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Rev. Anal. Num´er. Th´eor. Approx., vol. 34 (2005) no. 2, pp. 151–168 ictp.acad.ro/jnaat

SINGULAR INTEGRAL OPERATORS.

THE CASE OF AN UNLIMITED CONTOUR

V. NEAGA

Abstract. Let Γ be a closed or unclosed unlimited contour, a shiftα(t) maps homeomorphicly the contour Γ onto itself with preserving or reversing the direc- tion on Γ and also satisfies the conditions: for some naturaln2,αn(t)t, and αj(t) 6≡t for 1 j < n. In this work we study subalgebra Σ of algebra L(Lp(Γ, ρ)), which contains all operators of the form

(M ϕ) (t) =

n−1

X

k=0

ak(t)ϕ(αk(t)) +bkπi(t) Z

Γ ϕ(τ) τ−αk(t)

with piecewise-continuous coefficients. The existence of such an isomorphism between Σ and some algebra Aof singular operators with Cauchy kernel that an arbitrary operator from Σ and its image are Noetherian or not Noetherian simultaneously is proved. It allows to introduce the concept of a symbol for all operators from Σ and, using the known results for algebraA, in terms of a symbol to receive conditions of Noetherian property.

MSC 2000. 45E05.

Keywords. Lyapunov closed curves, Noetherian singular integral, piecewise Lyapunov contour, singular integral equations, singular operators with Cauchy kernel singular operators with shift.

1. INTRODUCTION

Let Γ be a closed or unclosed unlimited contour, a shiftα(t) maps homeo- morphicly the contour Γ onto itself with preserving or reversing the direction on Γ and also satisfies the conditions: for some natural1n≥2 ,αn(t)≡t, and for 1≤j < n,αj(t)6≡t(t∈Γ;αj(t) =α[αj−1(t)], j = 1,2, . . . , n−1, α0(t)≡t)

(1.1) α

0

i(t)(t−z0)2

i(t)−z0)2H(Γ) (z0 ∈C\Γ).

The class of such functions we shall designate by V(Γ). Obviously, V(Γ) does not depend on the choice of a pointz0 ∈C\Γ.

Universitatea de stat din Moldova, str. Mateevici 60, MD-2009 Chi¸sin˘au, Moldova, e-mail: [email protected].

1Ifαreverses the orientation on Γ,nis always equal to two.

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A singular integral operator with a Carleman shift is defined to be the operator of the form

(1.2) (M ϕ) (t) =

n−1

X

k=0

ak(t)ϕ(αk(t)) + bkπi(t) Z

Γ ϕ(τ) τ−αk(t)

,

whereak(t), bk(t) are functions given on the contour Γ. For a limited Lyapunov contour Γ the Noether theory of operators of the form (1.2), build in [10, 22], and the algebra, generated by operators of the form (1.2), are considered in papers [2, 3] and others. In the mentioned works the complete continuity of operatorsTk=WkSW−k−S, where (W ϕ)(t) =ϕ(α(t)) andS is the operator of singular integration along Γ

(Sϕ) (t) = πi1 Z

Γ ϕ(τ)

τ−tdt (t∈Γ), was essentially used.

The situation is absolutely different if Γ is an unlimited contour. In this case [6, 21] operatorW is unbounded, generally speaking, in spaces Lp; instead of the operatorW it is necessary to consider [13]–[16] the operator

(1.3) (V ϕ)(t) =α(t)−zt−z 0

0

λ

ϕ(α(t)) (1/p < λ <1 + 1/p)

and, as was shown in [16], for λ 6= 1 the operators VkSV−kS are not completely continuous. The difficulties of the research of operators (1.2) along an unlimited contour consist in these facts.

In the present paper on the base of results of works [13]–[17] the least subalgebra Σ of the algebra L(Lp(Γ, ρ)), containing all operators of the form (1.2) with piecewise continuous containing all operators of the form (1.2) with piecewise continuous containing all operators of the form (1.2) with piecewise continuous coefficients, is studied. It is necessary to consider separately the case, whenα preserves the orientation on Γ, and the case, whenαreverses the orientation. The algebra Σ contains the set Σ0 of all sums of compositions of operators of the form (1.2), and also operators, which are limits (in the sense of convergence by the norm of operators) of a sequence of operators from Σ0 The research of the set Σ0 is based on the suggested by I. Gohberg and N.Y.

Krupnik [4] method of the study of ”complicated” operators, which allows to receive necessary and sufficient conditions of Noetherian property of operators from Σ. In the paper the existence of such an isomorphism between Σ and some algebra A of singular integral operators with a Cauchy kernel that an arbitrary operator from Σ and its image are simultaneously Noetherian or not Noetherian is proved. It allows to introduce the concept of a symbol for all operators from Σ and, using known results for algebraA(see [4]), in terms of a symbol to receive conditions of Noetherian property for all operators from Σ, including for Σ\Σ0. Through the symbol the index of operatorsA∈Σ can be also expressed.

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The set of values of the determinant of a symbolA(t, µ) represents a closed continuous curve, which can be oriented in a natural way. The index of this curve (i.e. the number of turns about the origin), taken with the opposite sign, is equal to the index of the operatorA.

2. SYMBOL OF SINGULAR INTEGRAL OPERATORS WITHOUT A SHIFT ALONG AN UNLIMITED CONTOUR

Let Γ be an unlimited contour. We shall name Γ admissible, if the contour Γ =˜ γ(Γ), γ(t) = (t−z0)−1 (z06∈Γ) is a simple closed or unclosed Lyapunov contour. LetAbe the algebra, generated by singular integral operators of the form

(2.1) A=cI+dS,

wherec(t), d(t) are piecewise continuous matrix-functions of the ordernandS is a matrix operator of singular integration along Γ. The used by us methods of the research of operators from algebra Σ assume the use of the Noether theory of operators from A. In this connection, in this section we shall establish some results and formulate them so that it would be possible to use them conveniently in the case of an unlimited contour.

Let t1, t2, . . . , tm−1 be some various points on the unlimited curve Γ, tm =

∞; ρ, β1, β2. . . , βm−1, βm =Pm−1k=1 βk+p−2 be real numbers, satisfying the relations: −1< βk< p−1, k= 1,2, . . . , m, and

(2.2) ρ(t) =

m−1

Y

k=1

|t−tk|βk.

Let introduce the following notations: Lp(Γ, ρ) is the Banach spaceLp on the contour Γ with weight ρ(t) ; Lnp(Γ, ρ) is the Banach space of n-dimensional vector functions f = {fi}nj=1 with components fiLp(Γ, ρ), Λn(Γ) is the set of all matrix-functions F(t) of the ordern, continuous at each point of the contour Γ, except, possibly, a finite number of points, at which these functions are continuous from the left and have finite limits from the right. For further it is convenient to introduce the operatorsP = (I+S)/2 andQ=I−P. Then a usual singular operatorA=cI+dSin the spaceLnp(Γ, ρ) can be represented asA=aP +bQ, wherea=c+dand b=cd(a, b∈Λn(Γ)).

To the operator A its symbol will be assigned (see [4, 8]). Let introduce some notations, necessary for the definition of a symbol. Let us assume for the beginning that the contour is closed. By θ=θ(t), f(t, µ) and h(t, µ), t ∈ Γ,(0≤µ≤1) let us designate the following functions:

θ(t) =

π−2π(1 +βk)/p, ift=tk(k= 1,2, . . . , m), π−2π/p, ift∈Γ\{t1, t2, . . . , tm}.

f(t, µ) =

( sinθµexp(iθµ)

sinθexp(iθ) , ifθ6= 0,

µ, ifθ= 0

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and h(t, µ) =pf(t, µ)(1−f(t, µ)), respectively.

The symbol of an operator A is defined to be a matrix-functionA(t, µ)(t∈ Γ,0≤µ≤1) of order 2n, defined by the equality

(2.3) A(t, µ) =

=

f(t, µ)a(t+ 0) + (1−f(t, µ))a(t) h(t, µ)(b(t+ 0)−b(t)) h(t, µ)(a(t+ 0)−a(t)) f(t, µ)b(t) + (1−f(t, µ))b(t+ 0)

. If Γ is unclosed, we consider thatt1 is its beginning andtm=∞ is its end. If a∈Λn(Γ), then we set

a(t1) = lim

Γ3t→t1

a(t1) and a(∞) = lim

Γ3t→∞a(t).

At all points t∈ Γ\{t1,∞} the symbol A(t, µ) of the operatorA =aP +bQ will be defined by the equality (2.3). At pointst1 andt=∞ we set

(2.4)

A(t1, µ) =

=

f(t1, µ)a(t1) + (1−f(t1, µ)) h(t1, µ)(b(t1)−1) h(t1, µ)(a(t1)−1) f(t1, µ) + (1f(t1, µ))b(t1)

,

(2.5)

A(∞, µ) =

=

f(∞, µ) + (1−f(∞, µ))a(∞) h(∞, µ)(1b(∞)) h(∞, µ)(1a(∞)) f(∞, µ)b(∞) + (1−f(∞, µ))

, where, remind, in the definition of functionsf andhwitht=tm =∞the num- berβm=Pm−1k=1 βk. The symbolR(t, µ) of any operatorR∈Ais constructed from symbols of the operatorsA=aP +bQ.

This symbol will be written in the form

(2.6) R(t, µ) =

a11(t, µ) a12(t, µ) a21(t, µ) a22(t, µ)

,

whereaij(t, µ) are blocks of the dimensionn.

Theorem2.1. In order that the operatorR(∈A)be Noetherian in the space Lnp(Γ, ρ) it is necessary and sufficient that the condition:

(2.7) detR(t, µ)6= 0 (t∈Γ,0≤µ≤1) be satisfied. If condition (2.7)is satisfied, then

(2.8) IndA=− 1 2π

arg detR(t, µ)

deta22(t,0) det(a22(t,1))

0≤µ≤1t∈Γ

.

The number 1 {argg(t, µ)}(t,µ)∈Γ×[0,1]in the right-hand side of the equality (2.8) corresponds to the number of turns counter-clockwise of the curveg(t, µ) about the point z= 0 in the complex plane.

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The symbol of singular integrated operators can get a more perfect form in the case, when coefficients of operators are Noetherian. Really, if Γ is closed, then the symbol of the operatorA=aP +bQis determined by the equality

A(t, µ) =

a(t) 0 0 b(t)

Let Γ be an unclosed unlimited arc. As weight we shall take the function ρ(t) =|t−t1|β. Let us designate by Λ = Λ(Γ) the spatial closed unlimited curve (Λ⊂R3), consisting of all points (x,y,z), satisfying the relations x+ iy∈Γ,

−1≤z≤1, (1−z2)(x+ iy−t1) = 0.

By other words, the curve Λ(Γ) consists of two copies of the curve Γ, located in the planes z = 1 and z = −1, and a straight line segment, parallel to the axisz and passing through the beginning of the unclosed contour Γ.

The contour Γ is oriented so that in the plane z = 1 the direction along Λ(Γ) coincides with the direction along Γ (i.e. from the point t1), and in the planez=−1 is opposite. For the spaceLp(Γ,|t−t1|β) (−1< β < p−1) we shall designate by Ωp,ρthe function defined on Λ(Γ) by the following equalities (see [8]):

p,ρ(t, z) =

( z(1+d2)−iε(1−z2)d

1+z2d2 , fort=t1,∞, z, fort∈Γ\{t1,∞},

whered= ctgπ(1 +β)/p,ε= 1 fort=t1 andε=−1 for t=∞. The symbol of the operatorA=aI+bSwill be defined to be the following matrix-function (2.9) A(t, z) =a(t) + Ωp,ρ(t, z)b(t).

Theorem2.2. Letaandbbe continuous matrix-functions onΓof the order n. In order that the operatorA=aP+bQbe Noetherian in the spaceLnp(Γ, ρ) it is necessary and sufficient that the condition :

det(a(t) + Ωp,ρ(t, z)b(t)6= 0) (t, z∈Λ(Γ)) be satisfied. If this condition is satisfied, then

IndA=−inddet(a(t) + Ωp,ρ(t, z)b(t)).

The last formulas are in some sense more obvious and easier to check.

Let us consider the following example. Let Γ = [0,∞) and ρ(x) = xβ (−1< β < p−1). Let’s set z= ee2πξ2πξ−1+1 (ξ ∈R) assuming it to be continued on R by the continuity. Then (see [18, 9, 19]) the symbol of an operator A can be represented as

A(x, ξ) =a(x) +b(x)ψ(x, ξ), where

ψ(x, ξ) =

e2π(ξ+iγ)+1

e2π(ξ+iγ)−1, forx= 0,∞,

e2πξ−1

e2πξ+1, for 0≤x≤ ∞,

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and γ = 1+βp . Note that the domain of the symbolA(x, ξ) is the boundaryX of the set [0,∞]×[−∞,∞].

The following proposition plays an important role further on.

Theorem 2.3. Let δk be some real numbers and h(t) =

m−1

Q

k=1

(t−tk)δk. If

1+βp k < δk<1−1+βpk,

(2.10) −

1+

m−1

P

k=1

βk

p < δm=−

m−1

X

k=1

δk<1−

1+

m−1

P

k=1

βk

p

then the operator H = h(t)Sh−1(t)I belongs to the algebra A and its symbol H(t, µ) has the form

(2.11) H(t, µ) =

En U(t, µ)En

0 −En

, where En is a unity matrix of dimensionn and

U(t, µ) =

4ih(tk,µ) sin(πδk) exp(πiδk)

2if(tk,µ) sin(πδk) exp(πiδk)+1, for t=tk, k= 1,2, . . . , m, 0, for t∈Γ\{t1, . . . , tm}.

3. THE CONDITION OF NOETHERIAN PROPERTY OF OPERATORS FROMΣ. THE CASE OF A CLOSED CONTOUR

In this section we will construct such a homeomorphism ∆ of algebra Σ to algebraAthat the operatorsA∈Σ and ∆(A)∈Aare simultaneously Noether- ian or not Noetherian with the condition of Noether IndA= n1Ind∆(A). The symbol of operator A is defined as symbol of respective operator ∆(A) ∈ A.

We shall establish that an operator A is Noetherian if and only if the deter- minant of its symbol is not equal to zero. The index of the operatorAcan be also expressed through its symbol.

Let the shiftα(t) satisfy the conditions (0.1) andαn(t)≡t. We shall desig- natetn=∞and we shall calculate the iterationtj =αj(tn)(j= 1,2, . . . , n−1).

Let’s introduce the spaceLp(Γ, ρ) with weight

(3.1) ρ(t) =

n−1

Y

j=1

|t−tj|

−2 n ,

whereλis some real number, satisfying the condition (3.2) p(n−1)n−2 < λ < n(p+1)−2p(n+1) .

It is obvious that −1< pλn−2 < p−1 and −1< (pλ−2)(n−1)

n < p−1.

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The suppositions made about the weightρ(t) ensure (see [20, 7]) the bound- edness of the operatorsS and

(V ϕ)(t) =α(t)−zt−z 0

0

λ

ϕ(α(t))

in the space Lp(Γ, ρ). Let’s designate (Akϕ)(t) = ˜ak(t)ϕ(t) + ˜bk(t)(Sϕ)(t), where

a˜k(t) =αn−kt−z(t)−z0

0

λ

akn−k(t)), b˜k(t) =αn−kt−z(t)−z0

0

λ

bkn−k(t)), then the operator M, defined by equality (1.2), takes the form (3.3) M =A0+V A1+· · ·+Vn−1An−1.

Everywhere we shall consider that the functions ˜ak(t) and ˜bk(t)∈Λ1(Γ). The case when coefficients ˜ak,b˜k∈Λn(Γ) can be absolutely similarly investigated.

The following theorem thus is valid [16].

Theorem 3.1. The operator M is bounded in the space Lp(Γ, ρ).

We pass to the construction of the homomorphism ∆, specified above. Let’s designate byBnp,ρ(Γ) the Banach algebra of all operators, which operate in the spaceLnp(Γ, ρ), and bytnthe two-sided ideal of algebraBnp,ρ(Γ), which consists of all completely continuous operators.

Let’s take an arbitrary point τ0 on the contour Γ. Let’s calculate the it- erations τk = αk0), k = 1,2, . . . , n−1. As was established in [10], we can assume pointsτkordered in positive direction on Γ. The contour Γ is divided intonnot intersecting arcs: (τ0, τ1),(τ1, τ2), . . . ,(τn−2, τn−1),(τn−1, τ0,). Let’s define the functionω(t) on (τ0, τ1) as follows: the functionω(t) is equal to zero outside the arc [τ0, τ1] and ω(τ0) = 1, ω(τ1) =ε (ε=e2πin ); on [τ0, τ1], ω(t) is continuous, bounded and different from zero. By U(t) we shall designate the function

U(t) =

n−1

X

k=0

εkω(αn−k(t)) (t∈Γ).

It is clear that the function U(t) everywhere on Γ is continuous, bounded and different from zero, and everywhere on Γ it satisfies the conditionU(t)− ε−1U(α(t)) = 0.

Letδmj be the Kronecker delta. Let’s consider the operators R=δmjUm−1In

m,j=1, N =ε(m−1)(j−1)Vm−1n

m,j=1,

which belong to the algebra Bnp,ρ(Γ) (here and below m is the row number,j is the column number). The operatorsR and N are invertible, and

R−1=δmjU1−mIn

m,j=1, N−1 = n1ε(1−m)(j−1)V1−jn

m,j=1.

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Let’s define the homeomorphism δ : Bp,ρ(Γ) → Bnp,ρ(Γ) as follows. If A ∈ Bp,ρ(Γ), then we assume

(3.4) δ(A) =N RΘ(A)R−1N−1,

where Θ(A) = (δmjA)nm,j=1∈Bnp,ρ(Γ).

Lemma3.1. The operators A andδ(A) are Noetherian only simultaneously, and with the condition of Noether

IndA= n1Indδ(A).

For proof see [6].

Theorem3.2. The contractionof homeomorphismδ to algebra Σis also a homeomorphism

∆ : Σ→A, and, if M is the operator (3.3),then2

(3.5) ∆(M)'M ,˜

where M˜ = β(t)I + γ(t) ˜S, β(t) = ( ˜a−r+k+nk−1(t)))nr,k=1, γ(t) = ˜b−r+k+nk−1(t))n

r,k=1 and S˜ is defined in Lnp(Γ, ρ) by the equal- ity

S˜=

S 0

(t−t1)1−λS(tt1)λ−1

. ..

0 (t−tn−1)1−λS(ttn−1)λ−1

.

Proof. It is sufficient to find out the way in which the homomorphism ∆ acts on operators of the form (3.3). Let Mj =Pn−1r=0εrjVrAr (j = 0,1, . . . , n−1), Ar = ˜arI + ˜brS, L = (δmjMj−1)nm,j=1 (M0 = M) and Φ =Vk−1A−m+k+nV1−kn

m,k=1 (An+i=Ai)

Then, as is known, the following equality is satisfied:

(3.6) L=N−1ΦN.

The immediate check shows thatUjM 'MjUj. Then

(3.7) L'RΘ(M)R−1.

As (see [16]) VrSV−r ' (t−tr)1−λS(ttr)λ−1 (r = 1, . . . , m−1), then Φ'M. From here it follows that˜

∆(M)'M .˜

The theorem is proved.

2Here and further byK1 'K2 such operators that the operatorK1K2 is completely continuous are designated.

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The following theorem results from Lemma 3.1 and Theorem 3.2:

Theorem3.3. If A is an arbitrary operator from algebraΣ, then operators A and ∆(A) are Noetherian or not Noetherian simultaneously, with Noether condition

IndA= n1Ind∆(A).

As the operator ∆(A)∈Aand for operators from algebraAthe concept of symbol was introduced, then a symbol of an operator A∈Σ can be naturally defined to be the symbol of the operator ˜A= ∆(A). Thus, the symbol of an operator A ∈ Σ is the matrix-function A(t, µ) = ˜A(t, µ) of the order 2n. In particular, the symbol of the operatorM =A0+V A1+· · ·+Vn−1An−1 (Aj =

˜aI+ ˜bS) according to the formulas (2.3) and (2.10) has the form M(t, µ) =

=

f(t, µ)β(t+ 0) + (1−f(t, µ)β(t)) h(t, µ)(β(t+ 0)−β(t)) h(t, µ)(β(t+ 0)−β(t)) f(t, µ)β(t) + (1−f(t, µ)β(t+ 0))

+

f(t, µ)γ(t+ 0) + (1−f(t, µ))γ(t) (γ(t+ 0)−γ(t))

h(t, µ)(γ(t+ 0)−γ(t)) f(t, µ)γ(t) + (1−f(t, µ))γ(t+ 0)

·S(t, µ),˜ where

S(t, µ) =˜

1 0

. ..

0 1

. . . . . . . . . 0

0 0

U1(t, µ) . ..

0 Un−1(t, µ)

. . . . . . . . . . . .

−1 0

. ..

0 −1

and functionsUj(t, µ), because of formula (2.11), are defined by the equalities

Uj(t, µ) =

−4ih(tj,µ) sin(πλ) exp(−πiλ)

−2if(tj,µ) sin(πλ) exp(−πiλ)+1, ift=tj;

4ih(tn,µ) sin(πλ) exp(πiλ)

2if(tn,µ) sin(πλ) exp(πiλ)+1, ift=tn;

0, ift∈Γ\{t1, . . . , tn}.

If

(3.8) M =

m

X

j=1

Mj1Mj2. . . Mjr,

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whereMji are operators of the form (3.3), then

(3.9) M(t, µ) =

m

X

j=1

Mj1(t, µ)Mj2(t, µ). . . Mjr(t, µ),

whereMji(t, µ) is the symbol of operatorMji. The set of all operators of the form (3.8) we shall designate by Σ0. By the method described in [4] one can prove the following theorem:

Theorem 3.4. If the operator M belongs to algebra Σ0 and M(t, µ) =

=kajk(t, µ)k2j,k=1 is its symbol, then

(3.10) max

t∈Γ;0≤µ≤1|ajk(t, µ)| ≤ inf

T∈FkA+Tk,

where Tis the set of all completely continuous operators in the spaceLp(Γ, ρ).

Corollary 3.1. The symbol of operator M ∈Σ0 does not depend on rep- resentation of an operator in the form

M =

m

X

j=1

Mj1Mj2. . . Mjr.

Corollary3.2. The mapMM(t, µ)of algebraΣ0 to algebra of symbols is an algebraic homeomorphism, whose kernel contains the set of all completely continuous operators from algebra Σ0.

Let A ∈ Σ\Σ0 and A = limn→∞An, where An ∈ Σ0. By virtue of the relation (3.10), symbols of operators An are uniformly convergent to some continuous matrix-functionA(t, µ), which does not depend on the choice of a sequence {An}1 , and which we shall name the symbol of the operator A.

From Theorem 3.3 and from a property of the symbol of algebraA follows Theorem3.5. In order that the operatorA∈Σbe Noetherian in the space Lp(Γ, ρ) it is necessary and sufficient that its symbol

A(t, µ) =

a11(t, µ) a12(t, µ) a21(t, µ) a22(t, µ)

be non-degenerate: detA(t, µ) 6= 0 (t ∈ Γ,0 ≤ µ ≤ 1). If this condition is fulfilled, there exists a two-sided regulator of the operator,3 belonging to algebra Σ and

IndA=−2nπ1 narg detA(t, µ) deta22(t,0) deta22(t,1)

o

t∈Γ,0≤µ≤1.

3If A is Noetherian, then there exists such a linear bounded operatorB that operators IABandIBAare completely continuous;B is referred to as a two-sided regulator of the operatorA.

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4. THE CONDITION OF NOETHERIAN PROPERTY OF OPERATORS FROM Σ. CASE OF AN UNCLOSED CONTOUR

I. Continuous coefficients.

Let’s consider at first the case, when the algebra Σ is generated by operators S, V and operators of multiplication by continuous functions. Let Γ be an unclosed contour with the beginning t = t1 and the extremity t = ∞ and α: Γ→Γ satisfies the conditionα(α(t))t, then it is obvious that α(t1) =

∞ and α(∞) =t1. Let

ρ(t) =|t−t1|β β = pλ−22 , 1/p< λ <1 + 1/p

.

Lemma 4.1. Each operator A ∈Σ can be represented as a sum A =A1+ A2V (A1, A2 ∈A) and it is uniquely determined up to completely continuous summands.

Proof. Obviously, it is enough to show that the set of operators of the form

(4.1) A1+A2V,

where A1 and A2 run algebra A, is a Banach algebra and from the equality A1+A2V = 0, (A1, A2∈A) it follows thatA1, A2 are completely continuous operators. From the equalityV SV =−(t−t0)1−λS(tt0)λ−1+T (∈A) and V aV =a(α(t))I it follows that the sum and the product of operators of the form (4.1) have the same form. Let’s prove that this set is closed.

LetA(n)1 +A(n)2 V converge uniformly toA. LetU be a invertible operator, built in [13], satisfying the condition : U−1(A(n)1 +A(n)2 V)U =A(n)1A(n)2 V + Tn. Then 2 ˆA(n)1Aˆ+ ˆU−1AˆUˆ = 2 ˆA1 ∈ A, where ˆˆ A = A/T(Lp(Γ, ρ)) is the factor-algebra in all completely continuous operators, acting in the space Lp(Γ, ρ). Analogously, ˆA(n)2Aˆ2 ∈ A. So, ˆˆ A = ˆA1 + ˆA2V and, hence, A=A1+A2V (A1, A2∈A).

LetA1+A2V = 0, then ˆA1+ ˆA2Vˆ = 0 and ˆA1Aˆ2Vˆ = ˆU−1( ˆA1+ ˆA2Vˆ) ˆU,

i.e. A1, A2 ∈T(Lp(Γ, ρ)).

Lemma4.2. LetA, B, WL(B) andW2=I, then the equality is satisfied (see [10])

(4.2)

I W I −W

A B

W BW W AW

I I

W −W

=

= 2

A+BW 0

0 ABW

.

The proof is checked immediately (see [1]).

Theorem 4.1. Let A, B ∈ A and (V φ)(t) = α(t)−zt−z 0

0

λ

φ(α(t)). The op- erator R = A+BV is Noetherian in the space Lp(Γ, ρ) if and only if the

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operator

(4.3) RV =

A B V BV V AV

is Noetherian in the spaceL2p. IfA+BV is Noetherian, then (4.4) Ind(A+BV) = 12Ind

A B V BV V AV

.

Proof. Let’s write the equality (4.2) for operators A, B, V. Remark that U−1(A−BV)U =A+BV +T, hence, A+BV is Noetherian only simulta- neously with the operator ABV and Ind(A+BV) =Ind(A−BV). Since extreme multiplicands in the left-hand side of equalities (4.2) are invertible operators (their product is 2I), all the statements of theorem result from the

equality (4.2). Theorem is proved.

Remark4.1. The matrix operator (4.3) is a singular operator with matrix

coefficients (without shift!).

Corollary 4.1. Let A = aI +bS, B = cI +dS, a, b, c, dC(Γ), R = A+BV and(V ϕ)(t) = α(t)−zt−z 0

0 ϕ(α(t)). Then, (4.5) RV =

A B V BV V AV

= a c

˜c ˜(a)

!

+ b d

d˜ −d˜

! S+T, where f˜(t) =f(α(t)) and T is a completely continuous operator in L2p(Γ, ρ).

The symbol of the operator R =A+BV will be defined to be the matrix RV(t, µ) t∈Γ, 0≤µ≤1, defined by equalities (2.3), (2.4) and (2.5). From above reasonings and Theorems 2.1 and 4.1 results

Theorem 4.2. In order that the operator R = A+BV be Noetherian in the space Lp(Γ, ρ) it is necessary and sufficient that

detRV(t, µ)6= 0 (t∈Γ,0≤µ≤1).

If this condition is satisfied, then IndR=− 1

arg detRV(t, µ) deta22(t,0) deta22(t,1)

t∈Γ, 0≤µ≤1

.

Remark4.2. The symbol of the operatorRcan be defined by the equality (2.9) if substitute for the matrices a and b the respective matrices of the operator RV. In this case Theorem 2.2 for the operator R = A +BV is

true.

II. Discontinuous coefficients.

Theorems 4.1 and 4.2 generally speaking can not be transferred to the case, when the coefficients a, b, c, d ∈ Λ1(Γ). To show this it is enough to give an example of two singular operatorsAandBsuch thatA+BV is Noetherian, but

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ABV is not Noetherian. From equation (4.2) it results that the respective matrix operator is not Noetherian. The examples for Γ = [−1,1] or Γ = R and α(x) = −x are built in works [15,1]. After constructing the symbol of operators of the formA+BV we can construct such examples for every Γ and αV(Γ) (see§4).

Let introduce some notations. Let’s designate by Σ(Γ, ρ) the Banach sub- algebra of the algebra L(Lp(Γ, ρ)), ρ(t) = |t−tβ1|, β = pλ−22 , generated by an operator S, completely continuous operators and all operators of multi- plication by function a ∈ Λ1(Γ). By Σ(Γ, ρ;B)(BL(Lp(Γ, ρ))) we shall designate the Banach subalgebra generated by all operators from Σ(Γ, ρ) and by the operatorB. Let Σ be any subalgebra of algebraL(B). By Σnwe shall designate the subalgebra of the algebra L(Bnn) consisting of all operators of the form (Ajk)nj,k=1, whereAjk ∈Σ.

Definition4.1. Two algebrasA1(⊂L(B1))andA2(⊂L(B2))will be called equivalent [1] if there is an invertible operator ML(B1,B2) such that the set of operators of the form M AM−1(A∈A1) coincides with the algebra A2.

Theorem 4.3. Let Γ be some closed or unclosed contour in the complex plane and ρ(t) =

n

Q

k=1

|t−tk|βk,(−1< βk< p−1), then the algebra Σ(Γ, ρ) is equivalent with the algebra Σ(Γ) = Σ(Γ,1).

Proof. Let (M φ)(t) = ρ1/p(t)φ(t). As kM φkLρ(Γ) = kφkLρ(Γ,ρ), M maps isometrically the space Lp(Γ, ρ) on Lp(Γ). As M aM−1 =aI for any function a∈Λ1(Γ), therefore it is enough to show that M SM−1∈Σ(Γ). The operator M SM−1 can be represented as

M SM−1 =f(t)

n

Y

k=1

(t−tk)βk/pS

n

Y

k=1

(t−tk)−βkf−1(t)I

wheref(t) is different from zero and continuous everywhere on Γ with excep- tion, perhaps, the points tk. The numbers αk=βk/p satisfy the condition

−1/p < αk<1− 1p,

therefore (see [20]) the operatorQnk=1(t−tk)βk/pSQnk=1(t−tk)−βk/pI belongs

to the algebra Σ(Γ). Theorem is proved.

Remark4.3. If Γ is unlimited, it is necessary to require that numbersβkto- gether with the conditions−1< βk< p−1, (k= 1, . . . , n) satisfy the condition 1<

n

P

k=1

βk< p−1.

Corollary 4.2. Any two algebras Σ(Γ, ρ1) and Σ(Γ, ρ2) are equivalent.

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Further on we shall consider the algebra Σ(Γ, ρ;V), where ρ(t) =|t−t1|β (β = pλ−22 ,1/p < λ <1 + 1/p) and

(V φ)(t) =α(t)−zt−z 0

0

λ

(α ∈V(Γ)).

Remark that we can assume thatt1 = 0, as otherwise it can be achieved with the help of the invertible operator H : Lp(Γ, ρ) → Lp(˜Γ,ρ), defined by the˜ equality (Hφ)(z) = φ(z+t1)((H−1φ)(t) = φ(tt1)), where ˜ρ(t) = |z|β and Γ =˜ {z =tt1 :t∈ Γ}. For the same reason it is possible to consider that z06∈Γ. Besides it is enough to study the algebra Σ(R+, xβ;W), where

(W ϕ)(x) =α(x)+1x+1 λφ(α(x)), αV(R+).

and αV(R). Really, let γ :R+ → Γ be a homeomorphic map of R+ on Γ with the following properties: γ(0) = 1, γ(∞) =∞ and 06=γ0(x)∈H(R+).

Obviously (see [5]), such a function exists. The operator M, defined by the equality

(M ϕ)(x) =ϕ(γ(x)),

is a linear bounded invertible operator, acting fromLp(Γ,|t|β) intoLp(R+, xβ).

Let’s designate by ω the function ω(t) =γ−1(t). Then (M−1φ)(t) =φ(ω(t)), M SM−1 =S++T, where

(S+ϕ)(x) = πi1 Z

0

ϕ(y)dy y−x , (T ϕ)(x) = πi1

Z 0

γ0(y)

γ(y)−γ(x)y−x1 ϕ(y)dy

is completely continuous in Lp(R+, xβ). It is easy to see that M aM−1 = a(γ(x))IandM V M−1ϕ= α(γ(x))+1γ(x)+1 λϕ((ω◦α◦γ)(x)). Let’s designateν(x) = (ω◦αγ)(x). From properties of the functions γ, ω and α easily follows that νV(R+). The operator M V M−1 can be represented as M V M−1 = f W, where f(x) = α(γ(x))+1γ(x)+1 λ γ(x)+1x+1 λC(R+), is not equal to zero and (W ϕ)(x) = ν(x)+1x+1 λϕ(ν(x)). Thus MΣ(Γ,|t|β;V)M−1 = Σ(R+, xβ;W). So, further on we shall consider the algebra Σ(R+, xβ;W) and we shall show that it is equivalent to some algebra generated by singular integral operators without a shift.

Theorem 4.4. Let αV(R+),

(4.6) (W ϕ)(x) =α(x)+1x+1 λϕ(α(x)),

ρ(x) = xβ (β = pλ−22 , 1/p < λ < 1 + 1/p). The algebra Σ(R+, ρ;W) is equivalent to the algebra Σ(Γ0, ρ0, V0), where Γ0 = [−1,1], ρ0(t) = (1−t2)β and (V0ϕ)(t) =ϕ(−t).

To prove this theorem, we need the following lemma.

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Lemma 4.3. Let αV(R+) and α(x0) = x0, then there exists a homeo- morphic map µ:R+→R+ with the following properties:

(4.7) µ(0) = 0, µ(x0) = 1,06=µ0(x)∈H(R+),

(4.8) (µ−1αµ)(x) = 1x.

Proof. Let ˜µ(x) = xx

0µ−1(x) = x·x0), then, obviously, the map ν = µ˜◦αµ˜−1 belongs to the setV(R+), andν(1) = 1. The map

(4.9) ω(x) =

( x, ifx∈[0,1],

1

ν(x), ifx∈[1,+∞], satisfies the following conditions. There exists

(4.10) ω−1(x) =

x, ifx∈[0,1], γ(x1), ifx∈[1,∞],

where ω0(x) ∈ H([0,1]), ω0(x) ∈ H([1,∞]), ω0(x) 6= 0 and ω(1) = 1. Let’s show that ω0(x) is continuous at the point x = 1. This implies that ω0(x) ∈ H(R+). So, we have

ω0(1−0) = 1, ω0(1 + 0) = lim

x→1νν20(x)(x) = −νν20(1)(1) = 1,

since ν reverses the orientation on R+ and ν0(1) = −1. The function µ = µ˜−1ω−1 satisfies the conditions (4.7). Let show that (µ−1αµ)(x) = 1/x.

We have µ−1αµ = ωµ˜◦αµ˜−1ω−1 = ωνω−1. Let x ∈ [0,1], then (ω◦νω−1)(x) = ω(ν(ω−1(x))) =ω(ν(ν(1x))) =ω(x1) = 1x. Lemma is

proved.

Corollary 4.3. The algebra Σ(R+, xβ;W) is equivalent to the algebra Σ(R+, xβ;W0), where

(4.11) (W0ϕ)(x) = x1λϕ1x.

Indeed, let’s designate byM1 the operator (M ϕ)(x) =ϕ(µ(x)), whereµ is the function from Lemma 4.2. Then the operatorM1S+M1−1−S+is an integral operator with the kernel πi1 µ(y)−µ(x)µ0(y)x−y1 , and, according to property (4.7), it is completely continuous in Lp(R+, xβ). Besides, M1aM1−1 = a(µ(x))I ∈ Σ(R+, xβ;W0) for any function a ∈ Λ1(R+). Let’s show that M1W M1−1 ∈ Σ(R+, xβ;W0). So, we have

(M1W M1−1ϕ)(x) =M1W ϕ(µ−1(x)) =M1 α(x)+1 x+1

λ

ϕ(µ−1(α(x))) =

= α(µ(x))+1µ(x)+1 λϕ1x=f(x)(W0ϕ)(x), where the function f(x) = (α(µ(x))+1)x

µ(x)+1

λ

is continuous and different from zero on ˜R+. This fact results from properties of functions µand α (see [13]).

Besides thereby, M1W M1−1∈Σ(R+, xβ;W0).

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Proof of Theorem 4.4. Considering the above said, it remains to show that the algebra Σ(R+, xβ;W0) is equivalent with the algebra Σ(Γ0, ρ0;V0). LetM2 be the operator acting by the rule:

(4.12) (M2ϕ)(t) = 1−t1 ϕ 1+t1−t.

The operatorM2is linear, bounded and acting fromLp(R+, xβ) intoLp0, h), where Γ0 = [−1,1] and h(t) = |1 + t|β|1−t|p−β−2. The operator M2 is invertible, moreover,

(4.13) (M2−1Ψ)(x) = x+12 Ψ x−1x+1.

By simple calculations (we omit details) we make sure that M2S+M2−1 =S0, M2aM2−1 = ˜aI (a ∈ Λ1(R+)) and (M2W0M2−1ϕ)(t) = 1−t1+tλ−1ϕ(−t) where ˆa(t) = a 1+t1−t ∈Λ10) and (S0ϕ)(t) = πi1 R−11 ϕ(τ)τ−t (t∈ Γ0). Finally we’ll consider the operator

(M3ϕ)(t) = (1t)1−λϕ(t),

which isometrically maps the spaceLp0, h) into the spaceLp0, ρ0)(ρ0(t) = (1−t2)β). It is easy to see thatM3ˆaM3−1 = ˆaI,M2S0M2−1= (1−t)1−λS0(1− t)λ−1I and M3M2W0M2−1M3−1 = V0, where (V0ϕ)(t) = ϕ(−t). Considering the condition−1/p <1−λ <1−1/pandβ= pλ−22 , from Theorem 2.2 we con- clude that the operator (1−t)1−λS0(1−t)λ−1I belongs to the algebra Σ(Γ0, ρ0).

In particular its symbol can bee calculated with the help of (2.11). So, the algebra Σ(R+, xβ;W0) is equivalent to the algebra Σ(Γ0, ρ0;V0). Theorem is

proved.

Corollary 4.4. The algebra Σ(Γ, ρ;V) is equivalent with algebra Σ2(˜Γ0˜0), where ρ(t) = (1˜ −t)βt−1/2, Γ˜0 = [0,1].

Really, from the proved above it follows that the algebra Σ(Γ, ρ;V) is equiv- alent with the algebra Σ(Γ0, ρ0;V0). It remains to use results of work [1], in which it is proved that Σ(Γ0, ρ0;V0)∼Σ(˜Γ0˜0).

So, from Theorem 4.4 and results of work [1] we can conclude that there exists an invertible operatorXL(Lp(Γ, ρ), L2p( ˜Γ0˜0)) such thatXAX−1 ∈ Σ2( ˜Γ0˜0) for any operator A ∈ Σ(Γ, ρ, V). Let’s designate by A(t, µ) (0t, µ ≤ 1) the symbol of the operator XAX−1 ∈ Σ2( ˜Γ0˜0). The matrix- functionA(t, µ) (0t, µ≤1) of the fourth order can be naturally called the symbol of the operator A. The symbol A(t, µ) of the operatorA∈Σ(Γ, ρ;V) we shall denote bykajk(t, µ)k2j,k=1, whereajk(t, µ) are matrix-functions of the second order. From properties of the symbols of operators from the algebra Σ2( ˜Γ0˜0) (see [4]) and arguments given above, results the following theorem:

Theorem 4.5. In order that the operator A ∈Σ(Γ, ρ;V) be Noetherian in the space Lp(Γ, ρ) it is necessary and sufficient that the determinant of its

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