On a problem of Bharanedhar and Ponnusamy involving planar harmonic mappings

TL;DR

This paper disproves a previous conjecture about the univalence of certain harmonic mappings and introduces new results on a subclass of close-to-convex harmonic functions.

Contribution

It provides a negative answer to a longstanding problem on harmonic univalence and explores properties of a new subclass of close-to-convex harmonic mappings.

Findings

The class contains non-univalent functions for all parameter values.

New results on a subclass of close-to-convex harmonic mappings.

Disproof of a previous univalence conjecture.

Abstract

In this paper, we give a negative answer to a problem presented by Bharanedhar and Ponnusamy (Rocky Mountain J. Math. 44: 753--777, 2014) concerning univalency of a class of harmonic mappings. More precisely, we show that for all values of the involved parameter, this class contains a non-univalent function. Moreover, several results on a new subclass of close-to-convex harmonic mappings, which is motivated by work of Ponnusamy and Sairam Kaliraj (Mediterr. J. Math. 12: 647--665, 2015), are obtained.

Equations

$$f(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}+\overline{\sum_{k=1}^{\infty}b_{k}z^{k}}\quad(z\in\mbox{$\mathbb{D}$}).$$

$${\rm Re}\left(\frac{F^{\prime}(z)}{\phi^{\prime}(z)}\right)>0\quad(z\in\mbox{$\mathbb{D}$}).$$

$$\mathcal{M}=\left\{f=h+\overline{g}\in\mathcal{H}:\ g^{\prime}=zh^{\prime}\ {\it and}\ {\rm Re}\left(1+\frac{zh^{\prime\prime}(z)}{h^{\prime}(z)}\right)>-\frac{1}{2}\quad(z\in\mbox{$\mathbb{D}$})\right\}.$$

$$\mathrm{Re}\left(\frac{\lambda z{\omega }^{\prime }(z)}{1-\lambda \omega (z)}\right)>-\frac{1}{2}$$

$$\omega(z)=\lambda kz^{n}\quad\left(\left|\lambda\right|=1;\,0<k\leq\frac{1}{2n-1};\,n\in\mbox{$\mathbb{N}$}:=\{1,2,3,\ldots\}\right),$$

$$\mathrm{Re}\left(1+\frac{z{h}^{\prime \prime }(z)}{h^{\prime }(z)}\right)>-\frac{1}{2},$$

$$g^{\prime}(z)=\lambda kz^{n}h^{\prime}(z)\quad\left(n\in\mbox{$\mathbb{N}$};\,\left|\lambda\right|=1;\,0<k\leq\frac{1}{2n-1}\right).$$

$$\mathrm{Re}\left(1+\frac{z{h}^{\prime \prime }(z)}{h^{\prime }(z)}\right)>\alpha \left(-\frac{1}{2}\le \alpha <1\right),$$

$$g^{\prime}(z)=\zeta z^{n}h^{\prime}(z)\quad\left(\zeta\in\mbox{$\mathbb{C}$}\ {\rm with}\ \left|\zeta\right|\leq\frac{1}{2n-1};\,n\in\mbox{$\mathbb{N}$}\right).$$

$$\mathrm{Re}\left(1+\frac{z{F}^{\prime \prime }(z)}{F^{\prime }(z)}\right)<\beta \left(1<\beta \le \frac{3}{2}\right),$$

$$\mathcal{P}(\beta)=\left\{f=h+\overline{g}\in\mathcal{H}:\ g^{\prime}=zh^{\prime}\ {\it and}\ {\rm Re}\left(1+\frac{zh^{\prime\prime}(z)}{h^{\prime}(z)}\right)<\beta\quad(z\in\mbox{$\mathbb{D}$})\right\}.$$

$$h(z)=z-\lambda z^{2}andg(z)=\frac{z^2}{2}-\frac{2\lambda z^3}{3}.$$

$$\sup\limits_{z\in{\scriptsize\mbox{$\mathbb{D}$}}}\left\{{\rm Re}\left(1+\frac{zh^{\prime\prime}(z)}{h^{\prime}(z)}\right)\right\}=\frac{11}{8}$$

$$h(z)=z-\frac{3}{10}{z}^2.$$

$$h(z)=\frac{1}{\gamma }\left[1-(1-z{)}^{\gamma }\right]\left(1<\gamma \le \frac{7}{4}\right),$$

$$g(z)=\frac{1}{\gamma (1+\gamma )}\left[1-(1+\gamma z)(1-z{)}^{\gamma }\right]\left(1<\gamma \le \frac{7}{4}\right).$$

$$1+\frac{z{h}^{\prime \prime }(z)}{h^{\prime }(z)}=\frac{1-\gamma z}{1-z},$$

$$\mathrm{Re}\left(1+\frac{z{h}^{\prime \prime }(z)}{h^{\prime }(z)}\right)<\frac{1+\gamma }{2}\left(1<\frac{1+\gamma }{2}\le \frac{11}{8}\right).$$

$$f_{\gamma }=h+\overline{g}\in \mathcal{P}((1+\gamma )/2)\subset \mathcal{P}(\beta ).$$

$$\mathrm{Re}\left(f_{\gamma }\left(r{e}^{i\theta }\right)\right)=\mathrm{Re}\left(f_{\gamma }\left(r{e}^{-i\theta }\right)\right)$$

$$\mathrm{Im}\left(f_{\gamma }\left(r_0{e}^{i{\theta }_0}\right)\right)=\mathrm{Im}\left(f_{\gamma }\left(r_0{e}^{-i{\theta }_0}\right)\right)=0.$$

$$\mathrm{Im}\left(f_{\gamma }(z)\right)=\mathrm{Im}\left(h(z)-g(z)\right)=\mathrm{Im}\left(\frac{1-(1-z{)}^{\gamma +1}}{\gamma +1}\right)=-\mathrm{Im}\left(\frac{e^{(\gamma +1)\log (1-z)}}{\gamma +1}\right),$$

$$\begin{array}{rl}\mathrm{Im}\left(f_{\gamma }\left(r{e}^{i\theta }\right)\right)& =-\mathrm{Im}\left(\frac{e^{(\gamma +1)\log \left(1-r{e}^{i\theta }\right)}}{\gamma +1}\right)\\ & =-\frac{e^{(\gamma +1)\log |1-r{e}^{i\theta }|}}{\gamma +1}\sin \left[(\gamma +1)\arg \left(1-r{e}^{i\theta }\right)\right],\end{array}$$

$$\begin{array}{r}-\mathrm{Im}\left(f_{\gamma }\left(r{e}^{-i\theta }\right)\right)=\frac{e^{(\gamma +1)\log |1-r{e}^{-i\theta }|}}{\gamma +1}\sin \left[(\gamma +1)\arg \left(1-r{e}^{-i\theta }\right)\right]=\mathrm{Im}\left(f_{\gamma }\left(r{e}^{i\theta }\right)\right).\end{array}$$

$$\arg \left(1-r{e}^{i\theta }\right)\in \left(-\frac{\pi }{2},0\right)\cup \left(0,\frac{\pi }{2}\right),$$

$$\sin \left[(\gamma +1)\arg \left(1-r_0{e}^{i{\theta }_0}\right)\right]=0.$$

$$\mathrm{Im}\left(f_{\gamma }\left(r_0{e}^{i{\theta }_0}\right)\right)=\mathrm{Im}\left(f_{\gamma }\left(r_0{e}^{-i{\theta }_0}\right)\right)=0.$$

$$\left|a_{k}\right|\leq\frac{1}{k!}\prod_{j=2}^{k}(j-2\alpha)\quad(k\in\mbox{$\mathbb{N}$}\setminus\{1\}),$$

$$h(z)=\int_{0}^{z}\frac{dt}{(1-\delta t)^{2-2\alpha}}\quad(\left|\delta\right|=1;\,z\in\mbox{$\mathbb{D}$}).$$

$$\left|b_{n+1}\right|\leq\frac{\left|\zeta\right|}{n+1}\ \ \text{ and }\ \,\ \,\left|b_{k+n}\right|\leq\frac{\left|\zeta\right|}{(k+n)(k-1)!}\prod_{j=2}^{k}(j-2\alpha)\quad(k\in\mbox{$\mathbb{N}$}\setminus\{1\};\,n\in\mbox{$\mathbb{N}$}).$$

Full text

On a problem of Bharanedhar and Ponnusamy involvingplanar harmonic mappings

Zhi-Gang Wang, Zhi-Hong Liu, Antti Rasila and Yong Sun

Abstract. In this paper, we give a negative answer to a problem presented by Bharanedhar and Ponnusamy (Rocky Mountain J. Math. 44: 753–777, 2014) concerning univalency of a class of harmonic mappings. More precisely, we show that for all values of the involved parameter, this class contains a non-univalent function. Moreover, several results on a new subclass of close-to-convex harmonic mappings, which is motivated by work of Ponnusamy and Sairam Kaliraj (Mediterr. J. Math. 12: 647–665, 2015), are obtained.

††footnotetext:

Date: .

Key Words and Phrases. Planar harmonic mapping; univalent harmonic mapping; close-to-convex harmonic mapping.

2010 Mathematics Subject Classification. Primary 58E20; Secondary 30C55.

Z.-G. Wang was supported by the Natural Science Foundation of Hunan Province under Grant no. 2016JJ2036 and the National Natural Science Foundation under Grant no. 11301008. Z.-H. Liu was supported by the Foundation of Educational Committee of Yunnan Province under Grant no. 2015Y456 and the Young and Middle-aged Academic Training Object of Honghe University under Grant no. 2014GG0102. A. Rasila was partially supported by Academy of Finland under Grant no. 289576.

1 Introduction

In this paper, we consider univalency criteria for complex-valued harmonic functions $f$ in the open unit disk $\mathbb{D}$. It is well-known that such functions can be written as $f=h+\overline{g}$, where $h$ and $g$ are analytic functions in $\mathbb{D}$. We call $h$ the analytic part and $g$ the co-analytic part of $f$, respectively. Let $\mathcal{H}$ be the class of harmonic functions normalized by the conditions $f(0)=f_{z}(0)-1=0$, which have the form

[TABLE]

Since the Jacobian of $f$ is given by $\left|h^{\prime}\right|^{2}-\left|g^{\prime}\right|^{2}$, by Lewy’s theorem (see [10]), it is locally univalent and sense-preserving if and only if $|g^{\prime}|<|h^{\prime}|$, or equivalently, the dilatation $\omega={g^{\prime}}/{h^{\prime}}$ with $h^{\prime}(z)\neq 0$ has the property $|\omega|<1$ in $\mathbb{D}$. The subclass of $\mathcal{H}$ that are harmonic, univalent and sense-preserving in $\mathbb{D}$ is denoted by $\mathcal{S_{\mathcal{H}}}$. Univalent harmonic functions are also called harmonic mappings.

The classical family $\mathcal{S}$ of analytic univalent and normalized functions in $\mathbb{D}$ is a subclass of $\mathcal{S_{\mathcal{H}}}$ with $g(z)\equiv 0$. The family of all functions $f\in\mathcal{S}_{\mathcal{H}}$ with the additional property that $f_{\overline{z}}(0)=0$ is denoted by $\mathcal{S}^{0}_{\mathcal{H}}$. There exist reciprocal transformations between the classes $\mathcal{S}_{\mathcal{H}}$ and $\mathcal{S}^{0}_{\mathcal{H}}$ (see [5, 6]). Observe that the family $\mathcal{S}^{0}_{\mathcal{H}}$ is compact and normal, but the family $\mathcal{S}_{\mathcal{H}}$ is not compact. For recent results involving univalent harmonic mappings, we refer to [1, 2, 3, 4, 7, 8, 11, 12, 14, 15, 16, 17, 18, 20, 21], and the references therein.

A domain $\Omega$ is said to be close-to-convex if \mbox{\mathbb{C}}\backslash\Omega can be represented as a union of non-intersecting half-lines. Following the result due to Kaplan [9], an analytic function $F$ is called close-to-convex if there exists a univalent convex analytic function $\phi$ defined in $\mathbb{D}$ such that

[TABLE]

Furthermore, a planar harmonic mapping f:\mbox{\mathbb{D}}\rightarrow\mbox{\mathbb{C}} is close-to-convex if it is injective and f(\mbox{\mathbb{D}}) is a close-to-convex domain, we denote by $\mathcal{C}_{\mathcal{H}}^{0}$ the class of close-to-convex harmonic mappings.

This paper is organized as follows. In Section 2, we give a negative answer to a problem posed by Bharanedhar and Ponnusamy in [3]. In Section 3, we study a subclass of close-to-convex harmonic mappings, which is motivated by work of Ponnusamy and Sairam Kaliraj [16]. Coefficient estimates, a growth theorem, a covering theorem and an area theorem, for mappings of this class, are obtained.

2 A problem of Bharanedhar and Ponnusamy

Recently, Mocanu [11] proposed the following conjecture involving the univalency of planar harmonic mappings.

Conjecture 2.1

Let

[TABLE]

Then $\mathcal{M}\subset\mathcal{S}_{\mathcal{H}}^{0}$.

By applying the close-to-convexity criterion for analytic functions due to Kaplan [9], Bshouty and Lyzzaik [2] have solved the above conjecture by establishing the following stronger result: Theorem A  $\mathcal{M}\subset\mathcal{C}_{\mathcal{H}}^{0}$.

Later, Ponnusamy and Sairam Kaliraj [16, Theorem 4.1] generalized Theorem A, under the assumption that the analytic dilatation $\omega$ satisfies the condition

[TABLE]

for all $\lambda$ such that $\left|\lambda\right|=1$. In particular, for

[TABLE]

they gave the following result:

Theorem B  Suppose that $h$ and $g$ are analytic in $\mathbb{D}$ such that

[TABLE]

and

[TABLE]

Then $f=h+\overline{g}$ is univalent and close-to-convex in $\mathbb{D}$.

Motivated by Theorem B, we introduce the following natural class of close-to-convex harmonic mappings, which will be studied in Section 3. Note that for $n=1$, we have the class $\mathcal{M}(\alpha,\zeta)$, which was studied in [17].

Definition 2.1

A harmonic mapping $f=h+\overline{g}\in\mathcal{H}$ is said to be in the class $\mathcal{M}(\alpha,\zeta,n)$ if $h$ and $g$ satisfy the conditions

[TABLE]

and

[TABLE]

In 1995, Ponnusamy and Rajasekaran [13] derived the following starlikeness criterion for analytic functions. Theorem C  Suppose that $F$ is a normalized analytic function in $\mathbb{D}$. If $F$ satisfies the condition

[TABLE]

then $F$ is univalent and starlike in $\mathbb{D}$, i.e. F(\mbox{\mathbb{D}}) is a domain starlike with respect to the origin.

Motivated essentially by Theorems A and C, Bharanedhar and Ponnusamy [3, Problem 1, p. 763] posed the following problem, which we present here in a slightly modified form:

Problem 2.1

For $\beta\in(1,3/2)$, define

[TABLE]

Determine $\inf\left\{\beta\in(1,3/2):\,\mathcal{P}(\beta)\subset\mathcal{S}^{0}_{\mathcal{H}}\right\}$.

Let us recall the following result of Bshouty and Lyzzaik [2]:

Theorem D  Suppose that $0\leq\lambda<1/2$. Let $f=h+\overline{g}$ be the harmonic polynomial mapping with

[TABLE]

If  $0\leq\lambda\leq 3/10$, then $f$ is univalent in $\mathbb{D}$. But for $3/10<\lambda<1/2$, $f$ is not univalent in $\mathbb{D}$.

Remark 2.2

In view of Theorem D, we see that $\beta$ can be restricted to the value on the interval $(1,11/8]$, since

[TABLE]

for**

[TABLE]

Now, we are ready to give a counterexample, which shows that for all $\beta\in(1,11/8]$, the class $\mathcal{P}(\beta)$ of Problem 2.1 contains a non-univalent function.

Consider the harmonic function given by $f_{\gamma}=h+\overline{g}\in\mathcal{H}$, where

[TABLE]

and

[TABLE]

Clearly, we have $g^{\prime}=zh^{\prime}$. It follows that

[TABLE]

and therefore

[TABLE]

That is,

[TABLE]

In what follows, we shall prove that the function $f_{\gamma}$ is not univalent in $\mathbb{D}$. It is easy to verify that both the analytic and co-analytic parts of $f_{\gamma}$ have real coefficients, and thus, $f_{\gamma}(z)=\overline{f_{\gamma}(\overline{z})}$ for all z\in\mbox{\mathbb{D}}. In particular,

[TABLE]

for some $r\in(0,1)$ and $\theta\in(-\pi,0)\cup(0,\pi)$. It suffices to show that there exist $r_{0}\in(0,1)$ and $\theta_{0}\in(-\pi,0)\cup(0,\pi)$ such that

[TABLE]

In view of the relation

[TABLE]

we see that

[TABLE]

and

[TABLE]

By noting that

[TABLE]

we deduce that for each $1<\gamma\leq 7/4$, there exist $r_{0}\in(0,1)$ and $\theta_{0}\in(-\pi,0)\cup(0,\pi)$ such that

[TABLE]

It follows that

[TABLE]

Therefore, there exist two distinct points $z_{1}=r_{0}e^{i\theta_{0}}$ and $z_{2}=r_{0}e^{-i\theta_{0}}$ in $\mathbb{D}$ such that $f_{\gamma}(z_{1})=f_{\gamma}(z_{2})$, which shows that the function $f_{\gamma}(z)$ is not univalent in $\mathbb{D}$. Thus, we conclude that the conditions given in Problem 2.1 are not satisfied for any $\beta\in(1,11/8]$.

The image domain of $f_{\gamma}$ for $\gamma=5/4$ is given in Figures 1 and 2 to illustrate our counterexample.

3 The subclass $\mathcal{M}(\alpha,\zeta,n)$ of close-to-convex harmonic mappings

Recall the following lemma, due to Suffridge [19], which will be required in the proof of Theorem 3.2.

Lemma 3.1

If $h(z)=z+\sum_{k=2}^{\infty}a_{k}z^{k}$ satisfies the condition (2.1), then

[TABLE]

with the extremal function given by

[TABLE]

We now derive the coefficient estimates for the class $\mathcal{M}(\alpha,\zeta,n)$.

Theorem 3.2

Let $f=h+\overline{g}\in\mathcal{M}(\alpha,\zeta,n)$ be of the form (1.1). Then the coefficients a_{k}\ (k\in\mbox{\mathbb{N}}\setminus\{1\}) of $h$ satisfy (3.1). Moreover, the coefficients b_{k}\ (k=n+1,n+2,\cdots;n\in\mbox{\mathbb{N}}) of $g$ satisfy

[TABLE]

The bounds are sharp for the extremal function given by

[TABLE]

Proof. By equating the coefficients of $z^{k+n-1}$ in both sides of (2.2), we see that

[TABLE]

In view of Lemma 3.1 and (3.2), we get the desired result of Theorem 3.2.

Theorem 3.3

Let $f\in\mathcal{M}(\alpha,\zeta,n)$ with $0\leq\alpha<1$ and 0\leq\zeta<\frac{1}{2n-1}\ (n\in\mbox{\mathbb{N}}). Then

[TABLE]

where

[TABLE]

and

[TABLE]

All these bounds are sharp, the extremal function is $f_{\alpha,\zeta,n}=h_{\alpha}+\overline{g_{\alpha,\zeta,n}}$ or its rotations, where

[TABLE]

Proof. Assume that $f=h+\overline{g}\in\mathcal{M}(\alpha,\zeta,n)$. Also, let $\Gamma$ be the line segment joining [math] and $z$. Then

[TABLE]

Moreover, let $\widetilde{\Gamma}$ be the preimage under $f$ of the line segment joining [math] and $f(z)$. Then we obtain

[TABLE]

By observing that $h$ is a convex analytic function of order $\alpha\ (0\leq\alpha<1)$, it follows that

[TABLE]

By virtue of (3.5), (3.6) and (3.7), we see that

[TABLE]

which yields the desired inequalities (3.3).

Now, we shall prove the sharpness of the result. We only need to show that $f_{\alpha,\zeta,n}$ defined by (3.4) belongs to the class $\mathcal{M}(\alpha,\zeta,n)$ for each $\alpha\in[0,1)$. Suppose that

[TABLE]

Then, we find that $h_{\alpha}(z)$ satisfies the inequality (2.1) and the relation $g^{\prime}_{\alpha,\zeta,n}(z)=\zeta z^{n}h^{\prime}_{\alpha}(z)$ for each $\alpha\in[0,1)$. Moreover, for $0\leq\alpha<1$, $0<\zeta<1/(2n-1)$ with n\in\mbox{\mathbb{N}}, and $0<r<1$, it is easy to see that

[TABLE]

and therefore,

[TABLE]

This shows that the bounds are sharp.

Next, we consider a covering theorem for functions in the class $\mathcal{M}(\alpha,\zeta,n)$.

Theorem 3.4

Let $f\in\mathcal{M}(\alpha,\zeta,n)$ with $0\leq\alpha<1$ and 0\leq\zeta<\frac{1}{2n-1}\ (n\in\mbox{\mathbb{N}}). Then the range f(\mbox{\mathbb{D}}) contains the disk

[TABLE]

The bounds are sharp for the function $f_{\alpha,\zeta,n}=h_{\alpha}+\overline{g_{\alpha,\zeta,n}}$ given by (3.4) or its rotations.

Proof. By putting $r\rightarrow 1^{-}$ in the lower bound for $|f(z)|$ in Theorem 3.3, we get the desired result. The sharpness is similar to that of Theorem 3.3, we choose to omit the details.

Now, we consider the area theorem of the mappings belonging to the class $\mathcal{M}(\alpha,\zeta,n)$. Let us denote \mathcal{A}\left(f(\mbox{\mathbb{D}}_{r})\right) by the area of f(\mbox{\mathbb{D}}_{r}), where \mbox{\mathbb{D}}_{r}:=r\mbox{\mathbb{D}} for $0<r<1$.

Theorem 3.5

Let $f\in\mathcal{M}(\alpha,\zeta,n)$ with $0\leq\alpha<1$. Then, for $0<r<1$, \mathcal{A}\left(f(\mbox{\mathbb{D}}_{r})\right) satisfies the inequalities

[TABLE]

Proof. Let $f=h+\overline{g}\in\mathcal{M}(\alpha,\zeta,n)$. Then for $0<r<1$, we see that

[TABLE]

By observing that $h$ is a convex analytic function of order $\alpha\ (0\leq\alpha<1)$, in view of (3.7) and (3.9), we obtain the desired inequalities (3.8) of Theorem 3.5.

Remark 3.6

By setting $n=1$ in Theorems 3.2, 3.3, 3.4 and 3.5, respectively, we get the corresponding results obtained in [17].**

Finally, we discuss the radius of close-to-convexity of a certain class harmonic mappings related to the class $\mathcal{M}(\alpha,\zeta,n)$. The following lemma due to Clunie and Sheil-Small [5] will be required in the proof of Theorem 3.8.

Lemma 3.7

If $h,\,g$ are analytic in $\mathbb{D}$ with $\left|h^{\prime}(0)\right|>\left|g^{\prime}(0)\right|$, and $h+\lambda g$ is close-to-convex for each $\lambda\,(\left|\lambda\right|=1)$, then $f=h+\overline{g}$ is harmonic close-to-convex in $\mathbb{D}$.

Theorem 3.8

Suppose that $f=h+\overline{g}$ satisfies the inequality (2.1) with $-1/2<\alpha<0$. If $g^{\prime}(z)=z^{n}h^{\prime}(z)$ with n\in\mbox{\mathbb{N}}\setminus\{1\}, then $f$ is close-to-convex in the disk

[TABLE]

Proof. Suppose that $F_{\lambda}(z)=h(z)-\lambda g(z)$ with $\left|\lambda\right|=1$. It follows that

[TABLE]

For $z=re^{i\theta}\ (0<r<1)$, we see that

[TABLE]

Thus,

[TABLE]

for

[TABLE]

By Lemma 3.7 and Kaplan’s close-to-convexity criterion for analytic functions (see [9]), we deduce that $f$ is close-to-convex in the disk $\left|z\right|<r(\alpha,n)$.

Acknowledgments

The authors would like to thank the referees and Prof. S. Ponnusamy for their valuable comments and suggestions, which essentially improved the quality of this paper.