# k-Smooth Points in Some Banach Spaces

Authors: A. Saleh Hamarsheh
Journal: International Journal of Mathematics and Mathematical Sciences
 (2015)

---

## Abstract

We characterize thek-smooth points in some Banach spaces. We will deal with injective tensor product, the Bochner space L∞(μ,X) of (equivalence classes of) μ-essentially bounded measurable X-valued functions, and direct sums of Banach spaces.


---

## Body

## 1. Introduction

For a unit vectorx in a Banach space X, consider the state space Sx={x∗∈X∗:x∗=1=x∗(x)}. The point x is a smooth point if Sx consists exactly of one point. The set of all smooth points is denoted by smoothB(X). Smooth points are important tools in the study of the geometry of Banach spaces. For two Banach spaces X,Y Heinrich, [1], gave a description of smooth points of the unit ball in the space K(X,Y) of compact operators from X into Y. The research then turned to the space L(X,Y) of bounded operators. Kittaneh and Younis, [2], were the first to deal with this problem. They characterized smooth points in L(l2). Their result was then generalized in [3] to the space L(lp); 1<p<∞. For smooth points in L(lp,X) see [4, 5]. In [6] Werner gave a description of smooth points in L(X,Y) under some conditions on X and Y. Smooth points in certain vector valued function spaces were given in [7].In [8] the authors generalize the notion of smoothness by calling a unit vector x in a Banach space X a k-smoothpoint, ora multismooth point of orderk if Sx has exactly k linearly independent vectors, equivalently, if dim⁡(spSx)=k. For a natural number k, the set of k-smooth points in X is denoted by k-smoothB(X). Note that Sx is a weak∗-compact convex set and hence it is easy to see that x∈k-smoothB(X) if and only if dim⁡(spextSx)=k. Multismoothness in Banach spaces was extensively studied by Lin and Rao in [9]. In paricular, they showed that, in a Banach space of finite dimension k, any k-smooth point is unitary and hence a strongly extreme point. The aim of this paper is to characterize multismoothness in some Banach spaces. Indeed, we will deal with injective tensor product, the Bochner space L∞(μ,X) of (equivalence classes of) μ-essentially bounded measurable X-valued functions, and direct sums of Banach spaces.The set of all extreme points of the unit ball of a Banach spaceX is denoted by ext⁡B(X).


## 2. Multismoothness in Injective Tensor Products

In [9] the authors characterized multismoothness in the completed injective tensor product X⊗ϵY when X is an L1-predual space and Y is a smooth Banach space. We generalize their result toany Banach space Y. Note that X⊗ϵY=KW∗(Y∗,X) whenever either of X and Y has the approximation property. Here KW∗(Y∗,X) is the space of all compact and weak∗ to weakly continuous operators from Y∗ to X, endowed with usual operator norm.Recall that ifx∗∈X∗ and y∗∈Y∗ then x∗⊗y∗∈(KW∗(Y∗,X))∗ as follows:(1)ForT∈KW∗Y∗,X,x∗⊗y∗,T=y∗,T∗x∗.Theorem 1.
LetX be an L1-predual space and Y any Banach space. Let T∈X⊗ϵY=KW∗(Y∗,X) with T=1. Then T is a multismooth point of finite order in X⊗ϵY if and only if T∗ attains its norm at exactly finitely many independent vectors, say at x1∗,x2∗,…,xr∗∈ext⁡B(X∗) such that T∗xi∗ is a multismooth point of finite order in Y; i=1,2,…,r.
In this case the order of smoothness ofT is k=m1+m2+⋯+mr, where mi is the order of smoothness of T∗xi∗, i=1,2,…,r.Proof.
One can easily prove that ifT∗ attains its norm at infinitely many independent vectors in ext⁡B(X∗) then T∈X⊗ϵY is not a multismooth point of any finite order. The same conclusion will be obtained if T∗ attains its norm at some x∗∈ext⁡B(X∗) and T∗x∗ is not a multismooth point of finite order in Y. So, suppose T∗ attains its norm at exactly finitely many independent vectors, say at x1∗,x2∗,…,xr∗∈ext⁡B(X∗) such that T∗xi∗ is a multismooth point of finite order mi in Y; i=1,2,…,r and let k=m1+m2+⋯+mr. We will prove that T is a multismooth point of order k in X⊗ϵY.
For eachi there are exactly mi linearly independent functionals in ext⁡B(Y∗) attaining their norm at T∗xi∗, say yi,1∗,yi,2∗,…,yi,mi∗. Since X is an L1-predual space, then there are distinct atoms Ai with xi∗=±(1/μAi)χAi. Set Fi,t=xi∗⊗yi,t∗, where 1≤i≤r and 1≤t≤mi. These are k extreme functionals in (X⊗ϵY)∗ attaining their norms at T:(2)xi∗⊗yi,t∗,T=yi,t∗,T∗xi∗=1=T∗xi∗.
We claim that theFi,t’s are linearly independent. Indeed, if ∑i=1r‍∑t=1mi‍ai,tFi,t=0 for some scalars ai,t then ∑i=1r‍xi∗⊗(∑t=1mi‍ai,tyi,t∗)=0. But since the xi∗ correspond to distinct atoms then ∑t=1mi‍ai,tyi,t∗=0 for all i=1,2,…,r. Since {yi,t∗:t=1,2,…,mi} are linearly independent, then ai,t=0, ∀1≤i≤r,1≤t≤mi.
Finally, LetF∈(X⊗ϵY)∗ with F=1=F(T). We will show that F∈sp{Fi,t:1≤i≤r,1≤t≤mi}. We can suppose that F∈ext⁡B((X⊗ϵY)∗) (see Section 1). Now, by a result of Ruess and Stegall [10], F=x∗⊗y∗, where x∗∈ext⁡B(X∗) and y∗∈ext⁡B(Y∗). Then(3)FT=x∗⊗y∗,T=x∗,Ty∗=T∗x∗,y∗=1.
So,T∗x∗=1 and hence x∗∈sp{x1∗,x2∗,…,xr∗}. But since the xi∗ correspond to distinct atoms then x∗=xi∗ for some i=1,2,…,r and consequently y∗∈sp{yi,1∗,yi,2∗,…,yi,mi∗}. Therefore x∗⊗y∗=xi∗⊗y∗∈sp{xi∗⊗yi,1∗,xi∗⊗yi,2∗,…,xi∗⊗yi,mi∗}. Hence F∈sp{Fi,t:1≤i≤r,1≤t≤mi}. This proves that the Fi,t’s form a maximal linearly independent set in ST.
ThereforeT∈k-smoothB(X⊗ϵY).As a corollary, we get the following.Corollary 2 (see [9]).
LetX be an L1-predual space and Y a smooth Banach space. Let T∈X⊗ϵY=KW∗(Y∗,X) with T=1. Then T is a multismooth point of finite order k in X⊗ϵY if and only if  T∗ attains its norm at exactly k independent vectors in ext⁡B(X∗).Open Problem. For Banach spaces X and Y let T∈K(X,Y) with T=1. Is it true that T is a multismooth point of finite order k in K(X,Y) if and only if T∗ attains its norm at only finitely many independent vectors, say at y1∗,y2∗,…,yr∗∈ext⁡B(Y∗) such that each T∗yi∗ is a multismooth point of finite order, say mi, in X∗, where k=m1+m2+⋯+mr?Theorem1 above tells us that the answer is yes when Y is an L1-predual space, since in this case K(X,Y)=Y⊗ϵX∗=KW∗(X∗∗,Y).


## 3. Multismoothness in Bochner Spaces

LetX be a Banach space. In this section we discuss multismoothness in the Bochner space L∞(μ,X) of (equivalence classes of) μ-essentially bounded measurable X-valued functions. Recall that measurable functions are constants on the atoms.Lemma 3.
Letf∈L∞(μ,X) and suppose that there is no μ-atom A such that f(A)=1. Then f is not a multismooth point of any finite order.Proof.
Fixr∈N. We will prove that f is not a multismooth point of order r. Write Ω=∪i=1r+1Ei, where Ei’s are disjoint measurable sets of positive measure, with sup⁡{f(t):t∈Ei}=1. For 1≤i≤r+1, define(4)fit=ft,ift∈Ei0,ift∉Ei.Then f=f1+f2+⋯+fr+1 and fi=sup⁡{f(t):t∈Ei}=1. Moreover, fi±fj≤1 for all i≠j. This shows that f is not a smooth point of any order m≤r; see [8]. In particular, f is not a multismooth point of order r.Lemma 4.
Letf∈L∞(μ,X) and suppose that there are exactly nμ-atoms A1,A2,…,An such that f(Aj)=1. If sup⁡{f(t):t∉∪j=1nAj}=1, then f is not a multismooth point of any finite order.Proof.
Fixr∈N and write Ω∖∪i=1nAi as a disjoint union of r measurable sets Ei of positive measure and proceed as in the above proof.Theorem 5.
Letf∈L∞(μ,X) with f=1. Then f is a multismooth point of finite order if and only if there are exactly finitely many distinct atoms A1,A2,…,Ar such that f(Aj)=1,j=1,2,…,r and sup⁡{f(t):t∉∪j=1∞Aj}<1 and each f(Aj) is a multismooth point of finite order, say mj, in X. In this case the order of smoothness of f is k=m1+m2+⋯+mr.Proof.
The above two lemmas prove the “only if” part. For the converse, we choose, for anyj=1,2,…,r, linearly independent set {xj,1∗,xj,2∗,…,xj,mj∗}⊆Sf(Aj). So, xj,i∗=1=xj,i∗,f(Aj). For 1≤j≤r and 1≤i≤mj define Fj,i∈(L∞(μ,X))∗ by Fj,i(g)=xj,i∗,g(Aj). These are m1+m2+⋯+mr=k linear functionals attaining their norm at f. Suppose that ∑j=1r‍∑i=1mj‍aj,iFj,i=0 for some scalars aj,i. Then ∑j=1r‍∑i=1mj‍aj,ixj,i∗,g(Aj)=0, ∀g∈L∞(μ,X). Choosing gj(t)=x; if t∈Aj and gj(t)=0; if t∉Aj, where x∈X, we get (∑i=1mj‍aj,ixj,i∗)(x)=0 for all x∈X and j=1,2,…,r. Consequently, ∑i=1mj‍aj,ixj,i∗=0 for all j=1,2,…,r. Since {xj,1∗,xj,2∗,…,xj,mj∗}⊆Sf(Aj) is linearly independent, then aj,i=0 for all j=1,2,…,r and i=1,2,…,mj. Therefore, the Fj,i’s are linearly independent.
For an atomA and g∈L∞(μ,X) let gA=g(A)χA∈L∞(μ,X). We will prove that if F∈(L∞(μ,X))∗ with F=1=F(f) then F(g)=∑j=1r‍F(gAj) for all g∈L∞(μ,X). Without loss of generality, say g=1. We claim that there is ϵ>0 such that f±ϵg≤1. Indeed, if such ϵ does not exist, we would have a sequence (tk) outside ∪j=1rAj such that f(tk)±1/kg(tk)>1. But then(5)1<ftk±1kgtk≤ftk+1kgtk≤ftk+1k.
Hence,f(tk)>1-1/k for all k, a contradiction to our assumption. Thus, |F(f±ϵg)|≤1 and therefore F(g)=0 for all g∈L∞(μ,X) such that g(Aj)=0, ∀1≤j≤r. This proves that F(g)=∑j=1r‍F(gAj) for all g∈L∞(μ,X). Let Z=⊕j=1rX(l∞-sum) and let E={h∈Z∗:h=1=h(f(A1),f(A2),…,f(Ar))}. Then by Krein-Millman Theorem we have E=coext⁡E¯ with ext⁡E⊆ext⁡B(Z∗). So, any h∈ext⁡E has the form h=(0,0,…,0,x∗,0,…,0) for some x∗∈ext⁡B(X∗). Note that if, for example, h1,h2∈ext⁡E has the form h1=(x∗,0,0,…,0),h2=(y∗,0,0,…,0), then x∗,f(A1)=y∗,f(A1)=1. Since fA1∈m1-smoothB(X), then any m1+1h’s of the above form must be linearly dependent. Consequently, any m1+m2+⋯+mr+1=k+1 elements h∈E must be linearly dependent.
Now, letF1,F2,…,Fk+1∈L∞(μ,X)∗ such that Fi=1=Fi(f). We will prove that the Fi’s are linearly dependent. By the argument above we see that Fi(g)=∑j=1r‍Fi(gAj) for all g∈L∞(μ,X). For 1≤i≤k+1 define hi∈Z∗=⊕j=1rX∗(l1-sum) by hi(x1,x2,…,xr)=Fi(∑j=1r‍hxj), where hxj=xjχAj. Then hi∈Z∗ with hi=1=hi(f(A1),f(A2),…,f(Ar))=∑j=1r‍Fi(fAj)=Fi(f) so that hi∈E for all i=1,2,…,k+1 and therefore {h1,h2,…,hk+1} is linearly dependent. Hence, there are scalars a1,a2,…,ak+1, not all zeros, with ∑i=1k+1‍aihi=0. The proof will be complete if we show that ∑i=1k+1‍aiFi=0. Indeed, if g∈L∞(μ,X) then ∑i=1k+1‍aihi(g(A1),g(A2),…,g(Ar))=∑i=1k+1‍aiFi(∑j=1r‍gAj)=∑i=1k+1‍aiFi(g)=0. This shows that f∈k-smoothB(L∞(μ,X)) and completes the proof of the “if” part.


## 4. Multismoothness in Direct Sums of Banach Spaces

Lin and Rao characterized in [9] multisoothness in l∞-direct sums and proved the following theorem.Theorem 6 (see [9]).
Let{Xi:i∈I} be an infinite family of nonzero Banach spaces. Let X=⊕∞Xi and let x=(xi) be a unit vector in X. Let I1={i∈I:xi<1}, let I2={i∈I:xi=1andxiisamultismoothpointoffiniteorder}, and let I3=I∖(I1∪I2). Then, x is a multismooth point of finite order if and only if I3=∅,I2 is finite, and supi∈I1xi<1. In this case the order of smoothness of x is k=∑i∈I2‍mi, where mi is the order of smoothness of xi in Xi,i∈I2.In this section we deal withl1-direct sums. Indeed we prove the following result.Theorem 7.
Let{Xi:i∈I} be any family of nonzero Banach spaces. Let X=⊕1Xi and let x=(xi) be a unit vector in X. Let I1={i∈I:xi=0anddim⁡⁡Xi<∞}, I2={i∈I:xi=0anddim⁡⁡Xi=∞}, I3={i∈I:xi≠0andxi/xi∈smoothB(Xi)}, I4={i∈I:xi≠0andxi/xi∈r-smoothB(Xi)forsomenaturalnumberr≥2}, and I5=I∖∪i=14Ii. Then x is a multismooth point of finite order if and only if I2=I5=∅ and I1∪I4 is finite. In this case the order of smoothness of x is k=∑i∈I1‍dim⁡Xi+∑i∈I4‍mi-|I4|+1, where mi is the order of smoothness of xi in Xi,i∈I4 and |I4| is the number of elements in I4.For the sake of completeness, let us first state and prove the characterization of smoothness.Theorem 8.
Let{Xi:i∈I} be any family of nonzero Banach spaces. Let X=⊕1Xi and let x=(xi) be a unit vector in X. Then x is a smooth point if and only if for any i∈I,  xi≠0andxi/xi∈smoothB(Xi).Proof.
Supposej∈I and xj=0, or xj≠0 but xi/xi is not a smooth point in Xj. Then there are distinct unit functionals y0∗,z0∗∈Xj∗ with y0∗(xj)=z0∗(xj)=xj. For i≠j, choose any unit functional xi∗∈Xi∗ with xi∗(xi)=xi. Let y∗=(yi∗) and z∗=(zi∗) where yi∗=zi∗=xi∗,i≠j,yj∗=y0∗ and zj∗=z0∗. Clearly, y∗ and z∗ are distinct elements in Sx. So x is not a smooth point.
Conversely, supposexi∗∈Xi∗ is the unique element in Sxi/∥xi∥,i∈I. Let x∗=(xi∗). Then x∗∈X∗ with x∗=supi∈Ixi∗=1 and x∗(x)=∑i∈I‍xi∗(xi)=∑i∈I‍xi=x=1. Now if y∗=(yi∗)∈Sx then ∑i∈I‍yi∗(xi)=1=∑i∈I‍xi. Since yi∗(xi)≤xi then yi∗(xi/xi)=1 for all i∈I and hence y∗=x∗.Lemma 9.
LetY and Z be nonzero Banach spaces and X=Y⊕1Z. Let x=(0,z)∈X where z∈k-smoothB(Z). If dim⁡⁡Y=n<∞ then x∈(k+n)-smoothB(X).Proof.
Letz1∗,z2∗,…,zk∗∈Sz be linearly independent and choose n linearly independent unit functionals y1∗,y2∗,…,yn∗∈Y∗. Let (6)xi∗=0,zi∗,if1≤i≤k,yi-k∗,z1∗,ifk+1≤i≤k+n.Clearly xi∗∈Sx; 1≤i≤k+n. They are linearly independent. Indeed if ∑i=1k+n‍aixi∗=0 then ∑i=1n‍ak+iyi∗=0 and ∑i=1k‍aizi∗+(∑i=1n‍ak+i)z1∗=0. Since {y1∗,y2∗,…,yn∗} is lineraly independent then ak+i=0; 1≤i≤n and thus ∑i=1k‍aizi∗=0. But {z1∗,z2∗,…,zk∗} is also lineraly independent. So ai=0; 1≤i≤k. Hence {x1∗,x2∗,…,xk+n∗} is lineraly independent.
Now supposegi∗=(hi∗,ki∗)∈X∗=Y∗⊕∞Z∗; 1≤i≤k+n+1, where hi∗∈Y∗ and ki∗∈Z∗ with gi∗=1=gi∗(x)=ki∗(z). Thus gi∗=(hi∗,ki∗)∈Y∗⊕∞spSz. Since dim⁡⁡(Y∗⊕∞spSz)=k+n then we see that g1∗,g2∗,…,gk+n+1∗ must be linearly dependent. Therfore x∈k+n-smoothB(X).Lemma 10.
LetY and Z be nonzero Banach spaces and X=Y⊕1Z. Let x=(y,z)∈X where y,z≠0,y/y∈m-smoothB(Y) and z/z∈k-smoothB(Z). Then x∈(m+k-1)-smoothB(X).Proof.
Saym≥k. Let {y1∗,y2∗,…,ym∗}⊆Sy/∥y∥ and let {z1∗,z2∗,…,zk∗}⊆Sz/∥z∥ be linearly independent sets. Let (7)xi∗=yi∗,z1∗,if1≤i≤m,yi-m+1∗,zi-m+1∗,ifm+1≤i≤m+k-1.Clearly xi∗∈Sx; 1≤i≤m+k-1. They are linearly independent. Indeed if ∑i=1m+k-1‍aixi∗=0 then ∑i=1m‍aiyi∗+∑i=m+1m+k-1‍aiyi-m+1∗=0 and ∑i=m+1m+k-1‍aizi-m+1∗+(∑i=1m‍ai)z1∗=0. Since {y1∗,y2∗,…,ym∗} and {z1∗,z2∗,…,zk∗} are linearly independent sets then a1=0,ai=0;k+1≤i≤m,ai+am+i-1=0; 2≤i≤n and ai=0;m+1≤i≤m+k-1. This makes ai=0 for all 1≤i≤m+k-1. Hence {x1∗,x2∗,…,xm+k-1∗} is linearly independent.
Now to prove thatx∈(m+k-1)-smoothB(X) it is clearly enough to show that any x∗∈Sx must be in sp{x1∗,x2∗,…,xm+k-1∗}. So suppose x∗=(h∗,k∗)∈Sx, where h∗∈Y∗ and k∗∈Z∗. Then (8)1=x∗x=h∗y+k∗z≤h∗y+k∗z≤y+z=x=1.So h∗∈Sy/∥y∥ and k∗∈Sz/∥z∥ and hence there are scalars αi; 1≤i≤m and βj; 1≤j≤k such that h∗=∑i=1m‍αiyi∗ and k∗=∑j=1k‍βjzj∗. Thus 1=h∗(y/y)=∑i=1m‍αiyi∗(y/y)=∑i=1m‍αi. Similarly ∑j=1k‍βj=1.
letG=(y∗∗,z∗∗)∈X∗∗=Y∗∗⊕1Z∗∗ with G(xi∗)=0 for all 1≤i≤m+k-1. Then y∗∗(yi∗)+z∗∗(z1∗)=0; 1≤i≤m and y∗∗(yi-m+1∗)+z∗∗(zi-m+1∗)=0; m+1≤i≤m+k-1. Thus G(x∗)=y∗∗(h∗)+z∗∗(k∗) = y∗∗(∑i=1m‍αiyi∗)+z∗∗(∑j=1k‍βjzj∗)=-∑i=1m‍αiz∗∗(z1∗)+β1z∗∗(z1∗)-∑j=2k‍βjz∗∗(z1∗)=0 since ∑i=1m‍αi=1=∑j=1k‍βj. This proves that x∗∈sp{x1∗,x2∗,…,xm+k-1∗}. Therefore x∈(m+k-1)-smoothB(X).We now can easily prove Theorem7.Proof of Theorem7.
First note that ifI2≠∅, I5≠∅, I1 is infinite, or I4 is infinite, then one can easily construct an infinite linearly independent subset of Sx. On the other hand, if I2=I5=∅ and I1∪I4 is finite, then writing X=⊕1Xi as X=W1⊕1W2⊕1W3, where W1=⊕1{Xi:i∈I1}, W2=⊕1{Xi:i∈I3}, W3=⊕1{Xi:i∈I4} and applying Theorem 8 and Lemmas 9 and 10 we get the result.



---

*Source: 394282-2015-05-26.xml*