# On Kadison-Schwarz Type Quantum Quadratic Operators on𝕄2ℂ

Authors: Farrukh Mukhamedov; Abduaziz Abduganiev
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the description of Kadison-Schwarz type
quantum quadratic operators (q.q.o.) acting from𝕄2ℂ into 𝕄2(ℂ)⊗𝕄2(ℂ). Note that such kind of
operators is a generalization of quantum convolution. By means of
such a description we provide an example of q.q.o. which is not a
Kadison-Schwartz operator. Moreover, we study dynamics of an
associated nonlinear (i.e., quadratic) operators acting on the state
space of 𝕄2ℂ.


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## Body

## 1. Introduction

It is known that one of the main problems of quantum information is the characterization of positive and completely positive maps onC*-algebras. There are many papers devoted to this problem (see, e.g., [1–4]). In the literature the completely positive maps have proved to be of great importance in the structure theory of C*-algebras. However, general positive (order-preserving) linear maps are very intractable [2, 5]. It is therefore of interest to study conditions stronger than positivity, but weaker than complete positivity. Such a condition is called Kadison-Schwarz property, that is, a map ϕ satisfies the Kadison-Schwarz property if ϕ(a)*ϕ(a)≤ϕ(a*a) holds for every a. Note that every unital completely positive map satisfies this inequality, and a famous result of Kadison states that any positive unital map satisfies the inequality for self-adjoint elements a. In [6] relations between n-positivity of a map ϕ and the Kadison-Schwarz property of certain map is established. Certain relations between complete positivity, positive, and the Kadison-Schwarz property have been considered in [7–9]. Some spectral and ergodic properties of Kadison-Schwarz maps were investigated in [10–12].In [13] we have studied quantum quadratic operators (q.q.o.), that is, maps from 𝕄2(ℂ) into 𝕄2(ℂ)⊗𝕄2(ℂ), with the Kadison-Schwarz property. Some necessary conditions for the trace-preserving quadratic operators are found to be the Kadison-Schwarz ones. Since trace-preserving maps arise naturally in quantum information theory (see, e.g., [14]) and other situations in which one wishes to restrict attention to a quantum system that should properly be considered a subsystem of a larger system with which it interacts. Note that in [15, 16] quantum quadratic operators acting on a von Neumann algebra were defined and studied. Certain ergodic properties of such operators were studied in [17, 18] (see for review [19]). In the present paper we continue our investigation; that is, we are going to study further properties of q.q.o. with Kadison-Schwarz property. We will provide an example of q.q.o. which is not a Kadison-Schwarz operator and study its dynamics. We should stress that q.q.o. is a generalization of quantum convolution (see [20]). Some dynamical properties of quantum convolutions were investigated in [21].Note that a description of bistochastic Kadison-Schwarz mappings from𝕄2(ℂ) into 𝕄2(ℂ) has been provided in [22].


## 2. Preliminaries

In what follows, by𝕄2(ℂ) we denote an algebra of 2×2 matrices over complex filed ℂ. By 𝕄2(ℂ)⊗𝕄2(ℂ) we mean tensor product of 𝕄2(ℂ) into itself. We note that such a product can be considered as an algebra of 4×4 matrices 𝕄4(ℂ) over ℂ. In the sequel 1 means an identity matrix, that is, 1=(1001). By S(𝕄2(ℂ)) we denote the set of all states (i.e., linear positive functionals which take value 1 at 1) defined on 𝕄2(ℂ).Definition 1.
A linear operatorΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) is said to be  (a)
aquantum quadratic operator (q.q.o.) if it satisfies the following conditions:
(i)
unital, that is,Δ1=1⊗1;(ii)
Δ is positive, that is, Δx≥0 whenever x≥0;(b)
aKadison-Schwarz operator (KS) if it satisfies
(1)Δ(x*x)≥Δ(x*)Δ(x),∀x∈𝕄2(ℂ).One can see that ifΔ is unital and KS operator, then it is a q.q.o. A state h∈S(𝕄2(ℂ)) is called a Haar state for a q.q.o. Δ if for every x∈𝕄2(ℂ) one has
(2)(h⊗id)∘Δ(x)=(id⊗h)∘Δ(x)=h(x)1.Remark 2.
Note that if a quantum convolutionΔ on 𝕄2(ℂ) becomes a *-homomorphic map with a condition
(3)Lin¯((1⊗𝕄2(ℂ))Δ(𝕄2(ℂ)))=Lin¯((𝕄2(ℂ)⊗1)Δ(𝕄2(ℂ)))=𝕄2(ℂ)⊗𝕄2(ℂ),
then a pair (𝕄2(ℂ),Δ) is called a compact quantum group [20]. It is known [20] that for any given compact quantum group there exists a unique Haar state w.r.t. Δ.Remark 3.
LetU:𝕄2(ℂ)⊗𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a linear operator such that U(x⊗y)=y⊗x for all x,y∈𝕄2(ℂ). If a q.q.o. Δ satisfies UΔ=Δ, then Δ is called a  quantum quadratic stochastic operator. Such a kind of operators was studied and investigated in [17].
Each q.q.o.Δ defines a conjugate operator Δ*:(𝕄2(ℂ)⊗𝕄2(ℂ))*→𝕄2(ℂ)* by
(4)Δ*(f)(x)=f(Δx),f∈(𝕄2(ℂ)⊗𝕄2(ℂ))*,x∈𝕄2(ℂ).
One can define an operatorVΔ by
(5)VΔ(φ)=Δ*(φ⊗φ),φ∈S(𝕄2(ℂ)),
which is called a quadratic operator (q.c.). Thanks to conditions (a) (i), (ii) of Definition 1 the operator VΔ maps S(𝕄2(ℂ)) to S(𝕄2(ℂ)).


## 3. Quantum Quadratic Operators with Kadison-Schwarz Property on𝕄2(ℂ)

In this section we are going to describe quantum quadratic operators on𝕄2(ℂ) and find necessary conditions for such operators to satisfy the Kadison-Schwarz property.Recall [23] that the identity and Pauli matrices {1,σ1,σ2,σ3} form a basis for 𝕄2(ℂ), where
(6)σ1=(0110),σ2=(0-ii0),σ3=(100-1).In this basis every matrixx∈𝕄2(ℂ) can be written as x=w01+wσ with w0∈ℂ, w=(w1,w2,w3)∈ℂ3, here wσ=w1σ1+w2σ2+w3σ3.Lemma 4 (see [3]).
The following assertions hold true:(a)
x is self-adjoint if and only if w0,w are reals;(b)
Tr(x)=1 if and only if w0=0.5; here Tr is the trace of a matrix x;(c)
x>0 if and only if ∥w∥≤w0, where ∥w∥=|w1|2+|w2|2+|w3|2.Note that any stateφ∈S(𝕄2(ℂ)) can be represented by
(7)φ(w01+wσ)=w0+〈w,f〉,
where f=(f1,f2,f3)∈ℝ3 with ∥f∥≤1. Here as before 〈·,·〉 stands for the scalar product in ℂ3. Therefore, in the sequel we will identify a state φ with a vector f∈ℝ3.In what follows byτ we denote a normalized trace, that is, τ(x)=(1/2)Tr(x), x∈𝕄2(ℂ).LetΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a q.q.o. with a Haar state τ. Then one has
(8)τ⊗τ(Δx)=τ(τ⊗id)(Δ(x))=τ(x)τ(1)=τ(x),x∈𝕄2(ℂ),
which means that τ is an invariant state for Δ.Let us write the operatorΔ in terms of a basis in 𝕄2(ℂ)⊗𝕄2(ℂ) formed by the Pauli matrices, namely,
(9)Δ1=1⊗1,Δ(σi)=bi(1⊗1)+∑‍j=13bji(1)(1⊗σj)+∑‍j=13bji(2)(σj⊗1)+∑‍m,l=13bml,i(σm⊗σl),i=1,2,3,
where bi,bij(1),bij(2),bijk∈ℂ (i,j,k∈{1,2,3}).One can prove the following.Theorem 5 (see [13, Proposition 3.2]).
LetΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a q.q.o. with a Haar state τ, then it has the following form:
(10)Δ(x)=w01⊗1+∑m,l=13‍〈bml,w¯〉σm⊗σl,
where x=w0+wσ,  bml=(bml,1,bml,2,bml,3)∈ℝ3, m,n,k∈{1,2,3}.Let us turn to the positivity ofΔ. Given vector f=(f1,f2,f3)∈ℝ3 put
(11)β(f)ij=∑k=13‍bki,jfk.
Define a matrix 𝔹(f)=(β(f)ij)ij=13.By∥𝔹(f)∥ we denote a norm of the matrix 𝔹(f) associated with Euclidean norm in ℂ3. Put
(12)S={p=(p1,p2,p3)∈ℝ3:p12+p22+p32≤1}
and denote
(13)|∥𝔹∥|=supf∈S∥𝔹(f)∥.Proposition 6 (see [13, Proposition 3.3]).
LetΔ be a q.q.o. with a Haar state τ, then |∥𝔹∥|≤1.LetΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a liner operator with a Haar state τ. Then due to Theorem 5  Δ has the form (10). Take arbitrary states φ,ψ∈S(𝕄2(ℂ)) and let f,p∈S be the corresponding vectors (see (7)). Then one finds that
(14)Δ*(φ⊗ψ)(σk)=∑‍i,j=13bij,kfipj,k=1,2,3.Thanks to Lemma4 the functional Δ*(φ⊗ψ) is a state if and only if the vector
(15)fΔ*(φ,ψ)=(∑‍i,j=13bij,1fipj,∑‍i,j=13bij,2fipj,∑‍i,j=13bij,3fipj)
satisfies ∥fΔ*(φ,ψ)∥≤1.So, we have the following.Proposition 7 (see [13, Proposition 4.1]).
LetΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a liner operator with a Haar state τ. Then Δ*(φ⊗ψ)∈S(𝕄2(ℂ)) for any φ,ψ∈S(𝕄2(ℂ)) if and only if the following holds:
(16)∑‍k=13|∑‍i,j=13bij,kfipj|2≤1,∀f,p∈S.From the proof of Proposition6 and the last proposition we conclude that |∥𝔹∥|≤1 holds if and only if (16) is satisfied.Remark 8.
Note that characterizations of positive maps defined on𝕄2(ℂ) were considered in [24] (see also [25]). Characterization of completely positive mappings from 𝕄2(ℂ) into itself with invariant state τ was established in [3] (see also [26]).Next we would like to recall (see [13]) some conditions for q.q.o. to be the Kadison-Schwarz ones.LetΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a linear operator with a Haar state τ; then it has the form (10). Now we are going to find some conditions to the coefficients {bml,k} when Δ is a Kadison-Schwarz operator. Given x=w0+wσ and state φ∈S(𝕄2(ℂ)), let us denote
(17)xm=(〈bm1,w〉,〈bm2,w〉,〈bm3,w〉),fm=φ(σm),(18)αml=〈xm,xl〉-〈xl,xm〉,γml=[xm,x¯l]+[x¯m,xl],
where m,l=1,2,3. Here and in what follows [·,·] stands for the usual cross-product in ℂ3. Note that here the numbers αml are skew symmetric, that is, αml¯=-αml. By π we will denote mapping {1,2,3,4} to {1,2,3} defined by π(1)=2,π(2)=3,π(3)=1,π(4)=π(1).Denote(19)q(f,w)=(〈β(f)1,[w,w¯]〉,〈β(f)2,[w,w¯]〉,〈β(f)3,[w,w¯]〉),
where β(f)m=(β(f)m1,β(f)m2,β(f)m3) (see (11)).Theorem 9 (see [13, Theorem 3.6]).
LetΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a Kadison-Schwarz operator with a Haar state τ; then it has the form (10) and the coefficients {bml,k} satisfy the following conditions:
(20)∥w∥2≥i∑‍m=13fmαπ(m),π(m+1)+∑‍m=13∥xm∥2,(21)∥q(f,w)-i∑‍m=13fmγπ(m),π(m+1)-[xm,x¯m]∥≤∥w∥2-i∑‍k=13fkαπ(k),π(k+1)-∑‍m=13∥xm∥2
for all f∈S,w∈ℂ3. Here as before xm=(〈bm1,w〉,〈bm2,w〉,〈bm3,w〉); bml=(bml,1,bml,2,bml,3), and q(f,w), αml, and γml are defined in (19), (17), and (18), respectively.Remark 10.
The provided characterization with [2, 3] allows us to construct examples of positive or Kadison-Schwarz operators which are not completely positive (see next section).Now we are going to give a general characterization of KS operators. Let us first give some notations. For a given mappingΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ), by Δ(σ) we denote the vector (Δ(σ1),Δ(σ2),Δ(σ3)), and by wΔ(σ) we mean the following:
(22)wΔ(σ)=w1Δ(σ1)+w2Δ(σ2)+w3Δ(σ3),
where w∈ℂ3. Note that the last equality (22), due to the linearity of Δ, can also be written as wΔ(σ)=Δ(wσ).Theorem 11.
LetΔ:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be a unital *-preserving linear mapping. Then Δ is a KS operator if and only if one has
(23)i[w,w¯]Δ(σ)+(wΔ(σ))(w¯Δ(σ))≤1⊗1,
for all w∈ℂ3 with ∥w∥=1.Proof.
Letx∈M2(ℂ) be an arbitrary element, that is, x=w01+wσ. Then x*=w0¯1+w¯σ. Therefore
(24)x*x=(|w0|2+∥w∥2)1+(w0w¯+w0¯w-i[w,w¯])σ.
Consequently, we have
(25)Δ(x)=w01⊗1+wΔ(σ),Δ(x*)=w0¯1⊗1+w¯Δ(σ),(26)Δ(x*x)=(|w0|2+∥w∥2)1⊗1+(w0w¯+w0¯w-i[w,w¯])Δ(σ),(27)Δ(x)*Δ(x)=|w0|21⊗1+(w0w¯+w0¯w)Δ(σ)+(wΔ(σ))(w¯Δ(σ)).
From (26) and (27) one gets
(28)Δ(x*x)-Δ(x)*Δ(x)=∥w∥21⊗1-i[w,w¯]Δ(σ)-(wΔ(σ))(w¯Δ(σ)).
So, the positivity of the last equality implies that
(29)∥w∥21⊗1-i[w,w¯]Δ(σ)-(wΔ(σ))(w¯Δ(σ))≥0.
Now dividing both sides by ∥w∥2 we get the required inequality. Hence, this completes the proof.


## 4. An Example of Q.Q.O. Which Is Not Kadison-Schwarz One

In this section we are going to study dynamics of (57) for a special class of quadratic operators. Such class operators are associated with the following matrix {bij,k} given by
(30)b11,1=ε,b11,2=0,b11,3=0,b12,1=0,b12,2=0,b12,3=ε,b13,1=0,b13,2=ε,b13,3=0,b22,1=0,b22,2=ε,b22,3=0,b23,1=ε,b23,2=0,b23,3=0,b33,1=0,b33,2=0,b33,3=ε,
and bij,k=bji,k.Via (10) we define a liner operator Δε, for which τ is a Haar state. In the sequel we would like to find some conditions to ε which ensures positivity of Δε.It is easy that for given{bijk} one can find a form of Δε as follows.
(31)Δε(x)=w01⊗1+εω1σ1⊗σ1+εω3σ1⊗σ2+εω2σ1⊗σ3+εω3σ2⊗σ1+εω2σ2⊗σ2+εω1σ2⊗σ3+εω2σ3⊗σ1+εω1σ3⊗σ2+εω3σ3⊗σ3,
where, as before, x=w01+wσ.Theorem 12.
A linear operatorΔε given by (31) is a q.q.o. if and only if |ε|≤1/3.Proof.
Letx=w01+wσ be a positive element from 𝕄2(ℂ). Let us show positivity of the matrix Δε(x). To do it, we rewrite (31) as follows: Δε(x)=w01+εB; here(32)B=(ω3ω2-iω1ω2-iω1ω1-2iω3-ω2ω2+iω1-ω3ω1+ω2-ω2+iω1ω2+iω1ω1+ω2-ω3-ω2+iω1ω1+2iω3-ω2-ω2-iω1-ω2-iω1ω3),
where positivity ofx yields that w0,ω1,ω2,ω3 are real numbers. In what follows, without loss of generality, we may assume that w0=1, and therefore ∥w∥≤1. It is known that positivity of Δε(x) is equivalent to positivity of the eigenvalues of Δε(x).
Let us first examine eigenvalues ofB. Simple algebra shows us that all eigenvalues of B can be written as follows:
(33)λ1(w)=ω1+ω2+ω3+2ω12+ω22+ω32-ω1ω2-ω1ω3-ω2ω3,λ2(w)=ω1+ω2+ω3-2ω12+ω22+ω32-ω1ω2-ω1ω3-ω2ω3,λ3(w)=λ4(w)=-ω1-ω2-ω3.
Now examine maximum and minimum values of the functions λ1(w),λ2(w),λ3(w),λ4(w) on the ball ∥w∥≤1.
One can see that(34)|λ3(w)|=|λ4(w)|≤∑k=13‍|ωk|≤3∑k=13‍|ωk|2≤3.
Note that the functionsλ3,λ4 can reach values ±3 at ±(1/3,1/3,1/3).
Now let us rewriteλ1(w) and λ2(w) as follows:
(35)λ1(w)=ω1+ω2+ω3+223(ω12+ω22+ω32)-(ω1+ω2+ω3)2,(36)λ2(w)=ω1+ω2+ω3-223(ω12+ω22+ω32)-(ω1+ω2+ω3)2.
One can see that(37)λk(hω1,hω2,hω3)=hλk(ω1,ω2,w3),ifh≥0,(38)λ1(hω1,hω2,hω3)=hλ2(ω1,ω2,w3),ifh≤0.
where k=1,2. Therefore, the functions λk(w),  k=1,2 reach their maximum and minimum on the sphere ω12+ω22+ω32=1 (i.e., ∥w∥=1). Hence, denoting t=ω1+ω2+ω3 from (37) and (36) we introduce the following functions:
(39)g1(t)=t+223-t2,g2(t)=t-223-t2,
where |t|≤3.
One can find that the critical values ofg1 are t=±1, and the critical value of g2 is t=-1. Consequently, extremal values of g1 and g2 on |t|≤3 are the following:
(40)min|t|≤3g1(t)=-3,max|t|≤3g1(t)=3,min|t|≤3g2(t)=-3,max|t|≤3g2(t)=3.
Therefore, from (37) and (38) we conclude that
(41)-3≤λk(w)≤3,forany∥w∥≤1,k=1,2.
It is known that for the spectrum of1+εB one has
(42)Sp(1+εB)=1+εSp(B).
Therefore,
(43)Sp(1+εB)={1+ελk(w):k=1,4¯}.
So, if
(44)|ε|≤1max∥w∥≤1|λk(w)|,k=1,4¯,
then one can see 1+ελk(w)≥0 for all ∥w∥≤1,   k=1,4¯. This implies that the matrix 1+εB is positive for all w with ∥w∥≤1.
Now assume thatΔε is positive. Then Δε(x) is positive whenever x is positive. This means that 1+ελk(w)≥0 for all ∥w∥≤1(k=1,4¯). From (34) and (41) we conclude that |ε|≤1/3. This completes the proof.Theorem 13.
Letε=1/3 then the corresponding q.q.o. Δε is not KS operator.Proof.
It is enough to show the dissatisfaction of (21) at some values of w (∥w∥≤1) and f=(f1,f1,f2).
Assume thatf=(1,0,0); then a little algebra shows that (21) reduces to the following one:
(45)A+B+C≤D,
where
(46)A=|ε(ω¯2ω3-ω¯3ω2)-iε2(2ω¯2ω3-2|ω1|2-ω¯2ω1+ω¯1ω2-ω¯1ω3+ω¯3ω1)|2,B=|ε(ω¯1ω2-ω¯2ω1)-iε2(2ω¯1ω2-2|ω3|2-ω¯1ω3+ω¯3ω1-ω¯3ω2+ω¯2ω3)|2,C=|ε(ω¯3ω1-ω¯1ω3)-iε2(2ω¯3ω1-2|ω2|2-ω¯3ω2+ω¯2ω3-ω¯2ω1+ω¯1ω2)|2,D=(1-3|ε|2)(|ω1|2+|ω2|2+|ω3|2)-iε2(ω¯3w2-ω¯2ω3+ω¯2ω1-ω¯1ω2+ω¯1ω3-ω¯3ω1).
Now choosew as follows:
(47)ω1=-19,ω2=536,ω3=5i27.
Then calculations show that
(48)A=959419131876,B=1962586093442,C=16253779136,D=58917496.
Hence, we find
(49)959419131876+1962586093442+16253779136>58917496,
which means that (45) is not satisfied. Hence, Δε is not a KS operator at ε=1/3.Recall that a linear operatorT:𝕄k(ℂ)→𝕄m(ℂ) is completely positive if for any positive matrix (aij))i,j=1n∈𝕄k(𝕄n(ℂ)) the matrix (T(aij))i,j=1n is positive for all n∈ℕ. Now we are interested when the operator Δε is completely positive. It is known [1] that the complete positivity of Δε is equivalent to the positivity of the following matrix:
(50)Δ^ε=(Δε(e11)Δε(e12)Δε(e21)Δε(e22)),
here eij (i,j=1,2) are the standard matrix units in 𝕄2(ℂ).From (31) one can calculate that
(51)Δε(e11)=121⊗1+εB11,Δε(e22)=121⊗1-εB11,Δε(e12)=εB12,Δε(e21)=εB12*,
where
(52)B11=(1200-i0-120000-120i0012),B12=(0001-i2i01+i20i1+i2001-i2-i-i0).
Hence, we find that
(53)2Δ^ε=18+ε𝔹,
where 18 is the unit matrix in 𝕄8(ℂ) and
(54)𝔹=(100-2i0001-i0-1002i01+i000-102i1+i002i0011-i-2i-2i00-2i-2i1+i-1002i001-i2i010001-i02i00101+i000-2i00-1).
So, the matrix Δ^ε is positive if and only if
(55)|ε|≤1λmax(𝔹),
where λmax(𝔹)=maxλ∈Sp(𝔹)|λ|.One can easily calculate thatλmax(𝔹)=33. Therefore, we have the following.Theorem 14.
LetΔε:𝕄2(ℂ)→𝕄2(ℂ)⊗𝕄2(ℂ) be given by (31). Then Δε is completely positive if and only if |ε|≤1/33.


## 5. Dynamics ofΔε

LetΔ be a q.q.o. on 𝕄2(ℂ). Let us consider the corresponding quadratic operator defined by VΔ(φ)=Δ*(φ⊗φ), φ∈S(𝕄2(ℂ)). From Theorem 5 one can see that the defined operator VΔ maps S(𝕄2(ℂ)) into itself if and only if |∥𝔹∥|≤1 or equivalently (16) holds. From (14) we find that
(56)VΔ(φ)(σk)=∑‍i,j=13bij,kfifj,f∈S.
Here, as before, S={f=(f1,f2fp3)∈ℝ3:f12+f22+f32≤1}.So, (56) suggests that we consider the following nonlinear operator V:S→S defined by
(57)V(f)k=∑‍i,j=13bij,kfifj,k=1,2,3,
where f=(f1,f2,f3)∈S.It is worth to mention that uniqueness of the fixed point (i.e.,(0,0,0)) of the operator given by (57) was investigated in [13, Theorem 4.4].In this section, we are going to study dynamics of the quadratic operatorVε corresponding to Δε (see (31)), which has the following form
(58)Vε(f)1=ε(f12+2f2f3),Vε(f)2=ε(f22+2f1f3),Vε(f)3=ε(f32+2f1f2).Let us first find some condition onε which ensures (16).Lemma 15.
LetVε be given by (58). Then Vε maps S into itself if and only if |ε|≤1/3 is satisfied.Proof.
“If” Part. Assume thatVε maps S into itself. Then (16) is satisfied. Take f=(1/3,1/3,1/3), p=f. Then from (16) one finds that
(59)∑k=13‍|∑i,j=13‍bij,kfipj|2=3ε2≤1
which yields |ε|≤1/3.
“Only If” Part. Assume that|ε|≤1/3. Take any f=(f1,f2,f3),p=(p1,p2,p3)∈S. Then one finds that
(60)∑k=13‍|∑i,j=13‍bij,kfipj|2=ε2(|f1p1+f3p2+f2p3|2+|f3p1+f2p2+f1p3|2+|f2p1+f1p2+f3p3|2)≤ε2((f12+f22+f32)(p12+p22+p32)+(f32+f22+f12)(p12+p22+p32)+(p12+p22+p32)(f22+f12+f32))≤ε2(1+1+1)=3ε2≤1.
This completes the proof.Remark 16.
We stress that condition (16) is necessary for Δ to be a positive operator. Namely, from Theorem 12 and Lemma 15 we conclude that if ε∈(1/3,1/3] then the operator Δε is not positive, while (16) is satisfied.In what follows, to study dynamics ofVε we assume |ε|≤1/3. Recall that a vector f∈S is a fixed point of Vε if Vε(f)=f. Clearly (0,0,0) is a fixed point of Vε. Let us find others. To do it, we need to solve the following equation:
(61)ε(f12+2f2f3)=f1,ε(f22+2f1f3)=f2,ε(f32+2f1f2)=f3.We have the following.Proposition 17.
If|ε|<1/3 then Vε has a unique fixed point (0,0,0) in S. If  |ε|=1/3 then Vε has the following fixed points: (0,0,0) and (±1/3,±1/3,±1/3) in S.Proof.
It is clear that(0,0,0) is a fixed point of Vε. If fk=0, for some k∈{1,2,3} then due to |ε|≤1/3, one can see that the only solution of (61) belonging to S is f1=f2=f3=0. Therefore, we assume that fk≠0 (k=1,2,3). So, from (61) one finds
(62)f12+2f2f3f22+2f1f3=f1f2,f12+2f2f3f32+2f1f2=f1f3,f22+2f1f3f32+2f1f2=f2f3.
Denoting
(63)x=f1f2,y=f1f3,z=f2f3.
From (62) it follows that
(64)x(x(1+2/xy)1+2x/z-1)=0,y(y(1+2/xy)1+2yz-1)=0,z(z(1+2x/z)1+2yz-1)=0.
According to our assumptionx,y,z are nonzero, so from (64) one gets
(65)x(1+2/xy)1+2x/z=1,y(1+2/xy)1+2yz=1,z(1+2x/z)1+2yz=1,
where 2x≠-z and 2yz≠-1.
Dividing the second equality of (65) to the first one of (65) we find that
(66)y(1+2x/z)x(1+2yz)=1,
which with xz=y yields
(67)y+2x2=x+2y2.
Simplifying the last equality one gets
(68)(y-x)(1-2(y+x))=0.
This means that either y=x or x+y=1/2.
Assume thatx=y. Then from xz=y, one finds z=1. Moreover, from the second equality of (65) we have y+2/y=1+2y. So, y2+y-2=0; therefore, the solutions of the last one are y1=1,y2=-2. Hence, x1=1,x2=-2.
Now suppose thatx+y=1/2; then x=1/2-y. We note that y≠1/2, since x≠0. So, from the second equality of (65) we find
(69)y+41-2y=1+4y21-2y.
So, 2y2-y-1=0 which yields the solutions y3=-1/2,y4=1. Therefore, we obtain x3=1, z3=-1/2 and x4=-1/2, z4=-2.
Consequently, solutions of (65) are the following ones:
(70)(1,1,1),(1,-12,-12),(-12,1,-2),(-2,-2,1).
Now owing to (63) we need to solve the following equations:
(71)f1f2=xk,f2f3=zk.k=1,4¯,
According to our assumption fk≠0, we consider cases when xkzk≠0.
Now let us start to consider several cases.
Case  1. Let x2=1, z2=1. Then from (71) one gets f1=f2=f3. So, from (61) we find 3εf12=f1, that is, f1=1/3ε. Now taking into account f12+f22+f32≤1 one gets 1/3ε2≤1. From the last inequality we have |ε|≥1/3. Due to Lemma 15 the operator Vε is well defined if and only if |ε|≤1/3; therefore, one gets |ε|=1/3. Hence, in this case a solution is (±1/3;±1/3;±1/3).
Case  2. Let x2=1, z2=-1/2. Then from (71) one finds f1=f2,2f2=-f3. Substituting the last ones to (61) we get f1+3f12ε=0. Then, we have f1=-1/3ε,f2=-1/3ε,f3=2/3ε. Taking into account f12+f22+f32≤1 we find 1/9ε2+4/9ε2+1/9ε2≤1. This means |ε|≥2/3; due to Lemma 15 in this case the operator Vε is not well defined; therefore, we conclude that there is no fixed point of Vε belonging to S.
Using the same argument for the rest of the cases we conclude the absence of solutions. This shows that if|ε|<1/3 the operator Vε has unique fixed point in S. If |ε|=1/3, then Vε has three fixed points belonging to S. This completes the proof.Now we are going to study dynamics of operatorVε.Theorem 18.
LetVε be given by (58). Then the following assertions hold true:  (i)
if|ε|<1/3, then for any f∈S one has Vεn(f)→(0,0,0) as n→∞.(ii)
if|ε|=1/3, then for any f∈S with f∉{(±1/3,±1/3,±1/3)} one has Vεn(f)→(0,0,0) as n→∞.Proof.
Let us consider the following functionρ(f)=f12+f22+f32. Then we have
(72)ρ(Vε(f))=ε2((f12+2f2f3)2+(f22+2f1f3)2+(f32+2f1f2)2)≤ε2(f12+2|f2||f3|+f22+2|f1||f3|+f32+2|f1||f2|)≤ε2(f12+f22+f32+f22+f12+f32+f32+f12+f22)=3ε2(f12+f22+f32)=3ε2ρ(f).
This means(73)ρ(Vε(f))≤3ε2ρ(f).
Due toε2≤1/3 from (73) one finds that
(74)ρ(Vεn+1(f))≤ρ(Vεn(f)),
which yields that the sequence {ρ(Vεn(f))} is convergent. Next we would like to find the limit of {ρ(Vεn(f))}.(i)
First we assume that|ε|<1/3; then from (73) we obtain
(75)ρ(Vεn(f))≤3ε2ρ(Vεn-1(f))≤⋯≤(3ε2)nρ(f).
This yields that ρ(Vεn(f))→0 as n→∞, for all f∈S.(ii)
Now let|ε|=1/3. Then consider two distinct subcases.
Case A. Let f12+f22+f32<1 and denote d=f12+f22+f32. Then one gets
(76)ρ(Vε(f))≤ε2((f12+2|f2||f3|)2+(f22+2|f1||f3|)2+(f32+2|f1||f2|)2)≤ε2((f12+f22+f32)2+(f22+f12+f32)2+(f32+f12+f22)2)=3ε2d2=dd=dρ(f).
Hence, we have ρ(Vε(f))≤dρ(f). This means ρ(Vεn(f))≤dnρ(f)→0. Hence, Vεn(f)→0 as n→∞.
Case B. Now take f12+f22+f32=1 and assume that f is not a fixed point. Therefore, we may assume that fi≠fj for some i≠j, otherwise from Proposition 17 one concludes that f is a fixed point. Hence, from (58) one finds
(77)Vε(f)1=ε(f12+2f2f3)=ε(1-f22-f32+2f2f3)=ε(1-(f2-f3)2).
Similarly, one gets
(78)Vε(f)2=ε(1-(f1-f3)2),Vε(f)3=ε(1-(f1-f2)2).
It is clear that |Vε(f)k|≤|ε| (k=1,2,3). According to our assumption fi≠fj (i≠j) we conclude that one of |Vε(f)k| is strictly less than 1/3; this means Vε(f)12+Vε(f)22+Vε(f)32<1. Therefore, from Case A, one gets that Vεn(f)→0 as n→∞.



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*Source: 278606-2013-04-17.xml*