BOHR PHENOMENON FOR THE SPECIAL FAMILY OF ANALYTIC FUNCTIONS AND HARMONIC MAPPINGS Текст научной статьи по специальности «Математика»

Probl. Anal. Issues Anal. Vol. 9 (27), No3, 2020, pp. 3-13

3

DOI: 10.15393/j3.art.2020.7990 UDC 517.53, 517.54, 517.57

S. A. Alkhaleefah

BOHR PHENOMENON FOR THE SPECIAL FAMILY OF ANALYTIC FUNCTIONS AND HARMONIC MAPPINGS

Abstract. In this paper we obtain the sharp Bohr radius for a family of bounded analytic functions B' and for the family of sense-preserving K-quasiconformal harmonic mappings of the form f = h + g, where h e B'.

Key words: Bohr inequality, analytic functions, harmonic mappings, sense-preserving K-quasiconformal mappings 2010 Mathematical Subject Classification: Primary: 30A10, 30B10; 30C62, 30H05, 31A05, 41A58; Secondary: 30C75, 40A30

1. Introduction. Let D = {z e C : |z| < 1} denote the open unit disk, and let B denote the set of all bounded analytic functions defined on D, such that | f (z)| ^ 1 for all z e D. Then the classical inequality of Bohr [8] is the following:

ro

Theorem A. Let f e B, with Taylor series f (z) = Y1 anZn. Then we

n=0

have

ro

^ |an|rn ^ 1 for all r ^ ^, (1)

n=03

where r = |z|.

Actually, Herald Bohr obtained in 1914 inequality (1) for 0 ^ r ^ 1/6. The sharp version of the theorem in the above form was proved independently by M. Riesz, I. Schur, and F. Wiener. Other proofs were also given in [19]. Boas and Khavinson [7], and Aizenberg [3], [2] have extended the inequality to several complex variables. In [16], Kayumov et al. investigated the Bohr radius for complex-valued locally univalent harmonic mappings. Inspired by the work of [16], Evdoridis et al. [10] established several improved versions of Bohr's inequality for harmonic

© Petrozavodsk State University, 2020

mappings. In [13], Kayumov and Ponnusamy discussed the Bohr radius for bounded symmetric functions and settled, in particular, the conjecture of Ali et al. [4] on the Bohr radius for the class of bounded odd functions. In [5], Alkhaleefah et al. discussed Bohr's inequality with a fixed zero coefficient and introduced the study of the Bohr phenomenon for quasisubordination family of functions. We refer the reader to the recent survey by Abu Muhanna, et al. [1], and the references therein. More information about generalizations and extensions of Bohr's result can be found in [18], [19] and in the following recent papers [4], [5], [10], [12-17]. Moreover, in [20], [21], the authors considered the problem of determining the Bohr radius in an improved form, for functions having multiple zeros at the origin. For the point of view of function spaces, we refer to [6], [9] and the references therein. Bohr's phenomenon has been investigated in different contexts, e.g., holomorphic functions of several complex variables. Since this paper deals with the classical case only, we pay attention to and cite references related to this setting.

One of important proofs of Bohr's inequality was given by Wiener, (see also [18, p. 495] and [11, p. 162]), and the following coefficient inequality plays the key role.

<x

Theorem B. [8] Suppose that f e B, with Taylor series f (z) = Y1 anzn.

n=0

Then

K\ ^ 1 - |ao|2 for all n ^ 1. (2)

Note that, for all \a0\ ^ 1, we have 1 — \a0\ ^ 1 — \a0\2; so, we introduce the following subset B' of the set B:

Definition 1.

B' = {f : f e B with \an\ ^ 1 — K\ for all n E N}. (3)

There is another good reason to consider the subfamily B'. In 2004, Aizenberg and Vidras (see [3, p. 736]) considered a subclass of B and proved the following theorem:

Theorem C. Let f e B, such that the Taylor coefficients amn = 0 for a given m > 1 and all n ^ 1. Then

\an\ ^ 1 — \a0\ for all n ^ 1.

In this paper, we investigate and prove Bohr's inequality for functions f E B' and, also, for sense-preserving K-quasiconformal harmonic mappings of the form f = h + g, such that \g'(z)\ ^ k\h'(z)\ in the unit disc, for k = f-! E [0,1] and h E B'.

2. Bohr radius for a special family of analytic functions. In

this section, we find the Bohr radius for a family of bounded analytic functions B' and show its sharpness.

ro

Theorem 1. Let f (z) = E anZn E B'. Then

n=0

ro

^ \an\rn ^ 1 for all r ^ -, (4)

n=0

where r = \z\, and the constant 1/2 cannot be improved.

Proof. Suppose that f E B', \a0\ = a and \z\ = r. By the assumptions,

^ |a„|rra = a + ^ |ara|rn ^ a + ¿(1 - a)rn = a + , (5)

n=0 n=1 n=1

which is, clearly, less than or equal to 1, provided that

r1

^ 1, i. e. r ^ -,

1 — r 2

and the first part of the theorem is completed.

To show that the constant 2 is the best possible, let a = \a0\ E (0,1) and consider the function

/1 i 2\ ro

f (z) = a — ^-a + a)z = a — (1 — a) £ a"-1zn, z E D.

n=l

A simple calculation shows that

\f (z)\ ^ \f(ei0)\ ^ 1 for 0 ^ 0 ^ 2n,

whenever

2a cos 0 — (2 — a + a2) ^ 0,

which holds for all a e [0,1] and d e [0, 2n]. Thus, |f (z)| ^ 1 for all z e D. As for the function above, note that

|an| = (1 - a)an-1 ^ 1 - a, for all n ^ 1.

So f e B'. Now, for this function, we have

ro ro

^ |an|rn = a + (1 - a) ^ an-1rn = a + (1 - a)-

| n| - -

1 ar

n=0 n=1

which is less than or equal to 1 if ^ 1, i.e., if r ^ . By making a ^ 1, the proof of the theorem is completed. □

Remark 1. By substituting |a0|2 instead of |a0| in the left-hand side of

(5), we get

W2 + £|a,,r < 1 foraU r < =2+1001

Also, observe that

1 + |a0 | 1

lnf -:—- = -;

0<|«o|<i 2 + |a0| 2

however, the sharpness is still an open problem. Moreover, if we substitute |a0|m instead of |a0| in (5) for a fixed m e N\{1}, we get

m— 1

ro 1 I |m 1+ E Mfc

1 - | a0 | fc=1

|a0|m + V |an|rn ^ 1 for all r ^ -, 0|, , t

1 01 ^ n| 2 - |a0| - |a0|m m—1

n=1

2 + E Mk fc=1

Again, we see that

m— 1

1+ E Mfc • c fc=1 1 mi ---= -.

0<|ao|<1 m-\ Ifc 2

2 + E Mfc

fc=1

and, thus, the sharp Bohr radius remains open in this case, too.

N

E

n=0

Corollary 1. Suppose that f e B, and let SN(f, z) = E anzn denote the partial sum off, such that

|Sn(f, z)| ^ 1 for all N e N.

r

Then

^ |a„|rra ^ 1 for all r ^ -, (6)

n=0

Proof. The proof of this corollary follows from a result of Paulsen et al. [18]. For the reader's convenience, we include it in the form of a proposition below. □

N

Proposition 1. [18] For any complex polynomial p(z) = E anzn, with

n=0

|p(z)| ^ 1 for all |z| < 1 and |aN| = 0, the following inequality holds:

|«n| ^ 1 - K|. (7)

Example. Consider the function

f (z) = a - (1 - a9+ a)z = a - (1 - a)z V a2(n-1)zra-1, z e D, 1 — a2z

n=1

where a e [0,1]. This function f belongs to the class B, because |f(z)| ^ |f(e^)| ^ 1 for all 0 ^ Q ^ 2n,

if

2(a2 + a - 1) cosQ - (1 + a)(2 - 2a + a3) ^ 0,

which holds for all a e [0,1] and Q e [0, 2n]. The partial sum of this function is

N

Sn(f, z) = a - (1 - a)z ^ a2(n-1)zn-1 =

n=1

a - (1 - a + a3)z + a2N(1 - a)zN+1

1 - a2z

Note that the inequality |SN(f, ei0)| ^ 1 holds for N ^ 1 and Q e [0, 2n], since the following inequality holds:

a - (1 - a + a3)ei0 + a2N(1 - a)ei(N^ 1 - aV*; (8)

this means that

2(a2 + a - 1) cos 6 - (1 + a)(2 - 2a + a3) + a4N-1 (1 - a) +

+ 2a2N-1 (a cos ((N + 1)6) - (1 - a + a3) cos(N6)) ^ 0. (9)

Let (a, 6) be the expression in the left-hand side of the last inequality in (9). Note that a2 + a - 1 ^ 0 for a ^ £(V5 - 1) ^ 0.618034 and 1 - a + a3 ^ 0. Also note that when cos 6 ^ 1 and cos N6 ^ -1, we get cos(N + 1)6 ^ -1.

Now, using the observations above for a ^ 2(v^ - 1) and a ^ 1, we conclude the following:

(a,6) ^ 2(a2 + a - 1) - (1 + a)(2 - 2a + a3) + (1 - a) + +2(-a + (1 - a + a3)) = = -1 - 3a + 4a2 + a3 - a4,

which is less than or equal to zero for all 2 (v^5 - 1) ^ a ^ 1, so that (a, 6) ^ 0 for all 2 (v^ - 1) ^ a ^ 1; this means that inequality (9) holds for N ^ 1, 6 e [0, 2n] and a e [2(V5 - 1), 1]. Thus, |Sn(f, z)| ^ 1 holds for all N ^ 1, z e D, and a e [^2^5 - 1), 1].

Also, 2

ro

a +(1 - a)rVa2(fc-V-1 = a +(1 - a)—T— ^ 1 for all r ^ --,

^ 1 - a2r 1 + a2

n=1

and, thus,

inf = 0.5.

2 (^5-1)<a<1 1 + a2

ro

Corollary 2. Suppose that f e B with the Taylor series f (z) = anzn,

n=0

where amn = 0 for all n ^ 1 and a given m > 1. Then

ro

^ |an|rn ^ 1 for all r ^ -. (10)

n=0 2

Proof. We refer to Theorems 1 and C. □

3. Bohr radius for a special family of harmonic mappings. In

this section, we find the Bohr radius for the family of sense-preserving K-quasiconformal harmonic mappings of the form f = h + g, where h e B', and show its sharpness.

<x <x

Theorem 2. Suppose that f (z) = h(z) + g(z) = E anzn + E bnzn is

n=0 n=1

a harmonic mapping of the disk D, such that |g'(z)| ^ k|h'(z)| for some -k,

k e [0,1], and K = 1+k, where h e B'. Then

„n

y|a„|rn + y |bn|rn ^ 1 for all r ^

n=0 n=1

r ^ rK

K + 1 3K + 1

where r = |z|, and the constant rK cannot be improved.

To prove this theorem, we need the following Lemma proved in [16].

Lemma. Let h(z) = E anzn and g(z) = E bnzn be two analytic func-

n=0 n=0

tions defined in D, such that

|g'(z)| ^ k|h'(z)|, in D and for some k e [0,1]. Then

y |bn|2rn ^ k2 y |an|2rn for all |z| = r < 1.

n=1

n=1

Now we return to the proof of Theorem 2.

Proof. Consider the harmonic function f = h + g, such that |g'(z)| ^ k|h'(z)| in D, where k e [0,1) and h e B'. By Lemma and the Cauchy-Schwarz inequality,

y |bn|rn ^

n=1

\

y |bn|

n| 2rn

n=1

\

y rn ^

n=1

\

k2 y |an|2r

2n

n=1

\

y rn ^

n=1

\

k2(1 - |ao|)2 y rn

n=1

\

y rn = k(1 - |a0|) y r

n=1

n=1

Let m —y to to get

y |bn n=1

|bn|rn ^ k(1 -|ao|) y rn = k(1 -|ao|)

n=1

1r

r

so that

ro ro

|ao| + ^ |an|rn + ^ |bn|rn ^ |ao| + (1 + k)(1 - M)^, (12)

n=1 n=1

which is less than or equal to 1, provided that

r1

(1 + k)-^ 1, i.e., r ^ --.

V ;1 - r ' ' 2 + k

Substituting k = (K - 1)/(K + 1), we get

ro ro K + 1

]Clan|rn + J] |bn|rn ^ 1 for all r ^ 3K+T

n=0 n=1

This proves the first part of the theorem.

To show that the radius 3K+1 is the best possible, let a = |a0| e (0,1) and consider the function

(1 - a)z (1 - a)z

f (z) = h(z)+ g(z) = a ----k

1 az 1 az

= a - (1 - a) J] an-1zn - k(1 - a) J] an-1zn, z e D,

n=1 n=1

where k = (K - 1)/(K + 1). Then it is very simple to see that

00 00

J] |an|rn + |bn|rn = a + (1 - a) J] an-1rn + k(1 - a) J] an-1rn =

n=0 n=1 n=1 n=1

a - a2r + (1 + k)(1 - a)r 1 - ar

which is bigger than or equal to 1 if and only if

1 K + 1

r is -=-.

1 + a + k 2K + a(K + 1)

Since a could be chosen arbitrarily close to 1, we obtain that the radius 3KK+11 cannot be improved. The proof of the theorem is completed. □

Remark 2. By Substituting |a012 instead of |a01 in the left-hand side of (12), we get

00 00

|a012 + J] |an|rn + J] |bn|rn ^ 1

n=1 n=1

for all r with

1 — | ao |2 1 + |ao|

r ^

2 + k - (1 + fc)k|-|ao|2 2 + k + |ac|' We observe that

1 + |ao| 1 K + 1

r = mi -=-=-,

o<M<i 2 + k + |a0| 2 + k 3K + 1

which is an upper bound for the Bohr radius, but the sharpness case is still an open problem.

Moreover, if we substitute |a0|m instead of |a0| in (12) for a fixed m G N\{1}, we get

|ao|m + y |an|rn + y |bn|rn ^ 1,

n=1 n=1

for all

m—1

1 + E |ao|j 1 -|ao|m j=i | |

r ^

2 + k — (1 + k)|ao| — |ao|m 2 + , + m-11 j

2 + k + |ao |j j=i

Again, we see that

m—1

1+ E |aoj 1 K + -,

. , j=i 1 K + 1

ml

o<M<i , m—\ .. 2 + k 3K + 1 2 + k + E |ao|j j=i

which is an upper bound for the Bohr radius in this case, too. And also, the sharpness case is still an open problem.

Acknowledgment. I would like to thank my supervisor Prof. I. R. Kayu-mov for useful suggestions and valuable comments and would like to thank Prof. S. Ponnusamy for fruitful discussions and suggestions.

I am very grateful to the referees for the careful reading of the paper and for their comments and detailed suggestions, which helped me to improve the manuscript considerably.

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Received February 28, 2020. In revised form, August 04, 2020. Accepted August 11, 2020. Published online August 19, 2020.

Seraj A. Alkhaleefah Kazan Federal University, 420008, 18, ul. Kremlevskaya, Kazan, Russia

E-mail: s.alkhaleefah@gmail.com