1 IntroductionLet K be the field of real or complex numbers and let ( X, [ parallel ] * [ parallel ] ) be a normed space over K. Let ext [ S.sub.x * ] denote the set of all extreme points of [ S.sub.x * ], where Sx * is the unit sphere in [ X.sup. * ] .

For every x [member of] X we put

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E ( x ) = { f [ thành viên of ] ext [ S.sub.x * ] : f ( * ) = [ parallel ] x [ parallel ] }. ( 1 )By the Hahn-Banach and the Krein-Milman Theorems, E ( x ) [ not equal to ] [ empty set ]. Let for Y [ subset ] X ,[ P.sub.Y ] ( x ) = { y [ thành viên of ] Y : [ parallel ] x – y [ parallel ] = dist ( x, y ) } .A linear subspace Y [ subset ] X is called a Chebyshev subspace if for every x [ thành viên of ] X the set [ P.sub.Y ] ( x ) contains one and only one element .Theorem 1 ( see [ 3 ] ) Assume X is a normed space, Y [ subset ] X is a linear subspace, and let x [ thành viên of ] X \ Y. Then [ y.sub. o ] [ thành viên of ] [ P.sub.Y ] ( x ) if and only if for every y [ thành viên of ] Y there exists f [ membe of ] E ( x – [ y.sub. 0 ] ) with Rf ( y ) [ less than or equal to ] 0 .Definition ( see e. g. [ 8 ] ) An element [ y.sub. o ] [ thành viên of ] Y is caZZed a strongly unique best approximation for x [ thành viên of ] X if there exists r > 0 suc / z that for every y [ thành viên of ] Y ,[ parallel ] x – y [ parallel ] [ greater than or equal to ] [ paralle ] x – [ y.sub. 0 ] [ parallel ] + r [ parallel ] y – [ y.sub. 0 ] [ parallel ] .The biggest constant r satisfying the above inequality is called a strong unicity constant. There exist two main applications of a strong unicity constant : the error estimate of the Remez algorithm ( see e. g. [ 13 ] ), the Lipschitz continuity of the best approximation mapping at Xq ( assuming that there exists a strongly unique best approximation to Xq ) ( see e. g. [ 5,9,11 ] ) .Theorem 2 ( see [ 17 ] ) Let x [ thành viên of ] X \ Y and let Y be a linear subspace of X. Then [ y.sub. 0 ] [ thành viên of ] Y is a strongly unique best approximation for x with a constant r > 0 if and only if for euery y [ thành viên of ] Y there exists f [ thành viên of ] E ( x – [ y.sub. o ] ) with Rf ( y ) [ less than or equal to ] – r [ parallel ] y [ parallel ] .Recall that a k-dimensional subspace V of a normed space X is called an interpolating subspace if for any linearly independent [ f.sub. 1 ], [ f.sub. 2 ], …, [ f.sub. k ] [ thành viên of ] ext [ S.sub.x * ] and for every v [ thành viên of ] V the following holds :if [ f.sub. i ] ( v ) = 0, i = 1,2, …, k then v = 0 .Every interpolating subspace is a finite dimensional Chebyshev subspace. If V [ subset ] X is an interpolating subspace then every x [ thành viên of ] X has a strongly unique best approximation in V ( see [ 2 ] ) .In this paper we consider X = K ( [ c.sub. 0 ], [ c.sub. 0 ] ) ( the space of all compact operators from [ c.sub. 0 ] to [ c.sub. 0 ] equipped with the operator norm ). Here Cq denotes the space of all real sequences convergent to zero. For any x = ( [ x.sub. k ] ) [ thành viên of ] [ c.sub. 0 ] we put[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ]In [ 8, Theorem 3.1 ] it has been proved that if V [ subset ] K { [ c.sub. 0 ], [ c.sub. 0 ] ) is a finite-dimensional Chebyshev subspace then every A [ thành viên of ] K, { [ c.sub. 0 ], [ c.sub. 0 ] ) has a strongly unique best approximation in V .However, in [ 8 ] no example of a non-interpolating Chebyshev subspace has been proposed. If it were true that any finite-dimensional Chebyshev subspace of K ( [ c.sub. 0 ], [ c.sub. 0 ] ) is an interpolating subspace we would have obtained the proof of Theorem 3.1, [ 8 ] immediately ( see [ 2 ] for more details ) .The aim of this paper is to show that for every k < [ infinity ] there exists a k-dimensional non-interpolating Chebyshev subspace of K ( [ c.sub. 0 ], [ c.sub. 0 ] ) .This result is quite different from the result obtained in [ 7 ]. In the space L ( [ l.sup.n.sub. 1 ], [ c.sub. 0 ] ) any finite-dimensional Chebyshev subspace is an interpolating subspace .Additionally, we discuss the strong unicity of best approximation in some ( not necessarily Chebyshev ) subspaces of K { [ c.sub. 0 ], [ c.sub. 0 ] ) .2 k-dimensional Chebyshev subspaces of / C ( c0rc0 )Let A [ thành viên of ] K ( [ c.sub. 0 ], [ c.sub. 0 ] ) be represented by a matrix [ [ a.sub.ij ]. sub. i, j [ thành viên of ] N ] Note that( [ a.sub.ij ] ). sup. [ infinity ]. sub. j = 1 ] [ thành viên of ] [ c.sub. 0 ] for every ; [ thành viên of ] N .Since each row of a matrix [ ( [ a.sub.ij ] ) ]. sub. i, j [ thành viên of ] N corresponds to a linear functional on [ c.sub. 0 ] ,( [ a.sub.ij ] ). sup. [ infinity ]. sub. j = 1 ] [ thành viên of ] [ l.sup. 1 ] for every i [ thành viên of ] N .Moreover, by the Schur Theorem ( see [ 6 ] )[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ]Recall ( see [ 4 ] ) that ext [ S.sub.k * ] consists of functionals of the form [ e.sub. i ] [ cross product ] x, where x [ thành viên of ] ext [ S.sub.l [ infinity ] and[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ] ( 2 )It is easy to see that[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ]Remark 1 Let X be a Banach space and let V be a finite-dimensional subspace with [ V.sub. 1 ], [ V.sub. 2 ], ..., [ V.sub.k ] as a basis .V is an interpolating subspace if and only if for any linearly independent [ f.sub. 1 ], [ f.sub. 2 ], ..., [ f.sub. k ] [ thành viên of ] ext [ S.sub.x * ] the determinant of [ [ f.sub. i ] ] ( [ V.sub.j ] ) ]. sub. i, j = 1,2, ..., k ] is not equal to zero .Proof. We apply the definition of a k - dimensional interpolating subspace and the theory of linear equations. This completes the proof .In the sequel, we denote by lin { [ V.sub. 1 ], [ V.sub. 2 ], ..., [ V.sub.k ] } the k-dimensional subspace of K { [ c.sub. 0 ], [ c.sub. 0 ] ) with [ V.sub. 1 ], [ V.sub. 2 ], ..., [ V.sub.k ] as a basis .Example 1 Let V = [ [ v.sub.ij ] ]. sub. i, j [ thành viên of ] N ], where [ v.sub. i1 ] = 1/2, [ v.sub.ij ] = 0, i, j [ thành viên of ] N, j [ greater than or equal to ] 2 .It is obvious that V = lin { V } is a one-dimensional interpolating subspace of k ( [ c.sub. 0 ], [ c.sub. 0 ] ) .Theorem 3 Let V = lin { [ V.sub. 1 ], [ V.sub. 2 ], ... [ V.sub.n ] }. Let [ V.sub.m ] = [ { [ v.sub. m ] ). sub. i, j [ thành viên of ] N, m = 1,2, ..., n. If V is a Chebyshev subspace then[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ] ( 3 )for any [ f.sub. 1 ], ..., [ f.sub. n ] [ thành viên of ] [ extS. sub. k * ( [ c.sub. 0 ], [ c.sub. 0 ] ) ] such that [ f.sub. m ] = [ e.sub.im ] [ cross product ] [ x.sup. i ] m, m = 1,2, ..., n, where [ i.sub. m ] [ not equal to ] [ i.sub. k ] for m [ not equal to ] kProof. Assume ( 3 ) does not hold. Therefore there exist [ f.sub. 1 ], ..., [ f.sub. n ] [ thành viên of ] [ extS. sub. k * ( [ c.sub. 0 ], [ c.sub. 0 ] ), [ f.sub. m ] = [ e.sub.i.sub. m ] [ cross product ] [ x.sup.im ], m = 1,2, ..., n, where [ i.sub. m ] [ not equal to ] for m [ not equal to ] k such that detD = 0, where[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ]Since detD = det [ D.sup.T ], there exists y = { [ y.sub. 1 ], [ y.sub. 2 ], ..., [ y.sub. n ] ) [ not equal to ] 0 such that [ D.sup.T ] y = 0. Consequently ,[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ]. ( 4 )Since y [ not equal to ] 0, replacing [ f.sub. m ] by - [ f.sub. m ] if necessary, we may assume [ y.sub. m ] [ greater than or equal to ] 0 for m = 1,2, ..., n and[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ]Set C = { I [ thành viên of ] { 1,2, ..., n } : [ y.sub. l ] > 0 } .Fix ( [ d.sub. j ] ) j [ thành viên of ] N with the following properties :[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII. ]Define A = [ [ a.sub.ip ] [ i.sub. p ], j [ thành viên of ] N [ thành viên of ] k ( [ c.sub. 0 ], [ c.sub. 0 ] ) by[ ai.sub. p ] j = 0 for p [ ? ? ] C, j [ thành viên of ] N, [ a.sub.ip ] j = [ d.sub. j ] * sgn [ x.sup.ip ] ( j ) for p [ thành viên of ] C, j [ thành viên of ] N .Note that [ parallel ] A [ parallel ] = 1 andE ( A ) = { [ f.sub. p ] : p [ thành viên of ] C } .By ( 4 ) and Theorem 1, 0 [ thành viên of ] [ P.sub.v ] ( A ) .Since detD = 0, there exists x = { [ x.sub. 1 ], [ X.sub. 2 ], …, [ x.sub. n ] ) [ not equal to ] 0 such that Dx = 0. Put[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Note that V [ not equal to ] 0 and [ f.sub. m ] ( V ) = 0, m = 1,2, …, n. By Theorem 2, 0 is not a strongly unique best approximation for A in V. By [ 8, Theorem 3.1 ], V is not a Chebyshev subspace and the proof is complete .Theorem 4 Let V = lin { V }, V [ thành viên of ] k ( [ c.sub. 0 ], [ c.sub. 0 ] ), V [ not equal to ] 0. V is a Chebyshev subspace if and only ifV is an interpolating subspace .Proof. The classical work here is [ 12 ]. In [ l.sup. 1 ], the one-dimensional subspace lin { v } is Chebyshev iff for every x [ thành viên of ] ext [ S.sub.l [ infinity ] ] the following holds[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Note that for any x G Co we obtain = V ( x ) = [ [ f.sub. 1 ] ( x ), [ f.sub. 2 ] ( x ), … ], where the functional [ f.sub. i ] correspond to elements of [ l.sup. 1 ] .It is obvious that if for any j, lin { [ f.sub. j ] } is not a Chebyshev subspace of [ l.sup. 1 ], then lin { V } is not a Chebyshev subspace of k ( [ c.sub. 0 ], [ c.sub. 0 ] ) .This proves the theorem .Note that by a result of Malbrock ( see [ 10 ], Theorem 3.3 ) each one-dimensional subspace V = lin { V } [ subset ] { [ c.sub. 0 ], [ c.sub. 0 ] ) is a Chebyshev subspace iff there exists [ delta ] > 0 such that[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]where | x ( j ) | = 1, j [ thành viên of ] N .Corollary Let V [ subset ] K { [ c.sub. 0 ], [ c.sub. 0 ] ) be a one-dimensional Chebyshev subspace. Every operator A G k ( [ c.sub. 0 ], [ C.sub. 0 ] ) has a strongly unique best approximation in V .Proof. Obvious. For more details we refer the reader to [ 2 ] .It is clear that ( 3 ) is satisfied for any n-dimensional interpolating subspace. However, ( 3 ) is not sufficient for an n-dimensional ( n [ greater than or equal to ] 2 ) subspace to be Chebyshev .Example 2 Let V = lin { [ V.sub. 1 ], [ V.sub. 2 ] }, where[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Note that V satisfies ( 3 ). We claim that V is a non-Chebyshev subspace. Indeed, define A = [ [ a.sub.ij ] ]. sub. i, j [ thành viên of ] N ] by[ a.sub. 12 ] = 100, [ a.sub.ij ] = 0 for each ( i, j ) ( 1,2 ), i, j [ thành viên of ] N .It follows that[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Hence[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Theorem 5 Let [ V.sub. 1 ] [ V.sub. 2 ], [ V.sub.n ] be given by[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]where [ V.sub.ij ] [ not equal to ] 0 for each i [ thành viên of ] N, j [ thành viên of ] { 1,2, …, n } and[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] for each j [ thành viên of ] { 1,2, …, n } .The following statements are equivalent 🙁 i ) For every choice of distinct [ j.sub. i ], …, [ j.sub. k ] from { 1,2, …, n ), V ( [ j.sub. 1 ], …, [ j.sub. k ] : = lin { [ V.sub.j 1, …, [ V.sub.jk ] } is a Chebyshev subspace of k ( [ c.sub. 0 ], [ c.sub. 0 ] ) ,( ii ) [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Proof. First, we assume that ( ii ) holds .If k = 1 then V ( [ j.sub. 1 ] ) is an interpolating subspace for every [ j.sub. 1 ] [ thành viên of ] { 1,2, …, n }. Let 1 < k < n and assume that for any [ j.sub. 1 ], ..., [ j.sub. k ] [ thành viên of ] { 1,2, ..., n ), [ j.sub. p ] [ j.sub. q, p ] [ not equal to ] q, [ V.sub.k ] : = V ( [ j.sub. 1 ], ..., [ j.sub. k ] ) is a Chebyshev subspace .Suppose that there exist 1 [ less than or equal to ] [ j.sub. 1 ] < [ j.sub. 2 ] <, ..., < [ j.sub. k ] < [ j.sub. k + 1 ] [ less than or equal to ] n such that[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]is a non-Chebyshev subspace. Without loss of generality we can assume that for any k + 1 [ thành viên of ] { 1,2, n ), jm = m, m = 1,2, k + 1. This means precisely that [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] where[ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]for i [ thành viên of ] N, [ thành viên of ] { 1,2, ..., k, k + 1 } .Since [ V.sub.k + 1 ] is a non-Chebyshev subspace, there exists A = [ [ a.sub.ij ]. sub. i, j [ thành viên of ] N ] [ thành viên of ] k [ c.sub. 0 ], [ c.sub. 0 ] ) such that # [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] ( A ) > 1. We can assume that 0, W [ thành viên of ] [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] ), where W [ not equal to ] 0 .Let U = { i : [ parallel ] g ; oA [ parallel ] = [ parallel ] A [ parallel ] }. Since A [ thành viên of ] K, { [ c.sub. 0 ], [ c.sub. 0 ] ), # U < [ infinity ] For every i [ thành viên of ] U we put[ E.sub.i ] = { x [ thành viên of ] [ extS. sub. l [ infinity ] : ( [ e.sub. i ] [ cross product ] x ) ( A ) = [ parallel ] A [ parallel ] } .Since 0, W [ thành viên of ] [ P.sub.V.. sub. k + 1 ] ( ^ ) / we conclude that for all i [ thành viên of ] U and x [ thành viên of ] [ E.sub.i ]( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]. ( 5 )Let( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] .Since 0 [ thành viên of ] [ P.sub.v.sub.k + 1 ] ( A ), [ not equal to ] 0We will prove that for any i [ thành viên of ] [ U.sub. 1 ] and x, y [ thành viên of ] [ E.sub.i ] such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] ( 6 )On the contrary, suppose that ( 6 ) does not hold. Let x, y [ thành viên of ] [ E.sub.i ] be such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]andx ( l ) [ not equal to ] ( l ) for some l [ thành viên of ] { 1,2, ..., k + 1 } .Without loss of generality we can assumex ( j ) = y ( j ) for / = 50%, ..., p, p < k + 1andx ( j ) = - y ( j ) for ; = p + 1, p + 2, ..., k + 1 .Hence( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]. ( 7 )Asx ( j ) = - y ( j ) for j = p + 1, p + 2, ..., k + 1we obtain[ a.sub.ij ] = 0 for j = p + 1, p + 2, ..., k + 1 .By ( 5 ) ,( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Therefore( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]By ( 7 ), x ( k + 1 ) [ w.sub. i ] k + 1 = 0. Consequently, [ W.sub.i ] [ k + 1 ] = 0. Hence W [ thành viên of ] [ V.sub.k ] .Since 0 [ thành viên of ] [ V.sub.k ] and [ V.sub.k ] is a Chebyshev subspace, ( 6 ) is proved .We will show that there exists [ [ alpha ]. sub. 0 ] > 0 such that for every [ alpha ] [ thành viên of ] ( 0, [ [ alpha ]. sub. 0 ] ] ,( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]. ( 8 )We first prove that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] ( 9 )where A = [ [ a.sub.ij ] ]. sub. i, j [ thành viên of ] N ] .Let i [ thành viên of ] U, f = [ e.sub. i ] [ cross product ] x, f ( W ) < 0. Hence there exists ; ' o [ thành viên of ] { 1,2, ..., n } satisfying( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] .

Now, we will show

( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]and ( 9 ) is proved .We conclude from ( 9 ) that there exist [ [ alpha ]. sub. 0 ] > 0, b > 0 such that for every a [ thành viên of ] ( 0, [ [ alpha ]. sub. 0 ] ,( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]where f [ thành viên of ] [ extS. sub. k * ( [ c.sub. 0 ] / [ c.sub. 0 ] ) ], f ( W ) < 0 .Assume [ [ alpha ]. sub. 0 ] is so small that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Consequently, if f [ thành viên of ] E ( A - [ [ alpha ]. sub. 0 ] W ) then f = [ e.sub. i ] [ cross product ] x, where z [ thành viên of ] [ U.sub. 1 ] and f ( W ) = 0. Since[ parallel ] A - [ [ alpha ]. sub. 0 ] W [ parallel ] = [ parallel ] A [ parallel ] = dist ( A, [ V.sub.k + 1 ] ) ,( 8 ) is proved .Since [ [ alpha ]. sub. 0 ] W [ thành viên of ] [ P.sub.v.sub.k + 1 ] ( A ), we conclude ( see [ 16 ] ) that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ], ( 10 )where ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Let q be the smallest number having property ( 10 ). By ( 6 ), [ i.sub. j ] [ not equal to ] [ i.sub. l ] for ; j [ not equal to ] l, j, l [ thành viên of ] { 1,2, ..., q }. If q = k + 2 then ( see [ 18 ] ) [ [ alpha ]. sub. 0 ] W is the strongly unique best approximation for A in [ V.sub.k + 1 ], a contradiction .Suppose that 1 [ less than or equal to ] q [ less than or equal to ] k + 1. This contradicts ( ii ) .Let us assume that [ V.sub.k ] is a Chebyshev subspace of k ( [ c.sub. 0 ], [ c.sub. 0 ] ) for every 1 [ less than or equal to ] k [ less than or equal to ] n. Suppose that ( ii ) is false. Consequently, there exist( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]satisfying( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]It follows that there exist( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ], ( 11 )where [ x.sup.im ] = ( [ x.sup.im ] ( l ), [ x.sup.im ] ( 2 ), .... ), [ x.sup.im ] ( l ) = [ x.sub.ml ] .Without loss of generality we can assume( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]We define an operator B = [ [ b.sub.ij ]. sub. i, j [ memberof ] N by( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Hence ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] anddim span { [ e.sub.im ] [ cross product [ x.sup.im ] | [ V.sub.k ] } < k ,where dim [ V.sub.k ] = k. Therefore there exists V [ thành viên of ] [ V.sub.k ] \ { 0 } such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] .Note that ( see the proof of the formula ( 9 ) )( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Hence there exist olq > 0, b > 0 such thatf ( B – [ [ alpha ]. sub. 0 ] V ) – [ less than or equal to ] b, < [ parallel ] B [ parallel ], f [ thành viên of ] [ extS. sub. K * ], f ( V ) [ less than or equal to ] 0 .Consequently, [ parallel ] B - [ alpha ]. sub. 0 ] V [ parallel ] = [ parallel ] B [ parallel ], a contradiction .The proof is complete .Example 3 We will construct an n - dimensional Chebyshev subspace V [ subset ] K { [ c.sub. 0 ], [ c.sub. 0 ] ). Let 0 < [ t.sub. 1 ] < [ t.sub. 2 ] < ... < [ t.sub. n-1 ] be such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Define [ V.sub.m ] = [ ( [ V.sub.m ] ). sub.ij ] i, j [ thành viên of ] N ] by( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] .Hence [ V.sub.m ] [ thành viên of ] K ( [ c.sub. 0 ], [ c.sub. 0 ] ) for every m = 1, ..., 2, n - 1 .Let [ V.sub.n - 1 ] : = lin { [ V.sub. 1 ], [ V.sub. 2 ], [ V.sub.n - 1 ] ) satisfy the formula ( z'z ) for every 1 [ less than or equal to ] k [ less than or equal to ] n - 1. We will construct an operator [ V.sub.n ] [ thành viên of ] k ( [ c.sub. 0 ], [ c.sub. 0 ] ) such that [ V.sub.n ] : = lin { [ V.sub. 1 ], [ V.sub. 2 ], [ V.sub.n - 1 ], [ V.sub.n ] } satisfies the formula ( ii ) for every 1 [ less than or equal to ] k [ less than or equal to ] n. Our goal is to find x [ thành viên of ] R such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] ( 12 )and( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ], ( 13 )where k [ thành viên of ] { 1,2, ..., n ), [ i.sub. 1 ], [ i.sub. 2 ], ..., [ i.sub. k ] [ thành viên of ] N, [ y.sup. 1 ], ..., [ y.sup. k ] [ thành viên of ] { - 1,1 }. sup. k ] [ m.sub. 1 ], [ m.sub. 2 ], ..., [ m.sub. k-1 ] [ thành viên of ] { 1,2, ..., n - 1 }. Since W ( x, [ y.sub. 1 ], [ y.sub. k ], ..., [ i.sub. 1 ], ...., [ i.sub. k ] [ m.sub. 1 ], ... [ m.sub. k-1 ] ) is not totally equal to zero, we conclude that the set of roots of W ( x, [ y.sup. 1 ], ..., [ y.sub. k ], [ i.sub. 1 ], ..., [ i.sub. k ], [ m.sub. 1 ], ... [ m.sub. k-1 ] ) is finite for arbitrary but fixed ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] Therefore for all ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] as above, the set of roots of ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] is countable. Since IR is not countable we see that there exists x [ thành viên of ] IR satisfying ( 12 ) and ( 13 ) .Remark 2 An n-dimensional Chebyshev subspace proposed in Example 3 is a noninterpolating subspace of / C ( [ c.sub. 0 ], [ c.sub. 0 ] ) .Proof. Let us assume that [ V.sub.n ] = lin { [ V.sub. 1 ], [ V.sub. 2 ], [ V.sub.n ] } is an n-dimensional Chebyshev subspace, where [ V.sub.m ], m = 1,2, n are defined in Example 3 .Put ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]. Note that ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ], whereV = [ [ v.sub.ij ], j [ thành viên of ] N ]It is obvious that there exist [ x.sup. 1 ], [ x.sup. 2 ], ..., [ x.sup. n ] [ thành viên of ] [ extS. sub. l [ infinity ] ] such that [ x.sup. m ] ( 1 ) = [ x.sup. m ] ( 2 ) = 1, m = 1,2, ..., n and fm : = [ e.sub. 1 ] [ cross product ] [ x.sup. m ], m = 1,2, ..., n are linearly independent. Note that[ f.sub. m ] ( V ) = 0, m = l, 2, ..., n .This completes the proof .Lemma Let X be a normed space and let V be a finite-dimensional subspace of X. Let T [ thành viên of ] X. If 0 [ thành viên of ] [ P.sub.v ] ( T ) and 0 is not a strongly unique best approximation for T in V then( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Proof. Let us assume that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Set for any V [ thành viên of ] V, [ parallel ] V [ parallel ] = 1 ,( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]We show that r > 0 .If not, there exists ( [ V.sub.n ] ) [ subset ] [ S.sub.v ] such that – [ r.sub.vn ] [ greater than or equal to ]. – 1 / n. Since V is a finite-dimensional subspace, we may assume that [ V.sub.n ] [ left arrow ] V [ thành viên of ] [ S.sub.v ]. Take f [ thành viên of ] E ( T ), f ( V ) < 0. Hence for n [ greater than or equal to ] [ n.sub. 0 ] there exists d > 0 such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]a contradiction. Therefore( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]By the above ,( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Hence 0 is a strongly unique best approximation for T, a contradiction. This proves the lemma .Theorem 6 Let V [ subset ] k ( [ c.sub. 0 ], [ c.sub. 0 ] ) beann – dimensional subspace such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]where V = [ v.sub.ij ] ] i, j [ thành viên of ] N and let T [ thành viên of ] k ( [ c.sub. 0 ], [ c.sub. 0 ] ) .T has a unique best approximation in V if and only if T has a strongly unique best approximation in V .Proof. Let us assume that 0 is the unique best approximation for T in V. Suppose that 0 is not a strongly unique best approximation. Hence ( see Lemma )( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]where f = [ e.sub. i ] [ cross product ] [ x.sup. i ] for some [ x.sup. i ] [ thành viên of ] [ extS. sub. l [ infinity ]. Put( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] .Since T is compact, we conclude that # N < [ infinity ]. For every z [ thành viên of ] A / " we set( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Let i [ thành viên of ] N \ N. Hence there exists b > 0 such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] .Consequently, there exists [ [ alpha ]. sub. 0 ] > 0 such that for every [ alpha ] [ thành viên of ] ( 0, [ [ alpha ]. sub. 0 ] ] ,( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Therefore( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Let i [ thành viên of ] N and let [ x.sup. i ] [ ? ? ] [ E.sub.i ]. From this we conclude that there exists [ j.sub. 0 ] [ thành viên of ] N such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ] ,where T = [ [ t.sub.ij ] ] i, j, [ thành viên of ] N ]Set J = { j [ thành viên of ] N : [ v.sub.ij [ not equal to ] 0 } .If sgn [ x.sup. i ] ( ; ) = sgn ( [ t.sub.ij ] ) for any j [ thành viên of ] J, then there exists [ y.sub. i ] [ thành viên of ] [ E.sub.i ] such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]By the above ,( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Let sgn [ x.sup. i ] ( / o ) 7 ^ sgn ( fy0 ) for some ; ‘ o [ thành viên of ] /, where tjjQ 7 ^ 0. Since / is finite, there exists [ [ alpha ]. sub. 0 ] > 0 such that( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Let a [ thành viên of ] ( 0, [ [ alpha ]. sub. 0 ] ] – Hence( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]Finally ,[ parallel ] T – [ alpha ] V [ parallel ] = f ( T – [ alpha ] V ) ,where f = ( [ MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII ]. Hence[ parallel ] T – [ alpha ] V [ parallel ] = f ( T – [ alpha ] V ) [ less than or equal to ] [ parallel ] T [ parallel ] .The proof is complete .Acknowledgments The author wishes to thank Professor Lewicki for his remarks and suggestions concerning this article .Received by the editors July 2009 .References[ I ] A. ALEKSIEWICZ, Analiza funkcjonalna, PWN, Warszawa, 1969[ 2 ] D. A. AULT, F. R. DEUTSCH, P. D. MORRIS, J. E. OLSON, Interpolating Subspaces in Approximation Theory, J. Approx. Theory 3 ( 1970 ), 164 – 182[ 3 ] B. BROSOWSKI, R. 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WOJCIK, Characterization of strong unicity by tangent cones, Approximation and Function Spaces, Proceedings of the International Conference, Gdansk, ( 1979 ) ( Z. Ciesielski, Ed ), 854 – 866, PWN, Warszawa / North-Holland, Amsterdam, 1981[ 18 ] A. WOJCIK, J. SUDOLSKI, Some remarks on strong uniqueness of best approximation, J. Approx. Theory and its Appl. 6, No. 2, ( 1990 ), 44-78Research supported by local grant No. 10.420.03Communicated by R Bastin .

Joanna Kowynia

Department of Mathematics, AGH University of Science and Technology ,Al. Mickiewicza 30, 30-059 Cracow, Polande-mail : kowynia@wms.mat.agh.edu.pl

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