Uniqueness of Hahn-Banach extension and related norm- 1 projections in dual spaces
aa r X i v : . [ m a t h . F A ] S e p UNIQUENESS OF HAHN-BANACH EXTENSION ANDRELATED NORM- PROJECTIONS IN DUAL SPACES
SOUMITRA DAPTARI, TANMOY PAUL, AND T. S. S. R. K. RAO
Abstract.
In this paper we study two properties viz. property- U andproperty- SU of a subspace Y of a Banach space which correspond to theuniqueness of the Hahn-Banach extension of each linear functional in Y ∗ andin addition to that this association forms a linear operator of norm-1 from Y ∗ to X ∗ . It is proved that, under certain geometric assumptions on X, Y, Z these properties are stable with respect to the injective tensor product; Y hasproperty- U ( SU ) in Z if and only if X ⊗ ∨ ε Y has property- U ( SU ) in X ⊗ ∨ ε Z .We prove that when X ∗ has the Radon-Nikod´ y m Property for 1 < p < ∞ , L p ( µ, Y ) has property- U (property- SU ) in L p ( µ, X ) if and only if Y is so in X . We show that if Z ⊆ Y ⊆ X , where Y has property- U ( SU ) in X then Y /Z has property- U ( SU ) in X/Z . On the other hand Y has property- SU in X if Y /Z has property- SU in X/Z and Z ( ⊆ Y ) is an M-ideal in X . It isobserved that a smooth Banach space of dimension > Y, Z with property- SU in X , Y + Z hasproperty- SU in X whenever Y + Z is closed. We characterize all hyperplanesin c which have property- SU .Dedicated to the memory of Eve Oja. Introduction By X we mean a Banach space with real scalars and a subspace Y of X is always assumed to be a closed subspace of X except when we consider denselinear subspaces. The annihilator of Y in X ∗ is denoted by Y ⊥ and Y representsthe set { f ∈ X ∗ : k f k = k f | Y k} . B X and S X represent the closed unit ball andclosed unit sphere of X respectively. By a hyperplane in X we mean a subspaceof the form f − (0), for some f ∈ X ∗ . The classical Hahn-Banach Theoremensures that every f ∈ Y ∗ has a norm preserving extension ˜ f ∈ X ∗ . Uniquenessof such extension depends on some geometric structures of X ∗ . The authors in Mathematics Subject Classification.
Primary 46A22, 46B20; Secondary 46B22, 46M05.
Key words and phrases.
Hahn-Banach extension, property- U , property-( SU ), L -predual,Bochner integrable functions, tensor product spaces. [F, P, T] studied this property and characterize those Banach spaces where agiven subspace (all subspaces) has (have) this unique extension property. Wenow recall the following definitions. Definition 1.1. [P] A subspace Y of X said to have property- U if every f ∈ Y ∗ has unique norm preserving extension ˜ f to X ∗ .Also let us recall the following definitions. Definition 1.2. ( a ) [L] A subspace Y of X is said to be an ideal if thereexists a projection P : X ∗ → X ∗ where k P k = 1 and ker ( P ) = Y ⊥ .( b ) [O] A subspace Y of X is said to have property- SU if Y ⊥ has a com-plementary subspace G ⊆ X ∗ such that for all f ∈ X ∗ , f = g + hg ∈ G, h ∈ Y ⊥ and k f k > k g k whenever h = 0.It is well known that there exists a linear operator S : Y ∗ → X ∗ where S ( y ∗ )is a norm preserving extension of y ∗ if and only if Y is an ideal in X . Also recallthat, Y is an ideal in X if and only if Y ⊥⊥ is a range of a norm-1 projection in X ∗∗ ([Rao]). It is clear from the properties of a subspace Y having property- SU is that the canonical restriction map from Y to Y ∗ is an isometry. In [O] Ojaobserved, a subspace Y has property- SU in X if and only if Y has property- U and also an ideal in X .Let us recall that a subspace Y is said to have n.X.I.P. in X if n closedballs { B ( a i , r i ) } ni =1 in X with centres in Y and T ni =1 B ( a i , r i ) = ∅ , then Y ∩ T ni =1 B ( a i , r i + ε ) = ∅ , for all ε >
0. It follows from [L] that,
Theorem 1.3.
Let Y be a subspace of a Banach space X . If Y is an ideal in X then Y has n.X.I.P. , for all n ∈ N . Suppose Y has property- U in X , if Y has n.X.I.P. in X then Y is an ideal in X . Here it is enough to consider n = 3 . In this paper we discuss property- U and SU for various kinds of functionspaces, followed by their stability in quotient spaces. We also study these prop-erties for subspaces of Banach spaces which are of type L -preduals. Variousexamples are given at the end of the paper to illustrate the limitations of someof the results obtained here using the sequence space c and finite dimensionalspaces. NIQUENESS OF HAHN-BANACH EXTENSION 3 Notations and Definitions
Let Y be a subspace of X and let x ∈ X , define d ( x, Y ) = inf y ∈ Y k x − y k and P Y ( x ) = { y ∈ Y : k x − y k = d ( x, Y ) } . Note that P Y ( x ) may be empty for some x . Definition 2.1. ( a ) Y is said to be proximinal if P Y ( x ) = ∅ for all x .( b ) Y is said to be Chebyshev if P Y ( x ) is singleton for all x .[IS] is a standard reference for the notions defined above. Definition 2.2. [HWW]( a ) Y is said to be an M-summand (L-summand) in X if there existsa linear projection P : X → X such that Y = P ( X ) and X = P ( X ) L ℓ ∞ ( I − P )( X ) (cid:0) X = P ( X ) L ℓ ( I − P )( X ) (cid:1) .( b ) Y is said to be an M-ideal in X if Y ⊥ is an L-summand in X ∗ .( c ) Y is said to be a semi M-ideal in X if for any x ∗ ∈ X ∗ , we have k x ∗ k = k P x ∗ k + k x ∗ − P x ∗ k , where P : X ∗ → X ∗ is a projection satisfying P ( λx ∗ + P ( x ∗ )) = λP ( x ∗ ) + P ( x ∗ ), for x ∗ , x ∗ ∈ X ∗ and λ ∈ R ( quasiadditivity ) (see [HWW, Pg. 43]).Note that if Y is an M-ideal in X and X ∗ = Y ⊥ L ℓ Z then Z ∼ = Y , where Y is defined in Section 1. A Banach space X is said to be an M-embedded space if X is an M-ideal in X ∗∗ under the canonical embedding. An M-embedded space X satisfies many geometric properties, in particular this property is separablydetermined and X ∗ has Radon-Nikod ´ y m property (see below) [HWW, Pg. 126,Theorem 3.1]. c (Γ), for an arbitrary set Γ, K ( ℓ p ), for 1 < p < ∞ are someexamples of M-embedded spaces. If X is M-embedded then so is all its subspaces.Let (Ω , Σ , ν ) be a positive measure space, a measurable function f : Ω → X is said to be p -th Bochner-Integrable function if R Ω k f ( t ) k p dν ( t ) < ∞ . Thecorresponding p -th norm is defined by k f k p = (cid:0)R Ω k f ( t ) k p dν ( t ) (cid:1) p , where 1 ≤ p < ∞ . The corresponding Banach space consisting of all p -th Bochner Integrablefunctions is denoted by L p ( ν, X ). L ∞ ( ν, X ) represents the space of essentiallybounded, measurable functions from Ω to X . L ∞ ( ν, X ) forms a Banach spacewith respect to the essentially supremum norm.Let us recall the following Definition. Definition 2.3. [DU] A Banach space X is said to have Radon-Nikod´ y m prop-erty (RNP in short) if for any probability space (Ω , Σ , µ ) and any µ -continuous DAPTARI, PAUL, AND RAO vector measure G : Σ → X of bounded variation there exists g ∈ L ( µ, X ) suchthat G ( E ) = R E g ( t ) dµ ( t ) for all E ∈ Σ.Numerous characterizations are available in the literature for this property. Insome special cases the Banach spaces of vector valued functions can be expressedas the tensor product of classical spaces. The monographs [DU, LC] are somestandard references of tensor product of Banach spaces and related properties.We follow the notations from [DU], to define the injective ( projective ) tensorproduct of two Banach spaces X, Y , which will be denoted by X ⊗ ∨ ε Y ( X ⊗ ∧ π Y ).For z ∈ X ⊗ Y and let P ni =1 x i ⊗ y i be one such representation of z . Definethe following cross norms on X ⊗ Y , λ n X i =1 x i ⊗ y i ! = sup ((cid:13)(cid:13)(cid:13)(cid:13)(cid:13) n X i =1 φ ( x i ) y i (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) : φ ∈ S X ∗ ) and also, γ ( z ) = inf ( n X i =1 k x i kk y i k : z ≈ n X i =1 x i ⊗ y i ) . The completion of ( X ⊗ Y, λ ) (( X ⊗ Y, γ )) in B ( X × Y ) ∗ is said to be the injective(projective) tensor product of X and Y and is denoted by X ⊗ ∨ ε Y ( X ⊗ ∧ π Y ).It is well-known that (see [RR1, Pg. 114]) if one of the spaces X ∗ , Y ∗ has theRNP and one of them has Approximation property then ( X ⊗ ∨ ε Y ) ∗ ∼ = X ∗ ⊗ ∧ π Y ∗ and also L p ( ν, X ) ∗ ∼ = L q ( ν, X ∗ ), where p + q = 1, where 1 ≤ p < ∞ , q = ∞ when p = 1. Also L ( µ, X ) ∼ = L ( µ ) ⊗ ∧ π X and C ( K, X ) ∼ = C ( K ) ⊗ ∨ ε X , where X is a Banach space and K is a compact Hausdorff space. Here C ( K, X ) := { f : K → X : f is continuous on K } and the corresponding Banach space norm is k f k ∞ = sup K k f ( k ) k (see [LC]). Definition 2.4.
A Banach space X is said to be an L -predual space if X ∗ isisometrically isomorphic to L ( µ ) for some measure space (Ω , Σ , µ ).We refer Chapter 6 and 7 from the monograph [HEL] by Lacey for charac-terizations of these spaces and their properties. It is well-known that C ( K ) isan L -predual where K is compact Hausdorff. An L -predual space and its dualalways have the approximation property.If B is a closed, bounded, convex subset of a normed space, then ext ( B ) := { b ∈ B : b = x + x for some x , x ∈ B then b = x = x } , the set of all extremepoints of B . NIQUENESS OF HAHN-BANACH EXTENSION 5 Results on Banach spaces of vector valued functions
In this section we explore properties- U and SU in spaces of Bochner Integrablefunctions and Banach spaces of vector valued continuous functions. Both thesespaces can be interpreted as tensor products over suitable Banach spaces. The-orem 3.4, 3.6 are the main observations in this section. We first prove a formulafor the quotient of the spaces of Bochner Integrable functions. Proposition 3.1.
Let (Ω , Σ , µ ) be a probability space. Let X be a Banach spaceand Y be a subspace of X . Fix ≤ p < ∞ . Then L p ( µ, X ) /L p ( µ, Y ) is isometricto L p ( µ, X/Y ) .Proof. Let π : X → X/Y denote the quotient map.Define Φ : L p ( µ, X ) → L p ( µ, X/Y ) by Φ( f ) = π ◦ f .Φ is a bounded linear map and ker (Φ) = L p ( µ, Y ), which follows from R k π ( f ( w )) k p dµ ( w ) ≤ R k f ( w ) k p dµ ( w ).It remains to prove that for any f ∈ L p ( µ, X ), k f + L p ( µ, Y ) k = k Φ( f ) k . Itis clear that for any g ∈ L p ( µ, Y ), k f + g k p ≥ k Φ( f ) k p , we now prove the otherinequality.Using Bartle-Graves theorem (see [HR, Pg. 184]) we have a continuous crosssection map ρ : X/Y → X , such that ρ ( π ( x )) ∈ π ( x ). For g ∈ L p ( µ, X/Y ), wenow see that ρ ◦ g ∈ L p ( µ, X ) and Φ( ρ ◦ g ) = π ◦ ( ρ ◦ g ) = g . Thus Φ is onto.For any ε >
0, by taking λ = 1+ ε (as stated in [HR, Pg. 184]) we see that Φ is aquotient map, or in other words k f + L p ( µ, Y ) k = k Φ( f ) k i.e., L p ( µ, X ) /L p ( µ, Y )is isometric to L p ( µ, X/Y ). (cid:3) For the remaining part we may assume without loss of generality that µ is apurely non-atomic measure. Proposition 3.2.
Let (Ω , Σ , µ ) be a probability space. Let X be a Banach spaceand Y be a subspace of X . Suppose p, q ∈ (1 , ∞ ) with p + q = 1 and if p = 1 then q = ∞ . ( a ) If X ∗ has the RNP then L p ( µ, Y ) ⊥ = L q ( µ, Y ⊥ ) . ( b ) If L p ( µ, Y ) ⊥ = L q ( µ, Y ⊥ ) then Y ⊥ has the RNP.Proof. ( a ) . We always have L q ( µ, Y ⊥ ) ⊆ L p ( µ, Y ) ⊥ .To prove the other inclusion, we follow a similar arguments as stated in theproof of [DU, Theorem 1, Pg. 98]. DAPTARI, PAUL, AND RAO
RNP being hereditary property Y ⊥ has the RNP.Let g ∈ ( L p ( µ, X )) ∗ = L q ( µ, X ∗ ). Suppose g ∈ L p ( µ, Y ) ⊥ . Consider the vectormeasure G : Σ → X ∗ defined by G ( E )( x ) = R E g ( w )( x ) dµ ( w ). As in the proof of[DU, Theorem 1, Pg. 98], one has that G is a countably additive vector measureof bounded variation with respect to µ .We now claim that G is a Y ⊥ -valued measure.Let y ∈ Y and E ∈ Σ. Since yχ E ∈ L p ( µ, Y ), by the choice of g we have, Z E g ( w )( y ) dµ ( w ) = 0 . Hence G ( E )( y ) = 0 for all y . Thus G is a Y ⊥ -valued measure. Since Y ⊥ has theRNP, G has a Y ⊥ -valued derivative h . But by the uniqueness of the derivative g = h a.e. Thus g ∈ L q ( µ, Y ⊥ ).Identical arguments for p = 1 and the fact L ( µ, X ) ∗ = L ∞ ( µ, X ∗ ) lead to theproof for p = 1.( b ) . Now L q ( µ, Y ⊥ ) = L q ( µ, ( X/Y ) ∗ ) = L p ( µ, Y ) ⊥ .Also ( L p ( µ, X ) /L p ( µ, Y )) ∗ = L p ( µ, X/Y ) ∗ . Thus L p ( µ, X/Y ) ∗ = L q ( µ, Y ⊥ ).As µ is non atomic, by [DU, Theorem 1, Pg. 98], we have that ( X/Y ) ∗ = Y ⊥ has the RNP. (cid:3) We need the following observation in the subsequent discussion.
Remark 3.3.
Let µ be probability measure then for f ∈ X ∗ , k χ Ω f | L p ( µ,Y ) k = sup g ∈ B Lp ( µ,Y ) | Z Ω h g, χ Ω f i dµ | = sup g ∈ B Lp ( µ,Y ) | Z Ω h g, χ Ω f | Y i dµ |≤ sup g ∈ B Lp ( µ,Y ) Z Ω k f | Y kk g ( t ) k dµ ( t )= k f | Y k sup g ∈ B Lp ( µ,Y ) k g k ≤ k f | Y k . Again let ε > , there exist y ∈ B Y such that f | Y ( y ) ≥ k f | Y k − ε which implies R Ω h χ Ω y, χ Ω f i dµ ≥ R Ω χ Ω ( k f | Y k − ε ) dµ = k f | Y k − ε . Therefore k χ Ω f | L p ( µ,Y ) k ≥| R Ω h χ Ω y, χ Ω f i dµ | ≥ R Ω h χ Ω y, χ Ω f i dµ ≥ R Ω χ Ω ( k f | Y k − ε ) dµ = k f | Y k − ε and ε is arbitrary positive. So, we get k χ Ω f | L p ( µ,Y ) k ≥ k f | Y k . Hence k χ Ω f | L p ( µ,Y ) k = k f | Y k . NIQUENESS OF HAHN-BANACH EXTENSION 7
Theorem 3.4.
Let (Ω , Σ , µ ) be a probability space. Let Y be a subspace of aBanach space X such that X ∗ has RNP. Then, ( a ) Y has property- U in X if and only if L p ( µ, Y ) has property- U in L p ( µ, X ) , for < p < ∞ . ( b ) Y has property- SU in X if and only if L p ( µ, Y ) has property- SU in L p ( µ, X ) , for < p < ∞ .Proof. ( a ) . Let us assume that Y has property- U in X . It remains to showthat L p ( µ, Y ) ⊥ is Chebyshev in L p ( µ, X ) ∗ . Being a w ∗ -closed subspace of adual space, L p ( µ, Y ) ⊥ is proximinal. Now for any ϕ ∈ L p ( µ, X ) ∗ ∼ = L q ( µ, X ∗ ), d ( ϕ, L q ( µ, Y ⊥ )) = k d ( ϕ ( . ) , Y ⊥ ) k q , see [LC, Lemma 2.10]. Hence the uniquenessof the best approximation from ϕ to L q ( µ, Y ⊥ ) follows from the uniqueness ofbest approximation from ϕ ( ω ) to Y ⊥ for all ω ∈ Ω a.e. [ µ ].For the converse we follow the equivalence of (1) and (3) in [L, Theorem 2.1].Assume that f , f ∈ Y and f + f ∈ Y ⊥ . Since ( f + f )( y ) = 0 for all y ∈ Y we have χ Ω f + χ Ω f ∈ L p ( µ, Y ) ⊥ . It is clear that χ Ω f i ∈ L q ( µ, X ∗ ) and k χ Ω f i | L p ( µ,Y ) k = k χ Ω f i k = k f i k for i = 1 ,
2. Therefore χ Ω f , χ Ω f ∈ L p ( µ, Y ) .As L p ( µ, Y ) has property- U in L p ( µ, X ) we have χ Ω f + χ Ω f = 0, hence f + f =0. ( b ) . Let us recall the characterizations for property- SU in [O] (as stated inSection 1). Let us assume that Y has property- SU in X . It remains to provethat L q ( µ, Y ⊥ ) is an ideal in L q ( µ, X ∗ ). This follows from Proposition 3.2 andthe easy observation ˜ P : L q ( µ, X ∗ ) → L q ( µ, X ∗ ) defined by ˜ P ( g ) = P ◦ g , where P : X ∗ → X ∗ is a contractive projection with ker ( P ) = Y ⊥ .For the converse we follow the equivalence of (1) and (6) in [O, Pg. 1]. Assumethat f , f , f ∈ Y and f + f + f ∈ Y ⊥ . If L p ( µ, Y ) has property- SU in L p ( µ, X ) using similar arguments as stated above we can show f + f + f = 0,hence Y has property- SU in X . (cid:3) Remark 3.5. ( a ) Note that the converse of both the statements in Theo-rem 3.4 does not require the condition that the space X ∗ satisfy RNP. ( b ) We also note that the conclusion in Theorem 3.4 may not hold true for p = 1 . We now come to the spaces of type C ( K, X ), where X is a Banach space and K is a compact Hausdorff space. Let us recall C ( K, X ) ∼ = C ( K ) ⊗ ∨ ε X , the injectivetensor product of C ( K ) and X , as stated in Section 2. DAPTARI, PAUL, AND RAO
Theorem 3.6.
Let X be an L -predual space and Z be a Banach spaces suchthat Z ∗ has RNP. Let Y be a subspace of Z . Then, ( a ) Y have property- U in Z if and only if X ⊗ ∨ ε Y has property- U in X ⊗ ∨ ε Z . ( b ) Y have property- SU in Z if and only if X ⊗ ∨ ε Y has property- SU in X ⊗ ∨ ε Z .Proof. ( a ) . Let Y have property- U in Z .We now show that, ( X ⊗ ∨ ε Y ) ⊥ is Chebyshev in ( X ⊗ ∨ ε Z ) ∗ .Since Z ∗ has the RNP and X ∗ has approximation property (see [RR1,Pg. 73]), ( X ⊗ ∨ ε Z ) ∗ ∼ = X ∗ ⊗ ∧ π Z ∗ . On the other hand ( X ⊗ ∨ ε Y ) ⊥ ∼ =( X ⊗ ∨ ε Z/X ⊗ ∨ ε Y ) ∗ . Since X is an L -predual it follows from [MR, Corollary 18]that ( X ⊗ ∨ ε Z/X ⊗ ∨ ε Y ) ∼ = X ⊗ ∨ ε ( Z/Y ) and hence ( X ⊗ ∨ ε Y ) ⊥ ∼ = X ∗ ⊗ ∧ π Y ⊥ ∼ = L ( µ ) ⊗ ∧ π Y ⊥ ∼ = L ( µ, Y ⊥ ), the last identity follows from the fact that X ∗ ∼ = L ( µ )for some positive measure space (Ω , Σ , µ ) and the properties of projective ten-sor product. Now ( X ⊗ ∨ ε Z ) ∗ ∼ = L ( µ, Z ∗ ). Being a w ∗ -closed subspace of L ( µ, Z ∗ ), L ( µ, Y ⊥ ) is proximinal. It remains to prove that best approxima-tions in L ( µ, Y ⊥ ) are unique. This follows from similar arguments used in theproof of Theorem 3.4( a ).Conversely, let Y does not have property- U in Z . Let f ∈ Y ∗ and g , g be twonorm preserving extensions in Z ∗ . Let w ∈ S X ∗ , then w ⊗ g , w ⊗ g ∈ X ∗ ⊗ ∧ π Z ∗ (=( X ⊗ ∨ ε Z ) ∗ ) are two norm preserving extensions of w ⊗ f ∈ X ∗ ⊗ ∧ π Y ∗ (= ( X ⊗ ∨ ε Y ) ∗ ),which contradicts our assumption.( b ) . As the ideal property is stable under injective tensor product (see [RA,Lemma 2, pg-601]), the result is true for property- SU .Conversely assume that X ⊗ ∨ ε Y has property- SU in X ⊗ ∨ ε Z . Again we followthe equivalence for property- SU in [O].Let f , f , f ∈ Y such that f + f + f ∈ Y ⊥ . It remains to show that f + f + f = 0. Let g ∈ S X ∗ then g ⊗ f i ∈ ( X ⊗ ∨ ε Y ) , i = 1 , ,
3. In fact,for fixed i , k g ⊗ f i k = k g kk f i k = k f i k . Let ε >
0, there exist x ∈ S X suchthat | g ( x ) | > − ε . It is now clear that, k g ⊗ f i | X ⊗ ∨ ε Y k ≥ (1 − ε ) k f i k . Hence k g ⊗ f i | X ⊗ ∨ ε Y k = k g ⊗ f i k , i = 1 , , g ⊗ f + g ⊗ f + g ⊗ f ∈ ( X ⊗ ∨ ε Y ) ⊥ . In fact for simpletensor x ⊗ y , ( g ⊗ f + g ⊗ f + g ⊗ f )( x ⊗ y ) = 0. Hence, if D be the linear span NIQUENESS OF HAHN-BANACH EXTENSION 9 of all simple tensors in X ⊗ ∨ ε Y then ( g ⊗ f + g ⊗ f + g ⊗ f ) | D = 0. From thedensity of D we have f + f + f = 0. (cid:3) We do not know whether a similar conclusion derived in Theorem 3.6 alsoholds for the projective tensor product of the spaces.As an application of Theorem 3.6 we have the following for the subspaces oftype C ( K, Y ) in C ( K, X ). Corollary 3.7.
Let X be a Banach space such that X ∗ has RNP and Y bea subspace of X . Then Y has property- U (property- SU ) in X if and only if C ( K, Y ) has property- U (property- SU ) in C ( K, X ) . Property- ( U ) and ( SU ) in Quotient spaces The aim of this section is to discuss property- U and SU in the quotient spacesunder suitable assumptions on the underlying spaces. The first observation en-sures that if a subspace of a Banach space satisfies one of these properties thenit also transfers to the quotient spaces. Converse to this result is derived inTheorem 4.5 with suitable assumptions on the respective spaces. Theorem 4.1. ( a ) Let Z ⊆ Y ⊆ X be closed subspaces of X , where Y has property- SU in X , then Y /Z has property- SU in X/Z . ( b ) Let X be an L -predual. Let Y be a subspace of X which has property- SU and J be an M-ideal in Y . Then Y /J is an L -predual.Proof. ( a ) . We first show that
Y /Z has property- U in X/Z .Consider an element f ∈ Z ⊥ Y ∗ , with two norm preserving extensions, g, h ∈ Z ⊥ X ∗ .Now g | Y = h | Y , since Y has property- U we have g = h .For the remaining part, let P : X ∗∗ → Y ⊥⊥ be a contractive projection. Define P ′ : X ∗∗ /Z ⊥⊥ → Y ⊥⊥ /Z ⊥⊥ by P ′ ( π ( τ )) = π ( P ( τ )). For τ ∈ X ∗∗ , where π isthe quotient map on either quotient space. It is easy to see, P ′ is a contractiveprojection.( b ) . Suppose X ∗ = L ( µ ), for some measure µ and P : X ∗ → Y be a norm-1 projection for some subspace Y of X ∗ such that ker ( P ) = Y ⊥ . Being animage of a contraction of L ( µ ), Y is isometric with L ( ν ) for some measure ν ([HEL, Theorem 3, Chapter 6]). It is clear that Y is isomorphic with Y ∗ andlet P : Y ∗ → J ⊥ Y ∗ be an L -projection. Hence it follows that J ⊥ Y ∗ is isometric with L ( λ ). Now J ⊥ Y ∗ is isometrically isomorphic to ( Y /J ) ∗ , in other words Y /J is L -predual space. (cid:3) Proposition 4.2.
Let X be an L -predual. Let Y be a subspace of X which hasproperty- SU and J be an M-ideal in Y . Then J ⊥ Y ∗ is isometrically isomorphicwith J ⊥ X ∗ /Y ⊥ and J ⊥ X ∗ /Y ⊥ is isometric with L ( υ ) for some measure υ .Proof. Define φ : J ⊥ Y ∗ → J ⊥ X ∗ /Y ⊥ by the following. φ ( f ) = ˜ f + Y ⊥ , where ˜ f be theunique Hahn-Banach extension of f . Clearly k φ ( f ) k = k ˜ f + Y ⊥ k = k ˜ f | Y k = k f k and k φ − ( g + Y ⊥ ) k = k g | Y k = k g + Y ⊥ k . So J ⊥ Y ∗ isometrically isomorphic to J ⊥ X ∗ /Y ⊥ . And from above J ⊥ X ∗ /Y ⊥ is isometric with L ( υ ) for some measure υ . (cid:3) Proposition 4.3.
Let Z ⊆ Y ⊆ X . Suppose Y has property n.X.I.P. in X and Z is a proximal subspace of X . Then Y /Z has n.X/Z.I.P. in X/Z .Proof.
Let π denote the quotient map. Let { B ( π ( y i ) , r i ) } ni =1 be n -closed ballswith centers in Y /Z and let x ∈ X be such that k x + Z − π ( y i ) k = d ( x − y i , Z ) ≤ r i for 1 ≤ i ≤ n . Since Z is proximal in X , let z i ∈ Z be such that k x − y i − z i k = d ( x − y i , Z ) ≤ r i for 1 ≤ i ≤ n . Now for the collection { B ( y i − z i , r i ) } ni =1 of ballswith centers in Y , as Y has n.X.I.P. in X , for ε >
0, there is a y ∈ Y such that k y − y i − z i k ≤ r i + ε for 1 ≤ i ≤ n .Now π ( y ) ∈ Y /Z and k π ( y ) − π ( y i ) k = k π ( y − y i − z i ) k ≤ k y − y i − z i k ≤ r i + ε for 1 ≤ i ≤ n . Hence Y /Z has the n.X/Z.I.P in X/Z . (cid:3) Let Y , Z be subspaces of X such that Z ⊆ Y ⊆ X , the converse of Theorem 4.1is true when we are assuming Z is an M -ideal in X . Before going to proof of itwe first observe the following. Proposition 4.4.
Let Y , Z be subspaces of X such that Z ⊆ Y ⊆ X . Then ( Y /Z ) = { g ∈ Z ⊥ X ∗ : k g | Y k = k g k} .Proof. Let g ∈ Z ⊥ X ∗ and Λ ∈ ( X/Z ) ∗ (= Z ⊥ X ∗ ) is the corresponding element of g .Now k Λ | Y/Z k = sup k y + Z k≤ | Λ( y + Z ) | = sup k y + Z k≤ | g ( y ) | ≥ sup k y k≤ | g ( y ) | = k g | Y k . Again | Λ | Y/Z ( y + Z ) | = | Λ( y + Z ) | = | g ( y ) | = | g | Y ( y ) | = | g | Y ( y + z ) | ≤ k g | Y kk y + z k and this is true for all z ∈ Z , hence | Λ | Y/Z ( y + Z ) | ≤ k g | Y kk y + Z k . Therefore k Λ | Y/Z k = k g | Y k . As ( Y /Z ) = { Λ ∈ ( X/Z ) ∗ : k Λ | Y/Z k = k Λ k} and k Λ k = k g k we have ( Y /Z ) = { g ∈ Z ⊥ X ∗ : k g | Y k = k g k} . (cid:3) We now come to the main result of this section.
NIQUENESS OF HAHN-BANACH EXTENSION 11
Theorem 4.5.
Let
Y, Z be subspaces of X such that Z ⊆ Y ⊆ X . Suppose Z isan M-ideal in X and Y /Z has property- SU in X/Z . Then Y has property- SU in X .Proof. Let us consider the following cases to conclude the desired result. We firstshow that Y has property- U in X .Choose f , f ∈ Y with f + f ∈ Y ⊥ . Case 1: If f , f ∈ Z then as f + f ∈ Z ⊥ we have f + f = 0. Case 2:
Let f = p + q and f = p + q where p i ∈ Z and q i ∈ Z ⊥ .Now given that X ∗ = Z L ℓ Z ⊥ . Hence k f + f k = k p + p k + k q + q k . Claim 2.1: p + p = 0.In fact for any z ∈ Z , q i ( z ) = 0 and ( f + f )( z ) = 0, we have ( p + p )( z ) = 0.This follows that p + p ∈ Z ⊥ . Now from Case 1 it follows that p + p = 0. Claim 2.2: q + q = 0.Now 0 = k ( f + f ) | Y k = k ( p + p ) | Y k + k ( q + q ) | Y k = k ( q + q ) | Y k , thesecond identity follows from the fact that Z is an M-ideal in Y .Hence we have q + q ∈ Y ⊥ . We now show that q i ∈ ( Y /Z ) .Again since Z is an M-ideal in Y and k f i k = k f i | Y k , we have k p i k + k q i k = k p i | Y k + k q i | Y k , i = 1 ,
2. Now as k p i k = k p i | Z k ≤ k p i | Y k ≤ k p i k we have k q i k = k q i | Y k , which in other words q i ∈ Y Z ⊥ .Since Y /Z has property- SU in X/Z , we have q + q = 0 and hence f + f = 0. Case 3:
If one of f , f is in Z . Case 3.1:
Let f ∈ Z and f ∈ Z ⊥ . Then 0 = k ( f + f ) | Y k = k f | Y k + k f | Y k = k f k + k f k = k f + f k ⇒ f + f = 0. Case 3.2:
Let f ∈ Z and f = p + q where p ∈ Z and q ∈ Z ⊥ . Nowgiven that X ∗ = Z L ℓ Z ⊥ . Hence k f + f k = k f + p k + k q k . Claim 3.2.1: f + p = 0.In fact for any z ∈ Z , q ( z ) = 0 and ( f + f )( z ) = 0, we have ( f + p )( z ) = 0.This follows that f + p ∈ Z ⊥ . Now from Case 1 it follows that f + p = 0. Claim 3.2.2: q = 0.Now 0 = k ( f + f ) | Y k = k ( f + p ) | Y k + k q | Y k = k q | Y k , the second identityfollows from the fact that Z is an M-ideal in Y .Hence we have q ∈ Y ⊥ . We now show that q ∈ ( Y /Z ) . Again since Z is an M-ideal in Y and k f k = k f | Y k , we have k p k + k q k = k p | Y k + k q | Y k and k p k = k p | Y k . Now as k p k = k p | Z k ≤ k p | Y k ≤ k p k wehave k q k = k q | Y k , which in other words q ∈ Y Z ⊥ .Since Y /Z has property- SU in X/Z , we have q = 0 and hence f + f = f + p + q = 0.It remains to prove that Y is an ideal in X . Now by Theorem 1.3, as Y hasproperty- U , it is enough to check that Y has 3 .X.I.P. in X .Let { B ( y i , r ) } i =1 be 3 closed balls in X with centres in Y . Suppose T i B ( y i , r ) = ∅ . Choose x ∈ T i B ( y i , r ). Then it is clear that x + Z ∈ B ( y i + Z, r ),where the balls are now taken in the quotient space
X/Z .Now as
Y /Z has property- SU we have (cid:16)T i =1 B ( y i + Z, r + ε ) (cid:17) ∩ ( Y /Z ) = ∅ . Let us choose y + Z ∈ (cid:16)T i =1 B ( y i + Z, r + ε ) (cid:17) ∩ ( Y /Z ). In other words k y i − y + Z k ≤ r + ε for 1 ≤ i ≤
3. Get z i ∈ Z such that k y i − y − z i k < r + 2 ε .We now use a characterization of M-ideal stated in [HWW, Pg. 18]. Considerthe 3 balls { B ( y i − y , r + 2 ε ) } i =1 in X . Each ball intersects Z , as z i ∈ B ( y i − y , r + 2 ε ), and finally x − y ∈ T i =1 B ( y i − y , r + 2 ε ). Because Z is an M-idealin X , there must exists a z ∈ (cid:16)T i =1 B ( y i − y , r + 3 ε ) (cid:17) ∩ Z . Which concludesthat y + z ∈ T i =1 B ( y i , r + 3 ε ).As ε is arbitrary the result follows. (cid:3) Remark 4.6.
From the proof of Theorem 4.5, it is clear that if Z is a semi M-ideal in X and Y /Z has property- U in X/Z then Y has property- U in X . In fact,if Z is a semi M-ideal in X then k x ∗ k = k P x ∗ k + k x ∗ − P x ∗ k , where x ∗ ∈ X ∗ and P : X ∗ → Y ⊥ is a projection. This decomposition leads to that, Z has property- U in X and hence [L, Theorem 2.1, Pg. 99] applies for Z . Hence all the cases inTheorem 4.5 devoted to prove Y has property- U in X can be fitted for this casealso. Property- U and SU for spaces of type L -predual In this section we discuss property- U and SU for ideals and other subspaces of L -preduals. An easy argument concerning property- U and the 3 .X.I.P. ensurethe following. Theorem 5.1.
Let X be a Banach space. Y , Z are subspaces of X , and Z = X .If Y has property- U ( SU ) in span { Y ∪ { z }} , for all z ∈ Z then Y has property- U ( SU ) in X . NIQUENESS OF HAHN-BANACH EXTENSION 13
Using characterization of L -preduals discussed in [HEL, Chapter 7] we nowhave a similar result for ideals in Theorem 5.1. Theorem 5.2.
Let X be an L -predual space. Suppose Z ⊆ X is a dense subspaceand let Y ⊆ X be a closed subspace such that for any z ∈ Z \ Y , Y ⊆ span { z, Y } is an ideal if and only if Y is an ideal in X . Let us recall that a subspace W of a Banach space X is said to be 1-complemented if there exists a linear onto projection P : X → W with k P k = 1.It is clear that if X is a Banach space then X ⊥⊥ ( ∼ = X ∗∗ ) is 1-complemented in X ∗∗∗∗ . This leads to conclude that, Y is an ideal in X if and only if Y ⊥⊥ is anideal in X ∗∗ .Now observe that for a Banach space X of type L -predual or M-embedded, X ∗ is an L -summand in X ∗∗∗ . Hence we have the following. Theorem 5.3.
Let X be an L -predual (or an M -embedded) space, a finite co-dimensional subspace Y of X has property- U if and only if Y ⊥⊥ has property- U in X ∗∗ .Proof. The result follows from the above discussion and the fact that for any finitelinear functionals ( f i ) ni =1 in a dual space X ∗ , ( T i ker ( f i )) ⊥⊥ = T i ker ( ˜ f i ), where˜ f i ’s are the canonical images of f i ’s in X ∗∗∗ . Hence Y ⊥ = sp { f i : 1 ≤ i ≤ n } , asa Chebyshev subspace of X ∗ , continues to be Chebyshev in X ∗∗∗ . (cid:3) Hence we have the following.
Theorem 5.4.
Let X be an L -predual (or an M -embedded) space and Y be afinite co-dimensional subspace of X . Then Y has property- SU in X if and onlyif Y ⊥⊥ has property- SU in X ∗∗ . We now show that the condition of finite codimensionality can not be omittedin Theorem 5.4.
Example 5.5.
There exists a one-dimensional subspace of C [0 , which hasproperty- U but does not have property- U in C [0 , ∗∗ . Let f ( t ) = t if ≤ t ≤ − t ) if ≤ t ≤ ,f is the smooth point of C [0 , and δ ∈ M [0 , is the unique linear functionalwhich attains its norm at f ∈ C [0 , . Hence Y = span { f } has property- U in C [0 , . Now suppose Y has property- U in C [0 , ∗∗ , from the assumption, it is easy tosee that δ is a point of continuity of identity map I : ( B X ∗ , w ∗ ) → ( B X ∗ , w ) ,where X = C [0 , . Let t n → , t n = , for all n ∈ N , hence f ( t n ) → f ( ) , forall f ∈ C [0 , , i.e δ t n → δ in weak* topology. Since identity map is weak*-weakcontinuous at the point δ , δ t n → δ in weak topology, i.e δ ∈ conv w { δ t n : n ∈ N } , hence δ ∈ conv k . k { δ t n : n ∈ N } . Hence there exists a sequence ( µ n ) ⊆ conv { δ t n : n ∈ N } such that µ n → δ . Thus there exists n ∈ N such that k µ n − δ k < ε , hence k λδ t n + (1 − λ ) δ t n − δ k < ε , for some λ ∈ [0 , and n , n ∈ N , but k λδ t n + (1 − λ ) δ t n − δ k = 2 , hence contradiction. Let X be a Banach space. It is known that for any two M -ideals, Y, Z ⊆ X ,the sum space Y + Z is a closed M -ideal in X . See [HWW, Proposition I.1.11.].We next show that the validity of the same question for property- SU , determinesa Hilbert space in the class of Banach spaces which are smooth. Note that aHilbert space of dimension bigger than 1, has no non-trivial M -ideals, where as,in any Hilbert space, closed subspaces have property- SU Theorem 5.6.
Let X be a smooth, Banach space of dimension > . Supposefor every Y, Z ⊆ X subspaces having property- SU in X and with sum Y + Z isclosed, Y + Z also has property- SU . Then X is isometric to a Hilbert space.Proof. We note that since X is a smooth space, any one dimensional subspaceof X has property- SU in X . Thus by our hypothesis any two dimensional spacehas property- SU , so it has property- U and is the range of a projection of norm-1.Now by induction, every finite dimensional subspace has the property- SU . Thusany finite dimensional subspace is the range of a projection of norm one. Now byKakutani’s theorem (See [HEL, Pg.150]) we get that X is a Hilbert space. (cid:3) A few examples
In this section several examples are given satisfying properties U and SU . Ourfirst example characterizes finite co-dimensional subspaces of c with property- U .Let us recall that, f ∈ c ∗ where ker ( f ) is proximinal if and only if f has onlyfinite support and if Y ⊆ X is a proximinal subspace of finite co-dimensional and Y ⊆ Z ⊆ X then Z is proximinal in X (see [IS]). Proposition 6.1.
Let Y ⊆ c be a proximinal subspace of finite codimension. Y has property- U in c if and only if c /Y is isometric to a subspace, with property- U , of ℓ ∞ ( k ) for some integer k ≤ dim ( c /Y ) . NIQUENESS OF HAHN-BANACH EXTENSION 15
Proof.
Since Y is proximinal and finite codimensional, one has, Y ⊥ = span { f , ...f m } for some f i ∈ ℓ , each having only finitely many non-zero coordi-nates. Thus we may assume for some k > f i ( j ) = 0 for j > k and 1 ≤ i ≤ m .It is easy to see now, Y = F L ∞ { y ∈ c : y ( j ) = 0 for all j ≤ k } , for some finitedimensional subspace of c . By hypothesis F is a subspace with property- U of ℓ ∞ ( k ). Since c = ℓ ∞ ( k ) L ∞ { x ∈ c : x ( j ) = 0 for all j ≤ k } . Y is a subspacewith property- U of c . (cid:3) In the next result, we show the dimension n plays an important role for havingproperty- U of the subspace Y = ker { (1 , , . . . , } ⊆ ( R n , k . k ∞ ). Proposition 6.2.
Let Y = { ( y , y , . . . , y n ) : y + y + · · · + y n = 0 } be a subspaceof ( R n , k . k ∞ ) . Then ( a ) Y has property- U in ( R n , k . k ∞ ) , when n is odd. ( b ) Y does not have property- U in ( R n , k . k ∞ ) , when n is even.Proof. Case 1:
When n is odd. Let us assume n = 2 k + 1.It remains to prove that span { } is a Chebyshev subspace of ( R n , k . k ). Thenfor any ( x , x , . . . , x n ) ∈ R n , define α , α , . . . , α n such that { x i : 1 ≤ i ≤ n } = { α i : 1 ≤ i ≤ n } and α ≤ α ≤ . . . ≤ α n . Then ( α , α , . . . , α ) ∈ R n is theunique best approximation of ( x , x , . . . , x n ), where α = α k +1 . Case 2:
When n is even. Let us assume n = 2 k .Suppose that ( x , x , . . . , x n ) and ( α , α , . . . , α n ) stand with the similar mean-ing as Case 1 . Then for any β ∈ [ α k , α k +1 ], ( β, β, . . . , β ) ∈ R n is a best approx-imation from ( x , x , . . . , x n ) to span { } . (cid:3) An obvious question is to consider property- SU for the subspace Y in( R , k . k ∞ ). It was shown in [O] that when n = 3, Y fails to have property- SU in ( R , k . k ∞ ), where Y = { ( x, y, z ) ∈ R : x + y + z = 0 } ⊆ ( R , k . k ∞ ). Wederive a similar conclusion in ( R , k . k ). Proposition 6.3.
Let Y = { ( x, y, z ) ∈ R : x + y + z = 0 } ⊆ ( R , k . k ) , then Y does not have property- SU in ( R , k . k ) .Proof. It is clear that Y ⊥ = span { (1 , , } is a Chebyshev subspace of ( R , k . k ∞ ).Indeed, for ( x, y, z ) ∈ R , d (( x, y, z ) , Y ⊥ ) = x ∨ y ∨ z − x ∧ y ∧ z and unique nearest pointof ( x, y, z ) from Y ⊥ is x ∨ y ∨ z + x ∧ y ∧ z (1 , , It remains to prove that Y is not a linear subspace. one can check ext ( B Y ) = {± ( , − , , ± ( , , − ) , ± (0 , , − ) } .Let f = ( l, m, n ) ∈ Y . Then k f k = | l | ∨ | m | ∨ | n | = max ( x,y,z ) ∈ ext ( B Y ) | lx + my + nz | = k f | Y k . We consider the following to evaluate Y .It is known that | l | ∨ | m | = | l + m | + | l − m | , for any two real scalars l, m .Let the maximum is attained at ( x, y, z ) = ± ( , − , | l | ∨ | m | ∨ | n | = | l − m | = | l | ∨ | m | − | l + m | . Case 1: If | l | ∨ | m | ≤ | n | then | n | = | l | ∨ | m | − | l + m | ⇒ l + m = 0 . Case 2: If | l | ∨ | m | ≥ | n | then | l | ∨ | m | = | l | ∨ | m | − | l + m | ⇒ l + m = 0.Similarly, if the maximum is attained at ( ± ( , , − ) and ± (0 , , − ), we get l + n = 0 and m + n = 0 respectively.Hence Y = { ( a, b, z ) , ( a, z, b ) , ( z, a, b ) : a + b = 0 , z ∈ R } . Choose u =( − , , ∈ Y & v = (0 , , − ∈ Y but u + v = ( − , , − / ∈ Y . Therefore Y does not have property- SU in ( R , k . k ∞ ). (cid:3) Our next Theorem identifies hyperplanes Y in c with property- SU . Theorem 6.4.
Let ( a n ) ∈ S ℓ and sup n ∈ N | a n | > . Then ker { ( a n ) } ⊆ c hasproperty- SU .Proof. Let us first observe that ker ( a n ) is 1-complemented in c (see [BP, Theo-rem 6.1]) hence is an ideal in c .Let X = c , Y = ker { ( a n ) } . Since sup n ∈ N | a n | >
12 , there exist N ∈ R such that | a N | > . Note that, the N is unique, as ( a n ) ∈ S ℓ . Let G = ker { e N } . It is clearthat G is the complement of Y ⊥ = span { ( a n ) } in ℓ .It is enough to prove that G = Y .Let Y = ker( a n ) ⊆ c , where ( a n ) ∈ S ℓ , | a N | > and G = ker( e N ) ⊆ ℓ . Notethat for ( b n ) ∈ G , b N = 0. NIQUENESS OF HAHN-BANACH EXTENSION 17
Now k ( b n ) | Y k = sup ( x n ) ∈ B Y (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X n ∈ N b n x n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ X n ∈ N | b n | . For m > N , define x m by, x mn = sign ( b n ) if n = N, & n ≤ m, − P mn =1 ,n = N ( a n sign ( b n )) a N if n = N, n > m. It is clear that x m ∈ B Y and ( b n )( x m ) = m X n =1 | b n | .Now | ( b n )( x m ) | → X n ∈ N | b n | as m → ∞ and hence k ( b n ) | Y k = X n ∈ N | b n | = k ( b n ) k .Let ( b n ) ∈ Y and δ > X n ∈ N \{ N } | a n | < | a N | − δ . Claim :
If ( x n ) ∈ B Y then | x N | < − δ | a N | .Suppose not, i.e | x N | ≥ − δ | a N | , hence we have X n ∈ N \{ N } | a n | ≥ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X n ∈ N \{ N } a n x n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = | x N a N | ≥ | a N | (1 − δ | a N | ) = | a N | − δ contradiction. Now k ( b n ) k = X n ∈ N | b n | = k ( b n ) | Y k = sup ( x n ) ∈ B Y X n ∈ N | b n x n | ≤ X n ∈ N \{ N } | b n | + (1 − δ | a N | ) | b N | implies δ | a N | | b N | ≤ b N = 0. Hence ( b n ) ∈ G .This completes the proof. (cid:3) Example 6.5.
Let Y = { ( x, y, z ) : x + y + 6 z = 0 } and Y = { ( x, y,
0) : x, y ∈ R } be two subspaces of X = ( R , k . k ∞ ) . We know that both Y and Y have property- SU in X . Note that Y has property- SU in X and Y is an M-summand. It isclear that Y ∩ Y = { ( x, y,
0) : x + y = 0 } (= Z say) where Z ⊥ = { ( s, s, r ) : r, s ∈ R } , is not a Chebyshev subspace in X ∗ . Hence Y ∩ Y does not have property- U in X . We now conclude from the above example that, the assertion in Theo-rem 6.4 is not sufficient to conclude property- SU for finite codimensional ( >
1) subspaces of c . Note that c ∼ = ( R n , k . k ∞ ) L ℓ ∞ c and so its dual is( R n , k . k ) L ℓ ℓ , for any natural number n . Hence from Example 6.5 it is clearthat ker ( , , , , , . . . ) T ker ( e ) does not have property- U in c .The results in Theorem 6.4, [BP, Theorem 6.1] and the subsequent discussionlead to the following conclusion. Corollary 6.6.
Let ( a n ) ∈ S ℓ then ker ( a n ) has property- SU in c if and only if sup n ∈ N | a n | > .Proof. It remains to prove that the condition is necessary.Since c is an M -ideal in its bidual ℓ ∞ and Y = ker( a n ) is an ideal in c , by [RA,Proposition 2, Pg. 605] Y is an 1-complemented subspace. Hence sup n ∈ N | a n | ≥ n ∈ N | a n | = then there exist N ∈ N such that | a N | = . Case 1:
When a N = .Now d ( e N , Y ⊥ ) = inf α ∈ R {| − α | + | α | P n ∈ N ,n = N | a n |} = inf α ∈ R {| − α | + | α | } .Let d α = | − α | + | α | . Case 1.a:
Let 0 ≤ α ≤ d α = 1 − α + α = 1. Case 1.b:
Let α > d α = α − α = α − > Case 1.c:
Let α < d α = 1 − α − α = 1 − α > d ( e N , Y ⊥ ) = 1 and for any α ∈ [0 , α.a n ) ∞ n =1 is the best approxi-mation from e N to Y ⊥ . Case 2:
When a N = − .Now d ( e N , Y ⊥ ) = inf α ∈ R {| α | + | α | P n ∈ N ,n = N | a n |} = inf α ∈ R {| α | + | α | } .Let d α = | α | + | α | . Case 2.a:
Let α > d α = 1 + α + α = 1 + α > Case 2.b:
Let − ≤ α ≤ d α = α + 1 − α = 1. Case 2.c:
Let α < − d α = − − α − α = − − α > d ( e N , Y ⊥ ) = 1, ( α.a n ) ∞ n =1 is the best approximation of e N from Y ⊥ ,for α ∈ [ − , Y ⊥ can not be a Chebyshev subspace of ℓ . (cid:3) Acknowledgements.
The research of the second author is supported by Sci-ence and Engineering Research Board, India, Award No. MTR/ 2017/ 000061.Corresponding author would like to thank SERB for their financial support.
NIQUENESS OF HAHN-BANACH EXTENSION 19
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