Probl. Anal. Issues Anal. Vol. 4(22), No. 2, 2015, pp. 3-11
DOI: 10.15393/j3.art.2015.2950
3
UDC 517.51, 517.517
K. F. Amqzqva
NECESSARY AND SUFFICIENT CONDITIONS OF (a, P)-ACCESSIBILITY OF DOMAIN IN NONSMOOTH CASE
Abstract. In [1], [2] (a, 3)-accessible domains in C were defined and investigated, and the criterion of (a, 3)-accessibility in a smooth case was obtained. (a, 3)-accessible domains are starlike, Y-accessible (7 = min{a; 3}), and satisfy the so-called "cone condition" (i. e. the domains are conically accessible from the interior), which is important for applications, such as the theory of integral representations of functions, imbedding theorems, the questions of the boundary behavior of functions, the solvability of Dirichlet problem. In this paper the author obtains the necessary and some sufficient conditions of (a, 3)-accessibility of domain in nonsmooth case.
Keywords: a-accessible domain, (a, 3)-accessible domain, cone condition
2010 Mathematical Subject Classification: 52A30, 03E15
In [1, 2, 3] (a, 0)-accessible domains were studied as a generalization of a-accessible domains (see [4]).
Definition 1. [1, 2] Let a, 0 e [0,1), D C C, 0 e D. A domain D is called (a, 0)-accessible with respect to 0 if for each point p e dD there exists a number r = r(p) > 0 such that the cone
\ 0n an
K+(p, a, 0,r) = < z e C : —— < Arg(z — p) — Argp < —, \z — p| < r
is contained in C\D.
©Petrozavodsk State University, 2015 lM!iI!H|
In [2], it was shown that the domain D is (a, ^-accessible with respect to 0 if and only if, for each point p E dD, the unbounded cone
, mi fin . , N an
K+(p, a, fi) = < z E C : —— < Arg(z — p) — Argp < —
22
is contained in C\D.
If in the Definition^ a = fi, then we obtain the definition of a-accessible domain with respect to 0 (see [4]).
An (a, fi)-accessible domain is 7-accessible (7 = min{a, fi}), and starlike with respect to 0, as a consequence (see [4]). In the case a = fi = 0 the class of (0, 0)-accessible domains coincides with the class of starlike with respect to 0 domains (see [5]). Further, we assume that:
a) the function F(x) is defined and continuous in R2;
b) the open set D = {x E R2 : F(x) < 0} contains 0;
dF
c) there exist derivatives -— (p) at the points of the set level S = {p E
dl
E R2 : F(p) = 0} in all directions l E (K+ (p, a, a) — p) \{0}.
In [5] the following necessary condition of a-accessibility was obtained:
Theorem 1. [5] Let assumptions a), b), c) be satisfied. If D is an a-
dF
accessible domain for a certain a E [0; 1) then derivatives -— (p) are
dl
positive for any direction l E (K+(p,a,a) — p) \{0} and for any point p E S.
Regarding sufficiency the following theorems were received:
Theorem 2. [5] Let assumptions a), b), c) be satisfied. If for a certain dF
a E [0; 1) > 0 for any direction l E (K+ (p, a, a) — p) \{0} and for
any point p E dD then D is an a-accessible domain.
Theorem 3. [6] Let assumptions a), b) be satisfied and D be bounded set. If for a certain a E [0; 1), an arbitrarily small 5 > 0, and for a certain differentiable, strictly increasing numeric function ^ in R+ = {£ E R :
def
$ > 0}, such that lim = 0 = ^(0), we have
c^+o
f (p) > — F(p))) (—F(p)) cos an
for any direction l G (K+ (p, a, a) — p) \{0} and for any point p G D5 = = {x G D : p(x, dD) <5}, then D is an a-accessible domain.
In this article, analogs of theorems [1, 2 and 3 for (a,0)-accessible domain are obtained. In addition we have to replace the condition c) to the condition
dF
d) there exist derivatives (p) at the points p of the set level S in all
directions l G (K+(p, a, 0) — p) \{0}.
The following theorem gives a necessary condition of (a, 0)-accessibility.
Theorem 1'. Let assumptions a), b) and d) be satisfied. If D is an (a, 0)-accessible domain (with respect to 0) for certain a, 0 G [0; 1) then
dF
In (p) 2 0 (1)
for any p G S and for any direction l G (K+ (p, a, 0) — p) \{0}.
Proof. Let p G S. In the segment [0; p] there exists a point p0 G dD. Points p + pl, p > 0, l G (K+(p, a, 0) — p), lie in the cone K+(p, a, 0) Ç Ç K+(po,a,0). Since K+ (po,a,0) PlD = 0, we have K+ (p,a, 0) PlD = 0. So F(p + pl) > 0. Therefore
dF (p)= lim F(p + pl) — F(p) > 0.
dl p^o+ p
□
Theorem 2'. Let assumptions a), b) and d) be satisfied. If for certain a, 0 G [0; 1)
dF, ,
-Ql (p) > 0 (2)
for any direction l G (K+ (p, a, 0) — p) \{0} and for any point p G dD then D is an (a, 0)-accessible domain.
Proof. Note that by Theorem 2 the set D under the given conditions is a Y-accessible domain, where y = min{a,0}. It follows that D is starlike domain (see [4]).
Suppose that D is not an (a, 0)-accessible domain. Then there exists p G dD such that K+(p,a, 0,e) P D = 0 for any e > 0. So there exists a sequence of distinct points yn G D such that yn G K+ (p, a, 0) and
yn ^ p for n ^ œ. Notice, that these points yn do not lie on the ray {pt : t > 1}. Otherwise, by the condition of starlikeness of domain D, the segment [0, yn] C D, the point p G [0, yn] and the ray {pt : t > 1} lies in R2\D. This is a contradiction.
Draw a segment from each point yn to point 0. The segment [0, yn] intersects one of the sides of the cone K+(p, a, ft) at the point zn (but not the top of the cone). Choose the one of the two sides of the cone K+ (p, a, ft), which contains an infinite number of points zn. Since D is starlike (yn, 0 G
zn — p
G D), it follows that zn ^ p and zn G D. Let l0 = —"-r be a unit
|Zn — p|
vector. Note that l0 is independent of n. Let pn = |zn — p|. Since zn G D, then zn = p + pnl0 and F(zn) < 0. Therefore,
F (zn ) — F (p) = F (p + pnlo) — F (p) < 0
pn pn
and IF (p)= lim F(p + P"lo) — F (p) < 0.
dl0 o+ pn
dF
But this contradicts the condition > 0 for any direction l G
G (K+ (p,a,ft) — p) and for any point p G dD. Thus, D is an (a, ft)-accessible domain. □
Note that the necessary (Theorem 1') and sufficient (Theorem 2') conditions of (a, ft)-accessibility differ only by a sign of equality in inequalities (1) and (2). Below we represent another sufficient condition (an analog of Theorem 3), where in the right-hand side of an inequality there will be already negative expression, but it will be checked in the near-border zone D5, but not on the boundary of the domain D (compared to Theorem 2').
Theorem 3'. Let assumptions a), b) be satisfied and D be a bounded set. If for certain a, ft G [0; 1), an arbitrarily small 5 > 0, and for a certain differentiable, strictly increasing numeric function ^ in R+, such
def
that lim =0 = ^(0), derivatives
ç^+o
dF, , ______ „, _ 2 „_ ( l
(p) > — (—F(p)))^(—F(p)) ^-,pj (3)
for any direction l E (K+(p, a, fi) — p) \{0} and for any point p E D5, then D is (a, fi)-accessible domain.
Proof. For t > 0 denote Ft (x) = F(x) + and Dt = {x e
e R2 : Ft(x) < 0}. Suppose x e Dt. It follows from the inequality F(x) <
\x\2'
< — ^ \ J < 0 that x e D. Therefore Dt C D for each t > 0. Note
also that 0 e Dt, because 0 e D.
Let us show that there exists T > 0 such that for all 0 < t < T the level sets St = {x e Rn : Ft(x) = 0} lie in D5.
\x'2
At first, if x G St, then x = 0 and F(x) = — * ^t j < 0 for all
t > 0. Hence, St C D for all t > 0.
Secondly, for 0 < t1 < t2 the inclusion Dt2 C Dtl is valid. Indeed, if
|x|2'
x e Dt2, then F(x) + tf ^t2 —J < 0. Therefore, from a strict increasing
tf it follows that F (x) + tf ^ti ^^ < F (x) + tf ^2 < 0. Hence,
x e Dt 1.
Let us show that for a given 5 > 0 there exists a number T > 0 such that St C D5 for each t e (0, T). Suppose that it is not true. Then there exists a sequence yn e Stn such that the distance p(yn, S) >5 as tn ^ 0.
Since D is bounded, then from the sequence {yn} it is possible to select a converging subsequence (denote it the same way), such that yn ^ y0 e e R2.
( llyJ12
Passing to the limit as tn ^ 0 in the equality F(yn) = —tf I tn n
Jn I
'U
2
we see that F(y0) = — tf(0) = 0. Hence, y0 G S. However, p(yn,S) > 5. Therefore, p(y0 ,S) > 5. This is a contradiction. Thus, there exists a
number T > 0 such that St C D5 for each t G (0, T).
x2
Since t ^ = tf-1 (—F (x)) for all x G St, and St C D5 for 0 <t< T, 2
it follows from the condition of Theorem 3' that
f (x) > — n*-1 (—F(x)))^*-1(—F(x))( j x
V y (4)
= (¥) t (l,xN
for all x G St, 0 <t<T and l G (K+(x, a, 0 ) — x) \{0}.
Note that
f (4) = (t ^ , « =f (4)(* I). (5)
Relations (4) and (5) imply that for all t G (0, T) and x G St there exist derivatives in the directions l G (K+(x, a, ft) — x) \{0} such that
OFt^ dF, . dtf ( |x|2 ) 0
or (x) = « (x)+ « (' lr) > 0
Then by Theorem 2', the domain Dt is (a, ft)-accessible for each t G (0, T).
Let us show that for each x0 G D there exists such domain Dt, t G G (0, T), that xo G Dt.
Suppose xo G D, xo = 0 and F(xo) = — C < 0. Then, for all t such 2^-1(C )
that 0 < t < —-—TT2— = to, we have that xo G Dt. This is because
Ft(xo) = F(x„) + *(t J^^r) < — C + JXp) = — C + C = 0. Hence, xo E Dt for all t < t0. Therefore, D = (J Dt, where each
0<t<T
domain Dt is (a, fi)-accessible. In [2], it was shown that the union of (a, fi)-accessible domains is an (a, fi)-accessible domain. That proves the theorem. □
If in the conditions of the Theorem 3' we take = £n, n E N, then the right-hand side of (3) has the form
— *'(*-1(—F (p))) A (—F (p)) ( ||i =
= — n (*-1(—F(p)))n-1 A (—F(p))0 ( =
=—n (—F (p)) w(—F (p)) n (id = jfr-p
Then, in particular, we obtain the following sufficient condition of (a, fi)-accessibility of a domain
Corollary 1. Let assumptions a), b) be satisfied and D be a bounded set. If for certain a, fi E [0; 1), n E N, an arbitrarily small 5 > 0, and any point
p G D5 there exist derivatives in the directions l G (K+(p, a, 0) — p) \{0} such that
dF 2nF(p) f l \
dF(p) > -¡¡r( 1/|,p) (6)
for any point p G D5, then D is an (a, 0)--accessible domain. Note that inequality (6) is equivalent to
1 d (—F (p)) -n_ f J_
< 12 1 T7T,p
—F (p) dl |p|2 V |l|
and so (6) can be rewritten as
d ln(—F (p)) -n_ f J_
< 732 i Tn,p
dl |p|2 V |l|'
In the conditions of Theorem 3' let us consider the function n = =
= e-1/^. The function £ = tf-1 (n) = — ;— is an inverse function to n- In
ln n
this case the right part of (3) has the form
— F(p)))*-1 (—F(p)) (|l| =
1 _1 /AT/- T-f—^f^W - _1 , \ \ t l
1 <-f <p»> |-2 (—F (p))( Hi ,p) =
(tf-1(—F(p)))2 lpl2
- ( 1 \ ( l
= — (ln(—F(p)))2 eln(-FA (-,
2 \ ln(—F(p)) —-F (p) ln(—F (p)) f l
, p
, p =
|p|2
Thus, we obtain
Corollary 2. Let assumptions a), b) be satisfied and D be a bounded set. If for certain a, 0 G [0; 1) and an arbitrarily small 5 > 0, and for any point p G D5 there exist derivatives in the directions l G (K+(p, a, 0) — p) \{0} such that
d_F > —FMMzFM ( Lp) (7)
dl (p) > |p|2 V|l| /
for any point p G D5, then D is an (a, 0)-accessible domain.
Note that inequality (7) is equivalent to
d ln(— ln(—F(p))) —2 / J_
> i9 I T7T,p
dl |p|H |l|
Indeed, dividing inequality (7) by F(p), we get
1 d(—F(p)) < — 2ln(—F(p)) /1 p
—F (p) dl |p|
and so (7) can be rewritten as
d ln(—F(p)) — 2ln(—F(p)) /J_
< i__12 I 77T,p
dl |p |2
Dividing the last inequality by ln(—F(p)) and taking into account that ln(—F(p)) < 0 in D5 for sufficiently small 5 we obtain
1 d(— ln(—F(p))) > —2 /1 p
— ln(—F(p)) dl |p|2 V |l|
It follows that inequality (7) is equivalent to
d ln(— ln(—F(p))) —2 / J_
> 12 1 T7T,p
dl |p|2 V |l|'
Note also that if we introduce on the plane a complex structure everywhere in the text of the article, then the inner product ^can be
Re {I ■ p}
written as -r-,-, because
| l|
^p) = |p| cos(argp — argl) =
|p| cos(argp) cos(arg l) + |p| sin(argp) sin(arg l) =
t, T, l T t l Re {l ■ p}
= Re p Re — + Im p Im — =-—-.
|l| |l| |l|
Acknowledgment. This work was supported by RFBR 14-01-00510. The author would like to thank the referees for valuable comments on improving the paper.
References
[1] Anikiev A. N. Plane domains with .special cone condition. Russian Mathematics, 2014, vol. 58, no. 2, pp. 62-63. DOI: 10.3103/S1066369X14020108.
[2] Anikiev A. N. Plane domains with .special cone condition. Probl. Anal. Issues Anal., 2014, vol. 3 (21), no. 1, pp. 16-31. DOI: 10.15393/j3.art.2014.2609.
[3] Amozova K. F., Ganenkova E. G. About planar (a, ft)-accessible domains. Probl. Anal. Issues Anal., 2014, vol. 3 (21), no. 2, pp. 3-15. DOI: 10.15393/j3.art.2014.2689.
[4] Liczberski P., Starkov V. V. Domains in with conical accessible boundary. J. Math. Anal. Appl., 2013, vol. 408, no. 2, pp. 547-560. DOI: 10.1016/j.jmaa.2013.06.029.
[5] Amozova K. F., Starkov V. V. a-accessible domains, a nonsmooth case. Izv. Sarat. Univ. N. S. Ser. Math. Mech. Inform., 2013, vol. 13, iss. 3, pp. 3-8. (in Russian).
[6] Amozova K. F. Sufficient conditions of a-accessibility of domain in nons-mooth case. Probl. Anal. Issues Anal., 2013, vol. 2 (20), no. 1, pp. 3-13. DOI: 10.15393/j3.art.2013.2321 (in Russian).
Received September 1, 2015. In revised form, November 20, 2015.
Petrozavodsk State University
33, Lenina st., 185910 Petrozavodsk, Russia
E-mail: [email protected]