Isometries of a Generalized Numerical Radius on Compact Operators

Isometries of a Generalized Numerical Radius on Compact Operators

Isometries of a Generalized Numerical Radius on Compact Operators Maria Inez Cardoso Gon¸calves1,a, Vladimir G. Pestov2,a,b aDepartamento de Matem´atica - Universidade Federal de Santa Catarina - Trindade - Florian´opolis - SC - 88.040-900 - Brazil bDepartment of Mathematics and Statistics, University of Ottawa, Ottawa, ON, K1N 6N5, Canada Abstract We describe all isometries of the q-numerical radius on the space ( ) of K H compact operators on a (possibly infinite-dimensional) Hilbert space . H Keywords: q-numerical radius, isometries, compact operators 2010 MSC: 47A12, 15A60, 15A04, 47A30, 46B20, 46B28 1. Introduction Let 0 < q 1. The q-numerical radius of a bounded linear operator A on a Hilbert space≤ is given by H r (A) = sup Ax, y : x = y =1, x, y = q . q {h i k k k k h i } The q-numerical radius is a norm on ( ), equivalent to the uniform (spec- B H tral) norm. For q = 1, this reduces to the classical numerical radius: arXiv:1310.5155v1 [math.FA] 18 Oct 2013 r(A) = sup Ax, x : x =1 . {h i k k } In this article, we describe the isometries of the space ( ) of compact operators on an infinite-dimensional complex Hilbert spaceK withH regard to Email addresses: [email protected] (Maria Inez Cardoso Gon¸calves), [email protected] (Vladimir G. Pestov) 1Corresponding author. 2Special Visiting Researcher of the program Science Without Borders of CAPES (Brazil), processo 085/2012. Preprint submitted to Linear Algebra and Its Applications July 17, 2018 the q-numerical radius. In the finite-dimensional case, the description was obtained in [1], while the case of the classical numerical radius (also in finite dimensions) was previously treated by Leˇsnjak [6], Li and Semrlˇ [7] and Li et al. [8]. Here is the main result. Theorem 1.1. Let be an infinite-dimensional complex Hilbert space. Let 0 < q 1 and let Hφ: ( ) ( ). Then φ is an isometry of the q- ≤ K H → K H numerical radius if and only if there exist a compact operator S0 ( ), a unitary operator ( ) and a complex number µ with µ = 1∈KsuchH that for all A ( )U ∈ U H | | ∈K H ∗ † φ(A)= S0 + µU A U, where A† denotes either A ot At or A∗ or A¯. Here A∗ is the usual adjoint operator, given by x, Ay = A∗x, y , h i h i At is the traspose of A, x, Aty = y,¯ Ax¯ , h i h i and A¯ is the complex conjugate of A, x, Ay¯ = x,¯ Ay¯ . h i h i While the strategy of the proof is broadly similar to that in [1], the infinite dimensional case poses numerous difficulties and requires new techniques. For instance, one has to simultaneously deal with a number of different topologies on various spaces of operators, which all coincide between themselves in the finite-dimensional situation. The main challenge was the usage of extreme points, which was the key component of the proofs in [8, 1]. If one defines the operator 2 C = qE11 + 1 q E12 q − on = ℓ2, then the unit ball of the dualp norm to the generalized numerical radiusH is the convex hull of the saturated unitary orbit (C )= λU ∗C U : λ C, λ =1, U ( ) . SU q { q ∈ | | ∈ U H } It is easy to see that all points of (Cq) are extreme points, indeed exposed points of the dual unit ball. However,SU we were unable to prove that the set of 2 extreme (or exposed) points of the dual ball is exactly the set U(Cq), like it is in a finite-dimensional situation. Thus, a key component ofS the proofs in [8, 1] was missing, and we had to find an ingenious way around it, by using points of weak continuity with regard to the Hilbert-Schmidt norm topology combined with David Milman’s converse to the Krein-Milman theorem. We believe that this technique can be of interest in connection with the next natural open problem: that of characterizing isometries of the general- ized numerical radius on a large space ( ) of all bounded operators. Notice also that in order to simplifyB notation,H in the sequel we will often only give proofs for = ℓ2, but they remain valid for an arbitrary = ℓ2(Γ). H H 2. Generalized numerical radius 2.1. Duality Let be an infinite-dimensional separable Hilbert space. As usual, we H will identify the space 1( ) (the first Schatten class) of all trace class op- erators on , equippedC withH the trace-class norm T = tr( T ), with a H k k1 | | predual of the space B( ), equipped with the uniform operator norm. H Let us remind how to define a pairing , between 1( ) and B( ). If h· ·i C H H C is a trace-class operator and T is a bounded linear operator on , the value of C, T is given by H h i C, T = tr(CT ). (1) h i One can now prove that elements of B( ) are exactly the bounded linear ∗ H functionals on 1( ), and that 1( ) = B( ). C H C H ∼ H The space ( ) of all compact operators on , equipped with the uni- K H H form operator norm, is, under the above duality, the predual of 1( ). In particular, one has C H ( )∗∗ = ( ). K H B H For more on the duality and on the above classes of operators, see e.g. [11], 1.15 and 1.19. In the finite dimensional situation (dim < ), this duality allows to H ∞ conveniently identify all the spaces ( ), 1( ), ( ), as well as its dual K H C H B H space ( )∗, between themselves. In the infinite dimensions one has to treat them separately.B H 3 2.2. C-Numerical radius Fix an operator C 1( ). For any A ( ) the generalized C- numerical range of A is∈ defined C H by: ∈ B H W (A)= tr(CU ∗AU): U ( ) , C { ∈ U H } and the generalized C-numerical radius rC of A is defined by r (A) = sup tr(CU ∗AU) : U ( ) . C {| | ∈ U H } Theorem 2.1. r ( ) is a norm in ( ) if and only if C is non-scalar and C · B H tr(C) =0. 6 Proof. From the definition of r (A) we have that r (A) 0 for all A C C ≥ ∈ ( ), so we need to show that if rC (A) = 0 then A = 0. Suppose C is non-scalarB H and tr (C) = 0, then if r (A) = 0 we have that tr(CU ∗AU)=0 6 C for all unitary U, this implies that WC (A) is a singleton. By lemma 4.2 in [5], we have that A is a scalar operator, say A = λI, where λ is a scalar and I denotes the identity operator. Therefore we have that: 0 = tr (CU ∗AU)= λtr(C), since tr (C) = 0, this implies that λ = 0 and therefore A is the zero operator. 6 One can easily verify that rC (αA) = α rC(A), for all A ( ) and r (A + B) r (A)+ r (B), for all A, B | |( ). ∈ B H C ≤ C C ∈ B H Let us put the generalized numerical radius in the context of duality. Clearly, rC (A) = sup Re C, A , T h i where the supremum is taken over the saturated unitary orbit of A in 1( ): C H (A)= λU ∗AU : λ C, λ =1, U ( ) . SU { ∈ | | ∈ U H } That is to say, ∗ ∗ rC (A) = sup Re(tr (λUC U A)). λ =1 U| | ( ) ∈ U H Equivalently, r (A) is the supremum of the linear functional , A over the C h− i convex circled hull of the unitary orbit of C. In the case where C is non- scalar and tr(C) = 0, Theorem 2.1 implies that this hull is absorbing and 6 4 ∗ bounded in the vector space 1( ), so is the unit ball of a certain norm, rC . ∗ C H The dual norm to rC is the norm rC on ( ). B H ∗ We do not know if in this generality the norm rC is the dual norm to the restriction of rC to ( ), nor whether the norm rC is equivalent to the uniform operator (spectral)K H norm on ( ). In particular, we cannot assert ∗ B H that the norms rC and rC are complete. However, the answers to all of these questions are positive in the special case of a q-numerical radius. 2.3. q-Numerical radius 2 In the important particular case where = ℓ and C = E11 one recovers the classical numerical radius: H r(A) = sup Ax, x . kxk=1|h i| It is well known that the spectral radius is a norm on ( ) equivalent to the B H uniform norm, moreover r(A) A 2r(A). ≤k k ≤ (See e.g. Th. 2.14 in [4].) The case of interest for us is a more general case of the q-numerical radius, where 0 < q 1: ≤ rq(A) = sup Ax, y . kxk=kyk=1,hx,yi=q|h i| In this case, the matrix C assumes the form 2 C = qE11 + 1 q E12. q − Geometric considerations in a two-dimensionalp Euclidean space imply the following. Lemma 2.2. For all A ( ), ∈ B H r (A) qr(A). q ≥ Proof. Let ξ = 1 and Aξ,ξ = α, where α > 0. Identify ξ with k k h i ± 2 the second coordinate vector e2 of the plane R spanned by ξ,Aξ, so that Aξ belongs to either the first or the fourth quadrant ( Aξ, e1 0). Thus, Aξ = (√1 α2, α). Choose ζ with ζ = 1, ξ,ζ =h q, andi ≥ζ belonging − ± k k h i 5 to the first quadrant if so does Aξ, or the second quadrant if Aξ is in the fourth.

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