Plane domains with special cone condition Текст научной статьи по специальности «Философия, этика, религиоведение»

DOI: 10.15393/j3.art.2014.2609 Issues of Analysis. Vol. 3(21), No. 2, 2014

A. N. Anikiev

PLANE DOMAINS WITH SPECIAL CONE CONDITION

Abstract. The paper considers the domains with cone condition in C. We say that domain G satisfies the (weak) cone condition, if p + V(e(p), H) C G for all p € G, where V(e(p), H) denotes right-angled circular cone with vertex at the origin, a fixed solution e and a height H, 0 < H < to, and depending on the p vector e(p) axis direction.

Domains satisfying cone condition play an important role in various branches of mathematic (e.g. [1], [2], [3] (p. 1076), [4]).

In the paper of P. Liczberski and V. V. Starkov, a-accessible domains were considered, a € [0,1), — the domains, accessible at every boundary point by the cone with symmetry axis on {pt : t> 1}.

Unlike the paper of P. Liczberski and V. V. Starkov, here we consider domains, accessible outside by the cone, which symmetry axis inclined on fixed angle 0 to the {pt : t > 1}, 0 < ^^^ < < n/2.

In this paper we give criteria for this class of domains when the boundaries of domains are smooth, and also give a sufficient condition when boundary is arbitrary.

This article is the full variant of [5], published without proofs.

Key words: (a, 3)-accessible domain, cone condition. 2010 Mathematical Subject Classification: 26A21.

Introduction

In [6] (see also [7]) «-accessible domain, a E [0,1), were introduced and studied. A domain fi C Rn, 0 E fi, is called a-accessible, if for every point p E dfi there exists a number r = r(p) > 0 such that the cone

( ( p \ an 1

K+(p, a, r) = |x E Rn : ( x — p, 1 > \\x — p\\ cos(-^), \\x — p\\ < r >

1 This work was supported by Program of Strategic Development of Petrozavodsk State University.

© Anikiev A. N., 2014

is included in Q' = Rn \ Q.

In particular, in [6] the authors proved that «-accessible domains are bounded and satisfy cone condition when a E (0,1) and e(p)=—p. This condition of radiality axis of symmetry applies significant limitation on Q.

The paper consider the case, when the axis of cone symmetry is lies on ray, containing 0 and p, and crosses the cone.

Definition 1. A domain Q C C7 0 E Q, is called (a, fl)-accessible, a, /3 E E [0,1), if for every point p E dQ there exists a number r = r(p) > 0 so that the cone

K+ (p, a, fl, r) = |z E C : — fln — arg(z — p) — arg(p) — —, \z — p\ — r|

is included in Q' = C \ Q.

Let us denote a0 = min(a,fl), fl0 = max(a,fl). Note that the class of (a, fl)-accessible domains is intermediate between a0- and fl0-accessible classes.

The purpose of this paper is to discuss the failure of condition e(p) = = —p, when the angle (let us denote it by 0) of inclination axis of symmetry to the ray {pt : t > 0} is a constant.

It is interesting to figure out how the properties of domains with this inclination will be changed. This problem is very difficult for large values of 0 (0 > §) even in the case of permanent angle 0. In this case, the methods by which the results were obtained in [6] are no longer applicable.

This work does not provide a complete description of these areas - this task is too complex, but at this stage it's unable to get rid of condition e(p) = —p and replace it by the condition of the Def. 1, when 0 is constant. Let's introduce some other definitions.

Definition 2. We call a domain Q starlike with respect to 0 if for every point z E Q segment [0, z] is contained in Q.

Definition 3. We call a domain Q a strong-starlike with respect to 0 if [0, p] fl dQ = p for every point p E dQ.

Case of arbitrary boundary

Theorem 1. If the domain Q is (a, fl)-accessible, a, fl E (0,1), then for each point p E dQ and for every £ E (0, min(a, fl)) there exists a number

p such that p(p) > 0 and the cone K- (p, a — e,P — e, p) C Q, where

K- (p,a — e, P — e, p) = f _ (P — e)n (a — e)n 1

= s z e c :--2— < arg(z — p) — ars(—p) < —2— 'z — P' < P\ '

Proof. Suppose not. Then there exists a point p e dQ such that

K-(p,a — e,P — e,p) n Q' = 0

for p > 0 and e e (0, min(a,P)). This shows that there exists a sequence of points such that {wm} e K-(p,a — e,P — e,p) n Q', and zm ^ p as m ^ œ. Consider C(p, \ wm|) - circle with center of p and radius \wm\. This circle intersects the segment [0, p). Associate point wm with those, which are obtained as a result of intersection C (p, \zm\) n [0,p) with arc of circle, are placed in Int(K-(p,a — e,P — e,p)). As dQ is connected, this arc of circle intersects the bound of Q. Thus we get sequence of points lying on bound of Q, which converges to p. Let us denote this sequence {wm}.

Denote by l(0) the ray, starting from 0 and passing through the segment [0, p] with angle 0. In [6, proof of Theorem 1] was proved existence of le n dQ and a unique. Thus, Q is a strong-starlike domain.

Introduce a function r = r(0), the distance from 0 to the point of intersection of the ray l(0) with dQ. From [6, proof of Theorem 1] it follows that r(0) is continuous.

There exists n e N such that for all m > n

en

\ arg(wm) — arg(p)\ < —. (*)

Denote by 0m = arg(wm) — arg(p), 0m e (—n; n].

Now let us consider that L is part of dQ, lying between l(0) and l(0m). As wm e dQ, then for it exists cone K+ (wm, a, P, rm) C Q, rm > 0. Consider two ways:

1) Let 0m > 0. Draw a line through wm parallel those sides of cone K- (p, a — e, P — e, p), which intersect l(0m). This line intersects segment [0, p] at the point A : A < \p\. The same side of cone K+(wm, a, P) intersects the segment [0,p] at the point B : \B\ < \A\. This is true, when:

Pn (P — e)n . , . /1 \

^ > 2 + \0m\. (1)

2) Let now 0m — 0. By similar reasoning, we obtain:

an (a — £)n , , ,

-y > 3 + \0m\. (2)

From (*) it follows that for sufficiently large number m the inequalities (1) and (2) hold. By the fact, that Q is (a, fl)-accessible and

K+ (wm, a, fl, rm) f Q = 0

we have L f K+ (wm, a, fl, rm) = 0.

Consider L f [wm , B]. Let w0 is closest to B point of intersection L f [wm, B]. Denote by 00 = arg(w0) — arg(p), the angle between the ray l(00), going from 0 through point w0, and the segment [0,p]. As w0 E dQ, for it exists cone K+(w0, a,fl, S) such that

K+ (w0, a, fl, S) f Q = 0

for sufficiently small S > 0. The side of cone K+(w0 ,a,fl) intersects the segment [0,p] in point C in the way that \C\ < \B\. It follows from the fact that cone K+ (w0,a,fl) obtains from cone K+(wm,a,fl) by turning an angle (00 — 0m).

For L to connect w0 and p, it must either intersect (w0, B), or intersect segment [0,p]. None of both is possible. Indeed, by the definition of w0, L can't intersect the segment (w0,B). On the other hand, by virtue of an unambiguous definition r(0), dQ can't contain the radial segments [6, Theorem 1], so it doesn't contain the points from [0,p). Hence we get a contradiction with the fact, that theorem is wrong. The proof is complete now. □

Theorem 2. If Q is (a,fl)-accessible, then for every point p E dQ and for every fixed a, fl E [0,1) unbounded cone K+(p, a, fl, to) := K+ (p, a, fl) belongs to C \ Q = Q'.

Proof. Suppose that the theorem is wrong. Then there is a point p E dQ such, that z E K+ (p,a,fl) f Q, z E Q. Consequently there exists w E dQ such, that for every fixed R > 0, w E dK+(p, a, fl, R). Let us suppose, that point w is first, except p, contained in dK+ (p, a, fl, R), which means, that there were no other points from dQ on dK+(p, a, fl, R).

Suppose that w E dK+(p, a, fl). Then w E cM(p,R) and thus, there exists vicinity Uw C K+(p,a,fl). So, there is a point v E Q such that v E

E Uw. As Q is starlike, [0, v] is contained in Q. From the other hand

[0, v] f K+ (p,a,fl,R) = 0,

which contradicts the fact that K+ (p,a,fl,R) C Q'.

So w E dK+(p, a, fl). Through Theorem 1 there exists p = p(p) such that cone K-(w, a — e,fl — e, p) C Q for every e E (o, min(a, fl)). In C, we introduce polar coordinates 0 - pole, 0p - polar. Consider the points ax = p + (w — p)A, A E (0,1). We show, that ax E K-(w,a — e,fl — e,p) for sufficiently small p¿0, when A close to 1 and e close to 0. If this is true, then one the one hand ax E Q, which follows from Theorem 1, and on the other hand ax E dK+ (p,a,fl) since w E dK+(p, a, fl). This contradiction get us that the theorem is true.

To prove the inclusion ax E K- (p,a — e, fl — e, p) it is enough to show that

(fl — e)n (a — e)n

--2— < arg(ax— w) — arg(—w) <—2—' (3)

Since ax — w = p + (w — p)A — w = (p — w)(1 — A), (3) can be rewritten

as:

(fl — e)n , , , , (a — e)n

--2— < arg(p — w) — arg(—w) < —2—. (4)

Here we get two ways: 1) Suppose that

fln . . . .

—< arg(w) — arg(p) < 0;

this means that arg(w — p) — arg(p) = — .

We see that arg(p — w) — arg(—w) = arg(w — p) — arg(w), so

fln

arg(p — w) — arg(—w) = —— + arg(p) — arg(w) < 0.

2

As arg(w) — arg(p) < 0, for sufficiently small e > 0

arg(p) — arg(w) > Y,

and thus,

fln fln en

arg(p) — ~y — arg(w) > —y + ~2 . (5)

From inequality (5), it follows, that

(ß - e)n

< arg(p — w) — arg(-w) < 0. 2

2) Now let us suppose that

an

0 < arg(w) — arg(p) < —;

this means that arg(w — p) — arg(p) = .

We see that arg(p — w) — arg(—w) = arg(w — p) — arg(w) and so

an

0 < arg(p — w) — arg(—w) = — + arg(p) — arg(w).

As arg(w) — arg(p) > 0, for sufficiently e > 0, one has

arg(w) — arg(p) > en,

so that

an , , ,, an en — + arg(p) — arg(w) < —----• (6)

From (6), it follows, that

(a — e)n

0 < arg(p — w) — arg(—w) < ---.

2

Thus, from cases 1) and 2), it follows, that inequality (3) is true, and thus ax £ K- (w, a — e, [ — e, p) with A close enough to 1. Hence we get a contradiction. The proof is completed. □

Remark 1. Observe that (a, ()-accessible domains are bounded, if a, [ £ £ (0,1), since these domains are a0-accessible, a0 = (min(a, [)), and in [6] it was shown that a0-accessible domains are bounded for a0 > 0.

Theorem 3. If fi C C, 0 £ fi, a, [ £ (0,1), then the following assertions are equivalent:

(i) fi is (a,[)-accessible domain;

(ii) every unbounded cone K+ (p,a,[) C fi', p £ dfi;

(iii) every unbounded cone K+(p,a,[) C fi', p £ fi';

(iv) for every point p £ dfi and for every e £ (0, min(a, ()) there exists an r = r(p) > 0 such that the bounded cone K- (p, a — e, [ — e,r) C fi.

Proof. In view of Theorems 1 and 2, it is sufficient to prove the implications (iv) ^ (i) and (ii) ^ (iii).

Let w = I(z) be the mapping inversion, defined as:

w = 1. (7)

z

For the proof of (iv) ^ (i), under this mapping, consider the image of the cone K+ (p',fl,a) \ {p'} to K- (p,a,fl), where p = 1/p'. Indeed, (7) is a bilinear mapping, having a circular feature and the property of preserving angles, so that the boundary of K+ (p', a, fl) transfers into arcs, intersecting at points p and 0, and the angle of intersections of those circles at the point p is (a + fl)n/2, and the image will be lying inside intersection of these circles.

Now, let us consider the condition (iv). Denote G C C as the image I(Q \ 0). We will show that domain G' = C \ G is (fl, a)-accessible.

To show this, we note that 0 E G', and for every point p E dQ there exists cone K-(p,a — e,fl — e,p) C Q, for sufficient small p = = p(p,e) > 0. Obviously, there exists a number r = r(p,p) > 0 such that when z = 1/w follows the inclusion

I(K+ (p', fl — e,a — e,r) \ {p'}) C K-(p, a — e,fl — e,p),

which means that I(K+(p', fl — e,a — e,r) \ {p'}) C Q for every point p' E dG'. Thus G' is (fl — e,a — e)-accessible domain, e E (0, min(a, fl)). From Theorem 2 it follows, that K+ (p', fl — e,a — e) C G. Passing to the limit e ^ 0 we get that K+(p',fl,a) C G', so G' is (fl, a)-accessible. Hence, from Theorem 1, it follows that for every point p E dG' and for every e E (0, min(a, fl)) there exists an r = r(p', e) > 0 such that the cone K- (p', a — e, fl — e,r) belongs to G'.

Note that under the mapping (7) the image of cone K- (p', a — e, fl — — e, r) belongs to K+(p, a — e, fl — e, R) for some r = r(p, R) > 0, so that I(K- (p', a — e, fl — e,r)) C Q'. Hence and from definition we see that Q is (a — e, fl — e)-accessible domain. Using Theorem 1 and allowing e ^ 0 we get, that Q is (a,fl)-accessible. This proves the implication (iv) ^ (i).

We now show, that if Q satisfies the condition (ii), then Q satisfies the condition (iii). Take arbitrary point p E Q' \ dQ. The segment [0, p] intersects dQ. If this intersection has more than one point, then we take the closest to p and denote it as p', and the next one - as p''. Then the cone K+(p',a,fl) contains inside sufficient small surroundings

of point p'' and therefore points from fi. On the other hand, Theorem 2 says that K+(p', a, (3) C fi'. This is a contradiction the fact that [0,p] fl fidfi = p'. Hence, from Theorem 2, it follows that K+(p',a,() C fi'. We will now show, that K+(p,a,() C fi'. Indeed, since \p\ > |p'|, we have K+(p,a,()=K+(p',a,() + (p — p'), so that arg(p) = arg(p'). Then for every point z G K+(p', a, (3), one has z +(p—p') G K+ (p, a, (). Let us show that z + (p — p') belongs to K+ (p', a, (). Since z + (p — p') G K+ (p, a, (3), we see that

(n A an

— — < arg(z + (p — p ) — p) — arg(p) < —,

and so, as arg(p) = arg(p'),

—— < arg(z — p ) — arg(p ) < — •

Hence, from definition of K+(p',a,(), we obtain that z + (p — p') G G K+ (p',a,(). Thus K+(p,a,() C fi'. Since the point p G fi' \ dfi is arbitrary, we get the implication (ii) ^ (iii). □

Remark 2. If {fi7} is a family of (a, ()-accessible domains, then the union fi = (J {fiY} is also a (a,()-accessible domain. Actually, from Theorem 3, it follows that fi is (a,()-accessible domain if and only if K+(p,a,() f fi = 0 for every point p G fi . If p G fi, then p G fiY for every y. In this situation, K+(p,a,() f fiY = 0 for every y. Thus,

K+(p, a, (3) f (Ufi7) = 0.

Theorem 4. If fi is (a, (3 )-accessible domain, a, (3 G (0,1), then for every £ G (0, min(a,()) there exists an R = R(e) > 0 such that the cone K- (p,a — £, 3 — £, R) C fi for every point p G dfi.

Proof. From the implication (iv) ^ (i) in proof of Theorem 3, it follows that for (a, (3 )-accessible domains fi, the interior of complement I (fi') = = G', using z = I(w) = 1/w, is ((, a)-accessible domain. Therefore it is enough to show that for every fixed £ G (0, min(a, ()) there exists an R = R(£ > 0 such that for every point p G dfi, the image of every w G G K-(p,a — £,( — £,R) using z = I(w) considered inside K+(p', a), p' = 1 /p. Indeed, if it will be shown, then

I (K-(p,a — £,( — £, R)) C G = I (fi)

Hence, as I(w) is homeomorphism, we get K-(p, a — £, ß — £,R) C O.

Since w E K-(p,a - e,3 - e,R), w = p + rel^+arg(p)), r E (0, R],

0 E ((2 - / + e)n/2, (2 + a - e)n/2),

so that n - 0 E ((/ - e)n/2, (a + e)n/2).

By definition of the K+ (p,a), we get I(w) = 1/w E IntK+ (p',a) if and only if

an ( 11) (1) /n

—7T < arg = - = ) - arg = ^ ^ (8)

2 \w p J \p J 2

Now

arg (= - 1) - arg (1) = ar^p_w j - arg(p) = ar^p_w\ = \w pj \p J \ wp J \ w J

w p p

arg (= arg (ei(n-+-ag™) + arg (p + =

= n - 0 - arg(p) + arg (p + rel^+arg(p))^j . Since 0 E O, we have p = 0. Then there exists an Re (0, min

pEdO

such that

arg (p + - arg(p)

en

< t •

therefore, for every r E (0, R) and for every p E dO, the following inequality holds:

arg (p + - arg(p)

en

< t •

thus the inequality (8) holds.

Hence we get that there exists an R = R(e) > 0 such that for every p E E dO, the image of the cone K- (p, a-e, /-e, R) belongs to IntK+(p', ff, a). This proves the theorem. □

Theorem 5. If a domain O C C (O = C) is (a, 3)-accessible, a, 3 E E (0,1), then for every e E (0, min(a,/)) there exists an R = R(e) > 0 such that the cone K-(p, a - e, 3 - e,R) belongs to O for every p E O.

Proof. Assume that theorem is wrong. Then for some e E (0, min(a,3)) there exists sequence of points wk E O and a sequence of numbers rk such that the cone

K-(wk, a - e, 3 - e,rk) n O' = 0 (9)

for every number k G N, and rk — 0. Since fi is compact, there exists a convergent subsequence of sequence {wk}, that wk — w'0. Denote this subsequence as {wk}. If w0 G fi, then for sufficiently small p > 0 ball B(w0,p) C fi. Starting from some number k > N, points wk G B(w0,p), we have K-(wk,a — £,P — e)HB(w0,p) C fi. Since the last fact contradicts (9), we get that w0 G dfi.

Consider a sequence of points pk G dfi, pk = Akwk, Ak > 1- Note that Pk — w0 when k — to and lim pk = p0 = w0. Indeed, if it is wrong,

k—^^o

then p0 = Aow'0, Ao > 0 and A = 1. Since p0 G dfi, for every surroundings

Up0: Up0 H fi = 0. On the one hand fi is (a,0)-accessible domain and

w'0 G dfi, so the cone K+ (w'o,a,0) belongs to fi'. On the other hand,

since |p0| > |w01, the sufficient small surroundigs Up0 C K+(w'0), so

that K+(w'0) H fi = 0, but this can not be true (see Theorem 2).

Hence we get that p0=w'0.

Since lim wk = p0 = lim pk, pk = Akwk, Ak — 1+ as k — to. k—tt k—tt

Therefore for number R from Theorem 4 and for sufficient large number k, points wk G K-(pk, a — e, P — e,R) and

K-(wk, a — — e,r'k) C K-(pk, a — — e, R).

By Theorem 4, the cone K-(pk, a — e, P — e,R) C fi for some fixed R = = R(e) > 0, so that K-(wk,a — £,P — £,rk) C fi. The last contradicts the relation (9). Theorem 5 is proved. □

Case of domains with smooth boundary

Here we assume that the domain fi C R2 has smooth boundary dfi given by equation:

F (x,y)=0,

and

F(x,y) < 0.

is fi.

Smooth function F(x, y) can be set locally which means that F(x, y) = = Fp(x,y) in the neighborhood of each point p G dfi. Since dfi in the neighborhood of each point p G dfi can be defined by the equation:

x = f(y) or y = f(x), we can assume that gradF(p) = 0 for every point p G dfi.

Denote by n(p) = ||g™lF(p)||, the external unit normal vector at point p E dO.

The following lemma is a consequence of the lemma from [6].

Lemma 1. Let O C C with smooth boundary dO, and n(p) is external normal vector at point p E dO. Then for every fixed a, 3 E (0,1) there exists r > 0 such that

K + (p, a, 3, r) =

= |z E C : - 3n < arg(z - p) - arg(n(p)) < 02", \\z - p\\ < r| C O',

K-(p, a, 3, r) =

= |z E C : -3n < arg(z - p) - arg(-n(p)) < , \\z - p\\ < r| C O.

Theorem 6. Let O E C, dO be smooth boundary. Then for every fixed a, 3 E (0,1) domain O is (a,3)-accessible if and only if

(1 -T^ < arg(p) - arg(n(p)) < (1 - a) (10)

for every point p E dO.

Proof. Suppose that O is (a, [)-accessible. We will show that the inequality (10) holds. As O is (a, [)-accessible, it is starlike with respect to 0, and under our assumptions about F(z) it follows from [7] that O

starlike if and only if , — 0 for every p E dO. Indeed,

IgracLF(p)|| = n(p) is external normal vector at point p and

(à, in(M) — 0 "cos ^ — 0,

which means that \0p\ < n/2. Let = arg(p) — arg(n(p)), arg(p) E E [0, 2n]. arg(p) increases when crawling dO in positive direction, and arg(n(p)) changes continuously with a continuous changing of p E dO. Suppose that at point p the inequality (10) doesn't hold, then we get:

(1 — a)n / / w n

—2— < arg(p) — arg(n(p)) < ^' (11)

or

n . . , ... (1 — ¡3)n ,,

-^ < argp — arg(n(p)) <--^— • (12)

For simplicity, we assume that arg(p) = 0, p E R (this could be achieved by converting the rotation on which the domain Q is not sensitive). Thus

— 2 < arg(n(p)) < (13*)

or

(1 — 3)n , , n , _

v ^ < arg(n(p)) < 2• (14*)

As Q is (a, 3)-accessible, the cone K+ (p,a,3) C Q'. Let

K-(p, y, m) =

( Yn Yn 1

= < z E C : —— < arg(z — p) — arg(—n(p)) < —, \z — p\ < m> • 22

From a lemma proved in [6], it follows that for every fixed Y E (0,1) there exists an m > 0 such that K- (p,Y,m) C Q. Take a point z E E dK+(p, a, 3, r), z = p + pe^, 0 = {an/2, —¡n/2}, 0 < p < r. Separately consider the cases (13*), (14*).

1) Case (13*). Let z+ = p + peian/2. We will show that z+ belongs to K-(p,Y,m) if p < m. Choose arg(—n(p)) such that

arg(—n(p)) = n + arg(n(p))

Then

n ^ , , \ \ (1 + a)n 2 < arg(—n(p)) < ^—• (15)

Since arg(z + — p) = an/2, one has

n , + \ / / \\ (1 — a)n . .

—2 < arg(z — p) — arg(—n(p)) <--2— • (16)

From (16) it follows that for sufficiently small m > 0 there exists Y E (0,1) with

— f< arg(z+ — p) — arg(—n(p)) < ^

Last inequality means that z+ E K-(p,Y, m) with p < m^ A lemma from [6] guarantees that K- (p,Y,m) C Q and thus z+ also belongs Q, which contradicts the fact that z+ E dK+ (p, a, 3) C Q .

■ fn

2) Case (14*). Now let z- = p+pe-i~. We will show that z- belongs to K-(p,Y,m) if p < m. Choose arg(—n(p)) such that

arg(—n(p)) = arg(n(p)) — n.

Then

(1 +3)n , , n .

- g < arg(—n(p)) < — ^. (17)

Since arg(z- — p) = — 3n/2, we have

(1 — 3)n , _ x / / w n ,,

—2— < arg(z — p) — arg(—n(p)) < 2• (18)

From (18) it follows that for sufficiently small m > 0 there exists y G (0,1) with

—y < arg(z- — P) — arg(—n(P)) < y •

Last inequality means, that z- G K- (p, y, m) with p < m. A lemma from [6] guarantees that K- (p,j,m) C fi. Thus z- also belongs to fi, which contradicts the fact that z- G dK+ (p,a,3) C fi .

Contradictions in cases 1) and 2) mean that if fi is (a, 3)-accessible domain then inequality (10) holds.

Now let for every point p G dfi the inequality (10) hold. We show that fi is (a, 3)-accessible domain. At first show that fi is (n, 0)-accessible domain for n G (0,a),6 G (0,3), i.e. for every point p G dfi the cone K+(p, n,0,r) C fi , r = r(p) > 0. Fix p and take a point z G K+(p, n, 0, r) with sufficiently small r then

0n nn

— y < arg(z — p) — arg(p) < — • (19)

Compose (10) and (19):

(1 — 3 + 0)n (1 — a+n^

--2-< arg(z — p) — arg(n(p)) <-y-•

Last inequality means that z belongs to K + (p, 1 — a + n, 1 — 3 + 0, r). Denote ^ = max(1 — a + n, 1 — 3 + 0), ^ G (0,1). Then

K +(p, 1 — a + n, 1 — 3 + 0, r) C K +(p,^,r).

A lemma from [6] guarantees that K + (p,^,r) G O' for sufficiently small r > 0, and thus K + (p, 1 - a + n, 1 - 3 + 0,r) G O'.

We get that z G O' for every point z G K+(p, n, 0, r) with sufficiently small r > 0. Thus O is (n, 0)-accessible domain. Now, applying Theorem 2 and passing to a limit n a, 0 ^ 3 we get that O is a (a, 3)-accessible

domain. Theorem 6 proved. □

Corollary 1. Denote by e the symmetry axis of the cone K+ (p,a,3). Then a condition (10) is equivalent to

(m * -m-

Proof. Fix p G dfi. With rotation transformation, assume that arg(p)=0. The solution of the cone K+(p,a, /3,r) is . Note that arg(e) G

G (-4, 4). Then

( \ (a + 3)n -r / N ^ n

— - arg(e) = —4— if arg(e) >

or

In (a + 3 )n

— + arg(e) = —4—if arg(e) <

From the last inequalities we get arg(e) = . Thus from (10) we get:

(2 - a - 3)n (2 - a - 3)n --4- < arg(e) - arg(n(p)) < -4-,

and this is equivalent:

(^.*)) > sin (^) .

□

The following theorem gives a sufficient condition for (a, |)-accessible domains. Here A* denotes a matrix, conjugate to a matrix A. Let e, as in corollary to Theorem 6, be a vector lying on the symmetry axis of the cone K+ (p, a, |).

Theorem 7. Let fi C R2 be a bounded domain with 0 G fi, a, 3 G (0,1).

u

Let f = ^ v j be a diffeomorphism of a domain fi at the unit circle

centered at the point 0, f (0) = 0, and Df (x) is nonsingular differential in every point x G fi.

If for a number S > 0 the inequality:

f * (x)Df (x)e . ( (a + 3)n ' > sin '

. ((a + 3 )n \

||f*(x)Df(x)||||e|| " V 4

holds in fi(S) = {x G fi : dist(x, dfi) < S} then fi is (a,3)-accessible domain.

Proof. Denote by fir = {x G fi : u2 + v2 < r2} with r G (0,1). fir C fi and dfir - smooth boundary, given by equation:

F(x) = u2 + v2 - r2 = 0.

Since Df (x) is nonsingular for every x G fi, then f* (p)Df (p) = 0 for every point p G dfir. Note that

du du

2

/ du du \

f * (x)Df (x) = (u,v)[ dV 9JV_

\ Bxi dx_ J

du dv du dv /1 d 2 2 1 d 2 2 \ dxi dxi' dx2 dx2 \2 dxi '2 dx2 )

= i gradF (x).

Then gradF(p) = 2f*(p)Df (p) = 0 for every point p G dfi.

As fi is bounded, for fixed S > 0 dfir C fi(S) for r G (r0,1), with ro

sufficiently close to 1.

By the condition in Theorem 7 we get,

e gradF * (x) ^ gradF (x)e f *(x)Df (x)e

i e gradF * (x) \ \M' WgradF* (x)|| )

WefWgradF* (x)^ WgradF (x)||e| Wf* (x)Df (x)||e|

. f(a + 3)n \

y sin \—T- )

Now, from the corollary after Theorem 6 we get that fir is (a, 3)-accessible domain, and from remark after Theorem 3 it follows that fi = (JrE(r0 1) fir is (a, 3)-accessible. Thus, Theorem 7 is proved. □

References

[1] Besov O. V., Il'in V. P., Nikolskii S. M. Integral presentation of functions and theorem of embanding. M., Nauka, 1975.

[2] Dolzenko E. P. Boundary properties of arbitrary functions. (In Russian) Proc. of USSR. Ac. of Sci. Ser. Math., 1967, vol. 31, no. 1, pp. 3-14. DOI: 10.1070/IM1967v001n01ABEH000543.

[3] Math. encyclopedia. Moscow, 1979, vol. 2.

[4] Zaremba S. Sur le principe de Dirichlet. Acta Math., 1911, vol. 34, pp. 293-316. DOI: 10.1007/BF02393130.

[5] Anikiev A. N. Plane domains with special cone condition. Russian Mathematics, 2014, vol. 58, no. 2, pp. 62-63. DOI: 10.3103/S1066369X14020108.

[6] Liczberski P., Starkov V. V. Domains in Rn with conical accessible boundary. J. Math. Anal. Appl., 2013, vol. 408, no. 2, pp. 547-560.

[7] Liczberski P., Starkov V. V. Planar a-angularly starlike domains, a-angularly starlike functions and their generalizations to multidimensional case. International Conference "60 Years of analytic functions in Lublin", 2012, pp. 117-125.

The work is received on July 07, 2014.

Petrozavodsk State University,

Lenin Avenue, 33, 185910 Petrozavodsk, Russia.

E-mail: anikiev_a@mail.ru