ( [ [ 1 [ 1 ( [ [ [ x 1 x 1 s= s= 1 (1 q s x 1. (1 q s x 1 We ow split the fist of the above summatios ito two, witig say, whee x 1 s= 1 (1 q s x 1 = ψ (x+q 1 ψ +1 (x, ψ (x =x 1 s= (1 q s x 1. (7.60 O combiig the esultig thee summatios, the tems i ψ 0 (x ad ψ (x cacel, ad we obtai whee 1 (B 1 (f; x B (f; x (1 q s x 1 = a = [ [ f ( [ [ + q [ 1 [ f 1 [ =1 ( [ 1 f [ 1 a ψ (x, (7.61 ( [. (7.62 [

25 7.3 O the q-iteges 271 It is clea fom (7.60 that each ψ (x is oegative o [0, 1 fo 0 q 1, ad thus fom (7.61, it will suffice to show that each a is oegative. Let us wite [ λ =, x 1 = [ [ [ 1, ad x [ 1 2 = [ 1. It the follows that [ 1 λ = q [ ad λx 1 +(1 λx 2 = [ [, ad we see immediately, o compaig (7.62 ad (7.28, that a = λf(x 1 +(1 λf(x 2 f(λx 1 +(1 λx 2 0, ad so B 1 (f; x B (f; x. We obviously have equality at x = 0 ad x = 1, sice all Bestei polyomials itepolate f at these edpoits. The iequality will be stict fo 0 <x<1 uless evey a is zeo; this ca occu oly whe f is liea i each of the itevals betwee cosecutive poits [/[ 1, 0 1, whe we have B 1 (f; x =B (f; x fo 0 <x<1. This completes the poof. We ow give a algoithm, fist published i 1996 (see Phillips [42, fo evaluatig the geealized Bestei polyomials. Whe q = 1 it educes to the well-kow de Casteljau algoithm (see Hoschek ad Lasse [26 fo evaluatig the classical Bestei polyomials. Algoithm This algoithm begis with the value of q ad the values of f at the + 1 poits [/[, 0, ad computes B (f; x =f [ 0, which is the fial umbe geeated by the algoithm. iput: q; f([0/[,f([1/[,...,f([/[ fo =0to f [0 := f([/[ ext fo m =1to fo =0to m f [m ext ext m output: := (q q m 1 xf [m 1 f [ 0 = B (f; x + xf [m 1 +1 The followig theoem justifies the above algoithm. Theoem Fo 0 m ad 0 m, wehave f [m = m [ m f +t t t=0 m t 1 x t (q q s x, (7.63

26 Bestei Polyomials ad, i paticula, f [ 0 = B (f; x. (7.64 Poof. We use iductio o m. Fom the iitial coditios i the algoithm, f [0 := f([/[ = f, 0, it is clea that (7.63 holds fo m =0 ad 0. Note that a empty poduct i (7.63 deotes 1. Let us assume that (7.63 holds fo some m such that 0 m<, ad fo all such that 0 m. The, fo 0 m 1, it follows fom the algoithm that f [m+1 =(q q m xf [m + xf [m +1, ad usig (7.63, we obtai f [m+1 =(q q m x + x m [ m f +t t t=0 m [ m f +t+1 t t=0 m t 1 x t (q q s x m t 1 x t (q +1 q s x. The coefficiet of f o the ight of the latte equatio is m 1 (q q m x (q q s x= m (q q s x, ad the coefficiet of f +m+1 is x m+1. Fo 1 t m, we fid that the coefficiet of f +t is [ (q q m m x t m t 1 [ x t (q q s x+ m t 1 = a t x t (q q s x, m t 1 m t x t (q +1 q s x say. We see that [ a t =(q q m m x t [ + q m t (q +1 x m t 1 ad hece a t = q ([ m t + q m+1 t [ m t 1 ( [ q m t x q t m t [ + m t 1. It is easily veified (see (8.7 ad (8.8 that [ [ [ [ m + q m+1 t m = q t m + t t 1 t m t 1 = [ m +1 t

27 ad thus [ a t =(q q m t m +1 x t 7.3 O the q-iteges 273 Hece the coefficiet of f +t, fo 1 t m, i the above expessio fo f [m+1 is [ m t m +1 x t (q q s x, t ad we ote that this also holds fo t = 0 ad t = m + 1. Thus we obtai. m+1 [ f [m+1 m +1 = f +t t t=0 m t x t (q q s x, ad this completes the poof by iductio. The above algoithm fo evaluatig B (f; x is ot ulike Algoithm (Neville Aitke. I the latte algoithm, each quatity that is computed is, like the fial esult, a itepolatig polyomial o cetai abscissas. Similaly, i Algoithm 7.3.1, as we see i (7.63, each itemediate umbe f [m has a fom that esembles that of the fial umbe f [ 0 = B (f; x. We ow show that each f [m ca also be expessed simply i tems of q-diffeeces, as we have fo B (f; x i (7.53. Theoem Fo 0 m ad 0 m, wehave f [m = m [ q (m s m s s qf x s. (7.65 Poof. We may veify (7.65 by iductio o m, usig the ecuece elatio i Algoithm Alteatively, we ca deive (7.65 fom (7.63 as follows. Fist we wite m t 1 m t 1 (q q s x=q (m t (1 q s y, whee y = x/q, ad we fid with the aid of (7.54 that m t 1 m t [ (q q s x= ( 1 j q j(j 1/2+(m t j m t j j=0 x j. O substitutig this ito (7.63, we obtai f [m = m [ m f +t t t=0 m t [ x t ( 1 j q j(j 1/2+(m t j m t j j=0 x j.

28 Bestei Polyomials If we ow let s = j + t, we may ewite the above double summatio as m [ q (m s m s [ x s ( 1 j q j(j 1/2 s f s j +s j, j=0 which, i view of (1.121, gives m [ f [m = q (m s m s ad this completes the poof. s qf x s, Poblem Veify (7.65 diectly by iductio o m, usig the ecuece elatio i Algoithm Poblem Wok though Algoithm fo the case = 2, ad so veify diectly that f [2 0 = B 2 (f; x. 7.4 Total Positivity We begi this sectio by defiig a totally positive matix, ad discuss the atue of liea tasfomatios whe the matix is totally positive. We will apply these ideas i Sectio 7.5 to justify futhe popeties of the Bestei polyomials coceed with shape, such us covexity. Defiitio A eal matix A is called totally positive if all its mios ae oegative, that is, ( a i1,j 1 a i1,j k i1,i A 2,...,i k = det j 1,j 2,...,j k.. 0, (7.66 a ik,j 1 a ik,j k fo all i 1 <i 2 < <i k ad all j 1 <j 2 < <j k. We say that A is stictly totally positive if all its mios ae positive, so that is eplaced by > i (7.66. It follows, o puttig k = 1 i (7.66, that a ecessay coditio fo a matix to be totally positive is that all its elemets ae oegative. Theoem A eal matix A =(a ij is totally positive if ( i, i +1,...,i+ k A 0 fo all i, j, ad k. (7.67 j, j +1,...,j+ k Similaly, the matix A is stictly totally positive if the mios give i (7.67, which ae fomed fom cosecutive ows ad colums, ae all positive.

29 7.4 Total Positivity 275 Fo a poof, see Kali [28. I view of Theoem 7.4.1, we ca detemie whethe A is totally positive o stictly totally positive by testig the positivity of oly those mios that ae fomed fom cosecutive ows ad colums, athe tha havig to examie all mios. Example Let us coside the Vademode matix 1 x 0 x 2 0 x 0 1 x 1 x 2 1 x 1 V = V(x 0,...,x = ( x x 2 x As we showed i Chapte 1 (see Poblem 1.1.1, det V(x 0,...,x = i>j(x i x j. (7.69 Let 0 <x 0 <x 1 < <x. The we see fom (7.69 that det V > 0, ad we ow pove that V is stictly totally positive. Usig Theoem 7.4.1, it is sufficiet to show that the mios det x j i x j+1 i x j+k i x j i+1 x j+1 i+1 x j+k i+1... x j i+k x j+1 i+k x j+k i+k ae positive fo all oegative i, j, k such that i+k, j+k. O emovig commo factos fom its ows, the above detemiat may be expessed as (x i x i+k j det V(x i,...,x i+k > 0, sice det V(x i,...,x i+k = (x s x > 0. i <s i+k This completes the poof that V is stictly totally positive. If A = BC, whee A, B, ad C deote matices of odes m, m k, ad k, espectively, the ( ( ( i1,...,i A p = i1,...,i B p β1,...,β C p. (7.70 j 1,...,j p β 1< <β p β 1,...,β p j 1,,j p This is kow as the Cauchy Biet detemiat idetity. It follows immediately fom this most useful idetity that the poduct of totally positive matices is a totally positive matix, ad the poduct of stictly totally positive matices is a stictly totally positive matix.

30 Bestei Polyomials Defiitio Let v deote the sequece (v i, which may be fiite o ifiite. The we deote by S (v the umbe of stict sig chages i the sequece v. Fo example, S (1, 2, 3, 4, 5, 6 = 5, S (1, 0, 0, 1, 1 = 1, ad S (1, 1, 1, 1, 1, 1,...=, whee the last sequece alteates +1 ad 1 idefiitely. It is clea that isetig zeos ito a sequece, o deletig zeos fom a sequece, does ot alte the umbe of chages of sig. Also, deletig tems of a sequece does ot icease the umbe of chages of sig. We use the same otatio to deote sig chages i a fuctio. Defiitio Let v i = a ik u k, k=0 i =0, 1,...,m, whee the u k ad the a ik, ad thus the v i, ae all eal. This liea tasfomatio is said to be vaiatio-dimiishig if S (v S (u. Defiitio A matix A, which may be fiite o ifiite, is said to be m-baded if thee exists a itege l such that a ij 0 implies that l j i l + m. This is equivalet to sayig that all the ozeo elemets of A lie o m + 1 diagoals. We will say that a matix A is baded if it is m-baded fo some m. Note that evey fiite matix is baded. We have aleady met 1-baded ad 2-baded (tidiagoal matices i Chapte 6. I this sectio we will be paticulaly iteested i 1-baded matices, also called bidiagoal matices, because of Theoem below. We ow come to the fist of the mai esults of this sectio. Theoem If T is a totally positive baded matix ad v is ay vecto fo which Tv is defied, the S (Tv S (v. Fo a poof of this theoem see Goodma [22. Whe we fist ecoute it, the questio of whethe a liea tasfomatio is vaiatio-dimiishig may ot seem vey iteestig. Howeve, buildig o the cocept of a vaiatio-dimiishig liea tasfomatio, we will see i Sectio 7.5 that the umbe of sig chages i a fuctio f defied o [0, 1 is ot iceased if we apply a Bestei opeato, ad we say that Bestei opeatos ae shape-pesevig. This popety does ot always hold, fo example, fo itepolatig opeatos.

31 7.4 Total Positivity 277 Example Let v deote a ifiite eal sequece fo which S (v is fiite. Coside the sequece w =(w i i=0 defied by w i = v i + v i 1 fo i 1, ad w 0 = v 0. The w = Tv, whee T = Let us coside the mios of T costucted fom cosecutive ows ad colums. Ay such mio whose leadig (top left elemet is 0 has eithe a whole ow o a whole colum of zeos, ad so the mio is zeo. It is also ot had to see that ay mio costucted fom cosecutive ows ad colums whose leadig elemet is 1 has itself the value 1. Thus, by Theoem 7.4.1, the matix T is totally positive, ad so we deduce fom Theoem that S (w =S (Tv S (v. Theoem A fiite matix is totally positive if ad oly if it is a poduct of 1-baded matices with oegative elemets. Fo a poof of this theoem, see de Boo ad Pikus [13. A immediate cosequece of Theoem is that the poduct of totally positive matices is totally positive, as we have aleady deduced above fom the Cauchy Biet idetity. Example To illustate Theoem 7.4.3, coside the 1-baded factoizatio = The fou matices i the above poduct ae ideed all 1-baded matices with oegative elemets, ad thei poduct is totally positive. We ow state a theoem, ad give a elated example, coceig the factoizatio of a matix ito the poduct of lowe ad uppe tiagula matices. Theoem A matix A is stictly totally positive if ad oly if it ca be expessed i the fom A = LU whee L is a lowe tiagula matix, U is a uppe tiagula matix, ad both L ad U ae totally positive matices.

32 Bestei Polyomials Fo a poof, see Cye [9. Example To illustate Theoem 7.4.4, we cotiue Example 7.4.3, i which the matix A = is expessed as a poduct of fou 1-baded matices. If we multiply the fist two of these 1-baded matices, ad also multiply the thid ad fouth, we obtai the LU factoizatio A = = LU, ad it is easy to veify that L ad U ae both totally positive. The matix A i Example is the 3 3 Vademode matix V(1, 2, 3. I Sectio 1.2 we gave (see Theoem the LU factoizatio of the geeal Vademode matix. Example Let A = The we ca easily veify that A is totally positive, ad it is obviously ot stictly totally positive. We give two diffeet LU factoizatios of A: A = LU = , whee L is totally positive but U is ot, ad A = LU = whee both L ad U ae totally positive. This example shows that we caot eplace stictly totally positive by totally positive i the statemet of Theoem Defiitio Fo a eal-valued fuctio f o a iteval I, we defie S (f to be the umbe of sig chages of f, that is, S (f = sup S (f(x 0,...,f(x m, whee the supemum is take ove all iceasig sequeces (x 0,...,x m i I, fo all m.,

33 7.4 Total Positivity 279 Defiitio We say that a sequece (φ 0,...,φ of eal-valued fuctios o a iteval I is totally positive if fo ay poits 0 <x 0 < <x i I, the collocatio matix (φ j (x i i,j=0 is totally positive. Theoem Let ψ i (x =ω(xφ i (x, fo 0 i. The, if ω(x 0 o I ad the sequece of fuctios (φ 0,...,φ is totally positive o I, the sequece (ψ 0,...,ψ is also totally positive o I. Poof. This follows easily fom the defiitios. Theoem If (φ 0,...,φ is totally positive o I, the fo ay umbes a 0,...,a, S (a 0 φ + + a φ S (a 0,...,a. Fo a poof of this theoem see Goodma [22. Defiitio Let L deote a liea opeato that maps each fuctio f defied o a iteval [0, 1 oto Lf defied o [0, 1. The we say that L is vaiatio-dimiishig if S (Lf S (f. Poblem Show that a matix has (2 1 2 mios, of which ( +1 2 ae fomed fom cosecutive ows ad colums. How may mios ae thee i these two categoies fo a m matix? Poblem Show that the matix a 0 a 1 a 2 a 3 a 4 0 a 0 a 1 a 2 a a 0 a 1 a a 0 a a 0 is totally positive, whee a i 0 fo all i, ad a 2 i a i 1a i+1 0 fo 1 i 3. Poblem Let v deote a ifiite eal sequece fo which S (v is fiite. Coside the sequece w defied by w = Show that S (w S (v. v s fo 0.

34 Bestei Polyomials Poblem Repeat Poblem with the sequece w defied by ( w = v s s fo 0, agai showig that S (w S (v. Poblem Show that the matix (x 0 + t 0 1 (x 0 + t x 0 (x 1 + t 0 1 (x 1 + t x 1 (x 2 + t 0 1 (x 2 + t x 2 (x 3 + t 0 1 (x 3 + t x 3 is totally positive if 0 <x 0 <x 1 <x 2 <x 3 ad 0 <t 0 <t Futhe Results This sectio is based o the wok of Goodma, Ouç ad Phillips [23. We will use the theoy of total positivity, developed i the last sectio, to justify shape-pesevig popeties of the geealized Bestei polyomials. We will also show that if a fuctio f is covex o [0, 1, the fo each x i [0, 1 the geealized Bestei polyomial B (f; x appoaches f(x mootoically fom above as the paamete q iceases, fo 0 <q 1. I the last sectio we saw that fo 0 <x 0 <x 1 < <x, the Vademode matix V(x 0,...,x is stictly totally positive. It the follows fom Defiitio that the sequece of moomials (x i i=0 is totally positive o ay iteval [0,. We ow make the chage of vaiable t = x/(1 x, ad ote that t is a iceasig fuctio of x. Thus, if t i = x i /(1 x i, ad we ow let 0 <x 0 <x 1 < <x < 1, it follows that 0 <t 0 <t 1 < <t. Sice the Vademode matix V(t 0,...,t is stictly totally positive, it follows that the sequece of fuctios ( x 1, 1 x, x 2 (1 x 2,..., x (1 x is totally positive o [0, 1. We also see fom Theoem that the sequece of fuctios ( (1 x,x(1 x 1,x 2 (1 x 2,...,x 1 (1 x,x (7.71 is totally positive o [0, 1. Sice the + 1 fuctios i the sequece (7.71 ae a basis fo P, the subspace of polyomials of degee at most, they ae a basis fo all the classical Bestei polyomials of degee, defied by (7.1, ad we ca immediately deduce the followig poweful esult fom Theoem

35 7.5 Futhe Results 281 Theoem Let B (f; x deote the classical Bestei polyomial of degee fo the fuctio f. The S (B f S (f (7.72 fo all f defied o [0, 1, ad thus the classical Bestei opeato B is vaiatio-dimiishig. Poof. Usig Theoem 7.4.6, we have S (B f S (f 0,f 1,...,f S (f, whee f = f(/. Fo each q such that 0 <q 1, ad each 1, we ow defie j 1 P q,j (x =xj (1 q s x, 0 x 1, (7.73 fo 0 j. These fuctios ae a basis fo P, ad ae thus a basis fo all the geealized Bestei polyomials of degee, defied by (7.51. We have aleady see above that (P,0,P 1,1,...,P 1, 1 is totally positive o [0, 1, ad we will show below that the same holds fo (P q,0,pq,1,...,pq,, fo ay q such that 0 <q 1. Sice the fuctios defied i (7.73 ae a basis fo P, it follows that fo ay choice of q ad satisfyig 0 <q, 1, thee exists a osigula matix T,q, such that P q,0 (x. P,(x q P,0(x = T,q,. P,(x. (7.74 Theoem Fo 0 <q 1 all elemets of the matix T,q, ae oegative. Poof. We use iductio o. Sice T 1,q, is the 2 2 idetity matix, its elemets ae obviously oegative. Let us assume that the elemets of T,q, ae all oegative fo some 1. The, sice P q +1,j+1 (x =xp q,j (x, 0 j, (7.75 fo all q such that 0 <q 1, we see fom (7.75 ad (7.74 that P q +1,1 (x P +1,1(x. = T,q,.. (7.76 P q +1,+1 (x P+1,+1(x

36 Bestei Polyomials Also, we have P q +1,0 (x =(1 q xp q,0 (x =(1 q x j=0 t,q, 0,j P,j(x, (7.77 whee (t,q, 0,0,t,q, 0,1,...,t,q, 0, deotes the fist ow of the matix T,q,. If we ow substitute (1 q xp,j(x =P +1,j(x+( j q P +1,j+1(x i the ight side of (7.77 ad simplify, we obtai P q +1,0,q, (x =t 0,0 P +1,0(x+(1 q t,q, 0, P +1,+1(x ( + ( +1 j q t,q, 0,j 1 + t,q, 0,j P +1,j (x. (7.78 j=1 The, o combiig (7.76 ad (7.78, we fid that P q +1,0 (x P q +1,1 (x t,q, 0,0 v T +1. = P q +1,+1 (x 0 T,q, so that P +1,0(x P +1,1(x. P +1,+1(x, (7.79 t,q, 0,0 v T +1 T +1,q, =. ( T,q, I the block matix o the ight side of ( deotes the zeo vecto with + 1 elemets, ad v T +1 is the ow vecto whose elemets ae the coefficiets of P +1,1(x,...,P +1,+1(x, give by (7.78. O substitutig x = 0 i (7.80, it is clea that t,q, 0,0 = 1. We ca deduce fom (7.78 that if 0 <q 1, the elemets of v T +1 ae oegative, ad this completes the poof by iductio. It follows fom (7.80 ad the defiitio of v T +1 that t +1,q, 0,0 = t,q, 0,0 (7.81 ad t +1,q, 0,j =( +1 j q t,q, 0,j 1 + t,q, 0,j, 1 j. (7.82 We will equie this ecuece elatio, which expesses the elemets i the fist ow of T +1,q, i tems of those i the fist ow of T,q,,i the poof of ou ext theoem. This shows that the tasfomatio matix T,q, ca be factoized as a poduct of 1-baded matices. Fist we equie the followig lemma.

37 7.5 Futhe Results 283 Lemma Fo m 1 ad ay eal ad a, let A(m, a deote the m (m + 1 matix 1 m a 1 m 1 a 1 a. The A(m, aa(m +1, b=a(m, ba(m +1, a. (7.83 Poof. Fo i =0,...,m 1 the ith ow of each side of (7.83 is [ 0,...,0, 1, m+1 i + m i a b, ( m i a( m i b, 0,...,0. Fo 1 j 1, let B ( j deote the 1-baded ( +1 ( + 1 matix that has uits o the mai diagoal, ad has the elemets j q j, j 1 q j,..., q j, 0,...,0 o the diagoal above the mai diagoal, whee thee ae j zeos at the lowe ed of that diagoal. Thus, fo example, B ( 2 = The matix B (+1 j B ( j ad 1 2 q 2 1 q 2 ca be expessed i a block fom ivolvig the matix. We ca veify that B (+1 1 = B (+1 j+1 = 1 1 ct I ct j 0 B ( j 1. (7.84 (7.85 fo 1 j 1, whee each c T j is a ow vecto, 0 deotes the zeo vecto, ad I is the uit matix of ode +1.

38 Bestei Polyomials Theoem Fo 2 ad ay q,, wehave T,q, = B ( 1 B( 2 B ( 1, (7.86 whee B ( j is the 1-baded matix defied above. Poof. We use iductio o. Fo all 2 let It is easily veified that S,q, = B ( 1 B( 2 B ( 1. (7.87 T 2,q, = S 2,q, = B (2 1 = 1 q Let us assume that fo some 2, T,q, = S,q,. It follows fom (7.84 ad (7.85 that S +1,q, = 1 ct 0 1 ct 1 1 ct 1. ( I 0 B ( 1. 0 B ( 1 If we cay out the multiplicatio of the block matices o the ight of (7.88, the, usig (7.86, we see that S +1,q, = 1 dt, 0 T,q, whee d T is a ow vecto. Thus it emais oly to veify that the fist ows of T +1,q, ad S +1,q, ae equal. Let us deote the fist ow of S,q, by [ s,q, 0,0,s,q, 0,1,...,s,q, 0,. We will show that s,q, 0, = 0. Let us examie the poduct of the 1 matices o the ight of (7.87. We ca show by iductio o j that fo 1 j 1, the poduct B ( 1 B( 2 B ( j is j-baded, whee the ozeo elemets ae o the mai diagoal ad the j diagoals above the mai diagoal. (See Poblem Thus S,q, is ( 1-baded, ad so the last elemet i its fist ow, s,q, 0,, is zeo. Now let us wite the matix B (+1 j i the block fom B (+1 j = A(j, q+1 j O, C j D j

39 7.5 Futhe Results 285 whee A(j, q +1 j isthej (j + 1 matix defied i Lemma 7.5.1, O is the j ( +1 j zeo matix, C j is ( +2 j (j + 1, ad D j is ( +2 j ( +1 j. Thus B (+1 1 B (+1 2 B (+1 j = A(1,q A(j, q +1 j 0 T, (7.89 whee A(1,q A(j, q +1 j is1 (j + 1 ad 0 T is the zeo vecto with +1 j elemets. I paticula, o puttig j = i (7.89, we see fom (7.87 that the fist ow of S +1,q, is [ A(1,q A(2,q 1 A( 1,q 2 A(, q, 0 = [ w T, 0, (7.90 say, whee w T is a ow vecto with + 1 elemets. (We ote i passig that this cofims ou ealie obsevatio that the last elemet of the fist ow of S +1,q, is zeo. I view of Lemma 7.5.1, we may pemute the quatities q,q 1,...,q i (7.90, leavig w T uchaged. I paticula, we may wite w T = A(1,q 1 A(2,q 2 A( 1,qA(, q. (7.91 Now, by compaiso with (7.90, the poduct of the fist 1 matices i (7.91 is the ow vecto cotaiig the fist elemets i the fist ow of S,q,, ad thus w T =[s,q, 0,0,...,s,q, 0, 1 =[t,q, 0,0,...,t,q, 0, 1 F j G j 1 q q 1 q q This gives s +1,q, 0,0 = t,q, 0,0 ad s +1,q, 0,j =( +1 j q t,q, 0,j 1 + t,q, 0,j, 1 j, whee we ote that t,q, 0, = 0. It the follows fom (7.81 ad (7.82 that s +1,q, 0,j = t +1,q, 0,j, 0 j, ad sice s +1,q, 0,+1 =0=t +1,q, 0,+1, (7.86 holds fo + 1. This completes the poof. If 0 <q 1 1, all elemets i the 1-baded matices B ( j o the ight of (7.86 ae oegative. The, fom Theoem 7.4.3, we immediately have the followig esult coceig the total positivity of T,q,..