# On a Numerical Radius Preserving Onto Isometry onL(X)

Authors: Sun Kwang Kim
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We study a numerical radius preserving onto isometry onL(X). As a main result, when X is a complex Banach space having both uniform smoothness and uniform convexity, we show that an onto isometry T on L(X) is numerical radius preserving if and only if there exists a scalar cT of modulus 1 such that cTT is numerical range preserving. The examples of such spaces are Hilbert space and Lp spaces for 1<p<∞.


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## Body

## 1. Introduction

In this paper, we study a numerical radius preserving onto isometry on the set of bounded linear operators. We begin with some notation to present its definition. LetX be a Banach space over the field K=R or C. We use X∗ and L(X) for the dual of X and the space of bounded linear operators from X to X, respectively. We denote by BX (resp., SX) the closed unit ball of X (resp., the unit sphere of X).The numerical radius and numerical range of an operatorA∈L(X) are given by (1)vA=sup⁡x∗Ax:x,x∗∈ΠX,VA=convx∗Ax:x,x∗∈ΠX¯,where Π(X)={(x,x∗)∈SX×SX∗:x∗(x)=1}.Note that there are many different definitions of numerical range. The above definition ofV(A) is sometimes called Banach algebra numerical range to distinguish from others. It is worth mentioning that it is also usual to study spatial numerical range {x∗(Ax):(x,x∗)∈Π(X)}. From the definition, we see that v(A)≤A for every A∈L(X).The concept of numerical range was introduced by G. Lumer and F. Bauer in the sixties. Later this was extended to arbitrary continuous functions on a unit sphere of Banach spaces. On the other hand, the study of numerical radius was initiated by Harris for polynomials and holomorphic functions. We can find more details in [1–3].There has been many different types of research on these concepts. Among them, we can find a lot of works on linear maps which preserve numerical radius or numerical range [4].We say that an operatorT on L(X) preserves numerical radius (numerical range) when (2)vTA=vAVTA=VAfor every A∈L(X). It is clear that every numerical range preserving map is numerical radius preserving.In 1975, Pellegrini [5] studied numerical range preserving operators on a Banach algebra. Particularly, when H is a complex Hilbert space, it was shown that an isomorphism T on L(H) is C∗-isomorphism if and only if it is numerical range preserving. Later, Chan [6] showed that an isomorphism T on L(H) is numerical radius preserving if and only if cTT is a C∗-isomorphism for some scalar cT of modulus 1. These results say that for each numerical radius preserving isomorphism T on L(H) there exists a scalar cT of modulus 1 such that cTT is a numerical range preserving mapping.In Section2, we deduce a similar result for an onto isometry when a complex Banach space X has both uniform convexity and uniform smoothness. Indeed, we show that if T is a numerical radius preserving onto isometry on L(X) then there exists a scalar cT of modulus 1 such that cTT is numerical range preserving. It is known that Hilbert space and Lp for 1<p<∞ have both uniform convexity and uniform smoothness. However, we see that this does not hold when X=l∞2.


## 2. Results

We first recall the definition of uniform convexity.For everyϵ∈(0,2], themodulus of convexity of a Banach space (X,·) is defined by (3)δϵ=inf⁡1-x+y2:x,y∈BX,x-y≥ϵ.A Banach space(X,·) is said to be uniformly convex if δϵ>0 for all ϵ∈0,2.It is well known that every uniformly convex space is strictly convex andLp is uniformly convex when 1<p<∞.Very recently, the following characterization of uniform convexity was shown [7].Theorem 1.
A Banach spaceX is uniformly convex if and only if for each ε>0 there is η(ε)>0 such that for any (x,x∗)∈SX×SX∗ with x∗x>1-η(ε) there exists y∈SY satisfying (4)x∗y=1,y-x<ε.From this theorem, we show the following lemma.Lemma 2.
LetX be a uniformly convex space. For any ε>0 there exists 0<β(ε)<ε such that if two pairs (x,x∗),(y,y∗)∈Π(X) satisfy x∗-y∗<β(ε) then x-y<ε.Proof.
Letη(·) be the function in Theorem 1 and assume that (x,x∗),(y,y∗)∈Π(X) satisfy x∗-y∗<η(ε).
Note thatx is the unique element in BX such that x∗(x)=1 by the strict convexity of X. Since x∗y≥y∗y-y∗-x∗x>1-η(ε), we see that x-y<ε from Theorem 1. Hence we may take β(ε)=min⁡{η(ε),ε/2}.It is worth remarking that the converse of Lemma2 is true when the space X is finite dimensional. Indeed, if X is not uniformly convex, then there exist ε>0 and a sequence of pairs (xi,yi)∈BX×BX such that xi-yi>ε and xi+yi converge to 2. Since BX is compact, we may assume that (xi) and (yi) converge to x and y, respectively. It is clear that x+y=2 and so there exists a functional f∈SX∗ such that f(x+y)=2. Since (x,f),(y,f)∈Π(X) and x-y>ε, we get the desired contradiction.In paper [6] of Chan, it was shown that if T is a numerical radius preserving isomorphism on L(H), then T(I)=cTI for some constant cT of modulus 1, where I is the identity map on H. We show that the same result holds for an onto isometry on L(X) when X has uniform convexity and uniform smoothness. Before that, we first see the following.Lemma 3.
IfT is a numerical radius preserving map on L(X), then for each A∈L(X) there exists a scalar αA of modulus 1 such that v(T(I))+v(T(A))=v(T(I)+αAT(A)).Proof.
We first see that for anyA∈L(X) there is a scalar αA of modulus 1 satisfying v(I+αAA)=1+v(A). Indeed, for a sequence (xi,xi∗)∈Π(X) such that xi∗Axi converges to v(A), we may assume that xi∗(Axi) converges to some constant β. Let αA=β¯/β. Then, we have that xi∗((I+αAA)xi) converges to 1+β=1+v(A). Since it is clear that vI+αAA≤1+v(A), we deduce v(I+αAA)=1+v(A).
The fact thatT preserves numerical radius gives v(T(I))+v(T(A))=v(I)+v(A)=v(I+αAA)=v(T(I+αAA))=v(T(I)+αAT(A)).Lemma 4.
AssumeX is a strictly convex space and an operator A∈L(X) satisfies A=1. If absolute value of every element in V(A) is 1, then A=cAI for some constant cA of modulus 1.Proof.
From the assumption, we see thatAx=1=x∗Ax for arbitrary x∈SX and x∗ satisfying x∗(x)=1. Since X is strictly convex, x∗ attains its norm only at {λx:λ=1}. This implies that Ax=λxx for some constant λx of modulus 1.
Now assume that there exist two elementsx1,x2∈SX such that λx1≠λx2. Note that x1 and x2 are linearly independent. Consider z=(x1+x2)/x1+x2 and let Az=λzz. This means that (5)λzx1+λzx2x1+x2=λzz=Az=λx1x1+λx2x2x1+x2.
Hence, we have(λz-λx1)x1=(λx2-λz)x2 which is a contradiction.Theorem 5.
AssumeX has both uniform convexity and uniform smoothness. If T is a numerical radius preserving onto isometry on L(X), then T(I)=cTI for some constant cT of modulus 1.Proof.
From Lemma3, for each A∈L(X), there is αA so that v(T(I))+v(T(A))=v(T(I)+αAT(A)).
We now show thatx∗(TIx)=v(T(I))=v(I)=1 for any (x,x∗)∈Π(X). If this is not true, there exists (z0,z0∗)∈Π(X) such that z0∗TIz0<1. Take an operator B∈L(X) so that T(B)(x)=z0∗(x)z0 for each x∈X and take 0<γ<1 such that z0∗TIz0<γ<1.
Letδ>0 be a number satisfying that every (z,z∗)∈Π(X) with z-z0<δ and z∗-z0∗<δ implies z∗TIz<γ. Since X is uniformly convex, we may take β(·) in Lemma 2.
For a pair(y,y∗)∈Π(X), if y∗-z0∗<β(δ)<δ, then y-z0<δ. This gives that (6)y∗TI+αBBy≤y∗TIy+y∗TBy<γ+1.
On the other hand, there existsε>0 such that if (y,y∗)∈Π(X) satisfies y∗-z0∗≥β(δ), then y∗z0≤1-ε. To see that this is true, we note that dual of uniformly smooth space is uniformly convex. Hence X∗ is uniformly convex. We may use again Lemma 2, and then for some small enough ε>0 if (y,y∗)∈Π(X) satisfies y∗z0>1-ε then y∗-z0∗<β(δ).
Therefore we see that if(y,y∗)∈Π(X) satisfies that y∗-z0∗≥β(δ), then y∗TBy=y∗z0z0∗y≤1-ε. We have (7)y∗TI+αBBy≤y∗TIy+y∗TBy≤1+1-ε.
This shows thatv(T(I+αBB))≤max⁡{2-ε,γ+1}<2=v(T(I))+v(T(B)) and so we have desired contradiction.
From Lemma4, we see that T(I)=cTI for some modulus 1 constant cT.Remark 6.
From Theorem5, we can easily construct onto isometries which are not numerical radius preserving. For example, consider 2-dimensional Hilbert space R2; we define an operator B:R2→R2 by B(x1,x2)=(x1,-x2) for every (x1,x2)∈R2. It is clear that the operator T on L(R2) given by T(A)=B∘A for every A∈L(R2) is an onto isometry. However, this does not preserve numerical radius since T(I)≠cTI for any cT.We denote a state spaceS of a Banach algebra L(X) by (8)S=f∈LX∗:f=1=fI.WhenX is a complex Banach space, according to [5, Theorem 2.2], if the adjoint operator S∗ of operator S:L(X)→L(X) satisfies that S∗(S)⊂S, then V(S(A))⊂V(A). For T in Theorem 5, since T0=cT-1T is an onto isometry and so satisfies T0∗(S)=S, we see that V(T0(A))=V(A). Now, we deduce the following main result.Theorem 7.
Assume a complex Banach spaceX has both uniform convexity and uniform smoothness. An onto isometry T on L(X) is numerical radius preserving if and only if there exists a scalar cT of modulus 1 such that cTT is a numerical range preserving mapping.Remark 8.
Theorem7 does not hold for some Banach spaces. In order to see this, let us recall numerical index n(X) of a Banach space. (9)nX=inf⁡vT:T∈SLX.
We see thatn(X) is the greatest constant k≥0 such that kT≤v(T) for every T∈L(X), and so v is equivalent to the operator norm if and only if n(X)>0. For more information, we give [8–11].
From the definition of numerical index, ifn(X)=1, then every isometry on L(X) preserves numerical radius. There are many classes of Banach spaces having numerical index 1, like c0, l1, and l∞ [12]. Among them, one of the simplest examples having numerical index 1 is R2 as the typical subspace of l∞. Since the same operator T defined in Remark 6 is also an onto isometry on this space, T preserves numerical radius. However, this does not preserve numerical range. The reason is that we have V(I)={1} and -1=(0,1)T(I)(0,1)∈V(T(I)).Remark 9.
It is not possible to say that Theorem7 is true for every isometry. Consider shift operators B1 and B2 on lp (1<p<∞) defined by B1(xi)i=1∞=(xi)i=2∞ and B2(xi)i=1∞=(xi)i=0∞ for (xi)i=1∞∈lp, where x0=0. Then, it is easy to see that an operator T given by T(A)=B2∘A∘B1 for each A∈L(lp) is a numerical radius preserving isometry. However, T does not preserve numerical range since V(I)={1} and 0=(1,0,0,0,…)T(I)(1,0,0,0,…)∈V(T(I)).



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*Source: 9183135-2016-10-12.xml*