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Orthogonally Complemented Subspacesin Banach SpacesHenryk Hudzik a , Yuwen Wang b & Ruli Sha ca Faculty of Mathematics and Computer Science , Adam MickiewiczUniversity , Umultowska, Poznań, Polandb School of Mathematics and Computer Science, Harbin NormalUniversity , Harbin, People's Republic of Chinac Department of Mathematics , Hulunbeir College , Hailar, People'sRepublic of ChinaPublished online: 18 Sep 2008.

To cite this article: Henryk Hudzik , Yuwen Wang & Ruli Sha (2008) Orthogonally ComplementedSubspaces in Banach Spaces, Numerical Functional Analysis and Optimization, 29:7-8, 779-790, DOI:10.1080/01630560802279231

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Numerical Functional Analysis and Optimization, 29(7–8):779–790, 2008Copyright © Taylor & Francis Group, LLCISSN: 0163-0563 print/1532-2467 onlineDOI: 10.1080/01630560802279231

ORTHOGONALLY COMPLEMENTED SUBSPACESIN BANACH SPACES

Henryk Hudzik,1 Yuwen Wang,2 and Ruli Sha3

1Faculty of Mathematics and Computer Science, Adam Mickiewicz University,Umultowska, Poznan, Poland2School of Mathematics and Computer Science, Harbin Normal University, Harbin,People’s Republic of China3Department of Mathematics, Hulunbeir College, Hailar, People’s Republic of China

� In this paper, we extend the Moreau (Riesz) decomposition theorem from Hilbert spacesto Banach spaces. Criteria for a closed subspace to be (strongly) orthogonally complementedin a Banach space are given. We prove that every closed subspace of a Banach space Xwith dimX ≥ 3 (dimX ≤ 2) is strongly orthognally complemented if and only if the Banachspace X is isometric to a Hilbert space (resp. strictly convex), which is complementary tothe well-known result saying that every closed subspace of a Banach space X is topologicallycomplemented if and only if the Banach space X is isomorphic to a Hilbert space.

Keywords Banach space; Duality mapping; Hilbert space; Orthogonallycomplemented subspace; Strict convexity.

AMS Subject Classification 46B20.

1. INTRODUCTION

In 1971, Lindenstrauss and Tzafriri [9] proved that every closedsubspace of a Banach space X is topologically complemented in X if andonly if the space X is isomorphic to a Hilbert space, which is an importantresult in the geometry of Banach spaces.

In 1974, Nashed and Votruba [14] introduced the concept of theorthogonally complemented subspace in a normed linear space bythe definitions of the topologically complemented subspace and theorthogonality in normed spaces defined by James [8]. In the light

Address correspondence to Henryk Hudzik, Faculty of Mathematics and Computer Science,Adam Mickiewicz University, Umultowska 87, Poznan 61-614, Poland; E-mail: hudzik@amu.edu.pl

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of orthogonally complemented subspaces, they investigated orthogonallygeneralized inverses of linear operators in normed spaces [14]. Fordetails of various concepts and results, see [11, 13]. For a comprehensivetreatment of various aspects of generalized inverses of operators, see [12].It is well-known that each closed subspace of a Hilbert space has a uniqueorthogonal complement by means of the Riesz orthogonal decompositiontheorem. We may ask about criteria in order that every closed subspace ofa Banach space X is an orthogonally complemented subspace of X in theNashed and Votruba sense. To answer this problem, we should extend theRiesz orthogonal decomposition theorem from Hilbert spaces to Banachspaces.

In 1956, Moreau [10] extended the Riesz orthogonal decompositionfrom a closed subspace to a closed convex cone in a Hilbert space H . It isknown that if K is a closed convex cone in a Hilbert space H , we have thedecomposition

x = PK (x) + PK 0(x)

for every x ∈ H , where K 0 = �y ∈ H : (y, x) ≤ 0 ∀x ∈ K �, (·, ·) is a scalarproduct in H , and PK and PK 0 are the metric projectors of H onto K andK 0, respectively. This decomposition does not hold in arbitrary Banachspaces. Many authors have attempted to generalize this result.

In 1995, Wang and Li [17] obtained a decomposition by using themetric projector �K :

x = �K (x) + x2

for every x ∈ X , where K is a closed convex cone of a real reflexive strictlyconvex Banach space X , x2 ∈ F −1

X (K 0), FX : X → X ∗ is the duality mappingof X (see Definition 2.1 below), and K 0 = �x∗ ∈ X ∗ : 〈x∗, x〉 ≤ 0, ∀x ∈ K � isthe polar cone of K .

In 1998, Alber [1] obtained another decomposition in a reflexivestrictly convex and smooth Banach space X

x = J −1�K 0 J (x) + w

for every x ∈ X , where J = FX , K and K 0 are as above, �K 0 is thegeneralized projection of X ∗ onto K 0, and w ∈ K .

In this paper, we extend the Moreau (the Riesz) theorem from Hilbertspaces to general Banach spaces under some reasonable conditions, andwe get a generalized orthogonal decomposition theorem in Banach spaces.In view of this result, we prove that every closed subspace of a Banach spaceX with dimX ≥ 3 (dimX ≤ 2) is strongly orthogonally complemented ifand only if the Banach space X is isometric to a Hilbert space (resp. to astrictly convex Banach space), which is complementary to the well-knownresult given by Lindenstrauss and Tzafriri [9].

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2. GENERALIZED ORTHOGONAL DECOMPOSITIONIN BANACH SPACES

Let X ,Y be Banach spaces. For geometric properties of Banach spacessuch as strict convexity, smoothness, and also topologically complementedsubspaces, we refer to Diestel [4], Barbu and Precpuanu [2], and Holmes[5, 6].

Definition 2.1 (Barbu and Precpuanu [2]). The set-valued mappingFX : X → X ∗ defined by

FX (x) = �x∗ ∈ X ∗ | 〈x∗, x〉 = ‖x∗‖2 = ‖x‖2� (∀x ∈ X )

is called the duality mapping of X .

Here, 〈x∗, x〉 is the dual pair of x∗ ∈ X ∗ and x ∈ X , that is, 〈x∗, x〉 =x∗(x). From [2] we know that (i) FX is a homogeneous set-valued mapping;(ii) FX is surjective iff X is reflexive; (iii) FX is injective or strictly monotoneiff X is strictly convex; (iv) FX is single-valued iff X is smooth.

Definition 2.2 (Singer [16]). Let K ⊂ X . The set-valued mapping �K :X → K defined by

�K (x) = �y ∈ K | ‖x − y‖ = dK (x)� (∀x ∈ X ),

where dK (x) = infy∈K ‖x − y‖ is called the set-valued metric projection.Sometimes we also write �(K ; x) in place of �K (x)�

If �K (x) �= Ø for each x ∈ X , then K is said to be proximinal. If �k(x)is at most a singleton for each x ∈ X , K is said to be semi-Chebyshev. If Kis simultaneously a proximinal and semi-Chebyshev set, then K is said to beChebyshev set. We denote by �K any selection for the set-valued mapping�K , i.e., any single-valued mapping �K : �(�K ) → K , where �(��) = �x ∈X : �K (x) �= Ø�, and �K (x) ∈ �K (x) for any x ∈ �(�K )� In the particularcase when K is a Chebyshev set, �(�L) = X and �K (x) = ��K (x)� forany x ∈ X . The mapping �K (or �(K ; ·)) is called the metric projectionoperator (or metric projector) of X onto K .

Lemma 2.3 (Barbu and Precpuanu [2]). Let X be a normed space, C aconvex subset of X , x ∈ X \C, and x0 ∈ C. Then x0 ∈ �C(x) if and only if thereexists an element x∗ ∈ X ∗ such that:

(i) ‖x∗‖ = ‖x − x0‖;(ii) 〈x∗, y − x〉 ≥ ‖x0 − x‖2, ∀y ∈ C �

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Lemma 2.4 ([16]). Let X be a normed space, L a subspace of X . Then

(i) �2L(x) = �L(x) for any x ∈ �(�L), i.e., �L is idempotent;

(ii) ‖x − �L(x) ‖ ≤ ‖x‖ for any x ∈ �(�L).If L is a semi-Chebyshev subspace, then

(iii) �L(�x) = ��L(x) for any x ∈ X and � ∈ R, i.e., �L is homogeneous;(iv) �L(x + y)= �L(x) + �L(y) = �L(x) + y for any x ∈ �(�L) and y ∈ L, i.e.,

�L is quasi-additive.

Proof. See p. 40 in [16]. �

Theorem 2.5. Let X be a normed space, and C a nonempty convex cone in X .If C is proximinal, then for each x ∈ X , there exist x1 ∈ �C(x) and x2 ∈ F −1

X (C 0)such that

x = x1 + x2, (2.1)

where C 0 = �x∗ ∈ X ∗ | 〈x∗, x〉 ≤ 0,∀x ∈ C�. In other words, we have

X = �C(X ) + F −1X (C 0),

where F −1X (C 0) = �x ∈ X | FX (x) ∩ C 0 �= Ø�.

Proof. Let C be a proximinal convex cone in X . Then for each x ∈ X ,there exists x0 ∈ �K (x), and hence there exists x∗ ∈ X ∗ such that (i) ‖x∗‖ =‖x0 − x‖ and (ii) 〈x∗, y − x〉 ≥ ‖x0 − x‖2, ∀y ∈ C , by Lemma 2.3. Takingy = x0 in (ii) of Lemma 2.3, we get

〈x∗, x0 − x〉 = ‖x0 − x‖2 = ‖x∗‖2� (2.2)

The definition of FX yields that x∗ ∈ FX (x0 − x), i.e., x0 − x ∈ F −1X (x∗)�

Taking x1 = x − x0, we have −x1 ∈ F −1X (x∗) and x = x0 + x1. For any y ∈ C ,

(ii) in Lemma 2.3 and equality (2.2) yield that 〈x∗, y − x〉 ≥ 〈x∗, x0 − x 〉,and hence

〈x∗, y − x0〉 ≥ 0, ∀y ∈ C � (2.3)

Because C is a cone and x0 ∈ C , thus 2x0 ∈ C � By taking y = 2x0 in (2.3),we have 〈x∗, x0〉 ≥ 0, and hence

〈x∗, y〉 ≥ 〈x∗, x0〉 ≥ 0, ∀y ∈ C ,

so that −x∗ ∈ C 0� But the duality mapping FX is homogeneous, whenceF −1X (−x∗) = −F −1

X (x∗), so we have

x1 ∈ −F −1X (x∗) = F −1

X (−x∗) ⊂ F −1X (C 0),

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whence

x = x0 + x1 ∈ �C(x) + F −1X (C 0), i.e., X = �C(X ) + F −1

X (C 0)�

This completes the proof. �

Lemma 2.6. Let X be a normed space, L be a subspace of X , and x ∈X \L, x0 ∈ L. Then x0 ∈ �L(x) if and only if FX (x − x0) ∩ L⊥ �= Ø, where L⊥ =�x∗ ∈ X ∗ | 〈x∗, x〉 = 0,∀x ∈ L��

Proof. See [3]. �

Theorem 2.7. Let X be a normed space, and L be a subspace of X . Then thefollowing statements are equivalent:

(1) L is proximinal;(2) X admits the decomposition:

X = �L(X ) + F −1X (L⊥), i.e., x = x1 + x2, x1 ∈ �L(x), x2 ∈ F −1

X (L⊥)(2.4)

for any x ∈ X , where F −1X (L⊥) = �x ∈ X | FX (x) ∩ L⊥ �= Ø�, �L(X ) =⋃

��L(x) | x ∈ X ��(3) L is closed and for the canonical mapping �L : X → X /L, we have

�L(F −1X (L⊥)) = X /L, (2.5)

where X /L is the quotient space.

Proof. (1) ⇒ (2) Because L is a linear subspace, so L is a convex coneand the dual cone L0 = L⊥. But L is proximinal, so Theorem 2.5 yields that

X = �L(X ) + F −1X (L⊥)�

which finishes the proof of the implication (1) ⇒ (2).

(2) ⇒ (3) For any x ∈ L, identity (2.4) implies that there exist x1 ∈�L(x) and x2 ∈ F −1

X (L⊥) such that x = x1 + x2. Because x ∈ L, we have ‖x −x1‖ = infy∈L ‖x − y‖ = 0, whence x = x1 ∈ L. Hence L is closed.

For any [x] ∈ X /L, i.e., [x] = x + L, (2.4) implies that there exists x1 ∈�L(x) such that x − x1 ∈ F −1

X (L⊥).Because x − (x − x1) = x1 ∈ L, by the definition of the quotient space

X /L, we have x − x1 ∈ [x] and hence, by the definition of the canonicalmapping �L , we have �L(x − x1) = [x], whence

�L(F −1X (L⊥)) = X /L

and the implication (2) ⇒ (3) is proved.

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(3) ⇒ (1) For any x ∈ X , equality (2.5) implies that there exists anelement y ∈ F −1

X (L⊥) such that �L(y) = x + L. Taking y0 = x − y, by thedefinition of the canonical mapping �L , we know that y0 ∈ L and x − y0 =y ∈ F −1

X (L⊥), i.e.,

FX (x − y0) ∩ L⊥ �= Ø�

Lemma 2.6 implies that y0 ∈ �L(x) and hence L is proximinal. �

Theorem 2.8. For a linear subspace L of a normed space X , the followingstatements are equivalent:

(1) L is a Chebyshev subspace,(2) X can be decomposed in the form

X = �L(X )� F −1X (L⊥), i.e., x = �L(x) + x2, x2 ∈ F −1

X (L⊥), ∀x ∈ X ,(2.6)

(3) L is proximinal and for the canonical mapping �L : X → X /L, we have that�L : F −1

X (L⊥) → X /L is one-to-one and onto.

Proof. (1) ⇒ (2) Let L be a Chebyshev subspace. Theorem 2.7 impliesthat

X = �L(X ) + F −1X (L⊥)�

For any x0 ∈ �L(X ) ∩ F −1X (L⊥), there exists x ∈ X such that x0 = �L(x) and

FX (x0 − �) ∩ L⊥ �= Ø. By Lemma 2.6 and the idempotence of �L , we have� = �L(x0) = �2

L(x) = �L(x) = x0, and hence

X = �L(X )+·F −1

X (L⊥),

where the sign “+·” is used to indicate that the decomposition is unique.

Therefore, for any x ∈ X , x = x1 + x2, x1 ∈ �L(X ), x2 ∈ F −1X (L⊥). By quasi-

additivity of �L , we have �L(x) = x1 + �L(x2) = x1, and so x = �L(x) + x2,x2 ∈ F −1

X (L⊥)� The implication is proved.

(2) ⇒ (3) Equality (2.6) and Theorem 2.7 imply that L isproximinal and the canonical mapping �L : F −1

X (L⊥) → X /L is onto.Let x1, x2 ∈ F −1

X (L⊥) be such that

�L(x1) = �L(x2) = [x] ∈ X /L�

Then x1 ∈ x + L and x2 ∈ x + L, so there exist x ′1, x

′2 ∈ L such that x − x ′

i =xi ∈ F −1

X (L⊥) (i = 1, 2)� By the uniqueness of the decomposition (2.6), we

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have x ′1 = x ′

2, and so x1 = x2, i.e., the canonical mapping is one-to-one fromF −1X (L⊥) to X /L, which means that the implication (2) ⇒ (3) holds.

(3) ⇒ (1) Let L be proximinal and �L : F −1X (L⊥) → X /L be one-to-

one and onto. For any x ∈ X , we have �L(x) �= Ø. Taking x1, x2 ∈ �L(x),by Lemma 2.6, FX (x − xi) ∩ L⊥ �= Ø, i.e., x − xi ∈ F −1

X (L⊥), (i = 1, 2). Notethat x − (x − xi) = xi ∈ L, and so x − xi ∈ [x], i.e., �L(x − x1) = �L(x −x2) = [x]. Hence x − x1 = x − x2� Thus x1 = x2, i.e., L is Chebyshev. �

Definition 2.9. Let X be a normed space, L a subspace of X . L issaid to be metrically complemented in X , if L is proximinal and thereexists an operator �L : X → L such that (i) �L(x) ∈ �L(x), ∀x ∈ X ; (ii)�L(�x) = ��L(x), ∀x ∈ X , ∀� ∈ R ; (iii) �L(x + y) = �L(x) + �L(y) = �L(x)+y, ∀x ∈ X , ∀y ∈ L. In this case, �L is called the associated metric projector,� (�L) := �−1

L (�) is called the metrical complement of L.

Remark 2.10. Let X be a normed space, L a metrically complementedsubspace in X , �L be the associated metric projector. Then

(i) �2L = �L by Lemma 2.4,

(ii) X = L � �−1L (�) = �(�L)� � (�L),

(iii) � (�L) is a homogeneous set, i.e., ∀x ∈ � (�L), ∀ ∈ R ⇒ x ∈ � (�L),(iv) ‖x‖ = dL(x) = infy∈L ‖x − y‖, ∀x ∈ � (�L),(v) the metrical complement � (�L) of L is a subset of F −1

X (L⊥).

3. ORTHOGONALLY COMPLEMENTED SUBSPACESIN BANACH SPACES

Let X be a normed space. We write spA for the span of a set A andd(x ,A) or dA(x) for the distance from x to the set A. The following conceptof orthogonality has been defined by James [8]:

x⊥y means that d(x , sp�y�) = ‖x‖,A⊥B means that d(x ,B) = ‖x‖ ∀x ∈ A�

Note that this orthogonality in a normed space is not a symmetricrelation (see [12]), so we sometimes call it the generalized orthogonalityin a normed space (see [18]).

Next, we recall the concept of an orthogonally complementedsubspace in a normed space, which is equivalent to the definition given byNashed and Votruba (see [12], p. 34).

Definition 3.1. Let L be a closed subspace of a normed space X � If L hasa topological complement N in X , i.e., X = L ⊕ N , and N ⊥ L, we say that

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L is an orthogonally complemented subspace of X , and N is an orthogonalcomplement of L in X .

Note that orthogonal complements are not necessarily unique (see[12], p. 39).

We give the following definition.

Definition 3.2. Let L be a closed subspace of a normed space X � IfL(⊥) = �x ∈ X : ‖x‖ = dL(x)� is a closed linear space such that X = L ⊕L(⊥), then we say that L is a strongly orthogonally complemented subspaceof X .

Remark 3.3. (i) Let L be a closed subspace of a normed space X . L isan orthogonally complemented subspace of X if and only if L is bothtopologically complemented and metrically complemented subspace andL has the same complemented subspace (metrically and topologically.

(ii) By Lemma 2.6, we have

L(⊥) = F −1X (L⊥),

where L(⊥) = �x ∈ X | dL(x) = ‖x‖�, L⊥ = �x∗ ∈ X ∗ | 〈x∗, x〉 = 0,∀x ∈ L�,and

F −1X (L⊥) = �x ∈ X : FX (x) ∩ L⊥ �= Ø��

Lemma 3.4 (James [8], Rudin-Smith [15]). Let X be a normed linear spaceof dimension at least 3 such that for any subspace M of dimension n ∈ N satisfying1 ≤ n ≤ dim(X ) − 2, M is a Chebyshev subspace and �M is linear. Then X is aninner space.

For the proof of this result, see [5, p. 159].

Theorem 3.5. Let L be a closed linear subspace of a Banach space X � Then Lis strongly orthogonally complemented if and only if: (i) L is Chebyshev subspaceof X ; (ii) F −1

X (L⊥) is a closed linear subspace of X .

Proof. Necessity: Let L(⊥) be a closed subspace of X such that

X = L ⊕ L(⊥)� (3.1)

For any x ∈ X , by (3.1) and Remark 3.3, there exists x1 ∈ L such thatx − x1 ∈ L(⊥) = F −1

X (L⊥), and hence FX (x − x1) ∩ L⊥ �= Ø. By Lemma 2.6,we have that x1 ∈ �L(x), so L is proximinal.

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For any x ∈ X and x1, x2 ∈ �L(x), by Lemma 2.6, we have

FX (x − xi) ∩ L⊥ �= Ø, i = 1, 2�

Let x ′i = x − xi . Then x ′

i ∈ F −1X (L⊥) = L(⊥) and

x = xi + x ′i , xi ∈ L and x ′

i ∈ L(⊥), i = 1, 2�

Condition (3.1) implies that x1 = x2, so that L is a Chebyshev subspace.Also F −1

X (L⊥) = L(⊥) is a closed linear subspace.

Sufficiency: Because L is a Chebyshev subspace of X , it is closed.Noting that the duality mapping FX is homogeneous and F −1

X (L⊥) isadditive and closed, we get that L(⊥) = F −1

X (L⊥) is a closed linear subspaceof X .

It follows from Theorem 2.8 that for each x ∈ X , x has the uniquedecomposition

x = �L(x) + x2, x2 ∈ L(⊥) = F −1X (L⊥), (3.2)

and hence, we have

X = L � L(⊥), i.e., X = L + L(⊥) and L ∩ L(⊥) = Ø� (3.3)

Because L and L(⊥) are closed subspaces, thus

X = L ⊕ L(⊥),

i.e., L is topologically complemented in X (see [12], p. 24). Hence L isstrongly orthogonally complemented in X � �

Theorem 3.6. Let L be a closed subspace of a strictly convex Banach space X .Then the following statements are equivalent:

(i) L is a strongly orthogonally complemented subspace of X ,(ii) L is a orthogonally complemented subspace of X ,(iii) L is a Chebyshev subspace of X and F −1

X (L⊥) is an additive and closed set.

Proof. We only need to prove that (ii) ⇒ (iii)� Let Y be a closed subspaceof X with Y ⊂ L(⊥) such that

X = L ⊕ Y � (3.4)

By the same proof as those for the necessity in Theorem 3.5, we knowthat L is proximinal. Because X is strictly convex Banach space, L is aChebyshev subspace of X .

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Next, we intend to show that Y = F −1X (L⊥)� By Remark 3.3, Y ⊂ L(⊥) =

F −1X (L⊥)� Conversely, for any x ∈ F −1

X (L⊥), (3.4) implies that x has theunique decomposition

x = x1 + x2, x1 ∈ L and x2 ∈ Y ⊂ L(⊥),

and so, by Remark 3.3 and Lemma 2.6, we have that x1 ∈ �L(x)�On the other hand, x ∈ F −1

X (L⊥) yields that FX (x − �) ∩ L⊥ �= Ø� ByLemma 2.6, � ∈ �L(x), but L is a Chebyshev subspace, and so x1 = �� Thusx = x2 ∈ Y , whence F −1

X (L⊥) = Y is an additive and closed set. �

Theorem 3.7. Let X be a Banach space with dimX ≤ 2. Every closed subspaceL of X is strongly orthogonally complemented if and only if X is strictly convex.

Proof. Necessity: We may assume that dimX = 2, every closedlinear subspace L, (L = ��� or dimL = 1), is a strongly orthogonallycomplemented subspace in X . Then Theorem 3.5 yields that L is aChebyshev subspace. The fact that every closed subspace of X is aChebyshev subspace implies that X is a strictly convex Banach space (see[16], Theorem 3.16, p. 36).

Sufficiency: Let X be a strictly convex Banach space. Without loss ofgenerality, we may assume that L is a closed subspace of dimension oneand dimX = 2, so that L is a Chebyshev subspace of X �

Because L is a maximal subspace of X , i.e., codimL = 1, so we have anelement x∗

0 ∈ X ∗ such that L = �x ∈ X : 〈x∗0 , x〉 = 0,∀x ∈ L�� We have

X = L+·�tx0 | t ∈ R�

for some x0 ∈ F −1X (x∗

0 ). Next, we intend to prove that

L(⊥) = �tx0 | t ∈ R��

Indeed, for any t ∈ R and x = tx0, by the homogeneity of FX , we havethat tx∗

0 ∈ FX (tx0) = FX (x). Note that tx∗0 ∈ L⊥, and hence FX (x − �) ∩ L⊥ �=

Ø. Lemma 2.6 yields that � ∈ �L(x), i.e., ‖x‖ = dL(x). Thus �tx0 | t ∈ R� ⊂L(⊥). Conversely, for any x ∈ L(⊥), if x = �, then x ∈ �tx0 | t ∈ R�; if x �= �,then dL(x) = ‖x‖ > 0, whence it follows that x �L, and �L(x) = � by strictconvexity of X . On the other hand, for this x , we have the decomposition

x = x1 + tx0; x1 ∈ L, t ∈ R �

By Lemma 2.4, we have �L(x) = �L(tx0) + x1 = t�L(x0) + x1� Note thatx0 ∈ F −1

X (x∗0 ) ⊂ F −1

X (L⊥) = L(⊥), whence �L(x0) = �. Combining this fact

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Orthogonally Complemented Subspaces in Banach Spaces 789

with �L(x) = �, we have that x1 = �, i.e., x = tx0 ∈ �tx0 | t ∈ R�. So L(⊥) ⊂�tx0 | t ∈ R�. Thus

L(⊥) = �tx0 | t ∈ R��

By Remark 3.3, F −1X (L⊥) = L(⊥) = �tx0 | t ∈ R� is a closed linear subspace of

X . In view of Theorem 3.5, L is a strongly complemented subspace of X .�

Theorem 3.8. Let X be a Banach space with dimX ≥ 3. Then the followingstatements are equivalent:

(1) X is isometric to a Hilbert space,(2) Every closed subspace of X is strongly orthogonally complemented in X ,(3) X is reflexive and strictly convex, and every closed subspace of X is orthogonally

complemented in X .

Proof. The implication (1) ⇒ (2) is obvious.

(2) ⇒ (3) For any closed subspace L of X , (2) implies that Lis strongly orthogonally complemented in X , whence L is orthogonallycomplemented in X � By Theorem 3.5, L is a Chebyshev subspace, andhence, by Theorem 2.14 and Theorem 3.16 in [16], we know that X isreflexive and strictly convex.

(3) ⇒ (1) For any fixed positive integer n with 1 ≤ n ≤ dimX − 2,if M is an n-dimensional closed subspace, condition (3) and Theorem 3.6yield that M is a Chebyshev subspace of X and F −1

X (M⊥) is a closed linearsubspace of X .

Next, we want to show that

F −1X (M⊥) = �x − �M (x) : x ∈ X �� (3.5)

Indeed, for any x ∈ X , by Lemma 2.6, FX (x − �M (x)) ∩ M⊥ �= Ø,whence �x − �M (x) : x ∈ X � ⊂ F −1

X (M⊥). Conversely, for any x ∈ F −1X (M⊥),

by Lemma 2.6, � = �L(x), and hence x = x − �L(x) ∈ �x − �M (x) : x ∈ X ��Thus

F −1X (M⊥) ⊂ �x − �M (x) : x ∈ X �,

whence (3.5) holds.Because F −1

X (M⊥) is a closed linear subspace of X , �x − �M (x) : x ∈ X �is a closed linear subspace of X , too. Hence, for any x , y ∈ X , we have

(x − �L(x)) + (y − �L(y)) = (x + y) − �L(x + y)�

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790 H. Hudzik et al.

Subtracting x + y from both sides, we obtain

�L(x + y) = �L(x) + �L(y)�

Note that the metric projection operator �L is homogeneous, and thus �L

is linear. Lemma 3.4 implies that X is isometric to a Hilbert space. �

ACKNOWLEDGMENTS

The research was supported in part by the National ScienceFoundation (19971023) and the Science Foundation Grant ofHeilongjiang Province.

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