On a Numerical Radius Preserving Onto Isometry on
We study a numerical radius preserving onto isometry on . As a main result, when is a complex Banach space having both uniform smoothness and uniform convexity, we show that an onto isometry on is numerical radius preserving if and only if there exists a scalar of modulus 1 such that is numerical range preserving. The examples of such spaces are Hilbert space and spaces for .
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. Let be a Banach space over the field or . We use and for the dual of and the space of bounded linear operators from to , respectively. We denote by (resp., ) the closed unit ball of (resp., the unit sphere of ).
The numerical radius and numerical range of an operator are given by where .
Note that there are many different definitions of numerical range. The above definition of 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 . From the definition, we see that for every .
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 .
We say that an operator on preserves numerical radius (numerical range) when for every . It is clear that every numerical range preserving map is numerical radius preserving.
In 1975, Pellegrini  studied numerical range preserving operators on a Banach algebra. Particularly, when is a complex Hilbert space, it was shown that an isomorphism on is -isomorphism if and only if it is numerical range preserving. Later, Chan  showed that an isomorphism on is numerical radius preserving if and only if is a -isomorphism for some scalar of modulus 1. These results say that for each numerical radius preserving isomorphism on there exists a scalar of modulus 1 such that is a numerical range preserving mapping.
In Section 2, we deduce a similar result for an onto isometry when a complex Banach space has both uniform convexity and uniform smoothness. Indeed, we show that if is a numerical radius preserving onto isometry on then there exists a scalar of modulus 1 such that is numerical range preserving. It is known that Hilbert space and for have both uniform convexity and uniform smoothness. However, we see that this does not hold when .
We first recall the definition of uniform convexity.
For every , the modulus of convexity of a Banach space is defined by
A Banach space is said to be uniformly convex if for all .
It is well known that every uniformly convex space is strictly convex and is uniformly convex when .
Very recently, the following characterization of uniform convexity was shown .
Theorem 1. A Banach space is uniformly convex if and only if for each there is such that for any with there exists satisfying
From this theorem, we show the following lemma.
Lemma 2. Let be a uniformly convex space. For any there exists such that if two pairs satisfy then .
It is worth remarking that the converse of Lemma 2 is true when the space is finite dimensional. Indeed, if is not uniformly convex, then there exist and a sequence of pairs such that and converge to 2. Since is compact, we may assume that and converge to and , respectively. It is clear that and so there exists a functional such that . Since and , we get the desired contradiction.
In paper  of Chan, it was shown that if is a numerical radius preserving isomorphism on , then for some constant of modulus 1, where is the identity map on . We show that the same result holds for an onto isometry on when has uniform convexity and uniform smoothness. Before that, we first see the following.
Lemma 3. If is a numerical radius preserving map on , then for each there exists a scalar of modulus 1 such that .
Proof. We first see that for any there is a scalar of modulus 1 satisfying . Indeed, for a sequence such that converges to , we may assume that converges to some constant . Let . Then, we have that converges to . Since it is clear that , we deduce .
The fact that preserves numerical radius gives .
Lemma 4. Assume is a strictly convex space and an operator satisfies . If absolute value of every element in is 1, then for some constant of modulus 1.
Proof. From the assumption, we see that for arbitrary and satisfying . Since is strictly convex, attains its norm only at . This implies that for some constant of modulus 1.
Now assume that there exist two elements such that . Note that and are linearly independent. Consider and let . This means that Hence, we have which is a contradiction.
Theorem 5. Assume has both uniform convexity and uniform smoothness. If is a numerical radius preserving onto isometry on , then for some constant of modulus 1.
Proof. From Lemma 3, for each , there is so that .
We now show that for any . If this is not true, there exists such that . Take an operator so that for each and take such that .
Let be a number satisfying that every with and implies . Since is uniformly convex, we may take in Lemma 2.
For a pair , if , then . This gives that On the other hand, there exists such that if satisfies , then . To see that this is true, we note that dual of uniformly smooth space is uniformly convex. Hence is uniformly convex. We may use again Lemma 2, and then for some small enough if satisfies then .
Therefore we see that if satisfies that , then . We have This shows that and so we have desired contradiction.
From Lemma 4, we see that for some modulus 1 constant .
Remark 6. From Theorem 5, we can easily construct onto isometries which are not numerical radius preserving. For example, consider 2-dimensional Hilbert space ; we define an operator by for every . It is clear that the operator on given by for every is an onto isometry. However, this does not preserve numerical radius since for any .
We denote a state space of a Banach algebra by
When is a complex Banach space, according to [5, Theorem ], if the adjoint operator of operator satisfies that , then . For in Theorem 5, since is an onto isometry and so satisfies , we see that . Now, we deduce the following main result.
Theorem 7. Assume a complex Banach space has both uniform convexity and uniform smoothness. An onto isometry on is numerical radius preserving if and only if there exists a scalar of modulus 1 such that is a numerical range preserving mapping.
Remark 8. Theorem 7 does not hold for some Banach spaces. In order to see this, let us recall numerical index of a Banach space. We see that is the greatest constant such that for every , and so is equivalent to the operator norm if and only if . For more information, we give [8–11].
From the definition of numerical index, if , then every isometry on preserves numerical radius. There are many classes of Banach spaces having numerical index 1, like , , and . Among them, one of the simplest examples having numerical index 1 is as the typical subspace of . Since the same operator defined in Remark 6 is also an onto isometry on this space, preserves numerical radius. However, this does not preserve numerical range. The reason is that we have and .
Remark 9. It is not possible to say that Theorem 7 is true for every isometry. Consider shift operators and on ( defined by and for , where . Then, it is easy to see that an operator given by for each is a numerical radius preserving isometry. However, does not preserve numerical range since and .
The author declares that there is no conflict of interests regarding the publication of this paper.
This work was supported by Kyonggi University Research Grant 2015.
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