The Linearity of Metric Projection Operator for Subspaces of L p Spaces

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1 ISSN , Moscow University Mathematics Bulletin, 2010, Vol. 65, No. 1, pp c Allerton Press, Inc., Original Russian Text c Yu.Yu.Druzhinin, 2010, published in Vestni Mosovsogo Universiteta, Matematia. Mehania, 2010, Vol. 65, No. 1, pp The Linearity of Metric Projection Operator for Subspaces of L p Spaces Yu. Yu. Druzhinin Moscow State University, Faculty of Mechanics and Mathematics, Leninsie Gory, Moscow, Russia Received October 20, 2008 Abstract Let Y be a Chebyshev subspace of a Banach space X. Then the single-valued metric projection operator P Y : X Y taing each x X to the nearest element y Y is well defined. Let M be an arbitrary set and μ be a σ-finite measure on some σ-algebra Σ of subsets of M. Wegivea description of Chebyshev subspaces Y L p(m,σ,μ) with a finite dimension and a finite codimension which the operator P Y is linear for. DOI: /S INTRODUCTION Let X be a real Banach space, Y be its (closed linear) subspace, ρ(x, Y ):=inf{ x y : y Y } be the distance from some element x X to Y,andP Y (x) :={y Y : x y = ρ(x, Y )} be the set of elements from Y being nearest to x, i.e., the metric projection of the element x onto Y. A subspace Y is said to be Chebyshev if for any x X the set P Y (x) consists of exactly one element. Then the appearing metric projection operator P Y : X Y mapping each element x X to its best approximation element P Y (x) Y is, generally speaing, not linear. Reflexive strictly convex spaces are just those ones where each subspace is Chebyshev. In particular, the spaces L p are such for 1 <p<. The aim of this paper is to describe subspaces Y of finite dimension (Theorem 1) and finite codimension (Theorem 2) in the spaces L p (1 <p<, p 2) which the operator P Y is linear for. In addition, we present examples of subspaces having an infinite dimension and a codimension with linear and nonlinear metric projection operators. Note that Chebyshev subspaces of the space L 1 having a linear metric projection operator were described in papers of Morris [1] (for the classic Lebesgue measure on the segment [0, 1]) and of Borodin [2] (in the general case). The space L 2 is a Hilbert one. The operator P Y in this case is linear for any subspace Y because it coincides with the operator of orthogonal projection onto this subspace. For the case of a finite measure, Theorems 1 and 2 were proved in [3]. The proof is essentially based on Ando s theorem concerning the general form of a contracting projector in the space L p, see [4]. In this paper we consider the case of σ-finite measures and the proofs are based on elementary facts of the real analysis only. In what follows L p (M) always denotes the space L p (M,Σ,μ), where(m,σ,μ) is a measurable space with a σ-finite measure μ, and1 <p<, p QUASI-ORTHOGONAL SETS The main instrument of our study is the concept of so-called quasi-orthogonal sets. Definition. A quasi-orthogonal set Q(Y ) of a subspace Y of a space X is the family of all elements n X such that P Y (n) 0. A quasi-orthogonal set of a subspace is always closed and it is also a two-sided cone (i.e., if n Q(Y ), then λn Q(Y ) for all λ R). Lemma A [5]. AsubspaceY is Chebyshev if and only if the following two conditions are valid: (1) Q(Y )+Y = X and (2) P Y (q) =0for any q Q(Y ). Condition (1) is equivalent to the existence of the best approximation element and Condition (2) is equivalent to its uniqueness. Lemma B [6]. The metric projection operator P Y for a Chebyshev subspace Y is linear if and only if Q(Y ) is a linear subspace. 28

2 THE LINEARITY OF METRIC PROJECTION OPERATOR 29 Definition. The subspace Y = {f X : f(y) =0 y Y } is called the annihilator of a subspace Y. Lemma C [2]. The following equality is valid: Q(Y )={q X : there exists a nonzero functional f Y such that f(q) = f q }. This statement can be interpret as follows: in order to obtain the set Q(Y ), we have to tae all functionals from Y attaining their norm, and then all elements from X which this norm is attained at. The set of all these elements is Q(Y ). In the case of the space L p with 1 <p<, the adjoint space is the space L q (here 1 q + 1 p =1). Each nonzero functional g L q attains its norm at the element sgn g g q PRELIMINARIES Lemma 1. Let α>0, α 1, and let for nonzero numbers a and b the following equality hold: sgn (a + b) a + b α =sgna a α +sgnb b α Then a = b. Proof. Suppose the contrary. To be definite, assume a b. Dividing the given equality by sgn a a α, we get ( 1+ b ) α =1+sgn b a a b α a. It is not difficult to verify that for α>1 only t = 1 is the solution to the equation (1 + t) α (1 + sgn t t α )=0considered on the set [ 1, 0) (0, 1]. Thena = b, and the lemma is proved. Recall that a function f from the space L p (M) (1 p< ) is a class of functions pairwise equal to each other almost everywhere on the set M. Lemma 2. Let functions f 1,...,f n L p (M) be linearly independent and f 1,..., f n be some representatives of the classes f 1,...,f n, respectively. Then there exist points x 1,...,x n X such that Φ( f 1 ;...; f f 1 (x 1 )... f 1 (x n ) n ):= f n (x 1 )... f n (x n ) This lemma can be easily proved by induction over n. Let f 1,...,f n L p (M) be given functions and f 1,..., f n be their fixed representatives. Define the set { ( Ω=Ω( f 1 ;...; f n n ):= x M : ε >0 μ ) } f 1 i ( f i (x) ε, f i (x)+ε) > 0. Lemma 3. Let functions f 1,..., f n be some fixed representatives of function f 1,...,f n L p (M). Then μ(m \ Ω) = 0. Proof. We have: { ( n M \ Ω= x M : ε = ε(x) > 0 : μ ) } f 1 i ( f i (x) ε, f i (x)+ε) =0. For each point x M \ Ω we define a collection of intervals {(a i,x,b i,x )} n with rational ends so that f i (x) (a i,x,b i,x ) ( f i (x) ε, f i (x)+ε). 1 Assume A x := f 1 (a 1 1,x,b 1,x )... f n (a n,x,b n,x ). Note that the set of distinct interval collections {(a i,x,b i,x )} is at most countable, therefore, there are at most a countable number of the sets A x.dueto the construction, each set A x is of zero measure and x A x.thenm\ Ω A x, where the union is taen over all x M \ Ω. Hence μ(m \ Ω) Σμ(A x )=0. The lemma is proved.

3 30 DRUZHININ Lemma 4. Let f 1,...,f n be linearly independent functions from the space L p (M) and let M = M 1... M N,whereμ(M i ) > 0, i =1,...,N. Then there exists a system of pairwise disjoint measurable sets A 1,...,A n such that each A i lies in some M j, for all i =1,...,n the relation 0 <μ(a i ) < holds, and f 1 dμ... f 1 dμ A 1 A n I(f; A) :=..... f n dμ f n dμ A 1 A n Proof. Fix some representatives f 1,..., f n from the classes f 1,...,f n, respectively. Due to Lemma 3, the set Ω( f 1 ;...; f n ) coincides with M almost everywhere, therefore, the functions f 1,..., f n are linearly independent on Ω. Then by Lemma 2 there exist points x 1,...,x n Ω such that Φ( f 1 ;...; f n ) 0. For each function f i we introduce the set B i (ε) := 1 f i ( f i (x 1 ) ε, f i (x 1 )+ε)... 1 f i ( f i (x n ) ε, f i (x n )+ε). Show that for a sufficiently small ε the sets B 1 (ε),...,b n (ε) are pairwise disjoint. Tae the sets B (ε) and B l (ε). The determinant Φ is nonzero, therefore, each two its columns are distinct. For the th and lth columns there exists a number m such that the mth elements of these columns are distinct. Thus, f m (x ) f m (x l ). Therefore, for some ε the sets ( f m (x ) ε, f m (x )+ε) and ( f m (x l ) ε, f m (x l )+ε) are disjoint and hence the sets B (ε) and B l (ε) are also disjoint. We can assume that the sets B i (ε) have zero measure. For each B i (ε) we find M so that μ(b i (ε) M )>0. Then we assume A i := B i (ε) M. Thus, we have constructed the pairwise disjoint sets A 1,...,A n of finite positive measure. Further, for x A we have f i (x) f i (x ) <ε. Therefore, fi dμ μ(a ) f i (x ) <εμ(a ) fi dμ = μ(a )( f i (x )+α i ), A A where α ij are some numbers such that α ij <ε. The latter implies I(f; A) =Ψ n μ(a i), wherethe determinant Ψ obtained from Φ by small changes of its elements does not equal zero for sufficiently small ε. The lemma is proved. 4. SUBSPACES OF FINITE DIMENSION Theorem 1. Let Y be a subspace of finite dimension dim Y = n in L p (M) for p>1, p 2.Themetric projection operator P Y for Y is linear if and only if there exists a basis f 1,...,f n of the subspace Y such that the support of each function f i consists of one or two atoms. Here an atom is a measurable set A M such that the measure of any its measurable subset is either equal to zero, or equal to the measure of A itself. Proof. Prove the necessity. Let the operator P Y be linear. Then, in accordance with Lemma B, the set Q(Y ) is a linear subspace. Tae an arbitrary basis f 1,...,f n of the subspace Y. Assume M f := n supp f i. Partition the set M f into sets of zero measure as follows: M f = M 1... M N. We can assume that N n because f 1,...,f n are linearly independent. Due to Lemma 4, for M 1,...,M n there exist pairwise disjoint sets A 1,...,A n of finite positive measure such that each A i is entirely contained in some M j, j =1,...,n, and I(f; A) 0. Finally, in each M j, j = n +1,...,N, we find a set A j of finite nonzero measure. Let a function g Y tae constant values g i on the sets A i and equal zero outside of these sets. Then 0= gf i dμ = g 1 f i dμ g N f i dμ. M A 1 The latter implies g Y if and only if the value sequence (g 1,...,g N ) is a solution to the following system of linear equations: f 1 dμ... f 1 dμ A 1 A N g f n dμ.... =0. ( ) f n dμ g N A 1 A N A N

4 THE LINEARITY OF METRIC PROJECTION OPERATOR 31 Suppose N 2n +1. For any i = n +1,...,2n +1we have A i M f and hence there exists a number = (i) such that μ(supp f A i ) > 0. Then for those numbers i we choose A i to be equal to that of the pair of sets {x : f (x) > 0} A i and {x : f (x) < 0} A i which is of positive measure. Thus, we can assume that system ( ) has nonzero columns n +1,...,2n +1. Using system ( ), we construct a pair of functions ϕ, ψ Y as follows. Assume (ϕ 1,...,ϕ n )=(0,...,0). Then ( ) becomes a system of n equations on n +1 unnowns. Hence, there exists a nonzero solution (ϕ n+1,...,ϕ 2n+1 ). We may assume that ϕ 2n+1 =1(otherwise we can renumber the sets A n+1,..., ). For ψ we tae (ψ n+1,...,ψ 2n,ψ 2n+1 )=(0,...,0, 1). System( ) becomes nondegenerate, therefore, there exists a solution (ψ 1,...,ψ n ). Due to Lemma C, the functions sgn ϕ ϕ q 1 and sgn ψ ψ q 1 lie in the set Q(Y ), this set is linear; therefore, sgn ϕ ϕ q 1 +sgnψ ψ q 1 Q(Y ). Due to Lemma C again, there exists a function η Y such that sgn η η q 1 =sgnϕ ϕ q 1 +sgnψ ψ q 1. Calculating the values of η i,weget(η 1,...,η 2n+1 )= (ψ 1,...,ψ n,ϕ n+1,...,ϕ 2n, 2 1 q 1 ). Due to the construction, some th element of the last column of ( ) is nonzero. Then 0= ηf dμ = M = n ψ i f dμ A i f dμ + 2n i=n+1 f dμ +2 ϕ i A i f dμ q 1 1 q f dμ 0. We have got a contradiction. The latter means N 2n. Thus,M f cannot be partitioned into more than 2n sets of nonzero measure, therefore, M f consists of at most 2n atoms. Now prove that each supp f i consists of at most two atoms. Let M f be initially partitioned into atoms M 1,...,M N. The functions f i are constant on the atoms M j and tae the values f j i on them. Consider the matrix (f j i ), i = 1,...,n, j = 1,...,N, of these values. Using the Gauss method and elementary transformations of rows and permutations of columns only, this matrix can be transformed to the form (E B), where E is the (n n) identity matrix and B is some matrix. Row transformations of the matrix correspond to transformations of the basis f 1,...,f n and permutations of columns corresond to renumerations of the atoms M i. Thus, there exists a basis h 1,...,h n of the subspace Y whose matrix of values has the form (E B). Show that each row of the matrix (h j i )=(E B) contains at most two nonzero elements. Suppose the contrary. Then matrix B contains some row, say the th one, with two nonzero elements. After renumeration of the atoms M i we can assume that b,n+1 = h n+1 0and b,n+2 = h n+2 0. Construct a pair of new functions ϕ, ψ Y constant on M i and equal to zero outside of M f.forϕ we assume (ϕ n+1,ϕ n+2,...)=(1, 0,...). Then, solving system ( ) and taing into account the form of the matrix (h i j )=(E B), wegetϕi = h n+1 i m i,wherem i := μ(mn+1) μ(m i) for i =1,...,n.Forψ we assume (ψ n+1,ψ n+2,ψ n+3,...)=(0, 1, 0,...). Thenψ i = h n+2 i s i,wheres i := μ(mn+2) μ(m i) for i =1,...,n.SinceQ(Y ) is a linear subspace, then in accordance with Lemmas B and C, we get sgn ϕ ϕ q 1 +sgnεψ εψ q 1 Q(Y ). Due to Lemma C, there exists a function η Y such that sgn η η q 1 =sgnϕ ϕ q 1 +sgnεψ εψ q 1. We have (η n+1,η n+2,η n+3,...,η 2n )=(1,ε,0,...,0) and f dμ sgn η η q 1 =sgn( h n+1 m ) h n+1 m q 1 +sgn( εh n+2 s ) εh n+2 s q 1. Since η Y,then(η 1,...,η N ) is a solution to system ( ).Thenη = h n+1 m εh n+2 s. Due to Lemma 1, we obtain h n+1 m = εh n+2 s. Due to the arbitrariness of ε, we conclude that h n+1 =0. We have got a contradiction. The necessity is proved. Prove the sufficiency. Let f 1,...,f n be a basis in Y such that the support of each function f i consists of at most two atoms. Then each row of the matrix (f j i ) contains at most two nonzero elements. As before, transform this matrix to the form (E B) by the Gauss method. Note that the elementary row transformations used by the Gauss method do not increase the number of nonzero elements of each row. Therefore, there exists a basis h 1,...,h n with h j i =(E B) such that each row (hj i ) contains at most two nonzero elements. Let j {n +1,...,N} be the second subscript of the nonzero element h j = b,j of the th row; if such an element does not exist, then we put j to be equal to any value from {n +1,...,N}.

5 32 DRUZHININ Then the condition g Y is equivalent to the following one: gh dμ = g μ(m )+g j μ(m j )h j =0 g = a g j M for all =1,...,n,wherea = μ(mj ) μ(m ) hj. Using Lemma C, we derive from the latter relation that the quasi-orthogonal set Q(Y ) consists of the functions f L p (M) whose values {f } on the atoms M have the form {sgn g g q 1 }, respectively, and hence f =sgna a q 1 f j, =1,...,n. The latter implies that the set Q(Y ) is a linear subspace and so by Lemma B the operator P Y is linear. The theorem is proved. 5. SUBSPACES OF FINITE CODIMENSION Theorem 2. Let Y be a subspace of finite codimension codim Y = n in L p (M), p>1, p 2.Themetric projection operator P Y for Y is linear if and only if there exists a basis g 1,...,g n of the subspace Y L q such that μ(supp g i supp g j )=0for all i j. Proof. Prove the necessity. In accordance with Lemma B, we have dim Q(Y )=dimy = n. Tae some basis h 1,...,h n in Y. Using Lemmas 2 and 3, find points x 1,...,x n Ω( h 1 ;...; h n )=Ωsuch that the matrix ( h i (x j )) n i,j=1 is nondegenerate. Using elementary row transformations only, transform this matrix by the Gauss method to the identity matrix. At each step of the Gauss method we transform the functions h 1,..., h n in the same way as the corresponding rows are transformed. For example, if we subtract the jth row multiplied by a number λ from the ith row, then we have to change the function h i by h i λ h j. By g 1,..., g n we denote the resulting system of functions. It possesses the property g i (x i ) = 1 and g i (x j )=0for i j. Show that this is the required system. The functions g 1,...,g n form a basis in Y because they are obtained from the basis h 1,...,h n by a nondegenerate change of variables. Construct the functions f i := sgn g i g i q 1 for all i =1,...,n. In accordance with Lemma C, these functions belong to Q(Y ). Show that they are linearly independent. To do this, consider a linear combination λ 1 f λ n f n =0. Due to the definition of the set Ω, for the points x i and any ε>0 there exist sets M i = M i (ε) of positive measure such that x i M i and g 1 (x) ( ε, ε),...,g i (x) (1 ε, 1+ε),...,g n (x) ( ε, ε) for almost all x M i.thus,forapointx M i = M i (ε) we conclude 0=(λ 1 f λ n f n )(x) (λ i Dε, λ i + Dε), whered does not depend on ε. Due to the arbitrariness of ε, the latter implies λ i =0 for all i. Hence, the functions f 1,...,f n are linearly independent. Decompose the function sgn (g 1 + g 2 ) g 1 + g 2 q 1 Q(Y ) with respect to the basis f 1,...,f n : sgn (g 1 + g 2 ) g 1 + g 2 q 1 = 1 sgn g 1 g 1 q n sgn g n g n q 1. Calculating the values of the latter expression at the points x M i (ε), i =1,...,n,forε 0, weobtain 1 = 2 =1, 3 =...= n =0, i.e., Similarly, we get sgn (g 1 + g 2 ) g 1 + g 2 q 1 =sgng 1 g 1 q 1 +sgng 2 g 2 q 1. sgn (g 1 +2g 2 ) g 1 +2g 2 q 1 =sgng 1 g 1 q 1 +2 q 1 sgn g 2 g 2 q 1. Applying Lemma 1 to the previous two equalities, we have g 1 = g 2 and g 1 = 2g 2 almost everywhere on supp g 1 supp g 2. Thus, the supports of the functions g 1 and g 2 intersect over a set of zero measure. One can similarly verify that the measure of the set supp g i supp g j equals zero for all i j. The necessity is proved. Now prove the sufficiency. Tae the basis g 1,...,g n of the subspace Y from the conditions of the theorem. Then for any function g Y we have g = λ 1 g λ n g n and sgn g g q 1 = n sgn λ i λ i q 1 sgn g i g i q 1. (1)

6 THE LINEARITY OF METRIC PROJECTION OPERATOR 33 In accordance with Lemma C, for any function f Q(Y ) there exists a function g Y such that f =sgng g q 1. Equality (1) implies that the function f is a linear combination of the functions sgn g 1 g 1 q 1,...,sgn g n g n q 1 Q(Y ). Ifλ i varies over R, thensgn λ i λ i q 1 varies over the whole R too. Thus, each linear combination of the functions sgn g i g i q 1 lies in Q(Y ). Therefore, Q(Y ) is the linear hull of n functions and so it is a linear subspace. Lemma B implies that the operator P Y is linear. The theorem is proved. 6. EXAMPLES The theorems proved here give a complete description of all subspaces of finite dimension, or finite codimension in the spaces L p (M) with p>1, p 2, having linear metric projection operator. But these are not all the subspaces Y having a linear operator P Y. Consider L p [ 1, 1], p>1, p 2.Byf 1 we denote the restriction of an arbitrary function F L p [ 1, 1] onto [0, 1] and by f 2 we denote its restriction onto [ 1, 0]. Let an arbitrary bounded measurable function α(x) be fixed on [ 1, 0]. Construct the two following subspaces Y 1 := { F L p [ 1, 1] : f 1 L p [0, 1], f 2 = α(x)f( x) }, Y 2 := {F L p [ 1, 1] : f 1 L p [0, 1], f 2 (x) = x 1 f 1 ( t) dμ}. One can show that the operator P Y1 is linear, but the operator P Y2 is not linear. In conclusion we mention that Theorems 1 and 2 can be stated in terms of 1-complementable subspaces, see [7]. Definition. A subspace Z of an arbitrary Banach space X is said to be 1-complementable if there exists a projector of norm 1 onto this subspace. Theorem 3. Let Z L p be a subspace, p>1, p 2, codim Z = n. ThenZ is 1-complementable if and only if there exists a basis g 1,...,g n in Z such that the support of g j consists of one or two atoms for any j. Theorem 4. Let Z L p be a subspace, p>1, p 2, dim Z = n. ThenZ is 1-complementable if and only if Z = f 1,...,f n,whereμ(supp f i supp f j )=0for i j. ACKNOWLEDGMENTS The author expresses his gratitude to his scientific advisor P. A. Borodin for statement of the problem. The wor is supported by the Russian Foundation for Basic Research (project no a). REFERENCES 1. P. D. Morris, Chebyshev Subspaces of L 1 with Linear Metric Projection, J. Approx. Theory 29, 231 (1980). 2. P. A. Borodin, Linearity of Metric Projections on Chebyshev Subspaces in L 1 and C, Matem. Zameti 63 (6), 812 (1998) [Math. Notes 63 (6), 717 (1998)]. 3. T. Ando, Contractive Projections in L p Spaces, Pacif. J. Math. 17 (3), 391 (1966). 4. Lin. Pee-ee, Remars on Linear Selections for the Metric Projection, J. Approx. Theory 43, 64 (1985). 5. E. W. Cheney and D. E. Wulbert, The Existence and Unicity of Best Approximation, Math. scand. 24, 113 (1969). 6. I. Singer, Best Approximation in Normed Linear Spaces by Elements of Linear Subspaces, (Springer, Berlin, Heidelberg, New Yor, 1970). 7. B. Randrianantoanina, Contractive Projections in Nonatomic Function Spaces, Proc. Amer. Math. Soc. 123 (6), 1747 (1995). Translated by A. Ivanov