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### Article

Kyungpook Mathematical Journal 2022; 62(2): 243-256

Published online June 30, 2022

### Bohr’s Phenomenon for Some Univalent Harmonic Functions

Chinu Singla∗, Sushma Gupta and Sukhjit Singh

Department of Mathematics, Sant Longowal Institute of Engineering and Technology, Longowal-148106, India
e-mail : chinusingla204@gmail.com, sushmagupta1@yahoo.com and sukhjit_d@yahoo.com

Received: May 21, 2021; Revised: October 6, 2021; Accepted: October 7, 2021

### Abstract

In 1914, Bohr proved that there is an r0(0,1) such that if a power series m=0cmzm is convergent in the open unit disc and | m=0cmzm|<1 then, m=0|cmzm|<1 for |z|<r0. The largest value of such r0 is called the Bohr radius. In this article, we find Bohr radius for some univalent harmonic mappings having different dilatations. We also compute the Bohr radius for functions that are convex in one direction.

Keywords: Bohr radius, harmonic univalent functions, convex in one direction

### 1. Introduction

The Bohr inequality, first introduced in 1914 by Harald Bohr in his seminal work [3] and subsequently improved independently by M. Riesz, I. Shur and F. Wiener, essentially states that if f(z)= m=0amzm is an analytic function in the open unit disc D={z:|z|<1} and |f(z)|<1 for all zD, then

m=0|am|rm1

for all zD with |z|=rr0=1/3 and 1/3 is the largest possible value of r0, called the Bohr radius. Inequalities of type (1.1) have become famous by the name Bohr inequalities and the problems of finding the largest possible values of r0 in different setups are now called Bohr radius problems. For a glimpse of the ongoing current research in this area we refer the reader to some recent articles, e.g. [1, 2, 5, 8, 9] and references therein.

In 2010, Abu Muhanna [8] investigated some Bohr radius problems using the concept of subordination. For two analytic functions f and g in D, g is said to be subordinate to f (written, gf) if there exists a function ψ analytic in D with ψ(0)=0 and |ψ(z)|<1 such that g=fψ. In particular, when f is univalent, then gf is equivalent to g(0)=f(0) and g(D)f(D). We shall denote by S(f), the class of all functions g subordinate to a fixed function f. A class of analytic (harmonic) functions in the unit disc D is said to possess classical Bohr's phenomenon if an inequalty of the type (1.1) is satisfied in |z|<r0, for some r0,0<r01. It is known (see [8]) that not all classes of functions have classical Bohr's phenomenon. So, Abu Muhanna [8] reformulated classical Bohr's phenomenon and proved the following result:

Theorem 1.1. If f(z)= m=0amzm is a univalent function and g(z)= m=0bmzmS(f), then

m=1|bm|rmd(f(0),f(D))

for all |z|=rr0=38=0.17157.., where d(f(0),f(D)) is the Euclidean distance between f(0) and f(D), the boundary of f(D). The value of r0 is sharp for f(z)=z/(1z)2, the Koebe function. Further, if f is convex univalent in D, then r0=1/3.

In the recent years, a number of research articles (for example see [2, 6, 7]) are published and many hidden facts of this subject are brought into broad daylight. In particular, Bhowmik and Das [2] successfully extended the Bohr inequalities of type (1.2) for certain harmonic functions. A complex valued function f(z)=u(x,y)+iv(x,y) of z=x+iyD is said to be harmonic if both u(x,y) and v(x,y) are real harmonic in D. It is known that such an f can be uniquely represented as f=h+g¯, where h and g are analytic functions in D with f(0)=h(0). It immediately follows from this representation that f is locally univalent and sense preserving whenever its Jacobian Jf, defined by Jf(z):=|h(z)|2|g(z)|2, satisfies Jf(z)>0 for all zD; or equivalently if h0 in D and the (second complex) dilatation wf of f, defined by wf(z)=g(z)/h(z), satisfies the condition |wf(z)|<1 in D. A harmonic function f=h+g¯ defined in D is said to be K-quasiconformal if its dilatation wf satisfies |wf|k,k=(K1)/(K+1)[0,1). In view of the work of Schaubroeck in [10], aforesaid definitions and notations for subordination of analytic functions can be extended to harmonic functions without any change. This lead Bhowmik and Das [2] to extend Theorem 1.1. as under:

Theorem 1.2. Let f(z)=h(z)+g(z)¯= m=0amzm+ m=1bmzm¯ be a sense preserving K-quasiconformal harmonic mapping defined in D such that h is univalent in D, and let f1(z)=h1(z)+g1(z)¯= m=0cmzm+ m=1dm zm¯S(f). Then

m=1|cm|rm+ m=1|dm|rmd(h(0),h(D))

for |z|=rr0=(5K+18K(3K+1)/(K+1). This result is sharp for the function p(z)=z/(1z)2+kz/(1z)2¯, where k=(K1)/(K+1). Moreover, if we take h to be convex univalent then the inequality in (1.3) holds for |z|=rr0=(K+1)/(5K+1). This result is again sharp for the function q(z)=z/(1z)+kz/(1z)¯.

In this article, our aim is to establish the Bohr's phenomenon and compute Bohr radius for some subclasses of univalent harmonic functions. We also propose to improvise Theorem 1.1. and 1.2. stated above.

We close this section by setting certain notations for subsequent use in this paper. We denote by SH, the class of univalent harmonic functions f normalized by the conditions f(0)=0 and fz(0)=1. In addition, if fz¯(0)=0 also, then the class is denoted by SH0. Further, KH0 is the usual subclass of SH0 consisting of convex functions. A domain Ω is said to be convex in the direction θ,0θ<π, if the intersection of the straight line through the origin and the point eiθ in the complex plane is connected or empty. A function f mapping the open unit disc D onto such a domain is called convex in direction 𝜃.

### 2. Main Results

We begin this section by stating following lemma which immediately follows from the work of Bhowmik and Das [1].

Lemma 2.1. Let f(z)= m=0amzm and g(z)= m=0bmzm be two analytic functions in D and gf. Then

m=0|bm|rm m=0|am|rm

for |z|=r1/3.

Using this lemma, we now improvise Theorem 1.1 by taking univalent analytic function in D as f(z)=z+ m=2amzm. Making use of well known De Brange's theorem: |am|m,m=2,3,..., and after some simple calculations, we easily get:

Theorem 2.2. If f(z)=z+ m=2amzm is a univalent analytic function in D and g(z)= m=1bmzmS(f), then

m=1|bm|rm1

for all |z|=r1/3.

In a similar manner, we restate Theorem 1.2. as under;

Theorem 2.3. Let f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=1bmzm¯ be a sense preserving K-quasiconformal harmonic mapping in D, such that h is analytic univalent in D. Then

m=1|am|rm+ m=1|bm|rm1,a1=1

for |z|=rr0=(2K+1K(3K+2))/(K+1) and it is sharp for p(z)=z/(1z)2+kz/(1z)2¯. If we take h to be convex univalent then the inequality in (2.2) holds for |z|=rr0=(K+1)/(3K+1) and it is sharp for p(z)=z/(1z)+kz/(1z)¯, where k=(K1)/(K+1). Further, let f1(z)=h1(z)+g1(z)¯= m=1cmzm+ m=1dm zm¯S(f). Then

m=1|cm|rm+ m=1|dm|rm1

for |z|=rr0=min(1/3,(2K+1K(3K+2))/(K+1)). If we take h to be convex univalent then the inequality in (2.3) holds for |z|=rr0=min(1/3,(K+1)/(3K+1)).

In next theorem, we establish Bohr's phenomenon for univalent harmonic functions f=h+g¯SH whose dilatation g'/h' is suitably chosen.

Theorem 2.4. Let f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=1bmzm¯ be a univalent and K-quasiconformal harmonic mapping in D, where h is analytic univalent in D and g(z)/h(z)=keiθzn,k=(K1)/(K+1)(0,1),n,θ. Then

m=1|am|rm+ m=1|bm|rm1,a1=1

for |z|=rr0, where r0 is the only root of the equation

(k+1)r(1r)22nkr(1r)kn2log(1r)=1

in (0,1) and this r0 is best possible one.

Letting k1 (equivalently, K) we obtain the following result.

Corollary 2.5. Let f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=1bmzm¯ be a univalent harmonic mapping in D, where h is analytic univalent in D and g(z)/h(z)=eiθzn,n,θ. Then

m=1|am|rm+ m=1|bm|rm1,a1=1

for |z|=rr0, where r0 is the only root in (0,1) of the equation ϕ(r)=0, where

ϕ(r)=2r(1r)22nr(1r)n2log(1r)1.

This r0 is the best possible one.

By plotting the graph of ϕ(r) w.r.t r for different values of n, we observe that there is only one root of ϕ(r) in (0,1) which is the Bohr radius for that value of n in the dilatation function. Figure 1 illustrates the case when n=3 and in the following table we have listed values of r0 computed for n=1,2,3 and 4.

nr0
10.3485...
20.3121...
30.1794...
40.0959...

Figure 1. ϕ(r) w.r.t r for n=3.

We observe that if n, then r00.

Lemma 2.1 and Theorem 2.4 together lead us to the following result for the subordination class S(f).

Corollary 2.6. Let f1(z)=h1(z)+g1(z)¯= m=1cmzm+ m=1dm zm¯S(f) where f is as defined in Theorem 2.4. Then

m=1|cm|rm+ m=1|dm|rm1

for |z|=rr1=min(1/3,r0), where r0 is same as obtained in Theorem 2.4.

Next theorem shows the existence of Bohr's phenomenon for fSH with dilatation wf=(a+z)/(1+az),a(1,1).

Theorem 2.7. Let f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=1bmzm¯ be a univalent harmonic mapping in D, where h is univalent in D and g(z)/h(z)=a+z1+az,a(1,1) Then

m=1|am|rm+ m=1|bm|rm1+|a|,a1=1

for |z|=rr0, where r0=0.2291... is a unique root lying in (0,1) of r33r2+5r1=0.

Remark 2.8. We observe that if we take g/h=az1az,a(1,1), in Theorem 2.7, then we obtain the same value of r0.

In the following theorem we establish Bohr's phenomenon for univalent harmonic functions convex in one direction.

Theorem 2.9. Let f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=1bmzm¯ be a harmonic mapping in D, where h is analytic univalent in D and h(z)+eiθg(z) is convex univalent in D for some θ. Then

m=1|am|rm+ m=1|bm|rm1,a1=1

for |z|=rr0=0.2192....

We can drop the condition of univalency of h in Theorem 2.9 if we take b1=0.

Let f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=2bmzm¯SH0 be a harmonic mapping in D, where h(z)+eiθg(z) is convex univalent in D for some θ. Then

m=1|am|rm+ m=2|bm|rm1,a1=1

for |z|=rr0=0.3134..., where r0 is a unique root in (0,1) of 4r39r2+12r3=0. This result is sharp for Koebe function K(z)=z1/2z2+1/6z3(1z)3+1/2z2+1/6z3(1z)3¯.

Our last theorem gives Bohr radius for convex univalent harmonic functions in SH0.

Theorem 2.11. Let f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=2bmzm¯KH0,zD. Then

m=1|am|rm+ m=2|bm|rm1,a1=1

for |z|=rr0=(35)/2=0.3819.... This value of r0 is sharp for L(z)=12z1z+z(1z)2+z1zz(1z)2¯.

### 3. Proof of Theorems

We begin this section by stating a lemma which is easy to prove.

Lemma 3.1. Let h(z)= m=0amzm and g(z)= m=0bmzm be two holomorphic functions in D such that h(z)=g(z). Then

m=0|am|rm= m=0|bm|rm

for all |z|=r<1.

Proof of Theorem 2.4. From g(z)=keiθznh(z), we get

m=1mbmzm1=keiθ m=1mamzn+m1,zD,

where a1=1 and on integrating we obtain

m=1bmzm=keiθ m=1mm+namzm+n,zD.

Now, applying Lemma 3.1, we get

m=1|bm|rm=k m=1mm+n|am|rm+n

for all |z|=r<1. Since h is analytic univalent in D and according to De Brange's theorem, |am|m,m=2,3,..., therefore, from (3.2), we have

m=1|am|rm+m=1|bm|rm=m=1|am|rm+km=1m m+n|am|rm+n        m=1mrm+km=1 m2 m+nrm+n        =m=1mrm+km=n+1 (mn)2mrm        m=1mrm+km=1 (mn)2mrm        =(k+1)m=1mrm+kn2m=11mrm2knm=1rm        =(k+1)r (1r)2kn2log(1r)2knr1r.

Thus m=1|am|rm+ m=1|bm|rm1 if

(k+1)r(1r)2kn2log(1r)2knr1r1.

Now, we need to verify that inequality (3.3) holds for rr0, where r0 is the unique root of the equation (2.5) lying in (0,1). For this let

ϕ(r)=(k+1)r(1r)2kn2log(1r)2knr1r1.

Then ϕ(r) is continuous in (0,1),ϕ(0)=1<0 and limr1ϕ(r)>0 implies that there is atleast one root of ϕ(r)=0 in (0,1). But ϕ(r)>0 for all r(0,1),k(0,1) and for all n shows that ϕ is strictly increasing in (0,1). Hence ϕ(r)=0 has a unique root r0 in (0,1).

To see that this r0 is best possible one, we consider f0(z)=z/(1z)2+z/(1z)22z/(1z)log(1z)¯. f0 maps |z|<0.3485... onto the region given in the Figure 2 from which it is evident that r0 is sharp and can not be improved further.

Figure 2. Image of |z|<0.3485 under f0(z).

Proof of Theorem 2.7. From g(z)=a+z1+azh(z) we obtain

m=1mbmzm1=a+z1+az m=1mamzm1,zD

where a1=1 and this gives

m=1mbmzm1+ m=1mabmzm= m=1maamzm1+ m=1mamzm,zD.

Thus we have

m=1m|bm||z|m1 m=1m|a||bm||z|m m=1m|a||am||z|m1+ m=1m|am||z|m.

On integrating from 0 to r, we get

m=1|bm|rm m=1mm+1|a||bm|rm+1 m=1|a||am|rm+ m=1mm+1|am|rm+1,

and this implies

m=1|bm|( m1m)|a||b m1|rm m=1|a||am|+( m1m)|a m1|rm.

Now, we have

m=1(|am|+|bm|)rm=m=1(|am|+|bm|)rmm=1 m1m|a||bm1|rm1      +m=1 m1m|a||bm1|rm1      m=1(|am|+|bm|)rmm=1 m1m|a||bm1|rm      +m=1 m1m|a||bm1|rm1      =m=1|am|rm+m=1 |bm| m1m|a||b m1|rm      +m=1 m1m|a||bm1|rm1.

From (3.5), we get

m=1(|am|+|bm|)rmm=1|am|rm+m=1 |a||am|+ m1m|a m1|rm      +m=1 m1m|a||bm1|rm1      m=1(1+|a|)|am|rm+m=1 m1m|am1|rm1      +m=1 m1m|bm1|rm1      =m=1(1+|a|)|am|rm+m=1 m m+1|am|+|bm||rm.

Therefore, we get

m=11 m+1(|am|+|bm|)rm(1+|a|) m=1|am|rm.

Multiplying both sides with r and then differentiating w.r.t r, we get

m=1(|am|+|bm|)rm(1+|a|) m=1(m+1)|am|rm.

As h is univalent, so |am|m by De Branges's Theorem. From (3.6), we have

m=1(|am|+|bm|)rm(1+|a|)m=1(m+1)mrm      =(1+|a|) r(1+r) (1r)3+r (1r)2(1+|a|)

for r33r2+5r10 and this happens for r0.2291....

Proof of Theorem 2.9. Let h(z)+eiθg(z)=ψ(z), where ψ(z)= m=1Cmzm is a convex univalent function. So, we have |am+eiθbm|=|Cm|1 for all m. This implies |bm|1+|am|,m. We have with |a1|=1,

m=1|am|rm+m=1|bm|rmm=1|am|rm+m=1(1+|am|)rm        =2m=1|am|rm+m=1rm.

Since h is univalent in D, so by De Brange's Theorem, we have |am|m and hence we get

m=1|am|rm+ m=1|bm|rm1

for 2r(1r)2+r1r1 i.e. for 2r25r+10. This is true for rr0=5174=0.2192....

Now to prove Theorem 2.10., we first state the following result of Sheil-Small [11].

Lemma 3.2. If f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=2bmzm¯SH0 is convex in one direction, then

|am|(m+1)(2m+1)6|bm|(m1)(2m1)6.

Proof of Theorem 2.10. h+eiθg is convex univalent implies that f is convex in the direction θ/2, by the well known result of Clunie and Sheil-Small [4]. Therefore, from Lemma 3.2, we have

|am|(m+1)(2m+1)6|bm|(m1)(2m1)6

and so,

m=1|am|rm+m=2|bm|rmm=1 (m+1)(2m+1)6rm+m=2 (m1)(2m1)6rm        =m=1 2m2+13rm        =2r(r+1)3 (1r)3+r3(1r)        1

if 4r39r2+12r30. This inequality holds for rr0=0.3134..., where r0 is unique root of 4r39r2+12r3=0 in (0,1). This result is sharp for K(z)=z1/2z2+1/6z3(1z)3+1/2z2+1/6z3(1z)3¯,where K is harmonic mapping in D, which maps |z|<0.3134 onto region given in Figure 3. It is clear from Figure 3 that |K(z)|<1 for |z|<0.3134... and 0.3134... can not be improved. Hence this r0 is sharp for inequality (2.11) also.

Figure 3. Image of |z|<0.3134 under K(z).

To prove Theorem 2.11., we need following result of Clunie and Sheil-Small [4].

Lemma 3.3. If a harmonic function

f(z)=h(z)+g(z)¯=z+ m=2amzm+ m=2bmzm¯KH0,zD,

then

|am|m+12|bm|m12.

Proof of Theorem 2.11. In view of Lemma 3.3., f(z)KH0 implies that

amm+12bm    m12.

This gives for a1=1,

m=1|am|rm+m=2|bm|rmm=1 m+12rm+m=2 m12rm        =m=1mrm=r (1r)2        1.

for rr0=0.3819... This value of r0 is best possible, as the result is sharp for L(z)=12z1z+z(1z)2+z1zz(1z)2¯. For L(z) we have

m=1|am|rm+m=2|bm|rm=m=1 m+12rm+m=2 1m2rm        =m=1mrm        =r (1r)2        1

for r0.3819... Thus r0 is sharp.

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