Section II will contain the proof of the theorem of factorization of matrix-functions and present the theorem of singular operators that are Noetherian in weighted spaces

1(94), No. 4, 377-393

SINGULAR INTEGRAL OPERATORS ON MANIFOLDS WITH A BOUNDARY

R. KAPANADZE

Abstract. This paper deals with singular integral operators that are bounded, completely continuous, and Noetherian on manifolds with a boundary in weighted H¨older spaces.

We shall investigate the matrix singular operator A(u)(x) =a(x)u(x) +

Z

D

f€

x, x−y

|x−y|

|x−y|^−mu(y)dy, x∈D, D⊂R^m,

in weighted H¨older spaces and develop the results obtained in [1] for one- dimensional singular operators and in [2–8] for multidimensional singular operators in Lebesgue spaces.

This paper consists of two sections. In Section I we shall prove the theorems of integral operators that are bounded and completely continuous in H¨older spaces with weight. Section II will contain the proof of the theorem of factorization of matrix-functions and present the theorem of singular operators that are Noetherian in weighted spaces.

1. LetR^m(m≥2) be anm-dimensional Euclidean space,x= (x1, ..., xm), y= (y1, ..., ym) be points of the spaceR^m,

|x|=€X^m

i=1

x²_i¹₂

, Γ ={x: x∈R^m, xm= 0}, R^m+ ={x: x∈R^m, xm>0}, x⁰= (x1, . . . , xm−1),

B(x, a) ={y: y∈R^m, |y−x|< a}, S(x, a) ={y: y∈R^m, |y−x|=a}.

Definition 1. A function u defined on R^m\Γ belongs to the space H_α,β^ν (R^m\Γ) (0< ν,α <1,β ≥0,α+β < m) if

(i) ∀x∈R^m\Γ |u(x)| ≤c|xm|^−α(1 +|x|)^−β,

1991Mathematics Subject Classification. 45F15.

377

1072-947X/94/0700-0377$12.50/0 c1994 Plenum Publishing Corporation

(ii) ∀x∈R^m\Γ, ∀y∈B(x,1 2|xm|)

|u(x)−u(y)| ≤c|xm|^−(α+ν)(1 +|x|)^−β|x−y|^ν. The norm in the spaceH_α,β^ν (R^m\Γ) is defined by the equality

kuk= sup

x∈R^m\Γ|x|^α(1 +|x|)^β|u(x)|+

+ sup

x∈R^m\Γ y∈B(x,¹₂|xm|)

|xm|^α+ν(1 +|x|)^β|u(x)−u(y)|

|x−y|^ν . The spaceH_α,β^ν (R^m+) is defined similarly.

Note that ify∈B(x,¹₂|xm|), then

|y| ≤ 3

2|x|, |y| ≥ 1

2|x|; |ym| ≤ 3

2|xm|, |ym| ≥ 1

2xm|. (1) Thus fory∈B(x,¹₂xm) we have |x| ∼ |y|,|xm| ∼ |ym|.

Letx, y∈R^m\Γ and|x−y| ≥ ¹₂min(|xm|,|ym|). Then the condition (i) implies

|u(x)−u(y)| ≤c(min(|xm|,|ym|))^−α(min(1 +|x|,1 +|y|))^−β≤

≤c|x−y|^ν(min(|xm|,|ym|)^−α−ν(min(1 +|x|,1 +|y|)^−β; and therefore the condition (ii) can be replaced by the condition

∀x, y∈R^m\Γ

|u(x)−u(y)| ≤c|x−y|^ν(min(|xm|,|ym|)^−α−ν(min(1 +|x|,1 +|y|)^−β. One can easily prove that the space H_α,β^ν (R^m\Γ)[H_α,β^ν (R^m+)] is the Ba- nach one.

Consider the singular integral operator v(x) =A(u)(x) =

Z

R^mK(x, x−y)u(y)dy, whereK(x, z) =f€

x,_|^z_z|

|z|^−m.

Theorem 1. Let the characteristicf defined on (R^m\Γ)×(S(0,1)\Γ)sa- tisfy the conditions

(a) ∀x∈R^m\Γ Z

S(0,1)

f(x, z)dzS= 0;

(b) ∀x∈R^m\Γ, ∀z∈S(0,1)\Γ |f(x, z)| ≤c|zm|^−σ (0≤σ≤α);

(c) ∀x, y∈R^m\Γ, ∀z, θ, ω∈S(0,1)\Γ

|f(x, z)−f(y, z)| ≤c|x−y|^ν(min(|xm|,|ym|))^−ν|zm|^−σ

|f(x, θ)−f(x, ω)| ≤c|θ−ω|^ν1(min(|θm|,|ωm|))^−ν1−σ ν1> ν, ν1+σ <1.

Then the operator A is bounded in the spaceHα,β(R^m\Γ).

Proof. In the first place note that the second inequality of the condition (c) yields the inequality

∀x, y, z∈R^m\Γ

ŒŒf€ x, y

|y|

−f€ x, z

|z|ŒŒ≤c|y−z|^ν1 |y|^σ

|ym|^ν1+σ + |z|^σ

|zm|^ν1+σ

‘. (2)

Set

D1=B(x,¹₂|xm|), D2=B(x,¹₂(1 +|x|))\D1,

D3=B(0,¹₂(1 +|x|))\(D1∪D2), D4=R^m\(D1∪D2∪D3), (3) D={y: (y⁰, ym)∈R^m, |ym| ≤2(|y⁰|+ 1)}.

We have v(x) =

Z

D1

K(x, x−y)[u(y)−u(x)]dy+ X4 i=2

Z

Di

K(x, x−y)u(y)dy≡

≡ X4

i=1

Ii(x).

By virtue of the condition (ii) and the inequalities (1)

|I1(x)| ≤c|xm|^−(α+ν)(1 +|x|)^−β Z

D1

ŒŒf€ x−y

|x−y|ŒŒ|x−y|^ν−mdykuk ≤

≤c|xm|^−α(1 +|x|)^−β Z

S(0,1)|f(z)|dzSkuk. (4) Ify 6∈D1, then|x−y| ≥ ¹₆(|x⁰−y⁰|+|xm|+|ym|) and if y∈D2, then 1 +|y| ≥1 +|x|−|x−y| ≥ ¹⁺₂^|x|. Therefore due to the conditions (i) and (b)

|I2(x)| ≤c(1 +|x|)^−β Z

D2

|ym|^−α|ym−xm|^−σ

(|x⁰−y⁰|+|xm|+|ym|)^m−σdykuk. After performing the spherical transformation ofy⁰−x⁰, we obtain

|I2(x)| ≤c(1 +|x|)^−β Z _∞

0

r^m−2dr Z _∞

−∞

|ym|^−α|xm−ym|^−σ

(r+|xm|+|ym|)^m−σdymkuk.

The transformation ofr=|xm|er,ym=|xm|eym leads to

|I2(x)| ≤c|xm|^−α(1 +|x|)^−β Z _∞

−∞|ym|^−α|ym−signxm|^−σdym×

× Z _∞

0

(r+|ym|+ 1)^σ−2drkuk ≤c|xm|^−α(1 +|x|)^−β×

× Z _∞

−∞|ym|^−α|ym−signxm|^−σ(1 +|ym|)^σ−1dymkuk. (5) The term I4(x) is evaluated in the same manner, since in that case, too, 1 +|y| ≥ ¹₂(|x|+ 1). RepresentI3(x) in the form

I3(x) = Z

D3∩B(0,¹₂|xm|)

K(x, x−y)u(y)dy+ +

Z

D3\B(0,¹₂|xm|)

K(x, x−y)u(y)dy≡J1(x) +J2(x).

If y ∈ D3, then |x−y| ≥ ¹2(1 +|x|) ≥ |y|; if y 6∈ B(0,¹₂|xm|), then

|y| ≥ ^|x₂^m| and hence |y| ≥ ¹₄(|y⁰|+|xm|+|ym|). Therefore, in evaluating J2(x), we shall have

|J2(x)| ≤c(1 +|x|)^−β×

× Z

D3\B(0,¹₂|xm|)|ym|^−α|xm−ym|^−σ(|y⁰|+|xm|+|ym|)^σ−mdykuk. After performing the spherical transformation of y⁰, we obtain, as in the case of evaluatingI2(x),

|J2(x)| ≤c|xm|^−α(1 +|x|)^−βkuk. (6) WriteJ1(x) in the form

J1(x) = Z

D3∩B(0,¹₂|xm|)∩D

K(x, x−y)u(y)dy+ +

Z

D3∩B(0,¹₂|xm|)∩(R^m\D)

K(x, x−y)u(y)dy≡J₁⁰(x) +J₁⁰⁰(x).

Ify∈B(0,¹₂|xm|), then|xm−ym| ≥ |xm| − |ym| ≥ ¹2|xm|. We have

|J₁⁰(x)| ≤c(1 +|x|)^−m+σ|xm|^−σ Z

|y⁰|≤1 2 (1+|x|)

(1 +|y⁰|)^−βdy⁰×

× Z

|ym|≤2(|y⁰|+1)|ym|^−αdymkuk ≤

≤c|xm|^−σ(1 +|x|)^σ−m Z

|y0|≤1 2 (1+|x|)

(1 +|y⁰|)^1−β−αdy⁰kuk ≤

≤c|xm|^−α(1+|x|)^−β|xm|^α−σ(1+|x|)^β+σ−m€

c+(1+|x|)^m−(β+α) kuk≤

≤c|xm|^−α(1 +|x|)^−βkuk, (7) sinceσ≤α,α+β < m.

Ify∈R^m\D, then 1 +|y|<1 +|y⁰|+|ym|<³₂|ym|. Therefore

|J₁⁰⁰(x)| ≤c|xm|^−σ(1 +|x|)^σ−m Z

|y|≤1 2 (1+|x|)

(1 +|y|)^−β−αdykuk ≤

≤c|xm|^−α(1 +|x|)^−βkuk. (8) From the estimates (4)–(8) we obtain

|v(x)| ≤c|xm|^−α(1 +|x|)^−βkuk. (9) Let us evaluate the difference v(x)−v(z). It is assumed that|x−z| ≤

1

8|xm|. Then|z| ∼ |x|,|zm| ∼ |xm|.

We introduce the set De1 = B(x,2|x−z|), De2 = B(z,3|x−z|), De3 = B(z,¹₂|xm| − |x−z|). Clearly, De1 ⊂De2 ⊂De3 ⊂ B(x,¹₂|xm|) = D1. We have the representation

v(x)−v(z) = Z

R^m[K(x, x−y)−K(z, z−y)]u(y)dy=

= Z

R^m[K(x, x−y)−K(x, z−y)]u(y)dy+ +

Z

R^m[K(z, x−y)−K(z, z−y)]u(y)dy≡I1(x, z) +I2(x, z).

By virtue of the first inequality of the condition (c) the term I1(x, z) is evaluated exactly in the same manner asv(x) and we obtain the estimate

|I1(x, y)| ≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^νkuk. (10) Rewrite the termI2(x, z) in the form

I2(x, z) = Z

D1

[K(z, x−y)−K(z, z−y)]u(y)dy+

+ Z

R^m\D1

[K(z, x−y)−K(z, z−y)]u(y)dy=

= Z

D1

K(z, x−y)[u(y)−u(x)]dy− Z

D1\^De³[K(z, z−y)u(y)dy−

− Z

e

D3\^De²K(z, z−y)[u(y)−u(x)]dy− Z

e

D2

K(z, z−y)[u(y)−u(z)]dy+ +

Z

R^m\D1

[K(z, x−y)−K(z, z−y)]u(y)dy=

= Z e

D2

+ Z

D1\^De³

‘K(z, x−y)[u(y)−u(x)]dy−

− Z

e

D2

K(z, z−y)[u(y)−u(z)]dy−

− Z

e

D3\^De²[K(z, x−y)−K(z, z−y)][u(y)−u(x)]dy−

− Z

D1\^De³K(z, z−y)u(y)dy+ +

Z

R^m\D1

[K(z, x−y)−K(z, z−y)]u(y)dy≡ X5 i=1

Ji(x, z).

In evaluatingJ1(x, z), note thatDe2 ⊂B(x,4|x−z|), B(x,¹₂|xm| −2|x− z|)⊂De3. Therefore by virtue of the condition (b) and the inequalities (1) we obtain

|J1(x, z)| ≤ Z

B(x,4(x−z))|K(z, x−y)||u(y)−u(x)|dy+ +

Z

D1\B(x,1

2|xm|)−2|x−z|)|K(z, x−y)||u(y)−u(x)|dy≤

≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^νkuk, (11) since€₁

2|xm|ν

−€₁

2|xm| −2|z−z|ν

≤c|x−z|^ν. Similarly, ify∈De2, then

|y−z| ≤3|x−z| ≤ ^3|x8^m| ≤³7|zm|. Therefore

|J2(x, z)| ≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^νkuk. (12) It is clear thatB(x,¹₂|xm|)⊂B(z,¹₂|xm|+|x−z|) and hence

|J4(x, z)| ≤c|xm|^−α(1 +|x|)^−β Z

D1\^De³|K(z, z−y)|dykuk ≤

≤c|xm|^−α(1 +|x|)^−βln|xm|+ 2|x−z|

|xm| −2|x−z|kuk ≤

≤c|xm|^−α(1 +|x|)^−β |x−z|

|xm| −2|x−z|kuk ≤

≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^νkuk. (13) Note that ify6∈De2, then

|x−y|>2|x−z|, |z−y|>3|x−z|,

|x−y|<|x−z|+|z−y|<4

3|z−y|, |z−y|<3 2|x−y|.

Taking these inequalities into account, the inequality (2) readily implies that fory6∈De2

|K(z, x−y)−K(z, z−y)| ≤

≤c |x−z|^ν1

|x−y|^m+ν1

 |x−y|^ν1+σ

|xm−ym|^ν1+σ + |z−y|^ν1+σ

|zm−ym|^ν1+σ

‘

, (14) using which we obtain

|J3(x, z)| ≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^ν1×

× Z

e

D3\^De²

1

|x−y|^m+ν1−ν

 |x−y|^ν1+σ

|xm−ym|^ν1+σ + |z−y|^ν1+σ

|zm−ym|^ν1+σ

‘dykuk ≤

≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^ν1×

א Z

D1\^De¹

1

|x−y|^m+ν1−ν

 |x−y|

|xm−ym|

‘ν1+σ

dy+ +

Z e

D3\^De²

1

|z−y|^m+ν1−ν

 |z−y|

|zm−ym|

ν1+σ‘ dy‘

.

Passing to the spherical coordinates and keeping in mind that ^x_|^m_x⁻₋^y_y^m_| ,

zm−ym

|z−y| do not depend on the radius, we have

|J3(x, z)| ≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^ν1€

|xm|^ν−ν1+|x−z|^ν−ν1 kuk ≤

≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^νkuk. (15) In deriving the estimate, we took into account thatν < ν1,ν1+σ <1. By virtue of the inequality (14)

|J5(x, z)| ≤c|z−x|^ν1 Z

R^m\D1

|x−y|^σ−m|xm−ym|^−σ−ν1|u(y)|dy+ +

Z

R^m\B(z,¹₃|zm|)|z−y|^σ−m|zm−ym|^−σ−ν1|u(y)|dy‘ .

We evaluate the obtained integral expression by the same technique as was used to evaluate the integral expression

Z

R^m\D1

|K(x, x−y)||u(y)|dy (see the estimates (5)–(8)) and finally obtain

|J5(x, z)| ≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^νkuk. (16) The estimates (10)–(13), (15), (16) show that

|v(x)−v(z)| ≤c|xm|^−(α+ν)(1 +|x|)^−β|x−z|^νkuk, which, with the equality (9) taken into account, proves the theorem.

Corollary 1. Under the conditions of Theorem 1 the operator Z

R^m+

K(x, x−y)u(y)dy

is bounded when acting from the space H_α,β^ν (R^m+) into the space H_α,β^ν (R^m\Γ).

Definition 2. LetMbe a closed set inR^m. The setMis called an (m−1)- dimensional manifold without a boundary of the class C^1,δ (0≤δ≤1), if for eachx∈M there exist a positive numberrxand a neighborhoodQ(x) of the pointxin R^m, which is mapped by means of the orthogonal transform Tx onto the cylinder Ω0 ={ξ : ξ ∈ R^m,|ξ⁰| < rx,|ξm| < rx} and if the following conditions are fulfilled: Tx(x) = 0, the setTx(M ∩Q(x)) is given by the equationξm=ϕx(ξ⁰), |ξ⁰| < rx; ϕx ∈C^1,δ in the domain |ξ⁰|< rx

and∂ξiϕx(0) = 0,i= 1, . . . , m−1.

Clearly,Q(x) is the cylinder to be denoted byC(x, rx).

In what follows the manifoldM will be assumed compact.

We introduce the notation d(x)≡d(x, M) = inf

y∈M|x−y|, M(τ) ={x∈R^m, d(x)< τ}. Note some properties of the functiond(x):

d(x)≤c(1 +|x|), |d(x)−d(y)| ≤c|x−y|;

∀x∈R^m\M,∀y∈B(x,¹₂d(x)) d(y)≤ 3

2d(x)≤3d(y), 1 +|x| ∼1 +|y|;

∀x∈M,∀y∈C(x,¹₃rx)

(17)

d(y, M) =d(y, M∩C(x, rx)) and if y=T_x⁻¹(η), then

d(y)≤ |ηm−ϕx(η⁰)| ≤2(1 +ax)d(y), (18) whereax is the Lipshitz constant of the functionϕx.

Definition 3. A function u defined on R^m\M belongs to the space H_α,β^ν (R^m\M) (0< ν, α <1,β≥0,α+β < m), if:

(i) ∀x∈R^m\M |u(x)| ≤cd^−α(x)(1 +|x|)^−β, (ii) ∀x∈R^m\M, ∀y∈B(x,¹₂d(x))

|u(x)−u(y)| ≤cd^−(α+ν)(x)(1 +|x|)^−β|x−y|^ν. The norm in the spaceH_α,β^ν (R^m\M) is defined by the equality

kuk= sup

x∈R^m\M

d^α(x)(1 +|x|)^β|u(x)|+

+ sup

x∈R^m\M y∈B(x,¹₂d(x))

d^ν+α(x)(1 +|x|)^β|u(x)−u(y)|

|x−y|^ν . The spaceH_α,β^ν (R^m\M) is the Banach one.

Theorem 2. Let M ∈C^1,δ, and let the characteristic f of the singular operator Abe defined on (R^m\M)×S(0,1) and satisfy the conditions:

(a) ∀x∈R^m\M, ∀z∈S(0,1)

|f(x, z)| ≤c, Z

S(0,1)

f(x, z)dzS = 0;

(b) ∀x, y∈R^m\M, ∀θ, ω∈S(0,1)

|f(x, θ)−f(y, θ)| ≤c|x−y|^ν€

min(d(x), d(y))₋ν

,

|f(x, θ)−f(x, ω)| ≤c|θ−ω|^ν1, ν < ν1. Then the operator Ais bounded in the spaceH_α,β^ν (R^m\M).

Proof. LetM ⊂B(0, r0) ande1be an infinitely differentiable function such thate1(x) = 1 for|x| ≤r0+1,e1(x) = 0 for|x| ≥r0+2. Settinge2= 1−e1, we have

v(x) = Z

R^mK(x, x−y)e1(y)u(y)dy+ +

Z

R^mK(x, x−y)e2(y)u(y)dy≡v1(x) +v2(x).

Let us evaluate the integral v1(x) =

Z

R^mK(x, x−y)u1(y)dy (u1=e1u).

Choose a constant r^∗ (0 < r^∗ < 1) such that the system {C(x,^{i 1}₄r^∗)}^li=1

(x,ⁱ i= 1, . . . , l, are points of the manifoldM) covers the manifold M and C(x,ⁱ 4r^∗),i= 1, . . . , l, are again the coordinate neighborhoods.

We introduce the sets

D1=B(x,¹₂d(x)), D2= (B(x,¹₄r^∗)\D1)∩B(0, r0+ 2) D3=B(0, r0+ 2)\(D1∪D2), D4={y: y∈R^m, d(y)<¹₂r^∗} Now

v1(x) = Z

D1

K(x, x−y)[u1(y)−u1(x)]dy+ +

X3 i=2

Z

Di

K(x, x−y)u1(y)dy≡ X3

i=1

Ii(x).

By virtue of the inequality (17) and the condition (ii) we obtain

|I1(x)| ≤cd^−α(x)(1 +|x|)^−βku1k ≤cd^−α(x)(1 +|x|)^−βkuk. (19) Next,

|I2(x)| ≤ Z

D2∩D4

|K(x, x−y)||u1(y)|dy+ +

Z

D2\D4

|K(x, x−y)||u1(y)|dy≡I₂⁰(x) +I₂⁰⁰(x).

Ify∈D2, then 1 +|x| ∼1 +|y| ∼c. Moreover,d(y)≤¹₂r^∗ fory∈D2∩D4

and therefore there exists i (i = 1, . . . , e) such that x, y ∈ C(x,^{i 3+}₂^√2r^∗).

Lety=T⁻_i ¹

x (η),x=T⁻i ¹

x (ξ). By virtue of the inequality (18)

|ηm−ϕi

x(η⁰)| ≤2(1 +ai

x)d(y), |ξm−ϕi

x(ξ⁰)| ≤2(1 +ai

x)d(x).

Taking into account that|x−y|>¹₂d(x), we therefore obtain

|x−y|=|ξ−η| ≥ ¹4(|ξ⁰−η⁰|+|ξm−ηm|+d(x))≥

≥1

16(1 +ai

x)⁻¹€

4(1 +ai

x)|ξ⁰−η⁰|+|ξm−ηm|+ 4(1 +ai

x)d(x)

≥

≥1

16(1 +ai

x)⁻¹€

|ξ⁰−η⁰|+|ηm−ϕi

x(η⁰)|+d(x)

, (20)

since|ξm−ηm| ≥ |ηm−ϕi

x(η⁰)| − |ϕi

x(η⁰)−ϕi

x(ξ⁰)| − |ξm−ϕi

x(ξ⁰)|. Thus

|I₂⁰(x)| ≤c Z

|ξ⁰−η⁰|≤4r^∗

dη⁰ Z 2r^∗

−2r^∗|ηm−ϕi

x(η⁰)|^−α×

׀

|ξ⁰−η|+|ηm−ϕi

x(η⁰)|+d(x)₋m

dηmkuk. Using the transformηm−ϕⁱ

x(η⁰) =ηem, we obtain

|I₂⁰(x)| ≤c Z 4r^∗

0

r^m−2dr Z 4r^∗

−4r∗|ηm|^−α€

r+|ηm|+d(x)₋m

dηmkuk, which, upon applying the transformr=d(x)er,ηm=d(x)ηem, gives

|I₂⁰(x)| ≤cd^−α(x)kuk ≤cd^−α(x)(1 +|x|)^−βkuk. (21) Ify ∈D2\D4, thend(y)≥ ¹₂r^∗, ¹₂d(x)≤ |x−y|< ¹₄r^∗ by virtue of which d(x)≥d(y)− |x−y| ≥ ¹₄r^∗,|x−y| ≥ ¹₈r^∗. Therefore

|I₂⁰⁰(x)| ≤ckuk ≤cd^−α(x)(1 +|x|)^−βkuk. (22) Finally, ify∈D3, then 1 +|x| ≤1 +|y|+|x−y| ≤c|x−y|. Therefore

|I3(x)| ≤c(1 +|x|)^−mkuk ≤cd^−α(x)(1 +|x|)^−βkuk. (23) The inequalities (19), (21)–(23) show that

|v1(x)| ≤cd^−α(x)(1 +|x|)^−βkuk. (24) In evaluating the differencev1(x)−v1(z), it will be assumed that|x−z|<

1

8d(x). Then 1 +|z| ∼1 +|x|, d(x)∼d(y).

We introduce the set

De1=B(x,2|x−y|), De2=B(z,3|x−z|), De3=B(z,¹₂d(x)− |x−z|).

Proceeding as in proof of Theorem 1, we obtain

|v1(x)−v1(z)| ≤cd^−(α+ν)(x)(1 +|x|)^−β|x−z|^νkuk. (25) To evaluate the integral

v2(x) = Z

R^mK(x, x−y)u2(y)dy (u2=e2u)

note that the function u2 is defined on R^m and satisfies the conditions of Definition 3, if the function d(x) is replaced by the function 1 +|x|. Therefore, after introducing the sets

D1=B(x,¹₂(1 +|x|)), D2=B(0,2|x|+ 1)\D1, D3=R^m\(D1∪D2), we readily obtain the estimate

|v2(x)| ≤cd^−α(x)(1 +|x|)^−βkuk. (26) Now, considering the sets

De1=B(x,2|x−z|), De2=B(z,3|x−z|), De3=B(z,¹₂(1 +|x|)− |x−z|)

it is easy to show that

|v2(x)−v2(z)| ≤cd^−(α+ν)(x)(1 +|x|)^−β|x−z|^νkuk. (27) The estimates (24)–(27) prove the theorem.

A result close to the one presented here is obtained in [9] (see also [10]).

Definition 4. A function udefined onR^m belongs to the spaceH_λ^ν(R^m) (ν, λ >0), if

|u(x)| ≤c(1 +|x|)^−β, |u(x)−u(y)| ≤c|x−y|^νρ⁻_xy^ν−λ, whereρxy= min(1 +|x|,1 +|y|).

Theorem 3. Let the characteristicf of the singular operatorAsatisfy the conditions of Theorem 1, assuming that σ < α and the first inequality of the condition (c) is fulfilled in the strong form

|f(x, z)−f(y, z)| ≤c|x−z|^ν1(min(|xm|,|ym|))^ν1|zm|^−σ.

It is also assumed that a ∈ C( ˙R^m) ( ˙R^m = R^m∪ ∞) and (a−a(∞)) ∈ H_δ^ν1(R^m). Then the integral operator

v(x) =B(u)(x) = Z

R^m[a(x)−a(y)]K(x, x−y)u(y)dy is completely continuous in the spaceH_α,β^ν (R^m\Γ).

Proof. From the proof of Theorem 1 it follows that B is the bounded operator from the space H_α,β^ν (R^m\Γ) into the space H_α^ν+γ_−γ,β+2γ(R^m\Γ), whereγ is an arbitrary positive number satisfying the condition

γ <min{λ, ν1−ν, α−σ,¹₂(m−β−α)}. Indeed, it is clear that

|v(x)| ≤ Z

D1∪D2∪D4

|a(x)−a(y)||K(x, x−y)||u(y)|dy+ +

Z

D3

(|a(x)−a(∞)|+|a(y)−a(∞)|)|K(x, x−y)||u(y)|dy.

Taking into account that (a−a(∞))∈H_γ^γ(R^m) and repeating the proof of Theorem 1, we obtain

|v(x)| ≤c|xm|^γ−α(1 +|x|)^−β−2γkuk. (28)

Let us now assume that |x−z| ≤ ¹8|xm| and evaluate the difference v(x)−v(z). We have

|v(x)v(y)| ≤ Z

e

D2

|a(x)−a(y)||K(x, x−y)||u(y)|dy+ +

Z e

D2

|a(z)−a(y)||K(z, z−y)||u(y)|dy+ +

Z

R^m\^De¹|a(x)−a(z)||K(x, x−y)||u(y)|dy+ +

Z

R^m\^De²|a(z)−a(y)||K(x, x−y)−K(z, x−y)||u(y)|dy≡ X4 i=1

Ii(x, z).

Hence

|Ii(x, z)| ≤c|xm|^−α(1 +|x|)^−β(1 +|x|)^−ν1−λ|x−z|^ν1kuk ≤

≤c|xm|^−α−γ(1 +|x|)^−β−2γkuk (i= 1,2). (29) Write the termI3in the form

I3(x, z) =|a(x)−a(z)| Z

D1\^De¹|K(x, x−y)||u(y)|dy+ +

Z

R^m\^De¹|K(x, x−y)||u(y)|dy‘ . This representation gives

|I3(x, z)| ≤c|x−z|^ν1(1 +|x|)^−ν1−λ|xm|^−α(1 +|x|)^−β×

א

ln |xm|

|x−z|+c1

‘kuk ≤c|xm|^−α−ν(1 +|x|)^−β−2γ|x−z|^ν+γkuk. (30)

To evaluate the integral termI4 note that fory6∈De1we have

|K(x, x−y)−K(z, z−y)| ≤c|x−z|^ν1

|x−y|^m|xm|^−ν1 |x−y|^σ

|xm−ym|^σ + +c |x−z|^ν1

|x−y|^m+ν1

 |x−y|^ν1+σ

|xm−ym|^ν1+σ + |z−y|^ν1+σ

|zm−ym|^ν1+σ

‘ .

Using this estimate in the same manner as in proving Theorem 1, we obtain

|I4(x, z)| ≤c|xm|^−α−ν(1 +|x|)^−β−2γ|x−z|^ν+γkuk. (31) The estimates (28)–(31) show that the operatorBis bounded from the space H_α,β^ν (R^m\Γ) into the spaceH_α^ν+γ_−γ,β+2γ(R^m\Γ). The validity of the theorem now follows from the complete continuity of the operator of the embedding of the spaceH_α^ν+γ_−γ,β+2γ(R^m\Γ) into the spaceH_α,β^ν (R^m\Γ).

In a similar manner we prove

Theorem 4. Let m∈C^1,δ, let the characteristicf of the operatorA sat- isfy the conditions of Theorem 2, the first inequality of the condition (b) being replaced by a stronger inequality

|f(x, θ)−f(y, θ)| ≤c|x−y|^ν1€

min(d(x), d(y))₋ν1

,

and let the functionasatisfy the conditions of Theorem3. Then the operator B is completely continuous in the spaceH_α,β^ν (R^m\M).

2. We shall consider the matrix-functionA(ξ) =kAij(ξ)kⁿ×n. Let A(λξ) =A(ξ) (λ >0), Aij ∈C^∞(R^m\0), detA(ξ)6= 0 (ξ6= 0).

We set A0 = A⁻¹(0, . . . ,0,−1)A(0, . . . ,0,+1). It is assumed that λj

(j = 1, . . . , s) is the eigenvalue of the matrix A0 and rj is its multiplic- ity (Ps

j=1rj=n).

We introduce the matricesBr(α)≡ kBνk(α)kr×rwhere

Bνk(α) =







0, ν < k;

1, ν=k;

α^ν−k

(ν−k)!, ν > k, B(ri;α) = diag‚

Bri1(α), . . . , Br_ipi(α)ƒ

(ri1+· · ·+ripi=ri).

By the Jordan theorem the matrix A0 is representable in the form A0 = gBg⁻¹, where detg 6= 0, B is the modified Jordan form of the matrix A0, B= diag[λ1B(r1; 1), . . . , λsB(rs; 1)]. We introduce the notation

δ_j⁰ = 1

2πilnλj, δj=δ⁰_k for

k−1

X

ν=1

rν < j≤ Xk ν=1

rν, j= 1, . . . , n;

α_±(ξ) = 1

2πilnξm± |ξ⁰|

|ξ⁰| (ξ⁰= (ξ1, . . . , ξm−1));

by lnzwe denote a logarithm branch which is real on the positive semi-axis, i.e.,−π <argz≤π,

ξm±i|ξ⁰|

|ξ⁰|

‘δ

≡diaghξm±i|ξ⁰|

|ξ⁰|

‘δ1

, . . . ,ξm±i|ξ⁰|

|ξ⁰|

‘δni , B_±(ξ)≡diag‚

B(r1;α_±(ξ), . . . , B(rs;α_±(ξ))ƒ .

Theorem 5. Let the matrix A be strongly elliptic. Then A admits the factorizationA(ξ) =cgA₋(ξ⁰, ξm)D(ξ)A+(ξ⁰, ξm)g⁻¹, where

c=A(0, . . . ,+1), D(ξ) =B₋(ξ)(ξm−i|ξ⁰|)^δ(ξm+i|ξ⁰|)^−δB₊⁻¹(ξ), A_±(λξ) =A_±(ξ) (λ >0), detA_±(ξ)6= 0.