Selcuk Journal of
Applied Mathematics
Vol. 4, No. 1, pp. 25–41, 2003On location of the matrix spectrum inside an
ellipse
Ay¸se Bulgak1, Gennadi˘ı Demidenko2, and Inessa Matveeva2 1 Research Centre of Applied Mathematics, Selcuk University, Konya, Turkey;
e-mail: [email protected]
2 Sobolev Institute of Mathematics, SB RAS, 630090 Novosibirsk, Russia;
e-mail: [email protected], e-mail: [email protected] Received: July 3, 2002
Summary. In the present article we consider the problem on loca-tion of the spectrum of an arbitrary matrixA inside the ellipse
E = {λ ∈ C : (Rea2λ)2 +(Imλ)
2
b2 = 1}, a > b.
One of the authors (see [9]) established a connection of the problem with solvability of the matrix equation
H − 1 2a2 + 1 2b2 A∗HA − 1 4a2 − 1 4b2 (HA2+ (A∗)2H) = C. In this article we construct a Hermitian positive definite solution H to the equation in the form of a power series. We prove that the norm
H characterizes an immersion depth of eigenvalues of the matrix A
in the inside of the ellipseE. On the base of these results we propose an algorithm to determine whether the spectrum of the matrix A belongs to the inside of the ellipseE.
Key words: ellipse, matrix spectrum, matrix equations, matrix series, algorithm
2000 Mathematics Subject Classification: 15A18, 15A24, 65F30
The research was financially supported by the Scientific and Technical
Re-search Council of Turkey (TUBITAK) in the framework of the NATO-PC Fellow-ships Programme.
1. Introduction
Spectral problems of linear algebra are very important both for the-oretical and for applied investigations. However, solving eigenvalue problems for nonsymmetric matrices can lead to essential errors (see, for example, [1–4]). Therefore, obtaining of various criteria and algo-rithms to determine location of the matrix spectrum in the complex plane is a very important problem.
Essential criteria used in linear algebra to localize the matrix spectrum are the Lyapunov Criteria. These criteria indicate neces-sary and sufficient conditions for the matrix spectrum to belong to the unit disk Ci = {λ ∈ C : |λ| < 1} and the open left half-plane C−={λ ∈ C : Re λ < 0}. These conditions are formulated by using
the Lyapunov matrix equations
(1) H − A∗HA = C
and
HA − A∗H = −C
respectively. It is very important that these criteria enable to carry out numerical research. In particular, using the Lyapunov Criteria H. Bulgak and S. K. Godunov elaborated algorithms with guaranteed accuracy for numerical solving problems on location of the matrix spectrum in Ci and C−. The survey of results in this direction is contained in [3, 5]. The articles [6–9] are devoted to generalizations of the Lyapunov Criteria on location of the matrix spectrum.
In the present article we continue to research spectral problems of linear algebra by using criteria of Lyapunov type. We will consider the problem on location of the spectrum of an arbitrary (N × N) matrixA inside an ellipse, namely, in the domain
Ei={λ ∈ C : (Reλ)
2
a2 +
(Imλ)2
b2 < 1}, a > b.
Our aim is to elaborate an algorithm for numerical solving this prob-lem. In our opinion such algorithms have important applications to numerical research of spectral portraits of matrices. For this we will use the following matrix equation
(2) H − 1 2a2 + 1 2b2 A∗HA− 1 4a2 − 1 4b2 (HA2+ (A∗)2H) = C. The equation was introduced in the article [9]. In the case ofa = b = 1 the equation coincides with the discrete Lyapunov matrix equation (1). Using the equation (2) the following result was established [9].
Theorem 1. Let C be a Hermitian positive definite matrix. If there
exists a solutionH = H∗ > 0 to the equation (2), then all eigenvalues of the matrix A belong to Ei.
As was mentioned in [9], the inverse assertion holds too. Moreover, the solutionH = H∗ > 0 is uniquely defined, i. e. a criterion for the matrix spectrum to belong to Ei holds. The criterion is formulated by using the matrix equation (2). Obviously, in the case ofa = b = 1 this criterion coincides with the Lyapunov Criterion. Therefore, by analogy with the problem on location of the matrix spectrum inside the unit disk, using the matrix equation (2) one can try to elaborate an algorithm to determine whether the matrix spectrum belongs to
Ei. In the present article we propose such algorithm. Its description is
contained in Section 4. Sections 2 and 3 include auxiliary assertions on properties of a solution to the equation (2) and location of the matrix spectrum. Particularly, in Section 2 we show how one can construct a Hermitian positive definite solution to the equation (2) withC = I in the form of a power series. In Section 3 we prove that if the spectrum of a matrixA belongs to Ei, then, using the normH, one can point out a distance between the spectrum and the ellipse. Hence, the norm H can be considered as a value characterizing an immersion depth of eigenvalues of the matrix A in the inside of the ellipse. It should be noted that the norm of a solution to the discrete Lyapunov matrix equation (1) takes an analogous role in the problem on location of the spectrum ofA inside the unit disk Ci (see, for example, [3]).
Note that an another algorithm for the considerable problem is proposed in the articles [10].
2. A solution to the equation (2) and its properties
In this section we construct a solution to the equation (2) withC = I and establish some its properties.
Introduce the following notation:
α = 1 2a2 + 1 2b2, β = 1 4a2 − 1 4b2.
We will seek a solution to the equation (2) withC = I in the form
H = I + H1. Then H1 has to be a solution to the equation
H1− αA∗H1A − β(H1A2+ (A∗)2H1) =h1,
where
Similarly, we will seek a solution to the above equation in the form
H1=h1+H2. ThenH2 has to be a solution to the equation
H2− αA∗H2A − β(H2A2+ (A∗)2H2) =h2,
where
h2=αA∗h1A + β(h1A2+ (A∗)2h1).
Repeating the procedure we obtain that H can be represented as follows H = k−1 j=0 hj +Hk, where h0 =I, hj =αA∗hj−1A+β(hj−1A2+(A∗)2hj−1), j = 1, . . . , k−1,
and Hk is a solution to the equation
Hk− αA∗HkA − β(HkA2+ (A∗)2Hk) =hk with
hk=αA∗hk−1A + β(hk−1A2+ (A∗)2hk−1).
Thus, if the matrix series ∞
j=0hj converges, then the matrix
(3) H =
∞
j=0
hj
is a solution to the equation (2) with C = I. Here
h0 =I, hj =αA∗hj−1A + β(hj−1A2+ (A∗)2hj−1), j = 1, 2, . . . . Obviously, the matrix series (3) is Hermitian. Now we show that the matrix series (3) is positive definite. We will use two auxiliary theorems for this aim.
Theorem 2. The series (3) has the form
(4) H = ∞ j=0 j l=0 Cl jαj−l(A∗)j−l βll s=0 Cs l((A∗)2)s(A2)l−s Aj−l ,
Proof. Now we show that each termhj,j = 1, 2, . . ., of the series (3) can be written as follows
(5) hj = j l=0 Cjlαj−l(A∗)j−l βl l s=0 Cls((A∗)2)s(A2)l−s Aj−l.
We prove (5) by the method of mathematical induction. Obviously, the representation (5) holds for j = 1. Suppose that it holds for
j = k − 1. Consider the case of j = k. By definition, hk=αA∗hk−1A + β(hk−1A2+ (A∗)2hk−1). Consequently, hk=αA∗ k−1 l=0 Ck−1l αk−1−l(A∗)k−1−l × βll s=0 Cls((A∗)2)s(A2)l−s Ak−1−l A +β k−1 l=0 Ck−1l αk−1−l(A∗)k−1−l × βll s=0 Cls((A∗)2)s(A2)l−s Ak−1−l A2 +β(A∗)2 k−1 l=0 Ck−1l αk−1−l(A∗)k−1−l × βll s=0 Cls((A∗)2)s(A2)l−s Ak−1−l =αk(A∗)k(A)k +αA∗ k−1 l=1 Ck−1l αk−1−l(A∗)k−1−l × βll s=0 Cls((A∗)2)s(A2)l−s Ak−1−l A +β k−2 l=0 Ck−1l αk−1−l(A∗)k−1−l
× βl l s=0 Cls((A∗)2)s(A2)l−s Ak−1−l A2 +β(A∗)2 k−2 l=0 Cl k−1αk−1−l(A∗)k−1−l × βl l s=0 Cls((A∗)2)s(A2)l−s Ak−1−l +βk k−1 s=0 Cs k−1((A∗)2)s(A2)k−s+ k−1 s=0 Cs k−1((A∗)2)s+1(A2)k−1−s =J1+J2+J3+J4+J5. Firstly we consider the matrixJ5. Clearly,
J5 =βk k−1 s=0 Ck−1s ((A∗)2)s(A2)k−s+ k s=1 Ck−1s−1((A∗)2)s(A2)k−s =βk(A2)k+βk k−1 s=1 [Ck−1s +Ck−1s−1]((A∗)2)s(A2)k−s+βk((A∗)2)k =βk k s=0 Cks((A∗)2)s(A2)k−s.
It is easy to show thatJ2,J3 and J4 can be written as follows
J2 = k−1 l=1 Cl k−1αk−l(A∗)k−l βll s=0 Cs l((A∗)2)s(A2)l−s Ak−l, J3 = k−1 l=1 Ck−1l−1αk−l(A∗)k−l βll−1 s=0 Cl−1s ((A∗)2)s(A2)l−s Ak−l, J4 = k−2 l=0 Cl k−1αk−1−l(A∗)k−1−l × ⎡ ⎣βl+1l+1 η=1 Clη−1((A∗)2)η(A2)l−η+1 ⎤ ⎦ Ak−1−l = k−1 l=1 Ck−1l−1αk−l(A∗)k−l βl l s=1 Cl−1s−1((A∗)2)s(A2)l−s Ak−l.
Then, reasoning in the same way as for J5, we have J3+J4 = k−1 l=1 Ck−1l−1αk−l(A∗)k−l βll s=0 Cls((A∗)2)s(A2)l−s Ak−l. Consequently, J2+J3+J4= k−1 l=1 [Ck−1l +Ck−1l−1]αk−l(A∗)k−l × βll s=0 Cls((A∗)2)s(A2)l−s Ak−l = k−1 l=1 Cklαk−l(A∗)k−l βll s=0 Cls((A∗)2)s(A2)l−s Ak−l. Thus, J1+J2+J3+J4+J5 = k l=0 Cklαk−l(A∗)k−l βll s=0 Cls((A∗)2)s(A2)l−s Ak−l.
Hence, the representation (5) holds for j = k.
From (5) we obtain (4). The theorem is proved. Theorem 3. The series (3) can be written as follows (6) H = (1 − γ2) ∞ j=0 fj∗fj, where f0=I, f1 =σA, fj =σfj−1A + γfj−2, j = 2, 3, . . . γ = −a − b a + b, σ = 2 a + b.
Proof. To prove the representation (6) it is sufficient to calculate
coefficients at the products
(A∗)lAs, l, s = 0, 1, 2, . . . l + s = 2m, m = 0, 1, 2, . . . , in (4) and (6). Now we verify some coefficients. Denote the coefficients at the products (A∗)lAs in (4) and (6) by αls and ˜αls respectively.
The coefficients α00 at the unit matrix I in (4) equals 1. On the other hand,
˜
α00= (1− γ2)(1 +γ2+γ4+γ6+. . .).
By definition,|γ| < 1. Hence, ˜α00= (1− γ2)1−γ1 2 = 1.
Consider the coefficients α11 and ˜α11 at the matrix A∗A in (4) and (6) respectively. By definition,
α11=α = σ2(1 +γ2) 1 (1− γ2)2. Since |γ| < 1 it follows that
α11=σ2(1 +γ2)(1 +γ2+γ4+γ6+. . .)(1 + γ2+γ4+γ6+. . .) =σ2(1 +γ2)(1 + 2γ2+ 3γ4+ 4γ6+. . .) = σ2(1 + 3γ2+ 5γ4+ 7γ6+. . .). On the other hand,
˜
α11=σ2(1− γ2)(1 + 4γ2+ 9γ4+ 16γ6+. . .)
=σ2(1 + 3γ2+ 5γ4+ 7γ6+. . .). Thus,α11= ˜α11.
Consider the coefficientsα02 and ˜α02at the matrixA2 in (4) and (6) respectively. Note that α02=α20 and ˜α02 = ˜α20, where α20 and
˜
α20are the coefficients at the matrix (A∗)2in (4) and (6) respectively.
By definition,
α02=β = σ2γ(1− γ1 2)2.
Then,
α02=σ2γ(1 + γ2+γ4+γ6+. . .)(1 + γ2+γ4+γ6+. . .)
=σ2γ(1 + 2γ2+ 3γ4+ 4γ6+. . .). On the other hand,
˜
α02=σ2γ(1 − γ2)(1 + 3γ2+ 6γ4+ 10γ6+. . .)
=σ2γ(1 + 2γ2+ 3γ4+ 4γ6+. . .). Thus,α02= ˜α02.
The other coefficients in (4) and (6) are verified in the same way. The theorem is proved.
Corollary 1. The matrix series (3) is positive definite. Moreover,
the matrix H − I is nonnegative definite, H ≥ 1, and H = 1 if and only if A = 0.
3. On location of the matrix spectrum
According to Theorem 1, if the equation (2) for a matrix A has a solutionH that is Hermitian and positive definite, then the spectrum of A belongs to Ei. However, using the norm ofH, one can indicate more exact boundaries of a domain including the spectrum ofA. Theorem 4. Let H be a Hermitian positive definite solution to the
equation (2) withC = I. Then the spectrum of A belongs to EiH = λ ∈ C : (Rea2λ)2 +(Imλ) 2 b2 ≤ 1 − 1 H .
Proof. Let λk be an eigenvalue of the matrix A, and let vk be an eigenvector corresponding to the eigenvalueλk. Then, from the equa-tion (2) withC = I we have
Hvk, vk − αA∗HAvk, vk − β(HA2+ (A∗)2H)vk, vk = vk2.
Since the vectorvk is an eigenvector of A it follows that
vk2 =Hvk, vk − α|λk|2Hvk, vk − β(λ2k+ ¯λ2k)Hvk, vk = 1−(Reλk) 2 a2 − (Imλk)2 b2 Hvk, vk .
Taking into account the inequality
Hvk, vk ≤ Hvk2 we have (Reλk)2 a2 + (Imλk)2 b2 ≤ 1 − 1 H.
The theorem is proved.
Remark 1. In the case of a = b = 1 the obtained result gives the
known one [3, 11].
It should be noted that the series (6) can be written as follows
(7) H = 1− γ 2 2 I + 1− γ2 2 ∞ j=0 (f0∗f1∗)(L∗)jLj f0 f1 , where f0 =I, f1 =σA, γ = −a − b a + b, σ = 2 a + b,
and the matrix L = 0 I γI σA
was introduced in [9]. Indeed, it is easy to show that fj−1 fj =L fj−2 fj−1 , j = 2, 3, . . . . Then, fj−1∗ fj−1+fj∗fj = (fj−1∗ fj∗) fj−1 fj = (fj−2∗ fj−1∗ )L∗L fj−2 fj−1 = (f0∗f1∗)(L∗)j−1Lj−1 f0 f1 , j = 2, 3, . . . .
Hence, we obtain the representation (7).
As was established in [9], if the spectrum ofA belongs to Ei, then the spectrum ofL ones to the unit disk. Hence, the Lyapunov matrix equation
(8) H − L∗HL = I
has a unique solution H = H∗> 0. Write the matrix H as follows
H = H11H12 H∗ 12H22 ,
whereHij are (N × N) matrices. Note some interesting properties of the matrices Hij. From (8) it follows that
H11=I + γ2H22, H12= γ
abσH22A + γ2
abσA∗H22,
and the matrixH22 satisfies the equality
H22− 1 2a2 + 1 2b2 A∗H 22A − 1 4a2 − 1 4b2 (H22A2+ (A∗)2H22) = (a + b) 2 2ab I,
i. e. the matrix H22 is a solution to the equation (2) with the right-hand side (a+b)2ab2I. Obviously, the matrix (a+b)2ab2H22 = 1−γ
2
2 H22 is a
solution to the equation (2) with C = I and can be represented in the form of the convergent series (7).
In the following theorem we indicate a disk including the spectrum of L if the spectrum of A belongs to EiH.
Theorem 5. If the spectrum of A belongs to EiH, then the spectrum of L ones to the disk
λ ∈ C : |λ| ≤ a + b1 a2− b2 H +b 1− 1 H < 1 . Proof. According to results established in [9], to define boundaries of
the spectrum of the matrix L we need to estimate |z|, where
z = a + b1 (λ +λ2− (a2− b2)) and λ belongs to the boundary ∂EiH ofEiH:
∂EiH ={λ ∈ C : (Reλ) 2 a2 + (Imλ)2 b2 = 1− 1 H}. Let
Reλ = ˜a cos ϕ, Im λ = ˜b sin ϕ, where ˜ a = a 1− 1 H, ˜b = b 1− 1 H, 0 ≤ ϕ ≤ 2π.
Then z can be written as follows
z = a + b1
˜
a cos ϕ + i˜b sin ϕ +(˜b cos ϕ + i˜a sin ϕ)2− 2
,
where
2 = a2− b2
H .
Denote the expression
(˜b cos ϕ + i˜a sin ϕ)2− 2 by z1. By definition, |Re√z1| = |z1| + Re z1 2 , |Im √z 1| = |z1| − Re z1 2 . Obviously, |z1| + Re z1 2 = 1 2
(˜b2cos2ϕ − ˜a2sin2ϕ − 2)2+ 4˜b2a˜2cos2ϕ sin2ϕ +1
2(˜b
2cos2ϕ − ˜a2sin2ϕ − 2)≤ 1
2(˜b
+1 2(˜b
2cos2ϕ − ˜a2sin2ϕ − 2) = ˜b2cos2ϕ.
By analogy, |z1| − Re z1 2 = 1 2
(˜b2cos2ϕ − ˜a2sin2ϕ − 2)2+ 4˜b2a˜2cos2ϕ sin2ϕ
−1
2(˜b
2cos2ϕ − ˜a2sin2ϕ − 2)≤ 1
2(˜b
2cos2ϕ + ˜a2sin2ϕ + 2)
−1
2(˜b
2cos2ϕ − ˜a2sin2ϕ − 2) = ˜a2sin2ϕ + 2.
Hence,
|Re√z1| ≤ ˜b| cos ϕ|, |Im√z1| ≤
˜ a2sin2ϕ + 2. Then, |z| = a + b1 (˜a cos ϕ + Re√z1)2+ (˜b sin ϕ + Im√z1)2 ≤ a + b1
(˜a| cos ϕ| + ˜b| cos ϕ|)2+ (˜b| sin ϕ| + ˜ a2sin2ϕ + 2)2 = 1 a + b ˜
a2+ ˜b2+2+ 2˜b(˜a cos2ϕ +˜a2sin2ϕ + 2| sin ϕ|).
Consider the function
F (ϕ) = ˜a cos2ϕ +˜a2sin2ϕ + 2| sin ϕ|, 0 ≤ ϕ ≤ 2π. Define the maximum of the function. It is sufficient to investigate the function
f(ϕ) = ˜a cos2ϕ +˜a2sin2ϕ + 2sinϕ, for 0≤ ϕ ≤ π. It is easy to show that
f(ϕ) = cos ϕ−2˜a sin ϕ +˜a2sin2ϕ + 2+ ˜a2sin2ϕ
˜ a2sin2ϕ + 2 = cosϕ ˜ a2sin2ϕ + 2 ˜ a sin ϕ −˜a2sin2ϕ + 2 2 . Then, max ϕ∈[0,π]f(ϕ) = f(π/2) = ˜ a2+2, and max ϕ∈[0,2π]F (ϕ) = ˜ a2+2.
Consequently, |z| ≤ a + b1 ˜ a2+ ˜b2+2+ 2˜b˜a2+2 = 1 a + b( ˜ a2+2+ ˜b) = 1 a + b a2− b2 H+b 1− 1 H < 1.
Note that the obtained inequality is exact. Indeed,
|z| = 1 a + b a2− b2 H+b 1− 1 H forλ = ib 1−H1 ∈ ∂EiH.
Thus, if the spectrum of A belongs to EiH, then the spectrum of
L ones to the disk
λ ∈ C : |λ| ≤ a + b1 a2− b2 H +b 1− 1 H .
The theorem is proved.
4. Numerical research of the matrix spectrum
In this section we propose an algorithm to research by a computer whether the matrix spectrum belongs toEi:
Ei={λ ∈ C : (Reλ)
2
a2 +
(Imλ)2
b2 < 1}, a > b.
By Theorem 1, if the equation (2) withC = C∗> 0 has a unique solution H = H∗ > 0, then all eigenvalues of A belong to Ei. The solution to the equation (2) is represented in the form of a convergent series. In particular, in the case ofC = I the solution can be written in the form of the series (6). Obviously, for the partial sum
(9) Hk = (1− γ2) k−1 j=0 f∗ jfj we have (10) Hk− 1 2a2 + 1 2b2 A∗HkA − 1 4a2 − 1 4b2 (HkA2+ (A∗)2Hk)
=I − Hk+ 1 2a2 + 1 2b2 A∗HkA + 1 4a2 − 1 4b2 ( HkA2+ (A∗)2Hk), where Hk= (1− γ2) ∞ j=k fj∗fj.
Since the series (6) converges it follows that the right-hand side of (10) is Hermitian positive definite for sufficiently largek. Consequently, to find out whether the spectrum of A belongs to Ei one can calculate the Hermitian matrixHk by using (9) and research the matrix
Hk− 1 2a2 + 1 2b2 A∗H kA − 1 4a2 − 1 4b2 (HkA2+ (A∗)2Hk). If it is positive definite then, according to Theorem 1, the spectrum of A belongs to Ei. Therein lies the main idea of our algorithm. To indicate k we will use the following result.
Theorem 6. Let H < ∞, ε < √ 2ab (2a2b2+ (a2− b2)A2)H, δ = a + b1 a2− b2 H+b 1− 1 H , L = 0 I −a−b a+bI a+b2 A ,
and let k > 2N be a number such that the following inequality holds δk+δk−1k(L + δ) + . . . + δk−2N+1k2N−1(L + δ)2N−1≤ ε.
Then the matrix
I − Hk+ 1 2a2 + 1 2b2 A∗HkA + 1 4a2 − 1 4b2 ( HkA2+ (A∗)2Hk) is positive definite.
Proof. Consider the matrix
Ik= Hk− 1 2a2 + 1 2b2 A∗HkA − 1 4a2 − 1 4b2 ( HkA2+ (A∗)2Hk). Show that (11) Ik < 1.
From this the required result will follow.
As was noted in the previous section, the matricesHkand Hkcan be written as follows Hk= 1− γ 2 2 I + 1− γ2 2 k−1 j=0 (f0∗f1∗)(L∗)jLj f0 f1 , Hk= 1− γ 2 2 ∞ j=k (f0∗f1∗)(L∗)jLj f0 f1 .
Introduce the matrix
L =k−1 j=0 (L∗)jLj. Then, Hk= 1− γ 2 2 I + 1− γ2 2 (f ∗ 0f1∗)L f0 f1 and Hk= 1− γ 2 2 (f ∗ 0f1∗) ∞ l=1 (l+1)k−1 j=lk (L∗)jLj f0 f1 = 1− γ 2 2 (f ∗ 0f1∗) ∞ l=1 (L∗)lkLlk+ (L∗)lk+1Llk+1 +. . . + (L∗)(l+1)k−1L(l+1)k−1 f0 f1 = 1− γ2 2 (f ∗ 0f1∗) ∞ l=1 (L∗)lk I + L∗L + . . . + (L∗)k−1Lk−1Llkf0 f1 = 1− γ2 2 (f ∗ 0f1∗) ∞ l=1 (L∗)lkLLlk f0 f1 .
By the definition of the matrix L we have
Hk = sup v=1 Hkv, v ≤ ∞ l=1 Lk2l 1− γ2 2 v=1sup L f0 f1 v, f0 f1 v ≤ ∞ l=1 Lk2l Hk.
Consider the expression enclosed in round brackets. SinceH < ∞, according to Theorem 4, all eigenvalues of the matrixA belong to
EH i = λ ∈ C : (Reλ)2 a2 + (Imλ)2 b2 ≤ 1 − 1 H .
Consequently, by Theorem 5 all eigenvalues of the matrix L belong to the disk{λ ∈ C : |λ| ≤ δ < 1}, where
δ = a + b1 a2− b2 H+b 1− 1 H .
For anyj > 2N we have the estimate [11]
Lj ≤ δj+δj−1j(L + δ) + . . . + δj−2N+1j2N−1(L + δ)2N−1.
Hence, according to the conditions of the theorem, we obtain
Lk ≤ δk+δk−1k(L + δ) + . . . + δk−2N+1k2N−1(L + δ)2N−1≤ ε. Then ∞ l=1 Lk2l ≤∞ l=1 ε2l = ε2 1− ε2 and Hk ≤ ε 2 1− ε2Hk. Consequently, Ik ≤ 1 + 1 2b2 − 1 2a2 A2 H k ≤ ε2 1 + 1 2b2 − 1 2a2 A2 Hk ≤ ε2 1 + 1 2b2 − 1 2a2 A2 H.
Taking into account the condition onε, we have the inequality (11). The theorem is proved.
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