Georgian Mathematical Journal Volume 9 (2002), Number 3, 507–524

ON A PROBLEM OF LINEAR CONJUGATION IN THE CASE

OF NONSMOOTH LINES AND SOME MEASURABLE COEFFICIENTS

E. GORDADZE

Abstract. A boundary value problem of linear conjugation is considered for more general curves than those studied previously. A condition on the coefficient is found, under which the classical results are valid for these curves.

2000 Mathematics Subject Classification: 30E20, 30E25.

Key words and phrases: Boundary value problem, cusp, weight, factorization.

1^{0}. Let Γ be a rectifiable closed Jordan curve dividing the plane into to domains D+_{Γ} and D−_{Γ} (it is assumed that ∞ ∈ D−_{Γ}). The direction that leaves the domain D+_{Γ} lie on the left is assumed to be the positive direction on Γ. The line directed in this direction is denoted by Γ^{+}, whereas the line directed in the opposite direction is denoted by Γ^{−}. Occasionally, we will write Γ instead of Γ^{+}.

In the sequel, we will need classes of analytic functions E_{p}(D±_{Γ}) which are usually called Hardy–Smirnov classes. For boundary value problems the following definition of such classes is convenient.

We say that Φ(z) ∈ E_{p}(D_{Γ}), where p > 0 and D_{Γ} denotes either D+_{Γ} or D−_{Γ}, if:

a) Φ(z) is analytic in D_{Γ};

b) Φ(∞) = 0 if D_{Γ} = D−_{Γ};

c) there exists a sequence of curves \{Γ_{n}\}_{n=1}^{∞} ∈ D_{Γ} such that ∞\overline{∈}Γ_{n}, Γ_{n} → Γ as n → ∞ and

sup_{n} ∫_{Γ_{n}} |Φ(z)|^{p}ds < ∞  (1.1)

By the convergence Γ_{n} → Γ we understand the same convergence as in [1], p. 203.

For p ≥ 1, the above-defined class E_{p}(D−_{Γ}) coincides with the class which is

often denoted by E^{∘}_{p}(D−_{Γ}). We will not use this notation because we will need only functions Φ(z) vanishing at infinity.

ISSN 1072-947X / $8.00 / \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag www.heldermann.de

As usual, we denote by (S_{Γ} φ(t))(τ) and (K_{Γ} φ(t))(z) the integrals (whose contour can be both closed and open):

(S_{Γ} φ(t))(τ) = \frac{1}{πi} ∫_{Γ} φ(t)(t − τ)^{−1}dt, τ ∈ Γ,

(K_{Γ} φ(t))(z) = \frac{1}{2πi} ∫_{Γ} φ(t)(t − z)^{−1}dt, t\overline{∈}Γ.

It is assumed that φ ∈ L(Γ). The first integral is understood in the sense of the Cauchy principal value (see, e.g., [2], p. 13). For these integrals we sometimes use the notations S_{Γ} φ and K_{Γ} φ or Sφ and Kφ.

We say that Γ ∈ R if for any φ ∈ L_{p}(Γ), p > 1, we have

∥S_{Γ} φ∥_{L_{p}(Γ)} ≤ M_{p}∥φ∥_{L_{p}(Γ)}.

If Γ ∈ R, φ ∈ L_{p}(Γ), ψ ∈ L_{q}(Γ), q = p(p − 1)^{−1}, then

∫_{Γ} (S_{Γ} φ)(t)ψ(t)dt = − ∫_{Γ} φ(t)(S_{Γ} ψ)(t)dt (Riesz equality),  (1.2)

and if, in addition, Γ is a closed line, then

S2_{Γ} φ = φ  (1.3)

(with respect to (1.2) and (1.3) for Γ ∈ R, see [3], [4], [5]).

In the sequel we will need one more well-known definition. We say that the positive measurable function ρ(t) is a weight and write ρ ∈ W_{p}(Γ), p > 1, if

∥ρS_{Γ} ρ^{−1}φ∥_{L_{p}(Γ)} ≤ M_{p}∥φ∥_{L_{p}(Γ)}, ∀ φ ∈ L_{p}(Γ).

We will need the following properties of functions of the class E_{p}(D_{Γ}), where Γ ∈ R:

(i_{1}) {Φ(z) = (K_{Γ^{±}}Φ^{±})(z), Φ^{±} ∈ L_{1}(Γ)} ⟺ {Φ(z) ∈ E_{1}(D±_{Γ})}.

(i_{2}) {Φ(z) = E_{δ}(D±_{Γ}), δ > 0, Φ^{±} ∈ L_{p}(Γ), p > δ} ⟺ {Φ(z) ∈ E_{p}(D_{Γ})}.

(i_{3}) {φ ∈ L_{∞}} ⟹ {(exp(Kφ)(z)) − 1 ∈ E_{δ}(D_{Γ}) for some δ > 0}.

(i_{4}) In (1.1) we can take as Γ_{n} the images of the circumferences |w| = r_{n}, r_{n} → 1 for the conformal transformation of the circle |w| < 1 to D_{Γ}.

If D_{Γ} = D+_{Γ}, then (i_{1}) and (i_{4}) are valid for all rectifiable simple closed lines (see, e.g., [1], p. 208 and p. 203); (i_{2}) follows from Smirnov’s theorem ([1], p. 264) and Havin’s theorem ([6], p. 512); (i_{3}) is proved in [7], p. 68.

For D_{Γ} = D−_{Γ}, these statements or their close analogs are used by various authors, but E_{p}(D−_{Γ}) is defined in different ways, which, for p < 1, gives different classes. Therefore here we present the proofs the more so that they are very simple.

Take the conformal transformation z = ω_{0}(ζ) = (ζ − a)^{−1} + a and the inverse transformation ζ = (z − a)^{−1} + a, where a ∈ D+_{Γ}, z ∈ D−_{Γ}, ζ ∈ D+_{Γ_{0}}, where Γ_{0} is the image of Γ. (An analogous reasoning can be found in [5], p. 32.) We obviously obtain the equality

(K_{Γ^{±}}φ)(z) = (ζ − a)(K_{Γ{∓ \atop 0}}(φ(ω_{0}(τ)(τ − a)^{−1}))(ζ)  (1.4)

and the inclusion

Φ(z) ∈ E_{p}(D−_{Γ}), p > 0 ⟺ \frac{1}{ζ − a}Φ(ω_{0}(ζ)) ∈ E_{p}(D+_{Γ_{0}}).  (1.5)

Indeed,

(K_{Γ^{±}}φ)(ω_{0}(ζ)) = (2πi)^{−1} ∫_{Γ^{±}} φ(t)(t − ω_{0}(ζ))^{−1}dt

= −(2πi)^{−1} ∫_{Γ_{0}^{∓}} (\frac{1}{τ − a} + a − \frac{1}{ζ − a} − a)^{−1}(τ − a)^{−2}φ(ω_{0}(τ))dτ

= (ζ − a)(K_{Γ{∓ \atop 0}}φ(ω_{0}(τ))(τ − a)^{−1})(ζ).

We will prove (1.5). Since a\overline{∈}Γ_{n}^{0}, where Γ_{n}^{0} are the images of the lines Γ_{n} ∈ D−_{Γ}, from (1.1), for the mapping ζ = ω_{0}^{−1}(z), we have for any analytic function Φ(z) in D_{Γ}^{−}

{Φ(z) ∈ E_{p}(D−_{Γ}), p > 0}

⟺ \{ sup_{n} ∫_{Γ_{n}^{0}} |Φ(ω_{0}(ζ))|^{p}|dζ| < ∞, Γ_{n}^{0} ∈ D_{Γ{+ \atop 0}}, a\overline{∈}Γ_{n}^{0}, Φ(ω_{0}(a)) = 0\}

⟺ \{\frac{1}{ζ − a}Φ(ω_{0}(ζ)) ∈ E_{p}(D_{Γ{+ \atop 0}})\}.

Thus (1.4) and (1.5) are proved. Hence we easily obtain (i_{1}), (i_{2}) and (i_{3}) for D_{Γ} = D−_{Γ}. Indeed,

\{Φ(z) ∈ E_{1}(D−_{Γ})\} ⟺ \{\frac{1}{ζ − a}Φ(ω_{0}(ζ)) ∈ E_{1}(D+_{Γ_{0}})\}

⟺ \{\frac{1}{ζ − a}Φ(ω_{0}(ζ)) = (K_{Γ{+ \atop 0}} (\frac{1}{τ − a}φ(ω_{0}(τ)))(ζ)\}

⟺ \{Φ(z) = (ζ − a)(K_{Γ{+ \atop 0}}(τ − a)^{−1}φ(ω_{0}(ζ)))(ζ) = (K_{Γ^{−}}φ)(z)\}.

We have obtained (i_{1}). Let us show (i_{2}) for D_{Γ} = D−_{Γ}:

{Φ(z) ∈ E_{δ}(D−_{Γ}), Φ^{−} ∈ L_{p}(Γ), p > δ}

⟺ \{\frac{1}{ζ − a}Φ(ω_{0}(ζ)) ∈ E_{δ}(D+_{Γ_{0}}), \frac{1}{τ − a}Φ^{+}(ω_{0}(τ)) ∈ L_{p}(Γ_{0}), p > δ\}

⟺ \{\frac{1}{ζ − a}Φ(ω_{0}(ζ)) ∈ E_{p}(D+_{Γ_{0}})\} ⟺ \{Φ(z) ∈ E_{p}(D−_{Γ})\}.

Let us show (i_{3}) for D_{Γ} = D−_{Γ}:

{Φ(z) = (exp K_{Γ^{−}}φ)(z) − 1, z ∈ D−_{Γ}, φ ∈ L_{∞}(Γ)}

⟺ \{Φ(ω_{0}(ζ)) = ( exp(ζ − a)(K_{Γ{+ \atop 0}}φ(ω_{0}(τ)(τ − a)^{−1})(ζ)) − 1, a ∈ D+_{Γ_{0}},

(τ − a)^{−1}φ(ω_{0}(τ)) ∈ L_{∞}(Γ)} ⟹ {Φ(ω_{0}(ζ)) ∈ E_{δ}(D+_{Γ_{0}}), φ(ω_{0}(a)) = 0}

⟺ (ζ − a)^{−1}Φ(ω_{0}(ζ)) ∈ E_{δ}(D+_{Γ_{0}}) ⟺ Φ(z) ∈ E_{δ}(D−_{Γ}).

Next let us show (i_{4}) for D_{Γ} = D−_{Γ}. Let Γ_{0} be as above and Γ_{n}^{0} be the images of the circumferences |w| = r_{n}, r_{n} → 1, for the conformal transformation ζ = ω_{1}(w) of the unit circle |w| < 1 in D+_{Γ_{0}}. Let further Γ_{n} be the images of the circumferences |w| = r_{n} for the transformation z = ω_{0}(ω_{1}(w)), where ω_{0}(ζ) is the same as above. It is obvious that

sup_{n} ∫_{Γ_{n}} |Φ(z)|^{p}|dz| = sup_{n} ∫_{Γ_{n}^{0}} |Φ(ω_{0}(ζ))|^{p}|ω′_{0}(ζ)||dζ|

= sup_{n} ∫_{Γ_{n}} |Φ(ω_{0}(ζ))|^{p}|ζ − a|^{−2}|dζ| ≤ const ∫_{Γ_{n}^{0}} |\frac{1}{ζ − a}Φ(ω_{0}(ζ))|^{p}||dζ| < ∞

(here we take into account that a\overline{∈}Γ_{n}^{0}).

2^{0}. Our boundary value problem of linear conjugation is formulated as follows:

Find a function Φ(z) ∈ E_{p}(D±_{Γ}) whose angular boundary values satisfy the condition

Φ^{+}(t) = G(t)Φ^{−}(t) + f(t), t ∈ Γ,  (2.1)

where f ∈ L_{p}(Γ), p > 1, and G ∈ L_{∞}(Γ) are the known functions.

In the case, in which G(t) is piecewise-continuous and Γ is a Lyapunov curve, this problem was solved for the first time by B. Khvedelidze [9]. The results obtained subsequently are presented in [2] and other works of various authors. We will be concerned with the case of a measurable coefficient which was investigated by I. Simonenko [10] for Lyapunov lines.

The conditions of [10] can be written as followsin the form

(S_{1}) 0 < vrai inf_{t∈Γ}|G(t)| < vrai sup_{t∈Γ}|G(t)| < ∞  (2.2)

(S_{2}) the function arg G(t) = φ(t) can be chosen so that φ(t) = φ_{1}(t) + φ_{2}(t), where φ_{1}(t) is continuous on Γ except perhaps for one point at which it has a first order discontinuity φ_{1}(t_{0}−) − φ_{1}(t_{0}+) = 2πκ, where κ is an integer and φ_{1}(t_{0}∓) denotes the unilateral limits along the negative and the positive direction on Γ, respectively; φ_{2}(t) satisfies the inequality

vrai sup_{t∈Γ}|φ_{2}(t)| < \frac{π}{max(p, q)}, q = p(p − 1)^{−1}, p > 1.  (2.3)

The function, whose argument satisfies (2.3) and modulus (2.2), is called sectorial. In the sequel, for p and q it will always be assumed that p > 1 and q = p(p − 1)^{−1}.

For the briefness of our exposition we introduce yet another definition:

The statement (A_{κ}) or, simply, (A) is valid for problem (2.1) if:

1) all solutions of this problems, when they exist, are written in the traditional form

Φ(z) = X(z)(K_{Γ^{+}}(f/X^{+}))(z) + X(z)P_{κ−1}(z),  (2.4)

where P_{κ−1}(z) is a polynomial of degree κ−1 and, in addition to this, it assumed that P_{k}(z) ≡ 0 for k < 0. By X(z) we mean the following:

X(z) = \begin{cases}X_{0}(z) & for z ∈ D+_{Γ}, \\ (z − a)^{−κ}X_{0}(z) & for z ∈ D−_{Γ}, \end{cases}  (2.5)

X_{0}(z) = exp (K_{Γ} ln(t − a)^{−κ}G(t))(z),

X_{0}(z) − 1 ∈ E_{p}(D±_{Γ}), X_{0}^{−1}(z) − 1 ∈ E_{q}(D±_{Γ}),

(2.6)

where a ∈ D+_{Γ}, κ is the integer depending on G(t) and called the index of the problem;

2) for κ ≥ 0 the problem is solvable unconditionally, while for κ < 0 for the problem to be solvable it is necessary and sufficient that the condition

∫_{Γ} t^{k}f(t)(X^{+}(t))^{−1}dt = 0, k = 0, 1, …, −κ − 1.  (2.7)

be fulfilled.

I. Simonenko showed [10] that for Lyapunov lines, when the conditions (S_{1}) and (S_{2}) are fulfilled, the statement (A_{κ}) is true for problem (2.1). Later, other authors extended this result to sufficiently general lines (see, for example, [7], [8], [11]), but in that case the conditions imposed on the lines exclude the existence of cusps.

Our objective here is to modify condition (2.3) so that the classical result be valid for the lines with cusps.

Denote by Γ_{ab} a simple arc with ends a, b and directed from a to b. Denote by χ(Γ_{ab}) the characteristic function of the set {t : t ∈ Γ_{ab}}.

We say that Γ_{ab} ∈ R_{A} if it can be complemented to a closed contour Γ_{ab}^{0} ∈ R, Γ_{ab} ⊂ Γ_{ab}^{0}, so that the statement (A_{0}) be true for any function [G(t)]^{χ(Γ_{ab})} satisfying, on Γ_{ab}^{0}, the conditions (S_{1}) and (S_{2}) with the zero index.

Main Result. If in problem (2.1) the simple closed curve Γ can be represented as Γ = ∪_{k=1}^{n}Γ_{a_{k}a_{k+1}}, a_{k+1} = a_{1}, where the direction on Γ_{a_{k}a_{k+1}} coincides with the positive direction on Γ, Γ_{a_{k}a_{k+1}} ∈ R_{A} (k = 1, 2, …, n) and the function G(t) satisfies, for t ∈ Γ, the conditions (S_{1}) and (S_{2}) and, moreover, for the points a_{k} (k = 1, 2, …, n) there exist, on Γ, angular neighborhoods Γ_{b_{k}a_{k}} ⊂ Γ, Γ_{a_{k}c_{k}} ⊂ Γ, where the conditions

vrai sup_{t∈Γ_{b_{k}a_{k}}}|φ_{2}(t)| < α_{k} \frac{π}{max(p, q)} ,

vrai sup_{t∈Γ_{a_{k}c_{k}}}|φ_{2}(t)| < (1 − α_{k})\frac{π}{max(p, q)}

(2.8)

are fulfilled for some α ∈ [0, 1], then the statement (A_{κ}) is valid for problem (2.1).

It is obvious that in these conditions both cusps and other cases may occur at the points a_{k}.

3^{0}. The solution of a boundary value problem of linear conjugation is usually closely connected with a factorization problem. The factorization of the function G(t) in E_{p}(D±_{Γ}) implies that it can be represented as

\begin{matrix}G(t) = X^{+}(t)(t − a)^{κ}/X^{−}(t), t ∈ Γ, a ∈ D+_{Γ}, \\ X(z) − 1 ∈ E_{p}(D±_{Γ}), X^{−1}(z) − 1 ∈ E_{q}(D±_{Γ}), \end{matrix}  (3.1)

where κ is an integer, X^{±}(t) are boundary values of the function X(z). It is obvious that (3.1) can be rewritten as

\begin{matrix}X^{+}(z) = (t − a)^{−κ} · G(t) · X^{−}(t), t ∈ Γ, a ∈ D+_{Γ}, \\ X(z) − 1 ∈ E_{p}(D±_{Γ}), X^{−1}(z) − 1 ∈ E_{q}(D±_{Γ}), \end{matrix}  (3.1′)

The function X(z) is called the factor function, and the number κ the index of the problem.

Lemma 1. If the function G(t) admits a factorization in E_{p}(D_{Γ}), p > 1, then to it there corresponds the unique number κ which can be put in (3.1) or, which is the same, in (3.1′) and for fixed a ∈ D+_{Γ} the factorization is unique.

Proof. Suppose there are two factorizations

G(t) = (t − a_{1})^{κ_{1}} · X_{1}^{+}(t) · (X_{1}^{−}(t))^{−1}
= (t − a_{2})^{κ_{2}} · X_{2}^{+}(t) · (X_{2}^{−}(t))^{−1}, a_{1}, a_{2} ∈ D+_{Γ}.  (3.2)

Let κ_{1} > κ_{2}. From (3.2) we have

X_{1}^{+}(t) · (X_{2}^{+}(t))^{−1} = (t − a_{2})^{κ_{2}−κ_{1}}(\frac{t − a_{2}}{t − a_{1}} )^{κ_{1}}X_{1}^{−}(t) · (X_{2}^{−}(t))^{−1}. (3.3)

We introduce the function

X(z) = \begin{cases}X_{1}(t) · (X_{2}(t))^{−1} & for z ∈ D+_{Γ}, \\ (z − a_{2})^{κ_{2}−κ_{1}}((z − a_{2})/(z − a_{1}))^{κ_{1}}X_{1}(z) · (X_{2}(z))^{−1} & for z ∈ D−_{Γ}. \end{cases}

By the definition of classes, the property (i_{4}) and the Hölder inequality it obviously follows that X(z) ∈ E_{1}(D±_{Γ}). (3.3) clearly implies that X^{+}(t) = X^{−}(t) for t ∈ Γ. Therefore X(z) ≡ 0, which is impossible. Thus κ_{1} = κ_{2}. If now, along with κ_{1} = κ_{2}, it is assumed that a_{1} = a_{2}, then we obtain X(z) ≡ 1, i.e., X_{1}(z) = X_{2}(z).

Remark. Representation (3.1) and formula (2.6) clearly imply that if the condition (A_{κ}) is fulfilled, then the function X_{0}(z) = exp(K_{Γ} ln(t−a)^{−κ}G(t))(z) is the factor function of the function (t − a)^{−κ}G(t) with the zero index and the factor function of the function G(t) with the index κ.

In connection with Lemma 1 see also [2], p. 110.

4^{0}. We will give some auxiliary propositions.

Lemma 2. If f ∈ L_{p}(Γ), Γ ∈ R, p > 1, G(t) satisfies the condition (S_{1}) and Φ(z) ∈ E_{p}(D±_{Γ}), then the following statements are equivalent:

(1) the boundary value problem (2.1) has a unique solution for any f ∈ L_{p}(Γ);

(2) The operator N_{G} φ = \frac{1}{2}(1 + G)φ + \frac{1}{2}(1 − G)Sφ is invertible in L_{p}(Γ);

(3) The function G(t) admits a factorization with the index κ = 0 and the function Φ(z) given by the equality

Φ(z) = X_{0}(z)(K(f/X_{0}^{+}))(z),  (4.1)

where X_{0}(z) is the factorization of the function G(t), is a solution of problem (2.1);

(4) The invertible operator N_{G} in L_{p}(Γ) is written in the form

N −1_{G} f = \frac{1}{2} (1 + G^{−1})f + \frac{1}{2} (1 − G^{−1})X_{0}^{+} S_{Γ}(f/X_{0}^{+}),

where X_{0}(z) is the same as in (3); (5) |X_{0}^{+}| ∈ W_{p}(Γ), where X_{0}(z) is the same as in (3) and (4).

The statements of the lemma follow from the classical schemes and are encountered in this form or another in the works of many authors. Nevertheless, for the completeness of our exposition, we will give the proofs, the more so they are simple and some of their arguments make them different from the well known proofs.

By virtue of (i_{1}) and (i_{2}) the solution of the problem is Φ(z) = K_{Γ} φ, φ ∈ L_{p}(Γ). Applying Privalov’s basic lemma ([1], p. 184), we obviously obtain Φ^{+} − GΦ^{−} = \frac{1}{2}(1 + G)φ + \frac{1}{2}(1 − G)Sφ. Hence it clearly follows that (1) is equivalent to (2).

Let us show now that (2) implies (3).

Note preliminarily that if (1) holds for some G(t), then the same is true for G_{1}(t) = AG(t), where A = const, A ≠ 0. Therefore it can be assumed without loss of generality that |G(t)| ≠ 1 for t ∈ Γ.

Taking into account the invertibility of the operator N_{G} in L_{p}(Γ) and using (1.2), we conclude that the operator \frac{1}{2}(1 + G)ψ − \frac{1}{2}S(1 − G)ψ is invertible in L_{q}(Γ). Using the notation (1 − G)ψ = ψ_{1}, we have

\frac{1}{2} (1 + G)ψ − \frac{1}{2} S(1 − G)ψ = \frac{1}{2} \frac{1 + G}{1 − G}ψ_{1} − \frac{1}{2} Sψ_{1} = \frac{1}{2} \frac{G^{−1} + 1}{G^{−1} − 1}ψ_{1} − \frac{1}{2} Sψ_{1}

= (G^{−1} − 1)^{−1}[\frac{1}{2} (1 + G^{−1})ψ_{1} − \frac{1}{2} (G^{−1} − 1)Sψ_{1}]

= (G^{−1} − 1)^{−1}[\frac{1}{2} (1 + G^{−1})ψ_{1} + \frac{1}{2} (1 − G^{−1})Sψ_{1}] = (G^{−1} − 1)^{−1}N_{G^{−1}}ψ_{1}.

Therefore N_{G^{−1}} ψ is invertible in L_{q}(Γ) and the problem

X_{1}^{+}(t) = G^{−1}(t)X_{1}^{−}(t) + G^{−1}(t), X_{1} ∈ E_{q}(D±_{Γ})  (4.2)

has a unique solution.

If to this we add that the problem

X^{+}(t) = G(t)X^{−}(t) + G(t), X ∈ E_{p}(D±_{Γ}),  (4.3)

has a unique solution by virtue of (1), then we obtain

X_{1}^{+}X^{+} = G^{−1}(X_{1}^{−1} + 1) · G(X^{−1} + 1) = (X_{1}^{−1} + 1)(X^{−1} + 1).

Therefore

X_{1}^{+}X^{+} − 1 = (X_{1}^{−} + 1)(X^{−} + 1) − 1.  (4.4)

Now since X_{1}X ∈ E_{1}(D+_{Γ}) and (X_{1} + 1)(X + 1) − 1 ∈ E_{1}(D−_{Γ}), from (4.4) we have

X_{1}(z)X(z) − 1 = 0 for z ∈ D+_{Γ},
(X_{1}(z) + 1)(X(z) + 1) − 1 = 0 for z ∈ D−_{Γ},

which implies

\begin{matrix}(X(z))^{−1} = X_{1}(z) ∈ E_{q}(D+_{Γ}) and (X(z) + 1)^{−1} = X_{1}(z) + 1, \\ (X(z) + 1)^{−1} − 1 ∈ E_{q}(D−_{Γ}). \end{matrix} (4.5)

Using (4.5) together with (4.3), we conclude that the function

X_{0}(z) = \begin{cases}X(z) & for z ∈ D+_{Γ}, \\ X(z) + 1 & for z ∈ D−_{Γ}, \end{cases}

is the factor function of the function G(t) with the index κ = 0.

By the classical scheme it further clearly follows that if Φ(z) is a unique solution of problem (2.1) and X_{0}(z) is the facor function, then

Φ^{+}/X_{0}^{+} = Φ^{−}/X_{0}^{−} + f/X_{0}^{+}, Φ/X_{0} ∈ E_{1}(D±_{Γ}).

From this we obtain (4.1). Thus (1) implies (3).

To prove that the converse statement is true, we must show only that the solution is unique. Assume that this not so. Then

Φ_{1}^{+} = GΦ_{1}^{−} + f, Φ_{2}^{+} = GΦ_{2}^{−} + f Φ_{1}, Φ_{2} ∈ E_{p}(D±_{Γ}).

From this and (3.1) we have

(Φ_{1}^{+} − Φ_{2}^{+})/X_{0}^{+} = (Φ_{1}^{−} − Φ_{2}^{−})/X_{0}^{−}.

But as above (Φ_{1}(z) − Φ_{2}(z))/X_{0}(z) ∈ E_{1}(D±_{Γ}) and therefore Φ_{1}(z) = Φ_{2}(z). Thus the statements (1), (2) and (3) are equivalent.

Concurrently, we have shown that for κ = 0 the solution, if it exists for given (not necessarily arbitrary) f, is unique.

Let us next show that (3) implies (4). Since (3) is equivalent to (1), there exists a unique solution in E_{p}(D±_{Γ}). By the definition of classes for p > 1 we have E_{p} ⊂ E_{1} and by (i_{1}) and (i_{2}) the solution is Φ(z) = Kφ, φ ∈ L_{p}(Γ). Now it is clear that

φ=N_{0}f =Φ^{+} − Φ^{−} = \frac{1}{2}X_{0}^{+}(f/X_{0}^{+}+S(f/X_{0}^{+}))− \frac{X_{0}^{−}}{2} ( − f/X_{0}^{+}+S(f/X_{0}^{+}))
= \frac{1}{2} (1 + G^{−1})f + \frac{1}{2} (1 − G^{−1})X_{0}^{+}S(f/X_{0}^{+}) = N −1_{G}f.  (4.6)

Thus we have found that (3) implies (4). It is obvious that (4) implies (2) and therefore (1), (2), (3), (4) are equivalent.

Let us show that (4) implies (5). As said above, it can always be assumed that G(t) ≠ 1 for t ∈ Γ. Therefore (4) clearly implies |X_{0}^{+}| ∈ W_{p}(Γ), i.e., (5) is true.

Assume now that (5) is fulfilled. If f is continuous, then by (i_{4}), the definition of classes and the Hölder inequality we obtain Φ(z) = X_{0}(z)(K(f/X_{0}))(z) ∈ E_{1}(D±_{Γ}) and Φ(z) = Kφ. It is obvious that Φ(z) satisfies the boundary condition (2.1). Now on account of (4.6), for continuous f_{n} we have that φ_{n} = N_{0}f_{n} and Kφ_{n} is a solution. But since |X_{0}^{+}| ∈ W_{p}(Γ), we can pass to the limit and thus obtain φ = N_{0}f for any f ∈ L_{p}, p > 1. Moreover, φ ∈ L_{p}(Γ), Φ^{±} = \frac{1}{2}(±φ + Sφ) ∈ L_{p}(Γ) and because of the property of classes (i_{2}) we have Φ(z) ∈ E_{p}(D±_{Γ}). The solution is unique by virtue of κ = 0. Therefore N_{0} = N −1_{G}. The statements (1), (2), (3), (4) and (5) are equivalent.

Lemma 3. If one of the statements of Lemma 2 is fulfilled for the function G_{0}(t), then for the problem

Φ^{+}(t) = (t − a)^{κ}G_{0}(t)Φ^{−}(t) + f(t), f ∈ L_{p}(Γ), p > 1,

the statements (A_{κ}) hold except formula (2.6), where X_{0}(z) means in a general case the factor function G_{0}(t).

Proof. The statement (1) of Lemma 2 is fulfilled in the conditions of the lemma. The rest of the proof follows from the well verified arguments of many authors (see, for example, [10], p. 291). True, they consider the Lyapunov lines but here it is sufficient to require of the lines that the properties of the classes E_{p}(D±_{Γ}), which we have given for Γ ∈ R in the beginning of the paper, take place.

5^{0}. Let us consider a simple closed contour Γ ∈ R. If Γ_{ab} ⊂ Γ, then the direction from a to b is assumed to coincide with the positice direction on Γ. Here and in the sequel it is assumed that Γ = ∪_{k=1}^{n}Γ_{a_{k}a_{k+1}}, where a_{n+1} = a_{1}, Γ_{a_{k}a_{k+1}} ∈ R_{A}, k = 1, …, n, and do not have pairwise the common internal points. Occasionally, instead of Γ_{a_{k}a_{k+1}}, we write Γ_{k}. Also, we denote by Γ_{k}^{0} the contour participating the definition of R_{A}, i.e., Γ_{k}^{0} ≡ Γ_{a_{k}a_{k+1}}^{0}, and by χ_{k}(t) the characteristic function of the set {t : t ∈ Γ_{k}}.

Lemma 4. If G(t) is a measurable function satisfying condition (2.2) and G_{k}(t) ≡ [G(t)]^{χ_{k} (t)}, t ∈ Γ ∪ Γ_{k}^{0}, satisfies on Γ_{k}^{0} the condition (S_{2}) with κ = 0, then:

1) X_{k}(z) = exp (K_{Γ}χ_{k} ln G)(z) is the factor function of the function G_{k}(t) in

E_{p}(D±_{Γ});

2) The boundary value problem

Φ^{+}(t) = G_{k}(t)Φ^{−}(t) + f(t), t ∈ Γ, f ∈ L_{p}(Γ),  (5.1)

has a unique solution in E_{p}(D±_{Γ});

3) | exp \frac{1}{2}(S_{Γ}(χ_{k} ln G))(t)| ∈ W_{p}(Γ).  (5.2)

Proof. Since Γ_{k} ∈ R_{A}, the factor function of the function G_{k}(t) in E_{p}(D±_{Γ^{0}_{k}}) is

X_{k}(z) = exp (K_{Γ^{0}_{k}} (χ_{k} ln G))(z). But then X_{k}(z) − 1 = K_{Γ^{0}_{k}} ψ, ψ ∈ L_{p}(Γ_{k}^{0}) and

X_{k}^{−1}(z) − 1 = (K_{Γ^{0}_{k}} ψ_{1})(z), ψ_{1} ∈ L_{q}(Γ_{k}^{0}). Obviously, ψ = (K_{Γ^{0}_{k}} ψ)^{+} − (K_{Γ^{0}_{k}})^{−} =

X_{k}^{+} − X_{k}^{−} = X_{k}^{−}(X_{k}^{+}/X_{k}^{−} − 1) = X_{k}^{−}(G_{k}(t) − 1) = χ_{k}(t)ψ(t). Analogously,

ψ_{1} = χ_{k}(t)ψ_{1}(t). Therefore K_{Γ^{0}_{k}} ψ = K_{Γ^{0}_{k}} χ_{k}ψ = K_{Γ}χ_{k}ψ ∈ E_{p}(D±_{Γ}) and K_{Γ^{0}_{k}} ψ_{1} =

K_{Γ}χ_{k}ψ ∈ E_{q}(D±_{Γ}).

Thus

X_{k}(z) − 1 ∈ E_{p}(D±_{Γ}) and X_{k}^{−1}(z) ∈ E_{q}(D±_{Γ}).  (5.3)

Moreover, for t ∈ Γ − Γ_{k} we have (K_{Γ}χ_{k}ψ)^{+}/(K_{Γ}χ_{k}ψ)^{−} = 1. Therefore

X_{k}^{+}(t)/X_{k}^{−}(t) = G_{k}(t) for t ∈ Γ  (5.4)

(5.3) and (5.4) prove the statement 1).

Now we will show the validity of the statment 2).

A solution of problem (5.1) is to be sought in the form

Φ(z) = Φ_{1}(z) + Φ_{2}(z)  (5.5)

where Φ_{1}(z) = (K_{Γ}ψ)(z), ψ ∈ L_{p}(Γ) and Φ_{2}(z) = (K_{Γ}(χ_{Γ−Γ_{k}} f))(z). Equality (5.1) then takes the form

Φ_{1}^{+} + Φ_{2}^{+} = G_{k}(t)(Φ_{1}^{−} + Φ_{2}^{−}) + f(t), t ∈ Γ.  (5.6)

Denoting f_{0}(t) = f(t) − Φ_{2}^{+}(t) + G_{k}Φ_{2}^{−}(t), we rewrite (5.6) as

Φ_{1}^{+}(t) = G_{k}(t)Φ_{1}^{−}(t) + f_{0}(t), t ∈ Γ, Φ_{1} ∈ E_{p}(D±_{Γ}),  (5.7)

and (5.7) is equivalent to

\begin{cases}ϕ_{1}^{+} = G(t)Φ_{1}^{−}(t) + f_{0}(t) & for t ∈ Γ_{k}, Φ_{1} ∈ E_{p}(D±_{Γ}) \\ Φ_{1}^{+} = Φ_{1}^{−} & for t ∈ Γ − Γ_{k}. \end{cases}  (5.8)

From (5.8) it follows that if Φ_{1}(z) is a solution of (5.8), then

Φ_{1}(z) = (K_{Γ}ψ)(z) = K_{Γ}(Φ_{1}^{+} − Φ_{1}^{−}) = K_{Γ}χ_{k}ψ = K_{Γ^{0}_{k}} χ_{k}ψ ∈ E_{p}(D±_{Γ^{0}_{k}}).

Therefore Φ_{1}(z) simultaneously belongs both to E_{p}(D±_{Γ}) and to E_{p}(D±_{Γ^{0}_{k}}).

Now it is clear that between the solutions of problem (5.1) in E_{p}(D±_{Γ}) and the problem

Φ_{1}^{+}(t) = G_{k}(t)Φ_{1}^{−}(t) + f_{0}(t), t ∈ Γ_{k}^{0}, Φ_{1} ∈ E_{p}(D±_{Γ^{0}_{k}}),  (5.9)

there exists a one-to-one correspondence carried out by formula (5.5). But (5.9) has a unique solution. Therefore (5.1), too, has a unique solution.

Now, by virtue of statement (5) of Lemma 2, 1) and 2) imply |X_{k}^{+}(t)| ∈ W_{p}(Γ), from which it follows that

|X_{k}^{+}(t)| = | exp(K_{Γ}χ_{k} ln G)^{+}| = G(t)^{\frac{1}{2} χ_{k}} exp \frac{1}{2} S_{Γ}χ_{k} ln G| ∈ W_{p}(Γ)

⟹ | exp \frac{1}{2}S_{Γ}χ_{k} ln G| ∈ W_{p}(Γ).

Corollary 1. If we take G_{k}(t) = |G(t)|^{χ_{k} (t)}, then by virtue of Lemma 4:

1) the function

exp K_{Γ_{k}} ln |G|  (5.10)

is the factor function of the function |G|^{χ_{k} (t)} in ∩_{p>1}E_{p}(D±_{Γ});

2)  | exp \frac{1}{2}S_{Γ_{k}} ln |G|| ∈ ∩_{p>1}W_{p}(Γ).  (5.11)

Corollary 2. If we take G_{k}(t) = exp iχ_{k}φ_{2}, where φ_{2} satisfies condition (2.3), then:

1) exp(K_{Γ}iχ_{k}φ_{2})(z) is the factor function for exp iχ_{k}φ_{2} in E_{p}(D±_{Γ}), where p participates in condition (2.3);

2)  | exp \frac{i}{2}S_{Γ}χ_{k}φ_{2}| ∈ W_{p}(Γ) for the same p.  (5.12)

Lemma 5. If Γ_{bc} is a simple arc on Γ, Γ_{bc} ⊂ Γ, Γ_{bc} = Γ_{ba} + Γ_{ac}, Γ_{ba} ∈ R_{A}, Γ_{ac} ∈ R_{A} and φ_{2}(t) is a measurable function on Γ which satisfies the conditions

vrai sup_{t∈Γ_{ba}}|φ_{2}(t)| < α\frac{π}{max(p, q)},
vrai sup_{t∈Γ_{ac}}|φ_{2}(t)| < (1 − α)\frac{π}{max(p, q)},

(5.13)

where α is some number from [0, 1], then

ρ(t) = | exp \frac{i}{2}(S_{Γ_{bc}}φ_{2})(t)| ∈ W_{p}(Γ) ∩ W_{q}(t).  (5.14)

Proof. We make use of the particular case of the well-known Stein theorem (see, for example [10], p. 288):

If ∥(Sφ)ρ_{k}∥_{p} ≤ M_{p}∥φρ_{k}∥, k = 1, 2, p > 1, then

∥(Sφ)ρ∥ ≤ M_{p}∥φρ∥_{p} for ρ = ρ_{1}^{α} · ρ_{2}^{1−α}, α ∈ [0, 1].

This fact can also be written as follows:

If ρ_{k} ∈ W_{p}(Γ), k = 1, 2, p > 1 and α ∈ [0, 1], then

ρ = ρ_{1}^{α} · ρ_{2}^{1−α} ∈ W_{p}(Γ).  (5.15)

Denote by χ_{1}(t) = χ(Γ_{ba}) the characteristic function of the set {t : Γ_{ba}}, and by χ_{2}(t) that of the set {t : t ∈ Γ_{ac}}. On account of (5.12) and (5.13) we have

ρ_{1} = | exp \frac{i}{2}S_{Γ} \frac{1}{α}χ_{1}φ_{2}| ∈ W_{p}(Γ), ρ_{2} = | exp \frac{i}{2}S_{Γ} \frac{1}{1 − α}χ_{2}φ| ∈ W_{p}(Γ),

from which by virtue of (5.15) we obtain

ρ = ρ_{1}^{α} · ρ_{2}^{1−α} = | exp \frac{i}{2}S_{Γ}χ_{1}φ_{2}| · | exp \frac{i}{2}S_{Γ}χ_{2}φ_{2}|

= | exp \frac{i}{2}S_{Γ}(χ_{1} + χ_{2})φ_{2}| = | exp (\frac{i}{2}S_{Γ_{bc}}φ_{2})(t)| ∈ W_{p}(Γ).

Obviously, in the conditions (5.13) p and q play the same role and therefore (5.14), too, is valid.

Theorem 1. If Γ = ∪_{k=1}^{n}Γ_{a_{k}a_{k+1}}, a_{n+1} = a_{1}, Γ_{a_{k}a_{k+1}} ⊂ R_{A}, and, moreover, for each point a_{k} (k = 1, …, n) there exists an arc neighborhood Γ_{b_{k}a_{k}} ∪ Γ_{a_{k}c_{k}} ⊂ Γ, where conditions (2.8) are fulfilled for the function G, and conditions (2.2) and (2.3) are fulfilled on the entire contour Γ, then

ρ(t) = | exp \frac{i}{2}(S_{Γ}φ_{2})(t)| ∈ W_{p}(Γ) ∩ W_{q}(t).  (5.16)

Proof. To prove the theorem, we make use of Theorem 4.2 from [13], p. 52, by virtue of which the necessary and sufficient condition for the inequality

∫_{Γ} |S_{Γ}f|^{p_{0}}w dν ≤ c ∫_{Γ} |f|^{p_{0}}w dν,  (5.17)

to hold for p_{0} > 1, is w ∈ A_{p_{0}}.

In terms of the above-mentioned work this means that

\{ρ = w^{\frac{1}{p_{0}}} ∈ W_{p_{0}}(Γ), p_{0} > 1, Γ ∈ R\} ⟺ \{ρ^{p_{0}} = w ∈ A_{p_{0}}(Γ)\}. (5.18)

(for the definition of A_{p}, see [13], p. 42).

Let us write the condition A_{p_{0}} for ρ = w^{\frac{1}{ρ_{0}}} . Denote by ν the length of the arc Γ(z, r) which is cut out from Γ by the circle with center z ∈ Γ and radius r. (Here, like in (5.17), ν is the arc measure on Γ.) We obtain

ρ ∈ W_{p_{0}} ⟺ sup_{\begin{matrix}z∈Γ \\ 0<r<diam Γ \end{matrix}} (\frac{1}{νΓ(z; r)} ∫_{Γ(z;r)} ρ^{p_{0}}dν)^{\frac{1}{p_{0}}} (\frac{1}{ν(z; r)} ∫_{Γ(z;r)} ρ^{−q_{0}}dν)^{\frac{1}{q_{0}}} < ∞,

q_{0} = p_{0}(1 − p_{0})^{−1}. (5.19)

For brevity, we denote by B(Γ(z; r); ρ(t); p_{0}) the expression in (5.19) from which the supremum is taken. Now (5.19) can be rewritten as

ρ ∈ W_{p_{0}} ⟺ sup_{\begin{matrix}z∈Γ \\ 0<r<diam Γ \end{matrix}} B(Γ(z; r); ρ(t); p_{0}) < ∞.  (5.20)

Take now the points b_{k} and c_{k} (k = 1, 2, …, n) according to condition (2.8) and take the points e_{k} ∈ Γ_{b_{k}a_{k}} and d_{k} ∈ Γ_{a_{k}c_{k}} different from the end points (Fig. 1).

Fig. 1

Also assume that d_{n+1} = d_{1}, e_{n+1} = e_{1}, b_{n+1} = b_{1}, c_{n+1} = c_{1} and Γ_{b_{k}a_{k}} (k = 1, …, n) do not intersect pairwise. Denote δ_{k}^{(1)} = min_{\begin{matrix}t∈Γ_{e_{k}d_{k}} \\ τ∈Γ−Γ_{b_{k}c_{k}} \end{matrix}} |t − τ|,

δ_{k}^{(2)} = min_{\begin{matrix}t∈Γ_{d_{k}e_{k+1}} \\ t∈Γ−Γ_{a_{k}a_{k+1}} \end{matrix}} |t − τ|.

520 E. GORDADZE

Also denote δ = \frac{1}{2} min_{\begin{matrix}i=1,2 \\ i≤k≤n \end{matrix}} δ_{k}^{(i)}. Assume that z ∈ Γ_{e_{k}d_{k}}. Then Γ(z, δ) ⊂ Γ_{b_{k}c_{k}}. But by virtue of Lemma 5 we have | exp \frac{i}{2}(S_{Γ_{b_{k}c_{k}}} φ_{2})(t)| ∈ W_{p}(Γ) ∩ W_{q}(t). Then if p_{0} is equal to p or q, we have

sup_{\begin{matrix}z∈Γ_{e_{k}d_{k}} \\ 0<r<δ \end{matrix}} (B(Γ(z; r); | exp \frac{i}{2}(S_{Γ_{b_{k}c_{k}}} φ_{2})(t)|, p_{0}) < ∞.  (5.21)

Meanwhile, if z ∈ Γ_{e_{k}d_{k}} and t ∈ Γ(z; δ), then

| exp \frac{i}{2} (S_{Γ−Γ_{b_{k}c_{k}}} φ_{2})(t)| < const

and therefore along with (5.21) we have

sup_{\begin{matrix}z∈Γ_{e_{k}d_{k}} \\ 0<r<δ \end{matrix}} (B(Γ(z; r); | exp \frac{i}{2} (S_{Γ}φ_{2})(t)|, p_{0}) < ∞.  (5.22)

Let now z ∈ Γ_{d_{k}e_{k+1}}. In view of (5.12) and (5.18) we obtain

sup_{\begin{matrix}z∈Γ_{d_{k}e_{k+1}} \\ 0<r<δ \end{matrix}} (B(Γ(z; r); | exp \frac{i}{2} (S_{Γa_{k}a_{k+1}} φ_{2})(t)|, p_{0}) < ∞.  (5.23)

But, as above, for t∈Γ(z, r), z ∈Γ_{d_{k}e_{k+1}} we have |exp\frac{i}{2}(S_{Γ−Γa_{k}a_{k+1}} φ_{2})(t)|< const, which, together with (5.23), gives

sup_{\begin{matrix}z∈Γ_{d_{k}e_{k+1}} \\ 0<r<δ \end{matrix}} (B(Γ(z; r); | exp \frac{i}{2} (S_{Γ}φ_{2})(t)|, p_{0}) < ∞.  (5.24)

Since (5.22) and (5.24) are fulfilled for k = 1, …, n, we obtain

sup_{\begin{matrix}z∈Γ \\ 0<r<δ \end{matrix}} B(Γ(z; r); | exp \frac{i}{2} (S_{Γ}φ_{2})(t)|, p_{0}) < ∞.  (5.25)

Since | exp \frac{i}{2}(S_{Γ}φ_{2})(t)| ∈ L_{p_{0}}(Γ), it is obvious that

sup_{\begin{matrix}z∈Γ \\ δ<r<diam Γ \end{matrix}} B(Γ(z; r); | exp \frac{i}{2} (S_{Γ}φ_{2})(t)|, p_{0}) < ∞,

which, together with (5.25), gives

sup_{\begin{matrix}z∈Γ \\ 0<r<diam Γ \end{matrix}} B(Γ(z; r); | exp \frac{i}{2} (S_{Γ}φ_{2})(t)|, p_{0}) < ∞  (5.26)

which, together with (5.20), proves the theorem.

Corollary. If φ_{2} satisfies conditions (2.3) and (2.8), then there exists a number 0 < ε < 1 such that (1 − ε)^{−1}φ_{2} also satisfies these conditions. Hence from Theorem 1 it follows that

| exp \frac{i}{2} (S_{Γ} \frac{1}{1 − ε} φ_{2})(t)| ∈ W_{p}(Γ) ∩ W_{q}(Γ)  (5.27)

Remark. By a reasoning similar to that used in proving Theorem 1 one can prove a more general statement.

Let Γ be a simple (not necessarily closed) continuous line and Γ ∈ R. Let Γ_{t} be an arc neighborhood of the point Γ ∈ R. For a more exact definition of Γ_{t} we use, along with the notation Γ_{ab}, the notations Γ_{[a,b]}, Γ_{(a,b)}, Γ_{[a,b)}, Γ_{(a,b]} which indicate the belonging or not belonging of the end points to the arc Γ_{ab}.

Now if t is an internal point of the line Γ, then Γ_{t} = Γ_{(t_{1},t_{2})}, where t ∈ Γ_{(t_{1},t_{2})} and t_{1}, t_{2} are the points on Γ taken according to the positive direction on Γ. If however Γ = Γ_{ab} and t = a or t = b, then Γ_{a} ≡ Γ_{[0,t_{1})} and Γ_{b} ≡ Γ_{(t_{2},b]}, t_{1}, t_{2} ∈ Γ_{ab}.

For p > 1 the following statement holds true:

(ρ(t) ∈ W_{p}(Γ_{t}) for any t ∈ Γ) ⟹ ρ ∈ W_{p}(Γ).  (5.28)

The proof is published in [12]. Like Theorem 1, this result is based on Theorem 4.2 from [13], p. 52.

Theorem 2. If Γ is the same as in Theorem 1, G(t) = |G(t)| exp i(φ_{1} + φ_{2}), where φ_{1}(t) is a continuous function, |G(t)| and φ_{2}(t) satisfy conditions (2.2) and (2.3) as well as the additional condition (2.8), then

ρ(t) = | exp \frac{1}{2} (S_{Γ} ln G)(t)| ∈ W_{p}(Γ) ∩ W_{q}(Γ).  (5.29)

Proof. Denote

ρ_{1}(t) ≡ | exp \frac{1}{2}(S_{Γ} ln |G|)(t)|, ρ_{2}(t) ≡ | exp \frac{i}{2}(S_{Γ}φ_{1})(t)|,

ρ_{3}(t) ≡ | exp \frac{i}{2}(S_{Γ}φ_{2})(t)|.

It is obvious that ρ_{1}(t) = ∪_{k=1}^{n}| exp \frac{1}{2}(S_{Γ_{k}} ln |G|)(t)|. For 0 < α < 1 we have

| exp \frac{1}{2}(S_{Γ_{1}} \frac{1}{α} ln |G|)(t)| ∈ W_{p}(Γ) for any p > 1, and | exp \frac{1}{2}(S_{Γ_{2}} \frac{1}{1−α} ln |G|)(t)| ∈ W_{p}(Γ) for any p > 1. Hence, applying the Stein theorem (5.15), we find that | exp \frac{1}{2}(S_{Γ_{1}+Γ_{2}} ln |G|)(t)| ∈ W_{p}(Γ) for any p > 1, and for any G(t) satisfying (2.2). Performing the same n times we obtain ρ_{1} ∈ W_{p}(Γ) for any p > 1 and G(t) satisfying (2.2). Therefore

ρ1^{\frac{1}{α_{0}}} ∈ ∩_{p>1}W_{p}(Γ) for 0 < α_{0} < 1.  (5.30)

ρ_{2} also belongs to ∩_{p>1}W_{p}(Γ), which follows from the fact that the boundary

value problem for Γ ∈ R with a continuous coefficient is solved completely ([14], p. 547).

It is clear that

ρ2^{\frac{1}{1−α_{0}}} ∈ ∩_{p>1}W_{p}(Γ), 0 < α_{0} < 1.  (5.31)

By the same Stein theorem, (5.30) and (5.31) imply that ρ_{1} · ρ_{2} ∈ ∩_{p>1}W_{p}(Γ)

for any G(t) and φ_{1}(t) satisfying the conditions of the theorem. Therefore

(ρ_{1} · ρ_{2})^{\frac{1}{ε}} ∈ ∩_{p>1}W_{p}(Γ) for 0 < ε < 1.  (5.32)

Let p_{o} be any one of the numbers p and q participating in condition (2.8), and ε be the same as in (5.27). Then

(ρ_{3})^{\frac{1}{1−ε}} ∈ W_{p_{0}}(Γ).  (5.33)

By virtue of the Stein theorem, (5.32) and (5.33) give

((ρ_{1} · ρ_{2})^{\frac{1}{ε}} )^{ε} · (ρ3^{\frac{1}{1−ε}} )^{1−ε} = ρ_{1} · ρ_{2} · ρ_{3} ∈ W_{p_{0}},

which is equivalent to (5.29).

Theorem 3. If Γ is the same as everywhere in Section 5^{0}, G(t) is the measurable function G(t) = |G(t)| exp(iφ_{1}(t) + iφ_{2}(t)) on Γ, conditions (2.2), (2.3) are fulfilled, the function φ_{1}(t) is continuous except for one point at which it has a first kind jump equal to 2πκ, where κ and integer, and if, additionally, (2.8) is fulfilled φ_{2}, then the statement A_{(κ)} is valid for the boundary value problem (2.1) with the coefficient G(t).

Proof. First assume that κ = 0. By virtue of Lemma 2 of [8], p. 1654, the boundedness of ln |G_{0}| and formula (5.29) imply that X_{0}(z)=exp(K_{Γ} ln G_{0})(z) is the factor function of the function G_{0}(t). Since we have defined E_{δ}(D^{−}) for δ < 1 and the factor function in a somewhat different way, in our case we will repeat with an insignificant modification the reasoning of [8], p. 1654.

From (i_{3}) it follows that X_{0}(z) − 1 ∈ E_{δ}(D±_{Γ}) for some δ > 0. Further, by (5.29) and (5.18) we obtaon ρ^{±} ∈ L_{p}(Γ) ∩ L_{q}(Γ), which, by (i_{2}), implies that X_{0}^{±}(z)−1 ∈ E_{p}(D±_{Γ})∩E_{q}(D±_{Γ}), where X_{0}(z), which evidently satisfies conditions (3.1), is the factor function with κ = 0.

Thus statement (5) is fulfilled and so are the other statements of Lemma 2 and problem (2.1) has a unique solution in E_{p}(D±_{Γ}).

If κ ≠ 0, then, taking a ∈ D+_{Γ}, we can write

G_{0}(t) = (t − a)^{−κ}G(t) = |G_{0}(t)| exp(iφ_{1}^{0} + iφ_{2}),

where φ_{1}^{0}(t) is already continuous, and by virtue of the facts proved the problem with the coefficient G_{0}(t) has a unique solution, while the factor function of the function G_{0}(t) has form (2.6). Now due to Lemma 3 we conclude that the statement (A_{κ}) is true for problem (2.1) with the coefficient G(t) = (t − a)^{κ}G_{0}(t).

Remark. Theorem 3 is the main result of this paper. As an example of curves for which this theorem is valid we may name piecewise-smooth curves and curves with bounded rotation (Radon curves), where an additional condition is imposed on coefficient (2.8) at cusps. As Γ_{k} we can also take the curves considered in [8] and others. Cases may occur, in which unilateral tangent lines do not exist at the points a_{k}.

Acknowledgement

This work was supported by Grant 1.14 of the Georgian Academy of Sciences.

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(Received 28.09.2001; revised 27.02.2002)

Author’s address:

A. Razmadze Mathematical Institute Georgian Academy of Sciences 1, M. Aleksidze St., Tbilisi 380093, Georgia