RemarksonanarticleofJ.P.King

Transcription

1 Commet.Math.Uiv.Caroli. 46,4(2005) RemarksoaarticleofJ.P.Kig Heier Goska, Paula Piţul Abstract. The preset ote discusses a iterestig positive liear operator which was recetly itroduced by J.P. Kig. New estimates i terms of the first ad secod modulus of cotiuity are give, ad iterates of the operators are cosidered as well. For geeral Kig operators the secod momets are miimized. Keywords: positive liear operators, degree of approximatio, cotractio priciple, secod order modulus, secod momets Classificatio: 41A25, 41A36, 47H10 1. Itroductio I[4] J.P. Kig defied the followig iterestig(ad somewhat exotic) sequeceofliearadpositiveoperators V : C[0,1] C[0,1]whichgeeralizethe classicalbersteioperators B : (1) V (f; x)= k=0 ( ) (r (x)) k (1 r (x)) k f k ( ) k forall f C[0,1],0 x 1,where r :[0,1] [0,1]arecotiuousfuctios. We list some of their properties. Property1.1. If {V } N aretheoperatorsdefiedi(1)wehave (2) V (e 0 ; x)=e 0 (x) V (e 1 ; x)=r (x) ad V (e 2 ; x)= r (x) + 1 (r (x)) 2 where e i (x)=x i, i=0,1,2,aretheclassicaltestfuctiosforpositiveliear operator approximatio. Theequatio V (e 1 ; x)=r (x)showsthattheclassicalbersteioperator B, whichisobtaiedfor r (x)=x,istheuiquemappigoftheform(1)which reproduces liear fuctios.

2 646 H. Goska, P. Piţul Theorem1.2. Oehaslim V f(x)=f(x)foreach f C[0,1], x [0,1], ifadolyiflim r (x)=x. Choosigthe right r fuctio,j.p.kigprovedthefollowig: Theorem1.3. Let {V } N bethesequeceofoperatorsdefiedi(1)with r (3) r (x):= 1 (x)=x2, =1, r (x)= 2( 1) x2 + 1, =2,3,... 4( 1) 2 The: (i) V (e 2 ; x)=e 2 (x), N; x [0,1], (ii) V (e 1; x) e 1 (x), (iii) lim V (f; x)=f(x)foreach f C[0,1]. Remark1.4. Sice V e 1= r,itisclearthat V isotapolyomialoperator. J.P.Kigalsogavequatitativeestimatesfor V itermsoftheclassicalfirst ordermodulus ω 1 (f; )usigaresultofo.shishaadb.mod[8]. Theorem1.5. For {V } Ndefiedi(1)wehave (4) V(f; x) f(x) 2ω 1 (f; ) 2x(x V(e 1 ; x)), f C[0,1]; x [0,1]. Remark1.6. Fromthefactthat V (e 1; x)=r (x)ad x r (x)thesquare root i(4) ideed represets a real umber. FromTheorem1.5oecaeasilyobtaithat V iterpolates f attheedpoits: Propositio1.7. With {V } N from(1)wehave V (f;0)=f(0)ad V (f;1)=f(1),i.e., V iterpolatesattheedpoits0ad1. Proof:Weput α (x):= 2x(x V(e 1 ; x)).for x=0wehave α (0)=0,so ω 1 (f; α (0))=0. Thatmeas V (f;0)=f(0). For x=1wehave V (e 1;1)= r (1),adifweiserti(3)thevalue1,weobtai r (1)=1. Thatleadsus agaito ω 1 (f; α (1))=0ad V(f;1)=f(1). Remark1.8. Foraliearadpositiveoperator L:C[0,1] C[0,1]with Le i = e i, i=0,1,itiskowthat Literpolates fi0ad1.thisfollowseasily,ifwe isert x=0ad x=1i L(f; x) f(x) 2 ω 1 (f; L( t x ; x)). The latter iequality ca be foud i Mamedov s article[5]. We observe ow, with the help of the operators itroduced by J.P. Kig, that the above property isolyecessaryadotsufficiet. Ideed,the V, N,iterpolate f i0 ad1,theyareliearadpositive,but V e 1 e 1.

3 Remarks o a article of J.P. Kig Quatitativeestimateswith ω 2 From Păltăea s Theorem i[6, p. 28], the followig is kow: Theorem2.1. Let L:C[0,1] C[0,1]beapositiveadliearoperator.The we have L(f; x) f(x) L(e 0 ; x) e 0 (x) f(x) + L(e 1 x; x) 1 h ω 1(f; h) ( + L(e 0 ; x)+ 1 ) 2 1 h 2 L((e 1 x) 2 ; x) ω 2 (f; h); where h >0, f C[0,1], x [0,1],ad ω 2 istheclassicalsecodordermodulus defied by ω 2 (f; h):=sup { f(x+t) 2f(x)+f(x t) x, x ± t [0,1]}. t h For V thismeas: V(f; x) f(x) (x r(x)) 1 h ω 1(f; h) + (1+ 1h ) 2x(x r (x)) ω 2 (f; h), adfor h:= x r (x)wearriveat (5) V (f; x) f(x) x r (x) ω 1 (f; x r (x))+(1+x)ω 2 (f; x r (x)). If f C 1 [0,1]theduetothefactthat ω 1 (f; h)=o(h)adalso ω 2 (f; h)=o(h) wehavetheapproximatioordero( x r (x)),whe.for f C 2 [0,1] havigsimilarpropertiesforthemoduli ω 1 (f; h)=o(h)ad ω 2 (f; h)=o(h 2 )we obtaio(x r (x)),. 3. Iteratesof V ThissectioismotivatedbyrecetpapersofO.AgratiiadI.A.Rus([1],[7]) i which the cotractio priciple was used to show the followig result of Kelisky ad Rivli[3]. Theorem3.1. If Nisfixed,theforall f C[0,1], x [0,1] lim m Bm (f; x)=f(0)+[f(1) f(0)] x=b 1(f; x). For over-iterated Kigoperators V wehaveasimilarresult,butwitha differet limitig operator.

4 648 H. Goska, P. Piţul Theorem3.2. If Nisfixed,theforall f C[0,1], x [0,1] lim m (V ) m (f; x)=f(0)+[f(1) f(0)] x 2 = V1 (f; x). Proof:FollowigRuswecosidertheBaachspace(C[0,1], )where is the Chebyshev orm. Let Weotethat X α,β = {f C[0,1]:f(0)=α, f(1)=β}, α, β R. a) X α,β isaclosedsubsetof C[0,1]; b) X α,β isaivariatsubsetof V forall α, β R, N (see Propositio 1.7); c) C[0,1]= α,β R X α,βisapartitioof C[0,1]. Nowweshowthat isacotractioforall α, β R. Let f, g X α,β.from(1)wehave V X α,β : X α,β X α,β V (f; x) V (g; x) = V (f g; x) 1 ( ) ( ) = (r k (x)) k (1 r(x)) k k (f g) k=1 1 (r (x)) (1 r (x)) f g ( 1 1 ) 2 1 f g, recalligthat r:[0,1] [0,1]. ( ) Hece V f V g f g,adthus V Xα,β iscotractive. Otheotherhad α+(β α)e 2 X α,β isafixedpoitfor V. If f C[0,1]isarbitrarilygive,the f X f(0),f(1) adfromthecotractio priciple[2] we kow that lim (V m )m f= f(0)+(f(1) f(0))e 2, which cocludes the proof.

5 Remarks o a article of J.P. Kig Polyomial operators of Kig s type Cawefidpolyomial operatorsoftheform(1)thatreproduce e 2? The aswer is egative! Ideed,bythelasttwoequatiosof(2)adthecoditio V (e 2 ; x)=x 2, r mustbeapolyomialoffirstdegree.weput r (x)=ax+badweget: x 2 = 1 ( ) ( a2 x 2 a 1)ab b + +2( x+ + 1 ) b2. This leads to the equatios: 1= 1 a2, 0= a +2( 1)ab, 0= b + 1 b2. So a=± 1ad b=0or b= 1 1.Butforthesevaluesthesecodequatio isotsatisfied. Oeopequestioremais:Cawefidaothertypeofliear adpositivepolyomialoperators Lforwhich Le 2 = e 2? 5. Geeral case Ithissectiowewatto optimize thesecodmomets V ((e 1 x) 2 ; x), x [0,1],ofthegeeral V adstudyithiscasewhichpropertiesremai. Thesecodmometsareithegeeralcase (6) α 2 (x)=v ((e 1 x) 2 ; x)= r (x) + 1 (r (x)) 2 2xr (x)+x 2 = 1 r (x)(1 r (x))+(r (x) x) 2, where0 r (x) 1arecotiuousfuctios.Wewattofid r sothat α 2 is miimal. Wedefie g x :[0,1] [0,1], x [0,1]afixedparameter,by g x (y):= 1 y(1 y)+(y x) 2. Wecawrite g x (y)= ( 1 1 ) y 2 + ( 1 2x ) y+ x 2. Because 1 1 >0, =2,3,...,thefuctio g xadmitsamiimumpoit: 1 y mi = 2x 2 2 = 2x Weeed0 y mi 1,whichmeas 2 1 x 1 2 1, =2,3,... Wedefie r mi :[0,1] [0,1]by

6 650 H. Goska, P. Piţul (7) r mi (x):= 0, x [ 0, 1 2), 2x 1 2 2, x [ 1 2,1 1 2], 1, x ( 1 2 1,1]. Theorem5.1. Thefuctio r mi defiedi(7)yieldsthemiimumvaluefor α 2. Proof: For x [ 1 2,1 2] 1 thiswasprovebefore. Itremaistoshowthe aboveaffirmatiofor x [ 0, 2 1 ) ( ad x 1 1 2,1 ]. Firstcase: x [ 0, 2 1 ) r mi (x)=0adwehavetoprovethat g x (y) g x (0) foreach y [0,1]or 1y(1 y)+(y x)2 x 2 foreach x [0,1].Butthelatter isequivaletto y ( 1 2 2) 1 x,whichistrueduetoourchoiceof x. Secodcase: x ( 1 2 1,1] r mi (x)=1adwehavetoprovethat g x (y) g x (1)foreach y [0,1]or 1y(1 y)+(y x)2 (1 x) 2.Thismeas ( 1 2) 1 (1 y) ( 1 2 2) 1 x,whichisagaitrueduetoourchoiceof x. Theoperators V defiedvia r mi wedeoteby V mi. Property 5.2. Forthe(miimal)secodmomets α 2 of V mi represetatio α 2 (x)= x 2, x [ 0, 2) 1, ( ) [ 1 1 x(1 x) 1 4, x 1 2,1 2] 1, (1 x) 2, x ( 1 1 2,1]. Proof: This follows immediately from the geeral form 1 r (x)(1 r (x))+(r (x) x) 2 wehavethe adtheaboverepresetatioof r mi (x). Usig Păltăea s theorem agai we arrive at V mi (f; x) f(x) x r mi (x) 1 h ω 1(f; h) ( ) 2 1 h 2 α2 (x) ω 2 (f; h), h >0. For h= α (x) weobtai V mi (f; x) f(x) x rmi (x) ω 1 (f; α (x) )+ 3 α (x) 2 ω 2(f; α (x) ). Notethat x r mi (x) = V mi (e 1 x; x) V mi ( e 1 x ; x) V mi ((e 1 x) 2 ; x)= α (x),adthus x rmi (x) 1, x [0,1]. α (x)

7 Remarks o a article of J.P. Kig 651 Remark5.3. (i)fromthedefiitioof r mi wehavelim r mi(x)=xad fromtheorem1.2 lim V (f; x)=f(x). ThelatterfactisalsoacosequeceofoursecodapplicatioofTheorem2.1for V mi. (ii) V mi doesotreproduce e 2.Startigfrom(2)weseethat V mi (e 2 ; x)= 0 x 2, x ( 0, 2) 1. (iii) Theiterpolatiopropertiesattheedpoitsremai.Ideed, V mi (f;0) = ( ) 0 (1 r (0)) f(0)=f(0),ad V mi (f;1)= ( ) f( )=f(1). (iv) For f C 1 [0,1]wehave,withacostat cidepedetof x, V mi (f; x) f(x) c ( x r mi (x) + α (x) )= 2x, x [ 0, 1 ( 2), hece O 1 ), = c 1 2 x ( ) [ 1 x(1 x) 1 4, x 1 2,1 2 1 ] (, hece O 1 ), 2(1 x), x ( 1 2 1,1], hece O ( ) 1. Sothedegreeofapproximatioisbetterclosetotheedpoits,afact sharedbythebersteioperatorswhere r (x)=x. (v) If f C 2 [0,1],the V mi (f; x) f(x) c ( x r mi (x) +α2 (x))= x+x 2, x [ 0, 2) 1, = c 1 2 x ( ) [ x(1 x) 1 4, x 1 2,1 2] 1, (1 x)+(1 x) 2, x ( 1 2 1,1]. Sofor C 2 -fuctioswegetaglobaldegreeofapproximatiooforder O ( 1 ) which is also the case for the classical Berstei operators. Refereces [1] Agratii O., Rus I.A., Iterates of a class of discrete liear operators, Commet. Math. Uiv. Caroliae 44(2003), [2] Beresi I.S., Zhidkov N.P., Numerische Methode II, VEB Deutscher Verlag der Wisseschafte, Berli, [3] Kelisky R.P., Rivli T.J., Iterates of Berstei polyomials, Pacific J. Math. 21(1967), [4] KigP.J.,Positiveliear operators which preserve x 2,ActaMath.Hugar.99(2003), [5] Mamedov R.G., O the order of approximatio of fuctios by sequeces of liear positive operators(russia), Dokl. Akad. Nauk SSSR 128(1959),

8 652 H. Goska, P. Piţul [6] Păltăea R., Approximatio by liear positive operators: Estimates with secod order moduli, Ed. Uiv. Trasilvaia, Braşov, [7] Rus I.A., Iterates of Berstei operators, via cotractio priciple, J. Math. Aal. Appl. 292(2004), [8] Shisha O., Mod B., The degree of covergece of liear positive operators, Proc. Nat. Acad. Sci. U.S.A. 60(1968), Departmet of Mathematics, Uiversity of Duisburg-Esse, D Duisburg, Germay Colegiul Naţioal Samuel vo Brukethal, RO Sibiu, Romaia pitul (Received Jauary 28, 2005, revised July 4, 2005)