Georgian Mathematical Journal 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 Hölder spaces.

We shall investigate the matrix singular operator

A(u)(x) = a(x)u(x) + ∫_{D} f(x, \frac{x − y}{|x − y|})|x − y|^{−m}u(y)dy,

x ∈ D, D ⊂ ℝ^{m},

in weighted Hölder spaces and develop the results obtained in [1] for onedimensional 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 Hölder 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. Let ℝ^{m} (m ≥ 2) be an m-dimensional Euclidean space, x = (x_{1}, ..., x_{m}), y = (y_{1}, ..., y_{m}) be points of the space ℝ^{m},

|x| = ( ∑_{i=1}^{m} x_{i}^{2})^{\frac{1}{2}} , Γ = \{x : x ∈ ℝ^{m}, x_{m} = 0\},

ℝ_{+}^{m} = {x : x ∈ ℝ^{m}, x_{m} > 0}, x′ = (x_{1}, …, x_{m−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|x_{m}|^{−α}(1 + |x|)^{−β},

377

1991 Mathematics Subject Classification. 45F15.

1072-947X/94/0700-0377$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1994 Plenum Publishing Corporation

(ii) ∀x ∈ ℝ^{m}\Γ, ∀y ∈ B(x, \frac{1}{2}|x_{m}|)
|u(x) − u(y)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − y|^{ν}.

The norm in the space H_{α,β}^{ν}(ℝ^{m}\Γ) is defined by the equality

∥u∥ = sup_{x∈ℝ^{m}\Γ} |x|^{α}(1 + |x|)^{β}|u(x)| +

+ sup_{\begin{matrix}x∈ℝ^{m}\Γ \\ y∈B(x, \frac{1}{2} |x_{m}|) \end{matrix}} |x_{m}|^{α+ν}(1 + |x|)^{β} \frac{|u(x) − u(y)|}{|x − y|^{ν}}.

The space H_{α,β}^{ν}(ℝ_{+}^{m}) is defined similarly. Note that if y ∈ B(x, \frac{1}{2}|x_{m}|), then

|y| ≤ \frac{3}{2}|x|, |y| ≥ \frac{1}{2}|x|; |y_{m}| ≤ \frac{3}{2}|x_{m}|, |y_{m}| ≥ \frac{1}{2}x_{m}|.  (1)

Thus for y ∈ B(x, \frac{1}{2}x_{m}) we have |x| ∼ |y|, |x_{m}| ∼ |y_{m}|.

Let x, y ∈ ℝ^{m}\Γ and |x − y| ≥ \frac{1}{2} min(|x_{m}|, |y_{m}|). Then the condition (i) implies

|u(x) − u(y)| ≤ c(min(|x_{m}|, |y_{m}|))^{−α}(min(1 + |x|, 1 + |y|))^{−β} ≤

≤ c|x − y|^{ν}(min(|x_{m}|, |y_{m}|)^{−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(|x_{m}|, |y_{m}|)^{−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 Banach one.

Consider the singular integral operator

v(x) = A(u)(x) = ∫_{ℝ^{m}} K(x, x − y)u(y)dy,

where K(x, z) = f(x, \frac{z}{|z|})|z|^{−m}.

Theorem 1. Let the characteristic f defined on (ℝ^{m}\Γ) × (S(0, 1)\Γ) satisfy the conditions

(a) ∀x ∈ ℝ^{m}\Γ ∫_{S(0,1)} f(x, z)d_{z}S = 0;
(b) ∀x ∈ ℝ^{m}\Γ, ∀z ∈ S(0, 1)\Γ |f(x, z)| ≤ c|z_{m}|^{−σ} (0 ≤ σ ≤ α);
(c) ∀x, y ∈ ℝ^{m}\Γ, ∀z, θ, ω ∈ S(0, 1)\Γ

SINGULAR INTEGRAL OPERATORS ON MANIFOLDS 379

|f(x, z) − f(y, z)| ≤ c|x − y|^{ν}(min(|x_{m}|, |y_{m}|))^{−ν}|z_{m}|^{−σ} |f(x, θ) − f(x, ω)| ≤ c|θ − ω|^{ν_{1}}(min(|θ_{m}|, |ω_{m}|))^{−ν_{1}−σ}

ν_{1} > ν, ν_{1} + σ < 1.

Then the operator A is bounded in the space H_{α,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
|f(x, \frac{y}{|y|}) − f(x, \frac{z}{|z|})| ≤ c|y − z|^{ν_{1}}(\frac{|y|^{σ}}{|y_{m}|^{ν_{1}+σ}} + \frac{|z|^{σ}}{|z_{m}|^{ν_{1}+σ}} ).  (2)

Set

D_{1} = B(x, \frac{1}{2}|x_{m}|), D_{2} = B(x, \frac{1}{2}(1 + |x|))\D_{1},
D_{3} = B(0, \frac{1}{2}(1 + |x|))\(D_{1} ∪ D_{2}), D_{4} = ℝ^{m}\(D_{1} ∪ D_{2} ∪ D_{3}), (3)
D = {y : (y′, y_{m}) ∈ ℝ^{m}, |y_{m}| ≤ 2(|y′| + 1)}.

We have

v(x) = ∫_{D_{1}} K(x, x − y)[u(y) − u(x)]dy + ∑_{i=2}^{4} ∫_{D_{i}} K(x, x − y)u(y)dy ≡

≡ ∑_{i=1}^{4} I_{i}(x).

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

|I_{1}(x)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β} ∫_{D_{1}} |f(\frac{x − y}{|x − y|})||x − y|^{ν−m}dy∥u∥ ≤

≤ c|x_{m}|^{−α}(1 + |x|)^{−β} ∫_{S(0,1)} |f(z)|d_{z}S∥u∥.  (4)

If y ∉ D_{1}, then |x − y| ≥ \frac{1}{6}(|x′ − y′| + |x_{m}| + |y_{m}|) and if y ∈ D_{2}, then 1+|y| ≥ 1+|x|−|x−y| ≥ \frac{1+|x|}{2}. Therefore due to the conditions (i) and (b)

|I_{2}(x)| ≤ c(1 + |x|)^{−β} ∫_{D_{2}} \frac{|y_{m}|^{−α}|y_{m} − x_{m}|^{−σ}}{(|x′ − y′| + |x_{m}| + |y_{m}|)^{m−σ}} dy∥u∥.

After performing the spherical transformation of y′ − x′, we obtain

|I_{2}(x)| ≤ c(1 + |x|)^{−β} ∫_{0}^{∞} r^{m−2}dr ∫_{−∞}^{∞} \frac{|y_{m}|^{−α}|x_{m} − y_{m}|^{−σ}}{(r + |x_{m}| + |y_{m}|)^{m−σ}} dy_{m}∥u∥.

The transformation of r = |x_{m}|r̃, y_{m} = |x_{m}|ỹ_{m} leads to

|I_{2}(x)| ≤ c|x_{m}|^{−α}(1 + |x|)^{−β} ∫_{−∞}^{∞} |y_{m}|^{−α}|y_{m} − sign x_{m}|^{−σ}dy_{m} ×
× ∫_{0}^{∞}(r + |y_{m}| + 1)^{σ−2}dr ∥u∥ ≤ c|x_{m}|^{−α}(1 + |x|)^{−β} ×
× ∫_{−∞}^{∞} |y_{m}|^{−α}|y_{m} − sign x_{m}|^{−σ}(1 + |y_{m}|)^{σ−1}dy_{m}∥u∥.  (5)

The term I_{4}(x) is evaluated in the same manner, since in that case, too, 1 + |y| ≥ \frac{1}{2}(|x| + 1). Represent I_{3}(x) in the form

I_{3}(x) = ∫_{D_{3}∩B(0, \frac{1}{2} |x_{m}|)} K(x, x − y)u(y)dy +
+ ∫_{D_{3}\B(0, \frac{1}{2} |x_{m}|)} K(x, x − y)u(y)dy ≡ J_{1}(x) + J_{2}(x).

If y ∈ D_{3}, then |x − y| ≥ \frac{1}{2}(1 + |x|) ≥ |y|; if y ∉ B(0, \frac{1}{2}|x_{m}|), then |y| ≥ \frac{|x_{m}|}{2} and hence |y| ≥ \frac{1}{4}(|y′| + |x_{m}| + |y_{m}|). Therefore, in evaluating J_{2}(x), we shall have

|J_{2}(x)| ≤ c(1 + |x|)^{−β} ×
× ∫_{D_{3}\B(0, \frac{1}{2} |x_{m}|)}|y_{m}|^{−α}|x_{m} − y_{m}|^{−σ}(|y′| + |x_{m}| + |y_{m}|)^{σ−m}dy∥u∥.

After performing the spherical transformation of y′, we obtain, as in the case of evaluating I_{2}(x),

|J_{2}(x)| ≤ c|x_{m}|^{−α}(1 + |x|)^{−β}∥u∥.  (6)

Write J_{1}(x) in the form

J_{1}(x) = ∫_{D_{3}∩B(0, \frac{1}{2} |x_{m}|)∩D}K(x, x − y)u(y)dy +
+ ∫_{D_{3}∩B(0, \frac{1}{2} |x_{m}|)∩(ℝ^{m}\D)}K(x, x − y)u(y)dy ≡ J′_{1}(x) + J′′_{1}(x).

If y ∈ B(0, \frac{1}{2}|x_{m}|), then |x_{m} − y_{m}| ≥ |x_{m}| − |y_{m}| ≥ \frac{1}{2}|x_{m}|. We have

|J′_{1}(x)| ≤ c(1 + |x|)^{−m+σ}|x_{m}|^{−σ} ∫_{|y′|≤ \frac{1}{2} (1+|x|)}(1 + |y′|)^{−β}dy′ ×

× ∫_{|y_{m}|≤2(|y′|+1)} |y_{m}|^{−α}dy_{m}∥u∥ ≤
≤ c|x_{m}|^{−σ}(1 + |x|)^{σ−m} ∫_{|y′|≤ \frac{1}{2} (1+|x|)}(1 + |y′|)^{1−β−α}dy′∥u∥ ≤

≤c|x_{m}|^{−α}(1+|x|)^{−β}|x_{m}|^{α−σ}(1+|x|)^{β+σ−m}(c+(1+|x|)^{m−(β+α)})∥u∥≤

≤ c|x_{m}|^{−α}(1 + |x|)^{−β}∥u∥,  (7)

since σ ≤ α, α + β < m.

If y ∈ ℝ^{m}\D, then 1 + |y| < 1 + |y′| + |y_{m}| < \frac{3}{2}|y_{m}|. Therefore

|J′′_{1}(x)| ≤ c|x_{m}|^{−σ}(1 + |x|)^{σ−m} ∫_{|y|≤ \frac{1}{2} (1+|x|)}(1 + |y|)^{−β−α}dy∥u∥ ≤

≤ c|x_{m}|^{−α}(1 + |x|)^{−β}∥u∥.  (8)

From the estimates (4)–(8) we obtain

|v(x)| ≤ c|x_{m}|^{−α}(1 + |x|)^{−β}∥u∥.  (9)

Let us evaluate the difference v(x) − v(z). It is assumed that |x − z| ≤ \frac{1}{8}|x_{m}|. Then |z| ∼ |x|, |z_{m}| ∼ |x_{m}|.

We introduce the set D̃_{1} = B(x, 2|x − z|), D̃_{2} = B(z, 3|x − z|), D̃_{3} = B(z, \frac{1}{2}|x_{m}| − |x − z|). Clearly, D̃_{1} ⊂ D̃_{2} ⊂ D̃_{3} ⊂ B(x, \frac{1}{2}|x_{m}|) = D_{1}. We have the representation

v(x) − v(z) = ∫_{ℝ^{m}}[K(x, x − y) − K(z, z − y)]u(y)dy =
= ∫_{ℝ^{m}}[K(x, x − y) − K(x, z − y)]u(y)dy +
+ ∫_{ℝ^{m}}[K(z, x − y) − K(z, z − y)]u(y)dy ≡ I_{1}(x, z) + I_{2}(x, z).

By virtue of the first inequality of the condition (c) the term I_{1}(x, z) is evaluated exactly in the same manner as v(x) and we obtain the estimate

|I_{1}(x, y)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν}∥u∥.  (10)

Rewrite the term I_{2}(x, z) in the form

I_{2}(x, z) = ∫_{D_{1}}[K(z, x − y) − K(z, z − y)]u(y)dy +

+ ∫_{ℝ^{m}\D_{1}}[K(z, x − y) − K(z, z − y)]u(y)dy =

= ∫_{D_{1}} K(z, x − y)[u(y) − u(x)]dy − ∫_{D_{1}\_{D_{3}}}[K(z, z − y)u(y)dy −

− ∫_{_{D_{3}}\_{D_{2}}} K(z, z−y)[u(y)−u(x)]dy − ∫_{D_{2}} K(z, z−y)[u(y)−u(z)]dy +

+ ∫_{ℝ^{m}\D_{1}}[K(z, x − y) − K(z, z − y)]u(y)dy =

= ( ∫_{D_{2}} + ∫_{D_{1}\_{D_{3}}} )K(z, x − y)[u(y) − u(x)]dy −

− ∫_{D_{2}} K(z, z − y)[u(y) − u(z)]dy −

− ∫_{_{D_{3}}\_{D_{2}}}[K(z, x − y) − K(z, z − y)][u(y) − u(x)]dy −

− ∫_{D_{1}\_{D_{3}}} K(z, z − y)u(y)dy +

+ ∫_{ℝ^{m}\D_{1}}[K(z, x − y) − K(z, z − y)]u(y)dy ≡ ∑_{i=1}^{5} J_{i}(x, z).

In evaluating J_{1}(x, z), note that D̃_{2} ⊂ B(x, 4|x − z|), B(x, \frac{1}{2}|x_{m}| − 2|x − z|) ⊂ D̃_{3}. Therefore by virtue of the condition (b) and the inequalities (1) we obtain

|J_{1}(x, z)| ≤ ∫_{B(x,4(x−z))} |K(z, x − y)||u(y) − u(x)|dy +
+ ∫_{D_{1}\B(x, \frac{1}{2} |x_{m}|)−2|x−z|)} |K(z, x − y)||u(y) − u(x)|dy ≤
≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν}∥u∥,  (11)

since (\frac{1}{2}|x_{m}|)^{ν} − (\frac{1}{2}|x_{m}| − 2|z − z|)^{ν} ≤ c|x − z|^{ν}. Similarly, if y ∈ D̃_{2}, then |y − z| ≤ 3|x − z| ≤ \frac{3|x_{m}|}{8} ≤ \frac{3}{7}|z_{m}|. Therefore

|J_{2}(x, z)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν}∥u∥.  (12)

It is clear that B(x, \frac{1}{2}|x_{m}|) ⊂ B(z, \frac{1}{2}|x_{m}| + |x − z|) and hence

|J_{4}(x, z)| ≤ c|x_{m}|^{−α}(1 + |x|)^{−β} ∫_{D_{1}\_{D_{3}}} |K(z, z − y)|dy∥u∥ ≤

≤ c|x_{m}|^{−α}(1 + |x|)^{−β} ln \frac{|x_{m}| + 2|x − z|}{|x_{m}| − 2|x − z|}∥u∥ ≤

SINGULAR INTEGRAL OPERATORS ON MANIFOLDS 383

≤ c|x_{m}|^{−α}(1 + |x|)^{−β} \frac{|x − z|}{|x_{m}| − 2|x − z|}∥u∥ ≤

≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν}∥u∥.  (13)

Note that if y ∉ D̃_{2}, then

|x − y| > 2|x − z|, |z − y| > 3|x − z|,

|x − y| < |x − z| + |z − y| < \frac{4}{3}|z − y|, |z − y| < \frac{3}{2}|x − y|.

Taking these inequalities into account, the inequality (2) readily implies that for y ∉ D̃_{2}

|K(z, x − y) − K(z, z − y)| ≤
≤ c\frac{|x − z|^{ν_{1}}}{|x − y|^{m+ν_{1}}} (\frac{|x − y|^{ν_{1}+σ}}{|x_{m} − y_{m}|^{ν_{1}+σ}} + \frac{|z − y|^{ν_{1}+σ}}{|z_{m} − y_{m}|^{ν_{1}+σ}} ),  (14)

using which we obtain

|J_{3}(x, z)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν_{1}} ×

× ∫_{_{D_{3}}\_{D_{2}}} \frac{1}{|x − y|^{m+ν_{1}−ν}} (\frac{|x − y|^{ν_{1}+σ}}{|x_{m} − y_{m}|^{ν_{1}+σ}} + \frac{|z − y|^{ν_{1}+σ}}{|z_{m} − y_{m}|^{ν_{1}+σ}} )dy∥u∥ ≤

≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν_{1}} ×

×( ∫_{D_{1}\_{D_{1}}} \frac{1}{|x − y|^{m+ν_{1}−ν}} (\frac{|x − y|}{|x_{m} − y_{m}|})^{ν_{1}+σ}dy +

+ ∫_{_{D_{3}}\_{D_{2}}} \frac{1}{|z − y|^{m+ν_{1}−ν}} (\frac{|z − y|}{|z_{m} − y_{m}|}^{ν_{1}+σ})dy).

Passing to the spherical coordinates and keeping in mind that \frac{x_{m}−y_{m}}{|x−y|}, \frac{z_{m}−y_{m}}{|z−y|} do not depend on the radius, we have |J_{3}(x, z)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν_{1}}(|x_{m}|^{ν−ν_{1}} + |x − z|^{ν−ν_{1}})∥u∥ ≤

≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν}∥u∥.  (15)

In deriving the estimate, we took into account that ν < ν_{1}, ν_{1} + σ < 1. By virtue of the inequality (14)

|J_{5}(x, z)| ≤ c|z − x|^{ν_{1}}( ∫_{ℝ^{m}\D_{1}} |x − y|^{σ−m}|x_{m} − y_{m}|^{−σ−ν_{1}}|u(y)|dy + + ∫_{ℝ^{m}\B(z, \frac{1}{3} |z_{m}|)} |z − y|^{σ−m}|z_{m} − y_{m}|^{−σ−ν_{1}}|u(y)|dy).

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

∫_{ℝ^{m}\D_{1}} |K(x, x − y)||u(y)|dy

(see the estimates (5)–(8)) and finally obtain

|J_{5}(x, z)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν}∥u∥.  (16)

The estimates (10)–(13), (15), (16) show that

|v(x) − v(z)| ≤ c|x_{m}|^{−(α+ν)}(1 + |x|)^{−β}|x − z|^{ν}∥u∥,

which, with the equality (9) taken into account, proves the theorem.

Corollary 1. Under the conditions of Theorem 1 the operator

∫_{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. Let M be a closed set in ℝ^{m}. The set M is called an (m−1)- dimensional manifold without a boundary of the class C^{1,δ} (0 ≤ δ ≤ 1), if for each x ∈ M there exist a positive number r_{x} and a neighborhood Q(x) of the point x in ℝ^{m}, which is mapped by means of the orthogonal transform T_{x} onto the cylinder Ω_{0} = {ξ : ξ ∈ ℝ^{m}, |ξ′| < r_{x}, |ξ_{m}| < r_{x}} and if the following conditions are fulfilled: T_{x}(x) = 0, the set T_{x}(M ∩ Q(x)) is given by the equation ξ_{m} = φ_{x}(ξ′), |ξ′| < r_{x}; φ_{x} ∈ C^{1,δ} in the domain |ξ′| < r_{x} and ∂_{ξ_{i}}φ_{x}(0) = 0, i = 1, …, m − 1.

Clearly, Q(x) is the cylinder to be denoted by C(x, r_{x}).

In what follows the manifold M 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 function d(x):

\begin{matrix}d(x) & ≤ c(1 + |x|), |d(x) − d(y)| ≤ c|x − y|; \\ ∀x ∈ ℝ^{m}\M, & ∀y ∈ B(x, \frac{1}{2}d(x)) \\ d(y) & ≤ \frac{3}{2}d(x) ≤ 3d(y), 1 + |x| ∼ 1 + |y|; \\ ∀x ∈ M, & ∀y ∈ C(x, \frac{1}{3}r_{x}) \end{matrix}  (17)

d(y, M) = d(y, M ∩ C(x, r_{x})) and if y = T_{x}^{−1}(η), then

d(y) ≤ |η_{m} − φ_{x}(η′)| ≤ 2(1 + a_{x})d(y),  (18)

where a_{x} is the Lipshitz constant of the function φ_{x}.

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, \frac{1}{2}d(x))
|u(x) − u(y)| ≤ cd^{−(α+ν)}(x)(1 + |x|)^{−β}|x − y|^{ν}.

The norm in the space H_{α,β}^{ν}(ℝ^{m}\M) is defined by the equality

∥u∥ = sup_{x∈ℝ^{m}\M} d^{α}(x)(1 + |x|)^{β}|u(x)| +

+ sup_{\begin{matrix}x∈ℝ^{m}\M \\ y∈B(x, \frac{1}{2} d(x)) \end{matrix}} d^{ν+α}(x)(1 + |x|)^{β} \frac{|u(x) − u(y)|}{|x − y|^{ν}}.

The space H_{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 A be 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, ∫_{S(0,1)} f(x, z)d_{z}S = 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|^{ν_{1}}, ν < ν_{1}.

Then the operator A is bounded in the space H_{α,β}^{ν}(ℝ^{m}\M).

Proof. Let M ⊂ B(0, r_{0}) and e_{1} be an infinitely differentiable function such that e_{1}(x) = 1 for |x| ≤ r_{0}+1, e_{1}(x) = 0 for |x| ≥ r_{0}+2. Setting e_{2} = 1−e_{1}, we have

v(x) = ∫_{ℝ^{m}} K(x, x − y)e_{1}(y)u(y)dy +
+ ∫_{ℝ^{m}} K(x, x − y)e_{2}(y)u(y)dy ≡ v_{1}(x) + v_{2}(x).

Let us evaluate the integral

v_{1}(x) = ∫_{ℝ^{m}} K(x, x − y)u_{1}(y)dy (u_{1} = e_{1}u).

Choose a constant r^{∗} (0 < r^{∗} < 1) such that the system \{C(x^{i}, \frac{1}{4}r^{∗})\}_{i=1}^{l} (x^{i}, i = 1, …, l, are points of the manifold M) covers the manifold M and C(x^{i}, 4r^{∗}), i = 1, …, l, are again the coordinate neighborhoods.

We introduce the sets

D_{1} = B(x, \frac{1}{2}d(x)), D_{2} = (B(x, \frac{1}{4}r^{∗})\D_{1}) ∩ B(0, r_{0} + 2) D_{3} = B(0, r_{0} + 2)\(D_{1} ∪ D_{2}), D_{4} = \{y : y ∈ ℝ^{m}, d(y) < \frac{1}{2}r^{∗}\}

Now

v_{1}(x) = ∫_{D_{1}} K(x, x − y)[u_{1}(y) − u_{1}(x)]dy +

+ ∑_{i=2}^{3} ∫_{D_{i}} K(x, x − y)u_{1}(y)dy ≡ ∑_{i=1}^{3} I_{i}(x).

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

|I_{1}(x)| ≤ cd^{−α}(x)(1 + |x|)^{−β}∥u_{1}∥ ≤ cd^{−α}(x)(1 + |x|)^{−β}∥u∥. (19)

Next,

|I_{2}(x)| ≤ ∫_{D_{2}∩D_{4}} |K(x, x − y)||u_{1}(y)|dy +
+ ∫_{D_{2}\D_{4}} |K(x, x − y)||u_{1}(y)|dy ≡ I′_{2}(x) + I′′_{2}(x).

If y ∈ D_{2}, then 1 + |x| ∼ 1 + |y| ∼ c. Moreover, d(y) ≤ \frac{1}{2}r^{∗} for y ∈ D_{2} ∩ D_{4} and therefore there exists i (i = 1, …, e) such that x, y ∈ C(x^{i}, \frac{3+\sqrt{2}}{2}r^{∗}). Let y = T −1 ; x^{i}(η), x = T −1 ; x^{i}(ξ). By virtue of the inequality (18)

|η_{m} − φx^{i}(η′)| ≤ 2(1 + ax^{i})d(y), |ξ_{m} − φx^{i}(ξ′)| ≤ 2(1 + ax^{i})d(x).

Taking into account that |x − y| > \frac{1}{2}d(x), we therefore obtain

|x − y| = |ξ − η| ≥ \frac{1}{4}(|ξ′ − η′| + |ξ_{m} − η_{m}| + d(x)) ≥
≥\frac{1}{16}(1 + ax^{i})^{−1}(4(1 + ax^{i})|ξ′ − η′| + |ξ_{m} − η_{m}| + 4(1 + ax^{i})d(x)) ≥
≥\frac{1}{16}(1 + ax^{i})^{−1}(|ξ′ − η′| + |η_{m} − φx^{i}(η′)| + d(x)),  (20)

since |ξ_{m} − η_{m}| ≥ |η_{m} − φx^{i}(η′)| − |φx^{i}(η′) − φx^{i}(ξ′)| − |ξ_{m} − φx^{i}(ξ′)|. Thus

|I′_{2}(x)| ≤ c ∫_{|ξ′−η′|≤4r^{∗}} dη′ ∫_{−2r^{∗}}^{2r^{∗}} |η_{m} − φx^{i}(η′)|^{−α} ×

×(|ξ′ − η| + |η_{m} − φx^{i}(η′)| + d(x))^{−m}dη_{m}∥u∥.

Using the transform η_{m} − φx^{i}(η′) = η̃_{m}, we obtain

|I′_{2}(x)| ≤ c ∫_{0}^{4r^{∗}} r^{m−2}dr ∫_{−4r^{∗}}^{4r^{∗}} |η_{m}|^{−α}(r + |η_{m}| + d(x))^{−m}dη_{m}∥u∥,

which, upon applying the transform r = d(x)r̃, η_{m} = d(x)η̃_{m}, gives

|I′_{2}(x)| ≤ cd^{−α}(x)∥u∥ ≤ cd^{−α}(x)(1 + |x|)^{−β}∥u∥.  (21)

If y ∈ D_{2}\D_{4}, then d(y) ≥ \frac{1}{2}r^{∗}, \frac{1}{2}d(x) ≤ |x − y| < \frac{1}{4}r^{∗} by virtue of which d(x) ≥ d(y) − |x − y| ≥ \frac{1}{4}r^{∗}, |x − y| ≥ \frac{1}{8}r^{∗}. Therefore

|I′′_{2}(x)| ≤ c∥u∥ ≤ cd^{−α}(x)(1 + |x|)^{−β}∥u∥.  (22)

Finally, if y ∈ D_{3}, then 1 + |x| ≤ 1 + |y| + |x − y| ≤ c|x − y|. Therefore

|I_{3}(x)| ≤ c(1 + |x|)^{−m}∥u∥ ≤ cd^{−α}(x)(1 + |x|)^{−β}∥u∥.  (23)

The inequalities (19), (21)–(23) show that

|v_{1}(x)| ≤ cd^{−α}(x)(1 + |x|)^{−β}∥u∥.  (24)

In evaluating the difference v_{1}(x)−v_{1}(z), it will be assumed that |x−z| < \frac{1}{8}d(x). Then 1 + |z| ∼ 1 + |x|, d(x) ∼ d(y).

We introduce the set

D̃_{1} = B(x, 2|x − y|), D̃_{2} = B(z, 3|x − z|), D̃_{3} = B(z, \frac{1}{2}d(x) − |x − z|).

Proceeding as in proof of Theorem 1, we obtain

|v_{1}(x) − v_{1}(z)| ≤ cd^{−(α+ν)}(x)(1 + |x|)^{−β}|x − z|^{ν}∥u∥.  (25)

To evaluate the integral

v_{2}(x) = ∫_{ℝ^{m}} K(x, x − y)u_{2}(y)dy (u_{2} = e_{2}u)

note that the function u_{2} 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

D_{1} = B(x, \frac{1}{2}(1 + |x|)), D_{2} = B(0, 2|x| + 1)\D_{1}, D_{3} = ℝ^{m}\(D_{1} ∪ D_{2}),

we readily obtain the estimate

|v_{2}(x)| ≤ cd^{−α}(x)(1 + |x|)^{−β}∥u∥.  (26)

Now, considering the sets

D̃_{1} = B(x, 2|x − z|), D̃_{2} = B(z, 3|x − z|), D̃_{3} = B(z, \frac{1}{2}(1 + |x|) − |x − z|) it is easy to show that

|v_{2}(x) − v_{2}(z)| ≤ cd^{−(α+ν)}(x)(1 + |x|)^{−β}|x − z|^{ν}∥u∥.  (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 u defined on ℝ^{m} belongs to the space H_{λ}^{ν}(ℝ^{m}) (ν, λ > 0), if

|u(x)| ≤ c(1 + |x|)^{−β}, |u(x) − u(y)| ≤ c|x − y|^{ν}ρ_{xy}^{−ν−λ},

where ρ_{xy} = min(1 + |x|, 1 + |y|).

Theorem 3. Let the characteristic f of the singular operator A satisfy 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|^{ν_{1}}(min(|x_{m}|, |y_{m}|))^{ν_{1}}|z_{m}|^{−σ}.

It is also assumed that a ∈ C(ℝ̇^{m}) (ℝ̇^{m} = ℝ^{m} ∪ ∞) and (a − a(∞)) ∈ H_{δ}^{ν_{1}}(ℝ^{m}). Then the integral operator

v(x) = B(u)(x) = ∫_{ℝ^{m}}[a(x) − a(y)]K(x, x − y)u(y)dy

is completely continuous in the space H_{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−γ,β+2γ}^{ν+γ}(ℝ^{m}\Γ), where γ is an arbitrary positive number satisfying the condition

γ < min\{λ, ν_{1} − ν, α − σ, \frac{1}{2}(m − β − α)\}.

Indeed, it is clear that

|v(x)| ≤ ∫_{D_{1}∪D_{2}∪D_{4}} |a(x) − a(y)||K(x, x − y)||u(y)|dy +
+ ∫_{D_{3}}(|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|x_{m}|^{γ−α}(1 + |x|)^{−β−2γ}∥u∥.  (28)

Let us now assume that |x − z| ≤ \frac{1}{8}|x_{m}| and evaluate the difference v(x) − v(z). We have

|v(x)v(y)| ≤ ∫_{D_{2}} |a(x) − a(y)||K(x, x − y)||u(y)|dy +

+ ∫_{D_{2}} |a(z) − a(y)||K(z, z − y)||u(y)|dy +

+ ∫_{ℝ^{m}\_{D_{1}}} |a(x) − a(z)||K(x, x − y)||u(y)|dy +

+ ∫_{ℝ^{m}\_{D_{2}}} |a(z) − a(y)||K(x, x − y) − K(z, x − y)||u(y)|dy ≡ ∑_{i=1}^{4} I_{i}(x, z).

Hence

|I_{i}(x, z)| ≤ c|x_{m}|^{−α}(1 + |x|)^{−β}(1 + |x|)^{−ν_{1}−λ}|x − z|^{ν_{1}}∥u∥ ≤
≤ c|x_{m}|^{−α−γ}(1 + |x|)^{−β−2γ}∥u∥ (i = 1, 2).  (29)

Write the term I_{3} in the form

I_{3}(x, z) = |a(x) − a(z)|( ∫_{D_{1}\_{D_{1}}} |K(x, x − y)||u(y)|dy +

+ ∫_{ℝ^{m}\_{D_{1}}} |K(x, x − y)||u(y)|dy).

This representation gives

|I_{3}(x, z)| ≤ c|x − z|^{ν_{1}}(1 + |x|)^{−ν_{1}−λ}|x_{m}|^{−α}(1 + |x|)^{−β} ×
×( ln \frac{|x_{m}|}{|x − z|} + c_{1})∥u∥ ≤ c|x_{m}|^{−α−ν}(1 + |x|)^{−β−2γ}|x − z|^{ν+γ}∥u∥. (30)

To evaluate the integral term I_{4} note that for y ∉ D̃_{1} we have

|K(x, x − y) − K(z, z − y)| ≤ c\frac{|x − z|^{ν_{1}}}{|x − y|^{m}} |x_{m}|^{−ν_{1}} \frac{|x − y|^{σ}}{|x_{m} − y_{m}|^{σ}} +

+c\frac{|x − z|^{ν_{1}}}{|x − y|^{m+ν_{1}}} (\frac{|x − y|^{ν_{1}+σ}}{|x_{m} − y_{m}|^{ν_{1}+σ}} + \frac{|z − y|^{ν_{1}+σ}}{|z_{m} − y_{m}|^{ν_{1}+σ}} ).

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

|I_{4}(x, z)| ≤ c|x_{m}|^{−α−ν}(1 + |x|)^{−β−2γ}|x − z|^{ν+γ}∥u∥.  (31)

The estimates (28)–(31) show that the operator B is bounded from the space H_{α,β}^{ν}(ℝ^{m}\Γ) into the space H_{α−γ,β+2γ}^{ν+γ}(ℝ^{m}\Γ). The validity of the theorem now follows from the complete continuity of the operator of the embedding of the space H_{α−γ,β+2γ}^{ν+γ}(ℝ^{m}\Γ) into the space H_{α,β}^{ν}(ℝ^{m}\Γ).

In a similar manner we prove

Theorem 4. Let m ∈ C^{1,δ}, let the characteristic f of the operator A satisfy 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|^{ν_{1}}( min(d(x), d(y)))^{−ν_{1}},

and let the function a satisfy the conditions of Theorem 3. Then the operator B is completely continuous in the space H_{α,β}^{ν}(ℝ^{m}\M).

2. We shall consider the matrix-function A(ξ) = ∥A_{ij}(ξ)∥_{n×n}. Let

A(λξ) = A(ξ) (λ > 0), A_{ij} ∈ C^{∞}(ℝ^{m}\0), det A(ξ) ≠ 0 (ξ ≠ 0).

We set A_{0} = A^{−1}(0, …, 0, −1)A(0, …, 0, +1). It is assumed that λ_{j} (j = 1, …, s) is the eigenvalue of the matrix A_{0} and r_{j} is its multiplicity (∑_{j=1}^{s} r_{j} = n).

We introduce the matrices B_{r}(α) ≡ ∥B_{νk}(α)∥_{r×r} where

B_{νk}(α) = \begin{cases}0, & ν < k; \\ 1, & ν = k; \\ \frac{α^{ν−k}}{(ν−k)!}, & ν > k, \end{cases}

B(r_{i}; α) = diag [B_{r_{i1}}(α), …, B_{r_{ip_{i}}} (α)] (r_{i1} + ⋯ + r_{ip_{i}} = r_{i}).

By the Jordan theorem the matrix A_{0} is representable in the form A_{0} = gBg^{−1}, where det g ≠ 0, B is the modified Jordan form of the matrix A_{0}, B = diag[λ_{1}B(r_{1}; 1), …, λ_{s}B(r_{s}; 1)]. We introduce the notation

δ′_{j} = \frac{1}{2πi} ln λ_{j}, δ_{j} = δ′_{k} for ∑_{ν=1}^{k−1} r_{ν} < j ≤ ∑_{ν=1}^{k} r_{ν}, j = 1, …, n;

α_{±}(ξ) = \frac{1}{2πi} ln \frac{ξ_{m} ± |ξ′|}{|ξ′|} (ξ′ = (ξ_{1}, …, ξ_{m−1}));

by ln z we denote a logarithm branch which is real on the positive semi-axis, i.e., −π < arg z ≤ π,

(\frac{ξ_{m} ± i|ξ′|}{|ξ′|})^{δ} ≡ diag [(\frac{ξ_{m} ± i|ξ′|}{|ξ′|})^{δ_{1}}, …, (\frac{ξ_{m} ± i|ξ′|}{|ξ′|})^{δ_{n}}],

B_{±}(ξ) ≡ diag [B(r_{1}; α_{±}(ξ), …, B(r_{s}; α_{±}(ξ))].

Theorem 5. Let the matrix A be strongly elliptic. Then A admits the factorization A(ξ) = cgA_{−}(ξ′, ξ_{m})D(ξ)A_{+}(ξ′, ξ_{m})g^{−1}, where

c = A(0, …, +1), D(ξ) = B_{−}(ξ)(ξ_{m} − i|ξ′|)^{δ}(ξ_{m} + i|ξ′|)^{−δ}B_{+}^{−1}(ξ), A_{±}(λξ) = A_{±}(ξ) (λ > 0), det A_{±}(ξ) ≠ 0.

For |ξ′| ≠ 0 the matrices A_{+}, A_{+}^{−1} (accordinly, A_{−}, A_{−}^{−1}) admit analytic continuations with respect to ξ_{m} into the upper (lower) complex half-plane and these continuations are bounded.

Moreover, for any natural number k the matrix A_{±} admits the expansion

A_{±}(ξ′, ξ_{m}) = I +
+ ∑_{p=1}^{k} ∑_{q=0}^{(p+1)(2n−1)} c^{pq}(\frac{ξ′}{|ξ′|} ln^{q} \frac{ξ_{m} ± |ξ′|}{|ξ′|}(\frac{ξ_{m} ± |ξ′|}{|ξ′|})^{−p} + A^{k}(ξ), (32)

where c^{pq} ∈ C^{∞}(ℝ^{m−1}\0), A^{k}(λξ) = A^{k}(ξ) (λ > 0), A^{k} ∈ C^{k}(ℝ^{m}\0). Similar expansions also hold for the inverse matrices A_{±}^{−1}.

Let us outline a scheme for proving the theorem.

We set

A_{∗}(ξ) = (\frac{ξ_{m} − i|ξ′|}{|ξ′|})^{−δ}B_{−}^{−1}(ξ)g^{−1}A^{−1}(0, …, 0, +1) ×

×A(ξ)gB_{+}(ξ)(\frac{ξ_{m} + i|ξ′|}{|ξ′|})^{δ},

Z_{+} = {z = x_{1} + ix_{2}, x_{2} > 0}, Z_{−} = {z = x_{1} + ix_{2}, x_{2} < 0}.

Consider the homogeneous Hilbert problem: Find an analytic (in the domain Z_{+} ∪Z_{−}) matrix-function Φ(ξ′, z), which is left and right continuously extendable on ℝ, by the boundary condition

Φ^{−}(ξ′, t) = A_{∗}(ξ′, t)Φ^{+}(ξ′, t), lim_{Z̄_{+}∋z→∞} Φ(ξ′, z) = I, lim_{Z̄_{−}∋z→∞} Φ(ξ′, z) = I.

The solution is to be sought in the form

Φ(ξ′, z) = \frac{1}{2πi} ∫_{ℝ} \frac{φ(ξ′, z)}{t − z}dt + I.

To define the matrix φ we obtain the system of singular integral equations

(A_{∗}(ξ′, t_{0}) + I)φ(ξ′, t_{0}) + \frac{1}{πi}(A_{∗}(ξ′, t_{0}) − I) ∫_{ℝ} \frac{φ(ξ′, t)}{t − t_{0}}dt =

= 2(I − A_{∗}(ξ′, t_{0})).  (33)

One can prove that the system (33) is unconditionally and uniquely solvable and obtain, after a rather sophisticated reasoning, the desired result.

Remark 1. The fact that partial indices of the strongly elliptic matrix A are equal to zero is proved in [5]. An expansion of the form (32) when A is a scalar function is also obtained therein.

Let D be a finite or infinite domain in ℝ^{m} bounded by the compact manifold without a boundary M from the class C^{1,ν_{1}}.

Consider the matrix singular operator

A(u)(x) = a(x)u(x) + ∫_{D} f(x, \frac{x − y}{|x − y|})|x − y|^{−m}u(y)dy,  (34)
a(x) = ∥a_{ij}(x)∥_{n×n}, f(x, z) = ∥f_{ij}(x, z)∥_{n×n},
u = (u_{1}, …, u_{n})

in the spaces [H_{α,β}^{ν}(D)]^{n} (0 < α, ν < 1, ν < ν_{1}, β > 0, α + β < m) and [L_{p}(D, (1+|x|)^{γ})]^{n} (p > 1, −\frac{m}{p} < γ < \frac{m}{p′}, p′ = \frac{p}{p−1}), u ∈ L_{p}(D, (1+|x|^{γ}) ↔ ∫_{D} |u(x)^{p}(1 + |x|)^{pγ}dx < ∞.

Taking into account the character of the linear bounded operator acting in the spaces with two norms (see [11]), the proved theorems enable us to prove

Theorem 6. Let a ∈ H_{λ}^{ν_{1}}(D̄), f(x, ·) ∈ C^{∞}(ℝ^{m}\0), ∫_{S(0,1)} f(x, z)d_{z}S = 0; ∂_{z}^{p}f(·, z)∈H_{λ}^{ν_{1}}(D̄), |p| = 0, 1, …, if the domain D is bounded; lim_{x→∞} f(x, z)≡

f(∞, z) and ∂_{z}^{p}(f(·, z)−f(∞, z))∈H_{λ}^{ν_{1}}(D̄), if the domain is unbounded. The determinant of the symbol matrix Φ(A)(x, ξ) of the integral operator (34) is different from zero and either of the following two conditions is fulfilled:

(i) ∀x ∈ M the matrix Φ(A)(x, ξ) is strongly elliptic and Hermitian; (ii) ∀x ∈ M the matrix Φ(A)(x, ξ) is strongly elliptic and odd with respect to the variable ξ.

Then the operator A is the Noether operator both in the space [L_{p}(D, (1+ |x|)^{γ})]^{n} and in the space [H_{α,β}^{ν}(D)]^{n}. Any solution of the equation

A(u)(x) = g(x), g ∈ [L_{p}(D, (1 + |x|)^{γ})]^{n} ∩ [H_{α,β}^{ν}(D)]^{n}  (35)

from the space [L_{p}(D, (1+|x|)^{γ})]^{n} belongs to the space [L_{p}(D, (1+|x|)^{γ})]^{n} ∩ [H_{α,β}^{ν}(D)]^{n}. For equation (35) to be solvable it is necessary and sufficient that (g, v) = 0, where v is an arbitrary solution of the formally conjugate equation A′(v) = 0 from the space [L_{p′}(D, (1 + |x|)^{−γ})]^{n} into [H_{α,β}^{ν}(D)]^{n}.

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(Received 17.06.1993)

Author’s address: A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, Z. Rukhadze St., 380093 Tbilisi Republic of Georgia