On Copositive Approximation in Spaces ofContinuous Functions II*:The Uniqueness of Best Copositive Approximation

Aref K.Kamal

Department of Mathematics and Statistics,S.Q.University,P.O.Box 36 Al Khoudh 123 Muscat,Sultanate of Oman

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On Copositive Approximation in Spaces of
Continuous Functions II*:The Uniqueness of Best Copositive Approximation

Aref K.Kamal?

Department of Mathematics and Statistics,S.Q.University,P.O.Box 36 Al Khoudh 123 Muscat,Sultanate of Oman

Abstract.This paper is part II of”O(jiān)n Copositive Approximation in Spaces of Continuous Functions”.In this paper,the author shows that if Q is any compact subset of real numbers,and M is any finite dimensional strict Chebyshev subspace of C(Q),then for any admissible function f∈C(Q)M,the best copositive approximation to f from M is unique.

Strict Chebyshev spaces,best copositive approximation,change of sign.

AMS Subject Classifications:41A65

Analysis in Theory and Applications

Anal.Theory Appl.,Vol.32,No.1(2016),pp.20-26

1　Introduction

If Q is a compact Hausdorff space,then C(Q)denotes the Banach space of all continuous real valued functions on Q,together with the uniform norm,that is,‖f‖=max{|f(x)|: x∈Q}.If M is a subspace of C(Q),and f∈C(Q),then g∈M is said to be copositive with f on Q iff f(x)g(x)≥0 for all x∈Q.The element g0∈M is called a best copositive approximation to f from M iff g0is copositive with f on Q and‖f?g0‖=inf{‖f?g‖:g∈M,and g is copositive withf on Q}.The set{g∈M:g is copositive with f on Q}is closed, so if the dimension of M is finite,then the best copositive approximation to each f∈C(Q) from M is attained.If Q is a compact subset of real numbers,then the n-dimensional subspace M of C(Q)is called Chebyshev subspace of C(Q)if each g6=0 in M has at most n?1 zeros.The n-dimensional Chebyshev subspace M of C(Q)is called a”Strict Chebyshev subspace”of C(Q)if each g6=0 in M has at most n?1 changes of signs,that is,no g6=0 in M alternates strongly at n+1 points of Q,which means that there do not exist n+1 points,x0＜x2＜···＜xn+1in Q so that g(xi)g(xi+1)＜0 for all i=1,2,···,n.

This paper is a continuation of the author’s paper[1].In this paper the author investigates the uniqueness of the best copositive approximation by elements of finite dimensional subspaces of C(Q).Passow and Taylor[2]showed that when Q is any finite subset of real numbers,and M is a finite dimensional strict Chebyshev subspace of C(Q) then the best copositive approximation to each f∈C(Q)from M is unique.Zhong[3] proved the same result for the case when Q is a closed and bounded interval[a,b]of the real numbers,and f does not vanish on any subinterval of[a,b].In this paper it is shown that this fact is true for any compact subset of real numbers.

The rest of this section will be used to cover some notation and results that will be used later in Section 2.As in Kamal[1],If Q is a compact subset of real numbers,and x1＜x2in Q thenthe”intervals”(x1,x2),(x1,x2],[x1,x2),and[x1,x2]in Q are defined in the ordinary way,for example;(x1,x2)={x∈Q:x1＜x＜x2}.If Q is not connectedthennone of those intervals need to be connected.The point z0in Q is called”a limit point from both sides”in Q if z0is an accumulation for the set{x∈Q:x＜z0},and the set{x∈Q:x＞z0}. If z0is an accumulation point for the set{x∈Q:x＜z0}or the set{x∈Q:x＞z0}but not for both then z0is called”a limit point from one side”in Q.The function f∈C(Q)is said to have at”least k changes of sign in Q”if there are k+1 point t1＜t2＜···,tk+1in Q so that f(ti)f(ti+1)＜0 for all i=1,2,···,k.The”number of changes of sign of f”is defined to be the sup{k:f has at least k changes of sign}.Assume that f6=0 in C(Q),the point z∈Q is said to be a”double zero”for f in Q if f(z)=0,and there are x＜z＜y in Q so that f(α)f(β)＞0 for all α6=z,and β6=z in[x,y].If f(z)=0,and z is not a double zero then z is called a”single zero”in Q(see[4]).Finally the function f∈C(Q)is called admissible if f does not vanish on any infinite interval of Q.

The following Proposition presents some of the known propertis of strict Chebyshev subspaces.

Proposition 1.1.Assume that Q is a compact subset of real numbers containing at least n+1 points,and that M is an n-dimensional strict Chebyshev subspace of C(Q).The following facts hold;

i).If z1＜z2＜···＜zn?1are n?1 points in Q,then there is g∈M,such that g(x)=0 for all x∈{z1,z2,···,zn?1}and;

1).g(x)＞0,if x＜z1,

2).(?1)n?1g(x)＞0 if x＞zn?1,and;

3).(?1)ig(x)＞0 if x∈(zi,zi+1),and i=1,2,···,n?1.

ii).No g6=0 in M alternates weakly at n+1 points in Q,that is,there do not exist x1＜x2＜···＜xn+1in Q,and g6=0 in M such that(?1)ig(xi)≥0 for each i=1,2,···,n+1.

iii).If g6=0 in M and k is the number of single zeros of g,and m is the number of double zeros of g then k+2m≤n?1.

Part i)in Proposition 1.1 can be obtained from Lemma 6.5 in Zielke[4],part ii),is in[4,Lemma 3.1b],part iii)is[4,Lemma 6.2].

Lemma 1.1(see[1]).Assume that Q is an infinite compact subset of real numbers,M is an n-dimensional strict Chebyshev subspace of C(Q),and q is a limit point from both sides in Q.If g and h are two elements in M and h6=0,then limand limboth exist as extended real numbers.

Lemma 1.2.Assume that Q is an infinite compact subset of real numbers,and that M is an n-dimensional strict Chebyshev subspace of C(Q).Let q be a limit point from both sides in Q and let g and h be two nonzero elements in M,such that g(q)=h(q)=0.If the number of zeros of g is n?1,then

Proof.It will be shown that limWith the same method one can prove that limBy Lemma 1.1,limexists as an extended real number.Assume that lim=0,and let x1＜x2＜···＜xn?be the zeros of g.For each k=1,2,···,n?2,1let Ik=(xk,xk+1).Let I0={x∈Q:x＜x1},and In?1={x∈Q:x＞xn?1}.By Proposition 1.1, all the zeros of g are single zeros.Thus one can assume without loss of generality that (?1)kg(x)＞0 for all x∈Ik,and k=0,1,2,···,n?1.The proof will be given first for the case at which Ik6=φ for all k.In this case for each k,choose tk∈Ik.Then g alternates strongly at the n points t0＜t1＜···＜tn?1in Q.Since q=xi0for some i0,then ti0?1＜xi0＜ti0. It is clear that g does not changes sign in neithernor in[xi0,ti0].Let c＞0 be chosen so that c‖h‖＜min{|g(t0)|,|g(t1)|,···,|g(tn?1)|}.Since lim=0,then there is y06=xi0in,so that|g(y0)|＜c|h(y0)|.If g＞0,thenlet ψ=g?ch,and if g＜0,then let ψ=g+ch.In bothcases ψ6=0,and ψ＜0.Therefore, ψ alternates weakly at the n+2 points of the setwhich contradicts Proposition 1.1.

Second,assume that some of the intervals I0,I1,···,In?1are empty.The proof will be given by strong induction.Assume that the number of empty intervals among I0,I1,···,In?1is k.Then 0≤k＜n.The hypothesis is true for k=0.Now let k≥0,and assume that the hypothesis is true for all 0≤i≤k.It will be shown that it is true for k+1. Assume that the number of empty intervals is k+1,and let Ij=(xj,xj+1)be one of those empty intervals.Since xi0is a limit point from both sides in Q then Ij6=Ii0?1and Ij6=Ii0. Q is infinite,so one can find a natural number α∈{0,1,2,···,n?1},so that Iαis infinite. Let s be any point in Iαsuch that{x∈Iα:x＜s}6=φ and{x∈Iα:x＞s}6=φ,and let g0be a non zero element in M having n?1 zeros at[{x1,x2,···,xn?1}{xj+1}]∪{s}.The zeros of g0includes q=xi0,and if J0,J1,···,Jn?1are the intervals between its zeros then the number of empty intervals among them is no more than k.By induction lim6=0.But lim=0.So lim=0 .Let t1be any element inand t2be any element in Ii0,thenChoose c＞0 be so that c‖g0‖＜min{|g(t1)|,|g(t2)|}. Since lim=0,then there issuch that|g(y0)|＜c|g0(y0)|.Ifg(t1)g0(y0)＞0,then let ψ=g?cg0,and if g(t1)g0(y0)＜0,then let ψ=g+cg0.In both cases ψ6=0,and ψ(t1)ψ(y0)＜0,and since g(t1)g(t2)＜0,it follows that ψ(t2)ψ(y0)＞0. Therefore,ψ alternates weakly at the points t1＜y0＜xi0＜t2.But g(xk)=g0(xk)=0 for all k6=j+1.So ψ(xk)=0 for all k6=j+1.Thus ψ alternates weakly at the n+1 points of the set [{x1,x2,···,xn?1}{xj+1}]∪{t1,t2,y0},which contradicts Proposition 1.1.

2　The main results

This section is devoted to show that the best copositive approximation is unique.Let n be any natural number,Q be any compact subset of the real numbers containing more than n+1 points,and let M be any n-dimensional strict Chebyshev subspace of C(Q).

Let f be any element in C(Q).If f has more than n?1 changes of sign then there are n+1 points t1＜t2＜···＜tn+1in Q so that f(ti)f(ti+1)＜0 for all i=1,2,···,n.If g is any best copositive approximation to f from M then g(ti)g(ti+1)≤0 for all i=1,2,···,n. Therefore by Proposition 1.1.g must be zero.Hence g=0 is the unique best copositive approximation to f from M.So in this section the function f will have no more than n?1 changes of sign.

As in Kamal[1],if Q is a compact subset of real numbers containing at least n+1 points,and f is an admissible function in C(Q)having no more than n?1 changes of sign.Define X0(f)={z1,z2,···,zm}to be the set of all z∈Q such that z is a limit point from both sides in Q,and that f changes sign at z.If M is an n-dimensional strict Chebyshev subspace of C(Q),then for each g6=0 in M,copositive with f,define;

X1(f,g)={x∈Q:|f(x)?g(x)|=‖f?g‖}∪{x∈Q:f(x)6=0 and g(x)=0},

X2(f,g)={x∈Q:g(x)=0,f(x)=0,and x is not an isolated point inQ}.

Let X(f,g)=X1(f,g)∪X2(f,g),and define M0to be{g∈M:g(z)=0 for all z∈X0(f)}. It is clear that M0is an(n?m)-dimensional subspace of M,and that if g∈M is copositive with f on Q,then g∈M0.

The function θ6=0 in M is said to be”copositive with f around the elements of X0(f)”if for each z∈X0(f),there is a neighborhood Uzaround z such that f(x)θ(x)≥0 for all x∈Uz.It is clear that θ(z)=0 for all z∈X0(f).For such function,define X3(f,g,θ)to be

and X4(f,g,θ)to be

Lemma 2.1(see[1]).Assume that f is admissible function in C(Q)M having no more than n?1 changes of sign.If g is a best copositive approximation to f from M then there is a non-zero function ?∈M0copositive with f around the elements of X0(f),such that the numberof elements in[X(f,g)X0(f)]∪X3(f,g,?)∪X4(f,g,?)is more than or equal to n?m+1.

Lemma 2.2.Assume that f is admissible function in C(Q)M having no more than n?1 changes of sign,g is a best copositive approximation to f from M,and let ? be any element in M0copositive with f around the elements of X0(f).For any h0∈M0,if there are ξ1,ξ2,···,ξηin X(f,g)X0(f),and y1,y2,···,yrin X3(f,g,?)∪X4(f,g,?),such that η+r=n?m+1,h0(ξi)=0 for all i=1,2,···,η,and for all j=1,2,···,r,either

then h0=0.

Proof.By contradiction,assume that there is h0∈M0with the given properties,and that h06=0.Since h0has zeros at the points of the two distinct setsand {z1,z2,···,zm},then η+m≤n?1.If Q is finite then r=0,so η=n?m+1.Thus η+m=n+1. But then h0has more than n?1 zeros,which contradict the fact that M is a strict ndimensional Chebyshev space.So one may assume that Q is infinite,and that r＞0. By Proposition 1.1,let h1be any nonzero element in M having n?1 zeros,including,and choose the location of the extra zeros so that h1,and h0have the same sign in some neighborhood around yjfor all j=1,2,···r.This can be done by replacing each double zero of h0by two very close single zeros for h1.By Proposition 1.1,the number of zeros of h1may still less than n?1.To make this number equal n?1, one can add extra zeros after zmor before z1.For each j=1,2,···,r,choose ejin Q so that if limthen ej＜yj,and if limthen ej＞yj,and with the properties that,if Ijis the open interval between ejand yjin Q,then Ijdoes not intersectneither h0,nor h1change sign or have zeros in Ij,and h0(ej)6=0.Let λ0＞0, so that

and let h2=h0?λ0h1.It is clear that h26=0,and that h2(x)=0 for all x∈∪{z1,z2,···,zm},and that h2(ej)h0(ej)＞0 for all j.Foreach 1≤j≤r,eitherlim=0, orlim0.Assumefirstthat lim=0.Since h1has n?1 zerosand Q is infinite and yjis a limit point from both sides in Q,then by Lemma 1.2 lim= 0.So lim=0.Since h1and h0have the same sign at ej,and h2(ej)h0(ej)＞0, then h2(ej)h1(ej)＞0.On the other hand lim=?λ0.Thus there is a point ujin Q such that ej＜uj＜yjand that h2(uj)h1(uj)＜0.Since h1has a constant sign in [ej,yj)and h2(ej)h1(ej)＞0 then h2(ej)h2(uj)＜0.In the same manner,if lim= 0,then one can find point ujin Q such that yj＜uj＜ejand that h2(uj)h2(ej)＜0.Let {s1,s2,···,sη+m}={ξ1,ξ2,···,ξη}∪{z1,z2,···,zm},then h2(si)=0 for all i=1,2,···,η+m, and if si=yjfor some j,then the two points uj,ejlie between siand si?1or si,and si+1. Furthermore h2(uj)h2(ej)＜0.Thus one can choose tj∈{uj,ej}so that h2alternates weaklyat the points of{s1,s2,···,sη+m}∪{t1,t2,···,tr}.But η+m+r=(n?m+1)+m=n+1.So h2alternates weakly at n+1 points of Q.This is a contradiction.

Theorem 2.1.Assume that Q is a compact subset of real numbers having at least n+1 points, and that M is an n-dimensional strict Chebyshev subspace of C(Q).If f is an admissible function in C(Q)M,then the best copositive approximation to f from M is unique.

Proof.If f has more than n?1 changes of sign then as the argument at the start of this section,thebestcopositiveapproximation to f from M is unique.So onemay assumethat f have no more than n?1 changes of sign.By contradiction,assume that g1and g2are two distinct best copositive approximations to f from M.Let g?g0=g1?g2,then g06=0 and g?is another best copositive approximation to f from M.By Lemma 2.1,there is a non-zero function ?∈M0copositive with f around the elements of X0(f)such that the number of the elements in[X(f,g?)X0(f)]∪X3(f,g?,?)∪X4(f,g?,?)is more than or equal to n?m+1.Thus letbe elements in X(f,g?)X0(f),and let y1,y2,···,yrbe elements in X3(f,g?,?)∪X4(f,g?,?)such that η+r=n?m+1.

It will be shown that g0(ξi)=0 for all i=1,2,···,η,and for all j=1,2,···,r,either lim=0 or lim=0.If this is true,then by Lemma 2.2,g0=0,which is a contradiction.

For each i=1,2,···,η,ξi∈X(f,g?)X0(f)=[X1(f,g?)∪X2(f,g?)]X0(f).So either ξi∈X1(f,g?),or ξi∈X2(f,g?)X0(f).If ξi∈X1(f,g?),and|(f?g?)(ξi)|=‖f?g?‖,then since

it follows that

Therefore,

If ξi∈X1(f,g?),and f(ξi)6=0,but g?(ξi)=0,then since g?,g1,and g2are copositive with f on Q,andit follows that=0.ThereforeIfthenand ξiis a limit point either from both sides or from one side in Q.Since g?,g1and g2are all continuous on Q,and copositive with the admissible function f,then g1(ξi)=g2(ξi)=g?(ξi)=0.Thus g0(ξi)=0.

Finally,it will be shown that for each j=1,2,···,r,either limor limSince yj∈X3(f,g?,?)∪X4(f,g?,?),then either limor limAssume first that limSince g?,g1,g2are all continuous on Q,and copositive with the admissible function f,then limand lim

In the same method one can show that if limthen lim

Acknowledgments

The author want to thank Dr.AL.Brown for his patience in reading the manuscript and all his suggestions and corrections which made the paper readable.

References

[1]A.K.Kamal,On copositive approximation in spaces of continuous functions I,the alternation property of copositive approximation,Anal.Theory Appl.,31(2015),354–372.

[2]E.Passow and G.D,Taylor,An alternation theory for copositive approximation,J.Approx. Theory,19(1977),123–134.

[3]J.Zhong,Best copositive approximation,J.Approx.Theory,72(1993),210–233.

[4]R.Zielke,Discontinuous Chebyshev system,Lecture Notes in Mathematics,Vol.707, Spriger-Verlag,Berlin-Heidelberg-New York,1979.

10.4208/ata.2016.v32.n1.2

2 April 2015;Accepted(in revised version)25 May 2015

?Corresponding author.Email address:akamal@squ.edu.om(A.K.Kamal)