Научная статья на тему 'MULTIVALENT α CONVEX HARMONIC MAPPINGS'

MULTIVALENT α CONVEX HARMONIC MAPPINGS Текст научной статьи по специальности «Математика»

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Аннотация научной статьи по математике, автор научной работы — Ganczar A.

In this paper we give coefficient conditions for complex-valued harmonic functions that are ultivalent, sense-preserving and α-convex. We determine the extreme points, distortion and covering theorems for these mappings.

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Текст научной работы на тему «MULTIVALENT α CONVEX HARMONIC MAPPINGS»

Труды Петрозаводского государственного университета

Серия “Математика” Выпуск 9, 2002

YJIK 517.54

MULTIVALENT a — CONVEX HARMONIC MAPPINGS

A. Ganczar

In this paper we give coefficient conditions for complex-valued harmonic functions that are multivalent, sense-preserving and a -convex. We determine the extreme points, distortion and covering theorems for these mappings.

§ 1. Introduction

A continuous function / = u + iv is said to be a complex-valued harmonic function in a domain 0 C C if both u and v are real harmonic in 0. In any simply connected domain D such a mapping / can be written in the form

f(z) = h{z) + g(z), (1)

where h(z) and g(z) are analytic in D (see [1]). The Jacobian of / is then given by

Jf(z) = \h'(z)\2-\g'{z)\2.

A harmonic function / of the form (1) will be called sense-preserving at a point zo E D if hf(z) ^ 0 and the second dilatation function cj(z) = is analytic at zo (possibly with a removable singularity), and |cl;(^o)| < 1-Note that if Jf(z) > 0 for each z E D then / is sense-preserving in D.

Suppose that f(zo) = 0 at some zq E D where / is sense-preserving, then we may express the analytic functions h and g as

oo oo

h(z) = a0 + ^2 ttn(z ~ zo)ny 9(z) = bo + ^2 bn(z ~ zo)n■

71=1 71=1

© A. Ganczar, 2002

We see at once that bo = — ao, because f(zo) = 0. Since h'{z) / 0 in D, it follows that some an must be nonzero. We denote the first such a coefficient by am. Then we have bn = 0 for 1 < n < m and |6m| < |am|, because the second dilatation cj(z) is analytic at the point zo and |^(^o)| < 1- In this case we say that f(z) has a zero of order m at zo Є D.

Let D be the open unit disk A = {z : \z\ < 1}. We can certainly assume that an = bn = 0 for 0 < n < m and am = 1. Denote by *S#(m) the set of all m-valent harmonic functions / = h + g of the form

oo oo

f(z) =Zm + Y, an+m-lZn+m~1 + Y, bn+ra-lZn+™~\ (2)

71 = 2 71=1

that are sense-preserving in A. Since |6m| < |am|, we see that |6m| < 1. By the argument principle for harmonic functions [2] and the above arguments, if Jf(z) > 0 in A \ {0} then / of the form (2) belongs to the class *S#(m)-Note that *S#(1) is the familiar class Sh of harmonic univalent and sense-preserving functions (see [1]).

We say that / Є Бн(т) is harmonic convex of order a (e.g. see [4]),

0 < a < 1 in A if

w {arg a “• (3)

for each z, \z\ = r < 1.

Let us denote by Кн(тп,а) the subclass of Sh(тп) consisting of functions / that are convex of order a. In particular, we will denote by Кн{тп) the class Кн(т,0).

We further denote by ТКн{т, a) the subclass of Кн(т, a) consisting of functions / = h + g so that h and g are of the form

oo oo

h(z)=zm-J2K+m-i\zn+m~\ g(z) = -J2\bn+m-i\zn+m-1. (4)

71 = 2 71=1

§ 2. Coefficient conditions

In this section we proved a sufficient coefficient condition for the class Кн(т,а).It is also shown that those condition is necessary when / Є ТКн(т,а).Those results are a generalization of the theorems for the classes lf#(l,a) and Tlf#(l,a) of convex univalent harmonic mappings of order a and convex univalent harmonic mappings of order a with negative coefficients, respectively (see [3]).

Theorem 1. Let f(z) = h(z) + g(z) be of the form (2). If

E°° n + ra — 1/n + ra — a — 1 n + ra + a — 1 .

~ \ ~ ~ |^n+ra —1| H- ~ “ \bn-\-m —1| J < 2,

n=l

ra

m — a

m — a

where am = 1 and rn > 1. then f(z) G

(5)

Proof. We first prove that the coefficient condition (5) is sufficient for the function

f(z) =Zm + J2 an+m-lZn+m~1 + 5] bn+m-lZ

71 + 771 — 1

71 = 2

71=1

to be sense-preserving in A.

Let us first observe that for each pair of numbers ra, n E N and 0 < a < 1 we have

m — a

m — a

(6)

Set h{z) = zm + Y, an+m-izn+rn 1 and 5(2;) = £ &n+m-i£n+m

a=2

i=l

Obviously, the second dilatation function a;(z) = has the removable singularity at zo = 0.

From (5) and (6) we conclude that for 0 < \z\ < 1 we have

\h'(z)\ > m\z\m 1 - ^T(n + to - l)|an+m-i||^

71+771 — 2 ____

71 = 2

o° _ 1

= TO|z|m_1 1-^-------------™------|an+m_i||,z|n_1

71 = 2

ra

>

771 — 1

o° 1

Ti + ra — 1. .

1 — ------------|an+m-l|

71 = 2

ra

>

71 = 2

oo

(n + ra — l)(n + ra — a — 1) ra(ra — a)

(ra + n — l)(n + ra + a — 1)

| ^71 + 771 — 1 | j

>

ra(ra — a)

| ^71 + 771 — 1 | j

>

>

>

>m\zr~'\YJn + m~ \bn+m-A

L ^ m

n= 1

oo

> ^(n + m- l)|6n+m_i| \z\n+m~2 >

= W(z)\.

n= 1 oo

^2(n + TO - l)bn+m-izn+m~2

n= 1

Therefore the harmonic function / = h + g of the form (2) is sense-preserving in A.

We next show that / E lf#(ra, a). By the definition of the class KniTn^a) , it remains to prove that

¿{arg (¿/(re"’)) }=R*

or equivalently if

'zh'(z) + z2h"(z) + zg'(z) + z2gn(z)'

zhf(z) — zg'(z)

> a,

Re r-zfe'(-z) + ¿2fr"(z) + zg'(z) + z2ff"(>) _ 1 > 0

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I- zh'(z) — zg'{z) -I —

for each z = rei6>, \z\ < 1. For z E A we have

rzft'(z) + z2h"(z) + zg'(z) + z2g"(z) i

Re ------------------------------- i--------------------— cm —

I- zh'{z) — zg'(z) -I

= Re [((m — a)zm + U m—-(n + ra — a — l)an+m_i^n+m 1 L rn

OO 1

ETl + Til - 1 / x-----------1--7-

-----------(n + m + a - l)bn+m-1zn+m-1

m

71=1

00 I 1 00 I 1

n + m-1 v^n + TO-1-

m It' T HI ~ -L ^ i 11 “T HI ~ i-r.----------:----

?m + V-----------------an+m-iz 1 - V--------------------------6n+m_iz”+™-1

m rn

71 = 2 71=1

Let z = rei6>, 0 < r < 1, then by the above

I {»g(|/(re‘»))} - o = (m - a) Re [ J +

where

°° _L _ 1

p(reie) = (V -------------(n - l)on+m_1r"-1e<(n-1)9 +

m

n=2

+ T n (n + 2m - l)6n+m_1rn-1e-i(n+2m-1)°) /

m /

n=l

o° ^

/(2(m - a) + V'' -——-----------------(n + 2m - 2a - l)an+TO_irn_1e^"_1^+

/ z—' TO

n=2

00 _i_ _ 1

+ ^ ------(n + 2a -

z—* m

m

n= 1

as it is easy to check.

The proof will be complete if we can show that |p(rei6,)| < 1. We have

\p(rel(>) \ < (53 n + m—-(n - 1) |an+m—i|r" * +

I I m

n=2

+ 53 ~—--------------+ 2to - 1) |&n+m_i| rn_1) /

m /

71=1

0° ^

/(4(to - a) - 53 ~— ------------------(n + 2m - 2a - 1) |an+m_i | rn_1 -

/ ' TO

71=1

00 1 1

En m — 1 1 „ix

--------------(n + 2a - 1) \bn+m-i r )<

m

71=1

00 I 1 00 I 1

/ . 77 + 771 — 1 . , v-^v 77 + 771—1. \ i j i\

^ ( / [P' 1) 71 + 771 — 1 + / j (77 + 2777/ 1) ^71 + 771 — 1 ) ,

771 771 /

71 = 2 71=1

00 1 1

I f A f \ X----^ ^ "I” ^ - I / r» r» -1 \ I I

/ (4(771 - a) - > -------------(77 + 2ra - 2a - 1) an+m_i -

/ 771

71=1

E°° 77 + 771 - 1 , _l7 R{rfl)

n=1 to (n + " 1 |fe"+m-l|) - <2(TO,a) - ’

which is due to the fact that

Q(ra, a) — R(m) =

z . r v—> Tl Til — \ ,Tl Til — OL — 1. .

— 2(ra ex) 2 y ^ — y — - |cfcn_|_m_i| H-

rn

n= 1

n + m + a — 1

H------------------

m — a

by (5).

From this we conclude that Re conclusion. □

\K+m-l\)] > 0,

1 + p(relB)

1 — p(relB)

> 0, which is the desired

The restrictions in the above Theorem placed on the moduli of the coefficients enable us to conclude for arbitrary rotations of the coefficients of / that the resulting functions would still be in the class lf#(ra, a). Now we show that such coefficient bounds can’t be improved.

Theorem 2. Let f(z) = h(z) + g(z) be of the form (4). Then f G TKh(tti, a) if and only if

E°° n + m — 1 (n + m — a — 1 n + m + a — 1 A

~ \ ~ ~ |^n+ra —1| ~ “ \bn-\-m —1| ] < 2,

m

m — OL

n— 1

where am = 1 and m > 1.

m — a

(?)

Proof. In view of Theorem 1, we need only show that / ^ Tlf#(ra,a) if the coefficient condition (7) does not hold.We examine the required condition (3) for / = h + g £ TKh{tïi, a). By the above this is equivalent to

Re

zh'(z) + z2h"(z) + zg'(z) + z2gn(z)

— a

zh'(z) — zg'{z)

\if \ m (ra+ ra - l)(ra + ra - a - 1), . n+m_i

= Re ((m — a)z — >. ------------------------------------------------------

L ' m

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n=2

-E

(n + m — l)(n + m + a — 1)

71=1

m

- V ————— |«n+m-l|2n+m-1 + y^ |6

\bn+m-l\zn+m~1) j (zm — . n + m — 1-

71 = 2

71 + 771 — 1 |

71=1

m

_ ^71 + 771 —1 ^

> 0.

The above condition must hold for all z Є A.Upon choosing the values of z on positive real axis and such that 0 < ^ = r < 1 we must have

((m — a) — le 53----------(n + m - a - 1) |an+m_i | rn

n=2

o° 1

ETl ~\~ TYl — 1 . x і і „_і ч /

----------(n + m + a - 1) \bn+m-i\r )

m /

n= 1

oo

/ /-1 / Th TYl — 1 І Гї —1 Я + Ш — 1 і ч . _

/(1- < ^ ------------- a„+m_i r + \-----------------\bn+m-l \r ) > 0.

/ Ш Ш

n = 2 71=1

If the condition (7) does not hold then the numerator in (2) is negative for r sufficiently close to l.Thus there exists ro Є (0,1) for which the quotient in (2) is negative, and we arrive at a contradiction. □

Next theorem shows that class ТКн(т,а) is closed under forming convex combinations.

oo

Theorem 3. If fi(z) є TKH(m,a) for і = 1,2,..., and if =

2=1

OO

1, 0 < Ai < 1, theng(z) = ^2 xifi(z) is a member of the class TKH(m, a).

2=1

Proof. Since

oo oo

fi(z) = zm-Y K+^l/+”-1 - 53 Ibi+m_1\^+^ e TKH(m,a).

71 = 2 71=1

Theorem 2 shows that for each і Є N we have

oo

ЕП + ТП — 1 іь -\- m — ul — i . • . іь -\- m -r u: — і, 7,

--------------------------------K+m-l H-----------------------------&

m m — rv ' rn — rv

a=2

71 + 771 — 1 I

< 2. (8)

OO oo

For ^Ai = l, 0 < A* < 1, the convex combination g(z) = ^ Aifi(z) is

71=1 2=1

of the form

oo oo oo oo

g(z) = zm- 53(53XiK^Dz^-1 - 53(53

71 = 2 2=1 71=1 2=1

We check at once that

En + m- lrn + m- a-1 /V-^ л . • l4

—i“- [ Ш-v g B +

n=2

n + m + a — 1,

2=1

+ -

¿=i

< i,

which is clear from (8).Theorem 2 implies that g{z) Є ТКн(т,а). □

§ 3. Distortion bounds and extreme points

We now give the distortion bounds for functions in ТКн{т, a), which yield a covering result for this class.

Theorem 4. If f є ТКн(т,а), then and

In) \f(z)\>(l-\b I)rm т(т~а)-т(т + а)\ь™\гш+і

(II) \f(z)\>(l \t>m\)r (m + l)(m — a + 1)

where \z\ = r < 1.

Equalities are rendered by the function

№) = - |6m|^ +

(m + 1)(ш — а + 1)

and its rotations.

Proof.

We shall justify the (z) right hand inequality only. For \z\ = r, we have

- 53 |an+m_!| + 53 |bn+m_!| і

n=2

71=1

<

< r"* + 53 |an+m_].| rra+m_1 + 53 \Ьп+т-г\ r

71 + 771 — 1

rm+1

|^n+m-l | ~r | ^n+ra —11) ‘

n=2

Theorem 1 now shows that

m(m-a)+m(m+a) \bm\ + (m + l)(m-a + l) 53(lan+m-i| + |&n+m-i|) <

n=2

oo

< 53 (n + TO “ !) [ l“n+m-i I (n + m-a- 1) + |6n+m_i I (n + m + a-1)] <

n= 1

< 2rn(rn — a),

and hence

E°° m(m - a) - m(m + a) \bm\

n=2 Pn+m-il) < (m + l)(m — a +1) ’

which establishes the formula. □

Remark. Bounds given in Theorem 4 also are valid for f E Kn(rn^ a) if the coefficient condition (5) is satisfied.

Letting r —> 1~ in the left hand inequality in Theorem 4 we obtain a covering result for the class Tlf#(ra, a).

Corolary. If f e TKH(m,a), then

, : < 2m — Q + 1 - |6m| {2m + 1)(1 - a) j c

l (ra + 1 )(m — a + 1) J

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In particular, if / E TK^(m,0) = TKh then ju; : \w\ < (1 —

\bm\)} C/(A).

For any compact family, the maximum or minimum of the real part of any continuous linear functional occurs at one of the extreme points of the closed convex hull.Since TKnim, a) is a convex family, we will use the necessary and sufficient condition of Theorem 2 to determine the extreme points.

Theorem 5. Let

hm(z) = Zm, htfn+m-^z) = Z™ - —------------------------ z«+™-1

(n + m - 1 )(n + m - a - 1)

n = 2, 3,... and

m(m — a)

Qn+m — l (z) — Z / 1 V _i_ i 1 A ^n_^m 1 ? 77- — 1,2,....

(n + m — l)(n + m + a — 1)

Then f = h + g E TKff(m,a) if and only if it can be expressed in the form

oo

/(*) — ^ ^ (z) + /^n+ra —l^n+ra —1 (^) J ,

n= 1

oo

where An+m_i ^ 0, /^n+m—l ^ 0 and (An_|_m_i H- /^n+m—i) — 1*

n=l

In particular, the extreme points of Tlf#(ra,a) are {/in+m_i(z)} and

{^n+ra — 1 (^)}'

Proof. Suppose that

oo

f(z) — ^ ^ \^n+m-lhn+m-l (z) + /^n+ra —l^n+ra —1 (-¿Oj — n=l

____m m(m °0 \ „n+m-1

_ / v i i -, \ 71-)-771 1 ^

(n + m — l)(n + m — a — 1)

E°° m(m — a) —:-----r

-------------------------------- II , -, 7n + 771-l

/ , -|\/ , , -jxPn+m — 1*

(n + m — l)(ra + n + a — 1)

71=1 V V 7

Then

E°° n + m- 1 rn + m- cc - 1 / m(m — a) \

m V rri — a \n + m — a — 1 n+m-1/

<2,

71=1

n + m + a — 1/ m(m — a)

m — a \ (n + m — l)(n + m + a — 1) ^n+m 1 and by Theorem 2 / G Tlf#(ra,a). Conversely, if

oo oo

/ = *"*- £ |on+m_1|«n+m-1 - 53 |6n+m_1|z"+m-1 € TKH(m, a),

71 = 2 71=1

then by Theorem 2 we have

f^n-\-m — 1 —

(n + m — 1 )(n + m + a — 1)

7 \ |^n+:

m(m — a)

oo

oo

l-L(An+ m — 1 m —1) •

n=2

71=1

required. □

Remark. If the co-analytic part of f = h + g E S*H{m, a) is is zero, i.e. the function g(z) is identically zero, then we have analogous properties of m-valent analytic convex functions of order a in the unit disk.

[1] Clunie J. and Sheil-Small T. Harmonic univalent functions// Ann. Acad. Sci. Fenn Ser. A.I Math. 1984. V. 9. P. 3-25.

[2] Duren P.L., Hengartner W.,Laugesen R.S. The Argument principle for harmonic functions / / Amer. Math. Monthly. 1996. V. 103. № 5. P. 411-415.

[3] Jahangiri J. Coefficient bounds and univalence criteria for harmonic functions with negative coefficients// Ann. Univ. Mariae Curie-Sklodowska. Sect. A. 1998. V. 52. P. 57-66.

[4] Sheil-Small T. Constants for planar harmonic mappings// J. London Math. Soc. 1990. V. 42. № 2. P. 237-248.

Instytut Matematyki UMCS,

20-031 Lublin, Poland

E-mail: [email protected]

References

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