# Norm for Sums of Two Basic Elementary Operators

Authors: A. Bachir; F. Lombarkia; A. Segres
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

We give necessary and sufficient conditions under which the norm of basic elementary operators attains its optimal value in terms of the numerical range.


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## Body

## 1. Introduction

LetE be a normed space over 𝕂 (ℝ or ℂ), SE its unit sphere, and E* its dual topological space. Let D be the normalized duality mapping form E to E* given by
(1.1)D(x)={φ∈E*:φ(x)=∥x∥2,∥φ∥=∥x∥},∀x∈E.
Let B(E) be the normed space of all bounded linear operators acting on E. For any operator A∈B(E) and x∈E,
(1.2)Wx(A)={φ(Ax):φ∈D(x)},W(A)=∪{Wx(A):x∈SE}
is called the spatial numerical range of A, which may be defined as
(1.3)W(A)={φ(Ax):x∈SE;φ∈D(x)}.
This definition was extended to arbitrary elements of a normed algebra 𝒜 by Bonsall [1–3] who defined the numerical range of a∈𝒜 as
(1.4)V(a)=W(Aa),
where Aa is the left regular representation of A in B(𝒜), that is, Aa=ab for all b∈𝒜. V(a) is known as the algebra numerical range of a∈𝒜, and, according to the above definitions, V(a) is defined by
(1.5)V(a)={φ(ab):b∈S𝒜;φ∈D(b)}.
For an operator A∈B(E),  Bachir and Segres [4] have extended the usual definitions of numerical range from one operator to two operators in different ways as follows.The spatial numerical rangeW(A)B of A∈B(E) relative to B is
(1.6)W(A)B={φ(Ax):x∈SE;φ∈D(Bx)}.
The spatial numerical range G(A)B of A∈B(E) relative to B is
(1.7)G(A)B={φ(Ax):x∈E;∥Bx∥=1,φ∈D(Bx)}.
The maximal spatial numerical range of A∈B(E) relative to B is
(1.8)M(A)B={φ(Ax):x∈SE;∥Bx∥=∥B∥,φ∈D(Bx)}.
For A,B∈B(E), let SE(B)={(xn)n:xn∈SE,∥Bxn∥→∥B∥}, then the set
(1.9)ℳ(A)B={limφn(Axn):(xn)n∈SE(B),φn∈D(Bxn)}
is called the generalized maximal numerical range of A relative to B. It is known that ℳ(A)B is a nonempty closed subset of 𝕂 and M(A)B⊆ℳ(A)B⊆W(A)B¯. The definition of ℳ(A)B can be rewritten, with respect to the semi-inner product [·,·] as
(1.10)ℳ(A)B={lim[Axn,Bxn]:(xn)n∈SE(B)},
with respect to an inner product (·,·) as
(1.11)ℳ(A)B={lim(Axn,Bxn):(xn)n∈SE(B)}.
We shall be concerned to estimate the norm of the elementary operator MA1,B1+MA2,B2, where A1,A2,B1,B2 are bounded linear operators on a normed space E and MA1,B1 is the basic elementary operator defined on B(E) by
(1.12)MA1,B1(X)=A1XB1.
We also give necessary and sufficient conditions on the operators A1,A2,B1,B2 under which MA1,B1+MA2,B2 attaints its optimal value ∥A1∥∥B1∥+∥A2∥∥B2∥.


## 2. Equality of Norms

Our next aim is to give necessary and sufficient conditions on the set{A1,A2,B1,B2} of operators for which the norm of MA1,B1+MA2,B2 equals ∥A1∥∥B1∥+∥A2∥∥B2∥.Lemma 2.1.
For any of the operatorsA,B,C∈B(E) and all α,β∈𝕂, one has
(2.1)ℳ(αA+βB)B=αℳ(A)B+β∥B∥2;ℳ(αA+βC)B⊆αℳ(A)B+βℳ(C)B.Proof.
The proof is elementary.Theorem 2.2.
LetA1,A2,B1,B2 be operators in B(E).
If∥A1∥∥A2∥∈ℳ(A1)A2∪ℳ(A2)A1 and ∥B1∥∥B2∥∈ℳ(B1)B2∪ℳ(B2)B1, then
(2.2)∥MA1,B1+MA2,B2∥=∥A1∥∥B1∥+∥A2∥∥B2∥.Proof.
The proof will be done in four steps; we choose one and the others will be proved similarly. Suppose that∥A1∥∥A2∥∈ℳ(A1)A2 and ∥B1∥∥B2∥∈ℳ(B1)B2, then there exist (xn)n∈SE(A2),φn∈D(A2xn) such that ∥A1∥∥A2∥=limnφn(A1xn) and there exist (yn)n∈SE(B2),ψn∈D(B2yn) such that ∥B1∥∥B2∥=limnψn(B1yn). Define the operators Xn∈B(E) as follows:
(2.3)Xn(yn)=(ψn⊗xn)(yn)=ψn(yn)xn,∀n.
Then ∥Xn∥≤∥B2∥,foralln≥1, and
(2.4)∥(MA1+B1+MA2+B2)Xn(yn)∥=∥(A1XnB1+A2XnB2)yn∥=∥A1Xn(B1yn)+A2Xn(B2yn)∥=∥φn∥∥φn∥∥A1ψn(B1yn)xn+A2ψn(B2yn)xn∥≥1∥φn∥∥φn(ψn(B1yn)A1xn+ψn(B2yn)A2xn)∥=1∥φn∥∥ψn(B1yn)φn(A1xn)+∥B2yn∥2∥A2xn∥2∥.(2.5)∥MA1,B1+MA2,B2∥≥∥(MA1,B1+MA2,B2)Xn(yn)∥∥Xn∥,∀n≥1.
Hence
(2.6)∥MA1,B1+MA2,B2∥≥∥ψn(B1yn)φn(A1xn)+∥B2yn∥2∥A2xn∥2∥∥A2∥∥B2∥,∀n≥1.
Letting n→∞,
(2.7)∥MA1,B1+MA2,B2∥≥∥A1∥∥B1∥+∥A2∥∥B2∥.
Since
(2.8)∥MA1,B1+MA2,B2∥≤∥A1∥∥B1∥+∥A2∥∥B2∥,
therefore
(2.9)∥MA1,B1+MA2,B2∥=∥A1∥∥B1∥+∥A2∥∥B2∥.Corollary 2.3.
LetE be a normed space and A,B∈B(E). Then, the following assertions hold: (1)
if∥A∥∥B∥∈ℳ(A)B, then ∥A+B∥=∥A∥+∥B∥;(2)
if∥A∥∈ℳ(I)A and ∥B∥∈ℳ(I)B, then ∥MA,B+I∥=1+∥A∥∥B∥.Remark 2.4.
In the previous corollary, if we setB=I, then we obtain an important equation called the Daugavet equation:
(2.10)∥A+I∥=1+∥A∥.
It is well known that every compact operator on C[0,1] [5] or on L1[0,1] [6] satisfies (2.10).
A Banach spaceE is said to have the Daugavet property if every rank-one operator on E satisfies (2.10). So that from our Corollary 2.3 if 1∈ℳ(I)A or 1∈ℳ(A)I for every rank-one operator A, then E has the Daugavet property.
The reverse implication in the previous theorem is not true, in general, as shown in the following example which is a modification of that given by the authors Bachir and Segres [4, Example 3.17].Example 2.5.
Letc0 be the classical space of sequences (xn)n⊂ℂ:xn→0, equipped with the norm ∥(xn)n∥=maxn|xn| and let L be an infinite-dimensional Banach space. Taking the Banach space E=L⊕c0 equipped with the norm, for x=(x1+x2)∈E,∥x∥=∥x1+x2∥=max{∥x1∥,∥Tx1∥+∥x2∥}, where T is any norm-one operator from L to c0 which does not attain its norm (by Josefson-Nissenzweig’s theorem [7]), we can find a sequence (φn)n⊂SE* such that φn converges weakly to 0. Therefore we get the desired operator T:L→c0 defined by
(2.11)(Tx)n=nn+1φn(x).
Let A1,A2,B1,B2 be operators defined on E as follows:
(2.12)A1(x1+x2)=0+Tx1;A2x=A2(x1+x2)=x1+0;B1(x1+x2)=x1-x2;B2=I,∀x=(x1+x2)∈L×c0,
where I is the identity operator on E. It easy to check that A1,A2,B1 are linear bounded operators and ∥A1∥=∥A2∥=∥B1∥=∥B2∥=1. If we choose X0=I and x0=x1+0 such that 1=∥Tx1∥≥∥x1∥, then ∥X0∥=∥x0∥=1 and
(2.13)∥MA1,B1+MA2,B2∥≥∥(MA1,B1+MA2,B2)X0(x0)∥=∥(A1X0B1+A2X0B2)(x0)∥=∥0+Tx1+x1+0∥=max{∥x1∥,2∥Tx1∥}=2,
and from
(2.14)∥MA1,B1+MA2,B2∥≤∥A1∥∥A2∥+∥B1∥∥B2∥=2
we get
(2.15)∥MA1,B1+MA2,B2∥=2=∥A1∥∥A2∥+∥B1∥∥B2∥.
It is clear from the definitions of ℳ(A1)A2 and W(A1)A2 that
(2.16)ℳ(A1)A2⊆W(A1)A2¯
(for details, see [4]).The next result shows that the reverse is true under certain conditions, before that we recall the definition of Birkhoff-James orthogonality in normed spaces.Definition 2.6.
LetE be a normed space and x,y∈E. We say that x is orthogonal to y in the sense of Birkhoff-James ([8, 9]), in short x⊥B-Jy, iff
(2.17)∀λ∈𝕂:∥x+λy∥≥∥x∥.
IfF,G are linear subspaces of E, we say that F is orthogonal to G in the sense of ⊥B-J, written as F⊥B-JG iff x⊥B-Jy for all x∈F and all y∈G.
IfT∈B(E), we will denote by Ran(T) and T† the range and the dual adjoint, respectively, of the operator T.Theorem 2.7.
LetA1,A2,B1,B2 be operators in B(E).
If∥MA1,B1+MA2,B2∥=∥A1∥∥B1∥+∥A2∥∥B2∥,
(2.18)Ran(A2†)⊥B-JRan(A1†-∥A1∥∥A2∥A2†),Ran(B2)⊥B-JRan(B1-∥B1∥∥B2∥B2),
then
(2.19)∥A1∥∥A2∥∈ℳ(A1†)A2†,∥B1∥∥B2∥∈ℳ(B1)B2.
Moreover, if
(2.20)Ran(A1†)⊥B-JRan(A2†-∥A2∥∥A1∥A1†),Ran(B1)⊥B-JRan(B2-∥B2∥∥B1∥B1),
then
(2.21)∥A1∥∥A2∥∈ℳ(A1†)A2†∩ℳ(A2†)A1†,∥B1∥∥B2∥∈ℳ(B1)B2∩ℳ(B2)B1.Proof.
If∥MA1,B1+MA2,B2∥=∥A1∥∥B1∥+∥A2∥∥B2∥, then we can find two normalized sequences (Xn)n⊆B(E) and (xn)n⊆E such that
(2.22)limn∥A1XnB1xn+A2XnB2xn∥=∥A1∥∥B1∥+∥A2∥∥B2∥.
We have for all n≥1(2.23)∥A1XnB1xn∥≤∥A1∥∥B1xn∥≤∥A1∥∥B1∥∥A2XnB2xn∥≤∥A2∥∥B2xn∥≤∥A2∥∥B2∥,
so we can deduce from the above inequalities and (2.10) that limn∥B1xn∥=∥B1∥ and limn∥B2xn∥=∥B2∥. From the assumptions Ran(B2)⊥B-JRan(B1-(∥B1∥/∥B2∥)B2) we get
(2.24)Ran(B1-∥B1∥∥B2∥B2)¯∩Ran(B2)¯={0}.
Set χn=(B1-(∥B1∥/∥B2∥)B2)xn and yn=B2xn for all n and define the function ϕn on the closed subspace F spanned by {xn,yn} for all n as
(2.25)ϕn(aχn+byn)=b∥yn∥2=b∥B2xn∥,∀a,b∈𝕂.
It is clear that ϕn is linear for all n and
(2.26)|ϕn(aχn+byn)|=|b|∥B2xn∥2=∥aχn+byn∥∥B2xn∥∥byn∥∥aχn+byn∥.
From the assumptions Ran(B2)⊥B-JRan(B1-(∥B1∥/∥B2∥)B2) it follows that
(2.27)|ϕ(aχn+byn)|≤∥B2xn∥∥aχn+byn∥,∀a,b∈𝕂,∀n.
This means that ϕn is continuous for each n on the subspace F with ∥ϕn∥=∥B2xn∥ (by (2.27) and ϕn(yn)=∥yn∥∥B2xn∥). Then by Hahn-Banach theorem there is ϕn~∈E* with ϕn~|F=ϕn and ∥ϕn∥=∥ϕn~∥, for each n. So
(2.28)ϕn~(χn)=ϕn~((B1-∥B1∥∥B2∥B2)xn)=0,
hence
(2.29)limnϕn~(χn)=ϕn~((B1-∥B1∥∥B2∥B2)xn)=0,ϕn~(B2xn)=∥B2xn∥2,∥ϕn~∥=∥B2xn∥.
Thus, 0∈ℳ(B1-(∥B1∥/∥B2∥)B2)B2 and by Lemma 2.1(2.30)0∈(ℳ(B1)B2-∥B1∥∥B2∥∥B2∥2)=ℳ(B1)B2-∥B1∥∥B2∥.
Therefore,
(2.31)∥B1∥∥B2∥∈ℳ(B1)B2.From∥MA1,B1+MA2,B2∥=∥A1∥∥B1∥+∥A2∥∥B2∥ we can find a normalized sequences (Xn)n⊆B(E) such that
(2.32)limn∥A1XnB1+A2XnB2∥=∥A1∥∥B1∥+∥A2∥∥B2∥.
Since ∥A1XnB1+A2XnB2∥=∥B1†Xn†A1†+B2†Xn†A2†∥, for each n, then we can find a normalized ϕnk∈E† such that
(2.33)limk,n∥B1†Xn†A1†ϕnk+B2†Xn†A2†ϕnk∥=∥A1∥∥B1∥+∥A2∥∥B2∥.
We argue similarly and get
(2.34)limk,n∥A1†ϕnk∥=∥A1†∥,limk,n∥A2†ϕnk∥=∥A2†∥.
Following the same steps as in the previous case we obtain ∥A1∥∥A2∥∈ℳ(A1†)A2†.Moreover, if we haveRan(A1†)⊥B-JRan(A2†-(∥A2∥/∥A1∥)A1†) and Ran(B1)⊥B-JRan(B2-(∥B2∥/∥B1∥)B1), it suffices to reverse, in the proof of the previous case, the role of A1† into A2† and B1 into B2.For the completeness of the previous theorem we need to prove the following result which is very interesting.We recall that Phelps [10] has proved that, for a Banach space E, ∪{D(x):x∈E} is dense in E*; this property is called subreflexivity of the space E. Using this fact, Bonsall and Duncan [2] has proved that for any operator T∈B(E) we have W(T)¯=W(T†)¯. The following result generalizes the Bollobas result in the case ℳ(A)B, where A,B∈B(E).Proposition 2.8.
LetE be a Banach space with smooth dual and let A,B∈B(E) such that B is a surjective operator. Then ℳ(A†)B†⊆ℳ(A)B.Proof.
Leta∈ℳ(A†)B†, then there are ψn∈D(B†φn),(φn)n∈SE*(B†) such that a=limnψn(A†φn).
By the subreflexivity ofE there exist sequences (φnk)nk⊆E* and (xnk)⊆E such that φnk∈D(xnk) and ∥φnk-∥Bxnk∥φn∥ to 0. It follows that the sequence (x^nk)⊆E** has an E**-weak convergent subsequence (x^nm)nm, that is,
(2.35)x^nm(f)→Ψ(f),∀f∈E*,Ψ∈E**.
On the one hand, we have
(2.36)∥Bxnm∥2=[B†(φnm-∥Bxnm∥φn)](xnm)+∥Bxnm∥(B†φn)(xnm).
Then
(2.37)∥Bxnm∥2≤∥B†(φnm-∥Bxnm∥φn)∥+∥Bxnm∥∥B†φn∥.
Thus
(2.38)∥Bxnm∥|∥Bxnm∥-∥B†φn∥|≤∥B†∥∥φnm-∥Bxnm∥φn∥.
On the other hand,
(2.39)|x^nm(B†φn)-∥B†φn∥|≤|x^nm(B†φn)-x^nm(B†φnm∥Bxnm∥)|+|1∥Bxnm∥x^nm(B†φnm)-∥B†φn∥|=|x^(B†φn-1∥Bxnm∥B†φnm)|+|∥Bxnm-∥B†∥φn∥|→0asm→∞.
So limmx^nm(B†φn)=∥B†φn∥ and ∥B†φn∥Ψn∈D(B†φn). Then by smoothness of the space E* we get ∥B†φn∥Ψn=Ψn,foralln. Next,
(2.40)|x^nm(A†φnm)-∥Bxnm∥∥B†φn∥ψn(A†φn)|≤|x^nm(A†φnm)-x^nm(∥Bxnm∥A†φn)|+∥Bxnm∥|x^nm(A†φn)-1∥B†φn∥ψn(A†φn)|=|x^nm(A†φnm-A†φn)|+∥Bxnm∥|x^nm(A†φn)-ψn(A†φn)|→0asm→∞.
Thenlimmx^nm(A†φnm)=ψn(A†φn) or limmφnm(Axnm=ψ(A†φn) and therefore
(2.41)limn[limmφnm(Axnm)]=limnψn(A†φn)=a
which means that a∈ℳ(A)B.Corollary 2.9.
LetE be a Banach space with smooth dual and A1,A2,B1,B2∈B(E).
If∥MA1,B1+MA2,B2∥=∥A1∥∥B1∥+∥A2∥∥B2∥ and Ran(A2†)⊥B-JRan(A1†-(∥A1∥/∥A2∥)A2†) with A2 being surjective, and Ran(B2)⊥B-JRan(B1-(∥B1∥/∥B2∥)B2), then
(2.42)∥A1∥∥A2∥∈ℳ(A1)A2,∥B1∥∥B2∥∈ℳ(B1)B2.
Moreover, if Ran(A1†)⊥B-JRan(A2†-(∥A2∥/∥A1∥)A1†), A2 is surjective, and Ran(B1)⊥B-JRan(B2-(∥B2∥/∥B1∥)B1), then
(2.43)∥A1∥∥A2∥∈ℳ(A1)A2∩ℳ(A2)A1,∥B1∥∥B2∥∈ℳ(B1)B2∩ℳ(B2)B1.Corollary 2.10.
LetE be a Banach space with smooth dual and A1,A2,B1,B2∈B(E) such that A1,A2 are surjective operators. If Ran(Ai†)⊥B-JRan(Aj†-(∥Aj∥/∥Ai∥)Ai†) and Ran(Bi)⊥B-JRan(Bj-(∥Bj∥/∥Bi∥)Bi),(i,j=1,2 such that i≠j) then the following assertions are equivalent:(1)
∥MA1,B1+MA2,B2∥=∥A1∥∥B1∥+∥A2∥∥B2∥;(2)
∥A1∥∥A2∥∈ℳ(A1)A2∩ℳ(A2)A1and∥B1∥∥B2∥∈ℳ(B1)B2∩ℳ(B2)B1.As a particular case, we obtain the following.Corollary 2.11.
LetE be a Banach space with smooth dual and A,B are surjective operators in B(H). If
(2.44)Ran(B†)⊥B-JRan(A†-∥A∥∥B∥B†),Ran(A)⊥B-JRan(B-∥B∥∥A∥A),
then the following assertions are equivalent:(1)
∥A∥∥B∥∈ℳ(A)B∩ℳ(B)A;(2)
∥MA,B+MB,A∥=2∥A∥∥B∥.


## 3. Hilbert Space Case

LetE=ℋ be a complex Hilbert space and A∈B(ℋ). The maximal numerical range of A [11] denoted by W0(A) is defined by
(3.1){λ∈ℂ:∃(xn),∥xn∥=1,suchthatlim〈Axn,xn〉=λandlim∥Axn∥=∥A∥},
and its normalized maximal range, denoted by WN(A), is given by
(3.2)WN(A)={W0(A∥A∥)ifA≠00ifA=0.
The set W0(A) is nonempty, closed, convex, and contained in the closure of the numerical range of A.In this section we prove that ifE=ℋ, the conditions
(3.3)∥A1∥∥A2∥∈ℳ(A1)A2∩ℳ(A2)A1,∥B1∥∥B2∥∈ℳ(B1)B2∩ℳ(B2)B1
would imply that
(3.4)∥A2*A1∥=∥A1∥∥A2∥,∥B2B1*∥=∥B1∥∥B2∥,WN(A2*A1)∩WN(B2B1*)≠∅.Proposition 3.1.
Letℋ be a complex Hilbert space, A1,A2,B1,B2∈B(ℋ).
If∥A1∥∥A2∥∈ℳ(A1)A2∩ℳ(A2)A1and∥B1∥∥B2∥∈ℳ(B1)B2∩ℳ(B2)B1, then ∥A2*A1∥=∥A1∥∥A2∥ and ∥B2B1*∥=∥B1∥∥B2∥ and WN(A2*A1)∩WN(B2B1*)≠∅.Proof.
IfA1=0 or A2=0 and B1=0 or B2=0, the result is obvious.
The proof will be done in four steps, we choose one and the others will be proved similarly. Suppose thatA1≠0 and A2≠0, if ∥A1∥∥A2∥∈ℳ(A1)A2, then there exists a sequence (xn)n∈Sℋ(A2) such that
(3.5)∥A1∥∥A2∥=lim〈A1xn,A2xn〉.
We have |〈A2*A1xn,xn〉|≤∥A2*A1∥≤∥A1∥∥A2∥; this yields
(3.6)lim∥A2*A1xn∥=∥A2*A1∥=∥A1∥∥A2∥.
From (3.5) and (3.6) we get
(3.7)∥A2*A1∥=∥A1∥∥A2∥,1∈W0(A2*A1∥A2*A1∥).
Suppose now that B1≠0 and B2≠0, if ∥B1∥∥B2∥∈ℳ(B1)B2, then there exists a sequence (yn)n∈Sℋ(B2) such that
(3.8)∥B1∥∥B2∥=lim〈B1yn,B2yn〉.
Since limn∥B1yn∥=∥B1∥, then limn(B1*B1yn-∥B1∥2yn)=0.
Suppose thatwn=B1yn/∥B1∥, then yn=B1*wn/∥B1∥+zn such that limnzn=0.
Hence(3.9)〈B2yn,B1yn〉=〈B2(B1*wn∥B1∥),∥B1∥wn〉=〈B2B1*wn,wn〉+〈B2zn,∥B1∥wn〉.
From this, we derive that
(3.10)lim∥B2B1*wn∥=∥B2B1*∥=∥B1∥∥B2∥.
From (3.8) and (3.10) we have
(3.11)∥B2B1*∥=∥B1∥∥B2∥,1∈W0(B2B1*∥B2B1*∥).
From (3.7) and (3.11) we get ∥A2*A1∥=∥A1∥∥A2∥ and ∥B2B1*∥=∥B1∥∥B2∥ and WN(A2*A1)∩WN(B2B1*)≠∅.Remark 3.2.
We remark that in the caseE=ℋ we obtain an implication given by Boumazgour [12].



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*Source: 643568-2012-12-08.xml*