Multidimensional boundary analog of the Hartogs Theorem in circular domains Текст научной статьи по специальности «Математика»

УДК 517.55

Multidimensional Boundary Analog of the Hartogs Theorem in Circular Domains

Alexander M. Kytmanov* Simona G. Myslivets^

Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041

Russia

Received 06.06.2017, received in revised form 12.07.2017, accepted 17.10.2017

This paper presents some results related to the holomorphic extension of functions, defined on the boundary of a domain D C Cn, n > 1, into this domain. We study a functions with the one-dimensional holomorphic extension property along the complex lines.

Keywords: functions with the one-dimensional holomorphic extension property, circular domain. DOI: 10.17516/1997-1397-2018-11-1-79-90.

Introduction

This paper presents some results related to the holomorphic extension of functions, defined on the boundary of a domain D c Cn, n > 1, into this domain. We consider a functions with the one-dimensional holomorphic extension property along the complex lines.

The first result related to our subject was obtained M.L.,Agranovsky and R.E.Valsky in [1], who studied functions with the one-dimensional holomorphic continuation property into a ball. The proof was based on the properties of the automorphism group of a sphere.

E. L. Stout in [2] used the complex Radon transformation to generalize the Agranovsky and Valsky theorem for an arbitrary bounded domain with a smooth boundary. An alternative proof of the Stout theorem was obtained by A. M .Kytmanov in [3] by using the Bochner-Martinelli integral. The idea of using the integral representations (Bochner-Martinelli, Cauchy-Fantappie, logarithmic residue) has been useful in the study of functions with the one-dimensional holomorphic continuation property (see review [4]).

The question of finding different families of complex lines sufficient for holomorphic extension was put in [5]. As shown in [6], a family of complex lines passing through a finite number of points, generally speaking, is not sufficient. Thus, a simple analog of the Hartogs theorem should be not expected.

Various other families are given in [7-11]. In [12-16] it is shown that for holomorphic extension of continuous functions defined on the boundary of ball,there are enough n +1 points inside the bal, not lying on a complex hyperplane. This result was generalized by the authors n-circular domains.

* [email protected]

1 [email protected] © Siberian Federal University. All rights reserved

1. Main results

Let D be a bounded domain in Cn with a smooth boundary. Consider the complex line of the form

lzb = [C e Cn : Z = z + bt,t e C} = {(Zi, ■■■Cn): Cj = Zj + bj t,j =1, 2,...,n,t e C}, (1)

where z e Cn, b e CPn^1.

We will say that a function f e C(dD) has the one-dimensional holomorphic extension property along the complex line lz,b, if dDnlz,b = 0 and there exists a function Flzb with the following properties:

1) Fh b e C(B n iztb),

2) Fizbb = f on the set dD n ,

3) function Flzbb is holomorphic at the interior (with respect to the topology of lz,b) points of set D n lz,b.

Let r be a set in Cn. Denote by Lr the set of all complex lines lz,b such that z e r, and b e CPn_1, i.e., the set of all complex lines passing through z e r.

We will say that a function f e C(dD) has the one-dimensional holomorphic extension property along the family Lr, if it has the one-dimensional holomorphic extension property along any complex line lz,b e Lr.

We will call the set Lr sufficient for holomorphic extension, if the function f e C(dD) has the one-dimensional holomorphic extension property along all complex lines of the family Lr, and then the function f extends holomorphically into D (i.e., f is a Cfl-function on dD).

Theorem A. Let n = 2 and D be a bounded strictly convex circular domain with twice smooth boundary and a function f (C) e C(dD) have the one-dimensional holomorphic extension property along the family L{a,c,d}, and the points a,c,d e D do not lie on one complex line in C2, then the function f (C) extends holomorphically into D.

We denote by A the set of points ak e D c Cn, k = +1, which do not lie on a

complex hyperplane in Cn.

Theorem B. Let D be a bounded strictly convex circular domain with twice smooth boundary in Cn and the function f (C) e C(dD) have the one-dimensional holomorphic extension property along the family La, then the function f (C) extends holomorphically into D.

2. Construction of the Szego kernel

Let H(D) be the space of holomorphic functions in D with the topology of uniform convergence on compact subsets of D, and H(D) be the space of holomorphic functions in a neighborhood of D with the corresponding topology. Consider the measure dn = g(C)da, where g(C) e C1(dD), g(C) > 0, and da is the Lebesgue measure on dD. The space H(D) is the subspace in L2(dD) with the measure d/j, on dD. By the Maximum Modulus Theorem the mapping H(D) —> L2(dD) is injective. By H2 = H^dD) we denote the closure of H(D) in L2.

Consider a restriction mapping r : H(D) —> H(D). The mapping r extends by continuity from H2 in H(D).

Lemma 1 (Lemma 4.1. [17]). The restriction mapping r : H(D) —>• H(D) is continuous, if H(D) is considered in the topology induced by the space C?.

Therefore, the mapping r extends by continuity to the map i : H2 —> H(D). In this case, we say that for functions f e H2 there is a holomorphic continuation f = i(f) in D. Further on, this continuation will be denoted by the same symbol f.

In [17] as the measure considered by the Lebesgue measure da on the boundary of the domain, in our case, for the measure dn = g(Z)da the proof is similar.

Since the space H2 is a Hilbert separable space, then there exists an orthonormal basis

{fk }TO=i (2)

in the metric L2. Therefore, any function f G H2 extens in a Fourier series:

to

f (z ) = ck fk (z) (3)

k=i

with respect to the basis (2), which converges in the topology of L2, where ck = (f,fk) = f (u)Cpk(u) d^(u). Then

ISD

f (c HE f(u)LPk(u) d^(u)fk (o) = f (u)Y] fk (u)fk(z) d^(u). k=AJdD / JdD k=i

to

Denote K(Z,u) = ^^ fk (Z)fk (u) and K(Z,u) G H(D) on Z G D for a fixed u G D.

k=i

wc uan, uituuac un ui oitunui inw, ijubib {fk} k = 1 ln

2

in H(D).

Lemma 2. We can choose an orthonormal basis {fk }TO=i in H2 which consists of functions fk

Proof. Since the space H(D) is separable, then there exists a countable everywhere dense set. It will be the same in H2, since H2 is the closure of H(D). Using the process of Gram-Schmidt orthogonalization for the functions from this set, we get orthonormal basis in H2 consisting of functions fk G H(D). □

Lemma 3. If D is a bounded strictly convex domain with a smooth boundary, then we can choose a polynomials basis {fk}TO=i.

Proof. Since the domain D is strictly convex, the set D is polynomially convex and compact. On such sets functions, holomorphic in its neighborhood, are uniformly approximated by the polynomials [18]. Consequently, the polynomials are dense in the class of functions from H(D) and therefore from H2. Applying the Gram-Schmidt orthogonalization to this set we get an orthonormal basis in H2 consisting of polynomials. □

Let us call the function g(Z) invariant under rotations, if g(Zi, ■ Zn) = g(eiVZi, ■ ■ ■, eiVCn) for all f G [0, 2n).

Lemma 4. If D is a bounded strictly convex circular domain with a smooth boundary and a function g(Z) is invariant under rotations, we can choose a basis {fk}TO=i of homogeneous polynomials.

Proof. Indeed, in this case, the measure dn is also invariant under rotations, so the homogeneous polynomials of different degrees of homogeneity are orthogonal in H2. □

Further on, we assume that the basis is chosen in accordance with Theorem 5.1 [17]. According to this theorem the continuation of the kernel K(Z, u) has the property:

i(f )(z)= f f (Z)K(z,p) d^(Z), z G D,

JdD

where K(z,Z) = ^^i(fk)(z)i(fk)(Z) and the series converges uniformly on compact subsets of

k=i

D x D. This kernel we call the Szego kernel. Then

f (z)= I f (Z)K(z,Z) ), (4)

JdD

where f (z) is identified with f(z) = i(f)(z) and f € H2. We define the Poisson kernel

K(z,Z) ■ K(Z,z) = K(z,Z) ■ K(z,Z) = IK(z,Z)I2

( ) K(z, z) K(z, z) K(z, z) '

OO OO

and K(z,z) = ^2 ¥k(z)<fk(z) = ^2 Ifk(z)12 > 0. k=i k=i

Lemma 5. The kernel K(z, z) > 0 for any z € D.

Proof. Let k(z, z) = 0 for some z € D. Then fk(z) = 0 for all k = 1, 2,..., so

fk (z)= ! fk (Z )K (z,Z) dp(Z) = 0. (5)

JdD

O

Since any function f €H2 decomposes into the Fourier series (3), f (Z) = ^^ ck fk(Z). Applying

k=i

the mapping i, we get that f (z) = ^^cki(yk)(z) = 0 in virtue of (5), i.e. f(z) = 0 in D for all

k=i

functions f € H2, which is impossible. □

Lemma 6. A function f € H(D) admits the integral representation

f (z)= I f (Z)P(z,Z) dp(Z), (6)

JdD

for z € D.

Proof. By definition of the kernel P(z, Z) and from the integral representation (4) we have

J f (Z)P(z,Z) dp(Z) = j f (Z)K(z,K\zKKfz) d^(Z) =

dD dD

1 f f (Z) K(', -z))Kz Z) d,(Z) = = f (z)-

K(z, z ) K(z, z )

dD

Corollary 1. If the space H(D) is dense in the space H(D) n C(dD) = A(D), then a function f € A(D) admits the integral representation (6).

Suppose that the domain D satisfies the condition

(A): for any point Z € dD and any neighborhood U(Z) the Szego kernel K(z,Z) is uniformly bounded by z € D and z €U (Z).

Further, we assume that the domain D satisfies the condition (A).

Theorem 1. Let D be a strictly convex domain in Cn and the kernel K(z, Z) satisfies the Holder condition with exponent < a ^ 1 for Z € dD and a fixed z € D. Then the domain D and the kernel K(z,Z) satisfy the condition (A).

Proof. Let

D = {z € Cn : p(z) < 0},

(7)

where p £ C2(D) and gradp\dD = 0. For the proof we use Corollary 26.13 [3] for the Leray

ho

I 'n

f (z) =

integral representations for holomorphic functions f £ A(D) in strictly convex domains:

(n — 1)! [ f (C)£r=i 5k d([k] A d(

(2ni)n

dD

K( Ci - zi) + ••• + p'zn ( Cn - zn)Y

where

5k =

pd ••• pZn

p'U • • pI [k]

plCn • • pi Cn

k = 1,

dZ = dZi A ... A dZn, d,Z[k] = dZi A ... A dZik-i A dZik+i A ... A dC„.

The denominator of the kernel p'z1 (Zi — zi) + ... + pZn(Zn — zn) = 0 for Z £ dD, z £ D and Z = z. Indeed, the equality pZ1 (Zi — zi) + ... + p'n (Zn — zn) = 0 defines a complex tangent plane to dD at the point Z. If the domain D is strictly convex, then the tangent plane intersects the boundary of D only at a point Z.

For the domain D the Szego kernel K(z, Z) is the (generalized) Cauchy-Fantappie (Leray) kernel by Corollary 26.13 [3], so the same domain satisfy the condition (A).

□

Consider the restriction of the form

Er=i 4 dZ[k] A dZ

L(z,Z, C) =

to dD, then it would be

L(z, Z, C) =

M(C, C) da(Z)

[pZi(Ci - zi) + ••• + p'zn(Cn - zn)y

M(C, C) dKC)

K (Ci - zi) + •• • + p'zn (Cn - Zn)\n g(Z) pi (Ci - zi) + ••• + p'n (Cn - Zn)\

MC, C) dMC)

[p Zi ( Ci - zi) + ••• + p'z n ( Cn - zn)]

= L(z, Z, C) d^(C).

The proof of Theorem 1 shows that

K(z, Z) = L(z, Z, Z) (8)

for Z £ dD.

Lemma 7. The function K(z, Z) is unbounded as z ^ Z and Z £ dD, z £ D.

Proof. Consider the point z0 £ D, then the domain D is a strongly star-shaped with respect to z0, i.e. for any point Z0 £ dD the segment [z0, Z0] £ D. Let this segment have the form

{z £ D : z = Z0 + t(z0 — Z0), 0 < t < 1}. Then

pi ( z0 — zi) +... + pi (Zn — zn) = t(p'Cl ( Z0 — z0) +... + pi (Zn — zn ))■

If z ^ Z0, then t ^ 0 and (pZ1 (Zi° — z0) + ... + p'Z (Zn — z°n)) ^ 0. Then K(z, Z) ^ <x for z ^ Z, Z £ dD. □

. , n

3. Poisson kernel and its properties

For a function f € C(dD) we define the Poisson integral:

P[f](z) = F(z)= f f(Z)P(z,Z) dp(Z).

J8D

In strictly convex domain that satisfy the condition (A), from Equality (8) and the form of the kernel P(z, Z), it follows that this kernel is a continuous function for z € D and then the function F(z) is continuous in D.

Theorem 2. Let D be a bounded strictly convex domain in Cn satisfying the condition (A), and f € C(dD), then the function F(z) continuously extend onto D and F(z)\gD = f (z).

Proof. Theorem 1 and Lemma 7 show that the kernel P(yZ,t(z° — z)) tends uniformly to zero outside any neighborhood of the point Z for Z,z € dD, z0 € D, Z = z and t ^ 1. Moreover P(z,Z) > 0 and P[1](Z) = 1. Consequently, the Poisson kernel P(z,Z) is an approximative unit [19, Theorem 1.9]. □

Consider the differential form

n

w = ( — 1)k-1Zk dZ[k] A dZ, k=i

(n — 1)!

where c = —--—. Find the restriction of this form to dD for the domain D of the form (7).

(2ni)n

Then by Lemma 3.5 [20], we get

dZ[k] A dZ =( — 1)k-12n-1indP da

dZk I grad p[ Therefore, the restriction of w to dD is equal to

d = \ = (n — 1)! 'nZ dP da = w an = -^2=^ dZk ■ uradP"-

We denote

g(Z) = (n —1)! vz dt

2nn dZk I grad p[

Lemma 8. If D is a strictly convex circular domain, then g(Z) is a real-valued function that does not vanish on dD.

Proof. For circular domain p(Zl7 - - -, Zn) = p(Z1el9, - - -, Znei6), 0 < 0 < 2n, differentiating this equality with respect 0, we get

nn

0 = Y iZkeiedP —Y iCke-i9 dp.

k= dZk ¿1 dZk n d n d

Then we get = Y^ Z^tz for 0 = 0. The function g(Z) means being real that

k=1 dZk k=1 dZk

nnn YZk ^P = Y Zk ^P = Y Zk

2=1Zk dZk 2=1Zk dZk 2=1Zk dZk

1

The function g(Z) = 0 on dD, since the complex tangent plane does not pass through zero at the point Z. Therefore, the function g(Z) preserves sign on dD. □

Therefore, we can assume that g(Z) > 0 on dD. Therefore, da = gda is a measure and for it all previous constructions are true.

Lemma 9. Let D be a strictly convex (pi,... ,pn)-circular domain, i.e.

p( Zi,..., Zn)= p( Zieipie,..., Zueipnd), 0 < e < 2n,

where pi,... ,pn are positive rational numbers. Then the function

to dp

J2Zk pk dk

k=i Sk

is real-valued and not zero.

Proof repeats the proof of the previous Lemma 8. □

The function p can be chosen so that \gradp\\dD = 1, then

where ci =

dp

= ci Zk k=i

(n — 1)!

da = ci Zk —p da,

k= dZk ,

2nn

Consider the family of complex lines lzo b of the form (1) passing through the point z0 £ D, where b £ CPn-i. Calculate the form w in the variables b and t, we get

dZ = dZi A ... A dZn = d(z0 + bit) A ... A d(z0n + bnt)

idb[1] — b2

= tn-idt a£( — 1)k-ibk db[k] = tn-idt A v(b),

= d(bit) A ... A d(bnt) = tn-idt A (bidb[1] — b2db[2] + ... + ( — 1)n-ibndb[n] =

k=1

where v(b) = ^( — 1)k-ibk db[k]. Here we use the fact that b £ CPn-i.

k=1

Now we calculate

1)k-iZk dz [k]

k=1

= J2(z0 + bk t)d(z0 + bit) A ... A d(zk-i + bk-it) A d(z°0+i + bk+it) A ... A d(z°n + bnt) =

k=1

n

YJ( — 1)k-iz0 dZ[k] + J2( — 1)k-ibkt dZ[k] =

k=1 k=1

nn

= J2( — 1)k-iz°0tn-2 dt A x(b) +YJ( — 1)k-izltn-i db[k] + E bktn db[k],

k=1 k=1 k=1

where x(b) is a differential form of degree (n — 2). From here we get that

n

w\dD = c£( — 1)k-1Zk dZ[k] A dZ\dd=

k=1

c^("1)fc_1 z°kin-1tn-1 db[k] A dt A v(b) + c^("1)k-1 bktntn-1 db[k] A dt A v(b)

z kt

k=1 k=1

= (-1)ncdt A {yj{-1)k-1zl\t\2n-2 db[k] A v(b)+t\t\2n-2v(b) A v(b)} = \=1 '

= (-1)n-1c\t\2n-2 dt A (Ys-1)k-1 zk db[k] + tv(b)} A v(b).

Thus, we have Lemma:

Lemma 10. The form in the variables b and t has the form

"\öD = (-1)n-1 c\t\2n-2 dt A (±(-1)k-Z db[k] + tv(b )) A v(b).

k=1

Consider the modified Poisson kernel

K(z,Z) ■ K(Z,w)

Q(z,w,Z ) =

K(z, w)

For w = Z we obtain Q(z, z, Z) = P(z, Z) and K(z, z) > 0. Therefore, there exists a neighborhood U of the diagonal w = Z in Dz x Dw in which K(z, w) = 0. Consider the function

*(z,w)= f f (z)Q(z,w,Z) dp(Z), JdD

which is defined for (z, w) £ U. It is holomorphic in (z, w) £ U, and for w = z we have §(z,w) = F(z) and

df+Y&(z,w)

dzs dwY where

dS+Y$(z,w) = dfi + - + fn+7l + -+7n §(z,w)

dzsdwY = dzf1 ■ ■ ■ dztdwY1 ■ ■ ■ dw"t '

df+Y F (z

dzf dzY = dzf1 ■■■ dzt dZYl ■■■ dznn ' and S = (01,..., Sn), y = (Y1,.. .,Yn).

dS+Y F (z) (9)

dzf dZY ' (9)

4. Additional construction

Consider a mapping Z = x(n) : B —> D, where B is the unit ball in Cn centered at zero taking zero to a a € D. The mapping x is be constructed as follows: Consider the complex lines Xb = {n € Cn : n = br, b € CPn-1, T € C} and lab = {Z € Cn : Z = a + bt, b € CPn-1, t € C}. The intersection Da}b = D n la,b is a strictly convex domain in C; therefore, there exists a conformal mapping t = xb(T) of the unit disk B n Xb into Da b taking t = 0 to t = 0. By the

Caratheodory Theorem [21], this mapping extends to a homeomorphism of the closed domains. Then to a point n = br £ B n Xb there is assigned the point x(n) = a + bxb(r) £ Dab. Lemmas 11-14 are formulated and proved in the same way as in the paper [22].

Lemma 11. Let D be a bounded strictly convex circular domain with twice smooth boundary in Cn. Then x(n) is well defined and is a C1 -diffeomorphism from B onto D.

Henceforth, we assume that D is a bounded strictly convex circular domain with twice smooth boundary.

Lemma 12. The derivatives of x(n) are holomorphic functions in r for b fixed and where n = br.

Lemma 13. Let the function f £ C(dD) have the one-dimensional holomorphic extension property along complex lines passing through a £ D. Then the function f *(n) = f (x(n)) is continuous on dB and has the one-dimensional holomorphic extension property along complex lines passing through zero.

Performing a change of variables in integral for we obtain

$(z,w)= i f (z)Q(z,w,Z) d^(Z) = JdD

= f (x(n))Q(z,w,x(n)) d^(x(n))= f*(n)Q*(z,w,n) d^*(n).

JdB JdB

Consider the form

n

u*(v) = "(X(n)) = J2-1)k-1 Xk(n) dX(n)[k] A dx(n).

fc=i

By Lemma 12, the form dx(br) is holomorphic in t for b fixed, while the form dx(br)[k] is antiholomorphic in t for b fixed.

Lemma 14. The forms dx(bT)||rk = 1,... ,n, are forms with holomorphic coefficients with respect to t .

Theorem 3. Let D be a bounded strictly convex circular domain with twice smooth boundary and the function f £ C(dD) have the one-dimensional holomorphic extension property along complex lines passing through a £ D. Then

d Y §(z,w)

0

dwY

for II7II > 0, where 7 = (71, ...,Yn) and H7H = 71 + ... + Yn.

The proof of this Theorem is essentially as in the proof of Theorem 3 of [22].

Corollary 2. &(a,w) = const under the conditions of Theorem 3.

the same way as the previous theorem we prove the statement:

Theorem 4. Let D be a bounded strictly convex circular domain with twice smooth boundary and the function f £ C(dD) have the one-dimensional holomorphic extension property along complex

( , ) are polynomials in w of degree

lines passing through a £ D. Then the derivatives _ ,

dz°

at most .

Theorem 5. Let D be a bounded strictly convex circular domain with twice smooth boundary and the function f (Z) € C(dD), and a,c € D. Assume that &(z,w) satisfies the conditions ^ ^ d a$(a,w) d a$(c,w)

Q(a,w) = const, ®(c,w) = const and -dj~a-, -dT«- are polynomials in w of degree at

most ||a||. Then, for every fixed z on the complex line

la,c = {(z,w) : z = at + c(1 — t), w = at + c(1 — t),t € C}

dY $(z,w)

we have §(z,w) = const with respect to w; i.e., —d—Y— = 0 for ||71| > 0.

The proof of this Theorem is essentially the same as the proof of Theorem 5 of [22].

d Y F (z)

Corollary 3. Under the conditions of Theorem 5,

dcY

5. Proof of the main assertions

= 0 for ||7|| > 0.

z=at+(1-t)c

Theorem 6. Let n = 2 and D be a bounded strictly convex circular domain with twice smooth boundary and the function f € C(dD) have the one-dimensional holomorphic extension property along the family £{a,c,d} and the points a,c,d € D do not lie on one complex line in C2. Then d y $( z,w)

—d—y— = 0 for any z € D and ||7|| > 0, and f (Z) extends holomorphically into D.

Proof. Let 5 be an arbitrary point on lac. Then by Theorem 5, we have

d Y $(5,w)

dwY

0 (10)

for||71| > 0. Joining 5 with d by the line and again applying Theorem 5 with z € lz,d, we dY

conclude that —--=0 for ||^N > 0. Therefore, (10) is fulfilled for all 5 in some open set.

dwY

dY F (z)

Inserting w = z in (10), we have —d^Y— =0 in some open set in D. The real analiticity

dF (z)

of F(z) implies that ^^— = 0 for any z € D and j = 1,... ,n. Since by Theorem 2 we have dzj

F(Z)\sD = f(Z), the function f (Z) extends holomorphically into D. □

Denote by A the set of noncomplanar points ak € D c C", k = 1,... ,n +1.

Theorem 7. Let D be a bounded strictly convex circular domain with twice smooth boundary in C" and the function f € C(dD) have the one-dimensional holomorphic extension property

dY $(z,w)

along the family £3. Then --- = 0 for any z € D and H71| > 0, and f(Z) extends

dwY

holomorphically into D.

Proof. Proceed by induction on n. The induction base is Theorem 6 (n = 2). Suppose that the theorem holds for all k < n. Consider the complex plane r passing through ai,..., an, the dimension of r is by hypothesis equal to n — 1 and a„+1 € r. The intersection r n D is a strictly convex domain in C1.

The function f \ dD is continuous and has the property of holomorphic extension along the

dY $(z',w)

family £3l, where A1 = {a1,..., an}. By the induction assomption, -dj--= 0 for H71| > 0

for all z' £ m D.

dY §(z,w)

Joining z' € r with an+1, we find by Theorem 6 that --- =0 for INI > 0 for some

dwY

open set in D x D. In much the way as Theorem 6, this implies that F(z) is holomorphic in D, and so f(Z) extends holomorphically into D. □

Theorems 6 and 7 obviously imply Theorems A and B.

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Многомерные граничные аналоги теоремы Гартогса в круговых областях

Александр М. Кытманов Симона Г. Мысливец

Институт математики и фундаментальной информатики Сибирский федеральный университет Свободный, 79, Красноярск, 660041

Россия

В статье представлены некоторые результаты, связанные с голоморфным продолжением функций, определенных на границе области D С Cn, n > 1, в эту область. Речь идет о функциях с одномерным свойством голоморфного продолжения вдоль комплексных прямых.

Ключевые слова: функции с одномерным свойством голоморфного продолжения, круговые области.