n maximal area integral problem for analytic functions in the starlike famil



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`File: .tex, printed: --, .<100`

**Introduction, Preliminaries, and Main result**
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The univalent function has been the central object in the study of geometric function theory. Some of its natural geometric families act a prominent role in the theory of univalent functions @Dur83 [@Good83; @POM75] and their geometric properties. For instance, the classes of starlike, convex and close-to-convex, to name just a few. These classes have been familiarized and studied by many authors. It is interesting to observe that we can obtain many of their analytic properties by an integrated method. Study of various subclasses of the class of starlike functions have been appreciated by several authors. The class of starlike functions of order $\beta\,(0\leq \beta<1)$ was generated by Robertson @Rob36 and has been then studied by Schild @Sch58 and Merkes @Mer62. Marx @Mar33 and Strohacker @Str33 proved that if $f(z)$ maps the unit disk onto a convex domain, then $f(z)$ is starlike of order $1/2.$ Gabriel @Gab55 showed that the class of starlike functions of order $1/2$ played an important role in the solution of differential equations. In 1968, Padmanabhan @Pad68 discussed a different subfamily for the order of starlikeness. In this paper, we introduce a more general family than the family studied by Padmanabhan.

Define by $\mathbb{D}_r:=\{z\in\mathbb{C}:\,|z|<r\}$, the disk of radius $r$ centred at the origin. The unit disk is then defined by ${{\mathbb D}}:={{\mathbb D}}_1$. Let ${{\mathcal A}}$ denote the family of all functions $f(z)$ analytic in $\mathbb{D}$ and normalized so that $f(0) = 0 = f'(0)-1$, i.e. $f\in{{\mathcal A}}$ has the power series representation $f(z)=z+\sum_{n=2}^\infty a_n z^n$. Denote by ${{\mathcal S}}$, the class of univalent functions $f\in {{\mathcal A}}$. The Gaussian hypergeometric function ${}_2F_1(a,b;c;z)$ is defined by the series $$1+\sum_{n=1}^{\infty}\frac{(a)_n(b)_n}{(c)_n(1)_n}z^n, \quad |z|<1,$$ where $a,b$ and $c$ are complex numbers with $c$ is neither zero nor a negative integer. Clearly, the shifted function $z{}_2F_1(a,b;c;z)$ belongs to ${{\mathcal A}}$. The notation $(a)_n$ denotes the shifted factorial and it is defined by $$(a)_0 = 1, (a)_n = a(a + 1) \cdots (a + n-1) =\frac{\Gamma (a+n)}{\Gamma (a)},\quad n\geq 1.$$ Here, $\Gamma$ stands for the usual gamma function. If either (or both) of $a$ and $b$ is (are) zero or a negative integer(s), then the series terminates.

For two analytic functions $f$ and $g$ in ${{\mathbb D}}$, we say that $f$ is <span>*subordinate*</span> to $g$ if $$f(z)=g(w(z)),\quad |z|<1,$$ for some analytic function $w$ in ${{\mathbb D}}$ with $w(0)=0$ and $|w(z)|<1.$ We express this symbolically by $f \prec g$. In particular, if $g$ is univalent in ${{\mathbb D}}$, $f(0)=g(0)$ and $f({{\mathbb D}})\subset g({{\mathbb D}})$ then $f \prec g$. For instance, one can easily see that $1/(1+z)\prec (1+z)/(1-z)$, $z\in {{\mathbb D}}$.

We denote by ${{\mathcal S}}t(\beta)$, the well-known class of starlike functions of order $\beta$. Analytically, for $f\in {{\mathcal S}}, $ the starlike functions are characterized by the condition $ {\rm Re}\,\left(zf'(z)/f(z)\right)>\beta,$ where $ 0\leq \beta <1,$ i.e. $f$ has the subordination property, $$\frac{zf'(z)}{f(z)} \prec \frac{1+(1-2\beta) z}{1- z},\quad z\in {{\mathbb D}},\, 0\leq \beta<1.$$ The class ${{\mathcal S}}t:={{\mathcal S}}t(0)$ is the class of starlike functions, i.e. $f\in {{\mathcal S}}$ is starlike with respect to the origin, i.e. $tw\in f({{\mathbb D}})$ whenever $t\in [0, 1]$ and $w\in f({{\mathbb D}})$.

Suppose that $f(z)$ is a function analytic in the unit disk ${{\mathbb D}}$. For $0<r\le 1$, we denote by $\Delta(r,f)$, the area of the image of the disk $\mathbb{D}_r$ under $f(z)$. Thus, $$\Delta(r,f)=\iint_{\mathbb{D}_r}|f'(z)|^2 \, dxdy\quad (z=x+iy).$$ Computing this area is known as the <span>*area problem for the function of type $f$.*</span> The classical Parseval-Gutzmer formula for a function $f(z)=\sum_{n=0}^\infty a_nz^n$ analytic in $\overline{\mathbb{D}}_r$ states that $$\frac{1}{2\pi} \int_0^{2\pi}|f(re^{i\theta})|^2\,d\theta = \sum_{n=0}^\infty |a_n|^2r^{2n}.$$ By means of this formula, since $f'(z)=\sum_{n=1}^\infty n a_nz^{n-1}$, we find $$\Delta(r,f)=\pi \sum_{n=1}^\infty n |a_n|^2r^{2n}.$$ We call $f$ a Dirichlet-finite function if $\Delta(1,f)$, the area covered by the mapping $z \to f(z)$ for $|z| < 1$, is finite. Our interest in this paper was originated by the work of Yamashita @Yam90 and Ponnusamy et. al. @OPW13 [@OPW14; @PW13]. Yamashita @Yam90 initially conjectured that $$\displaystyle \max_{f\in {{\mathcal C}}}\Delta \left(r,\frac{z}{f}\right)=\pi r^2$$ for each $r,\, 0<r\leq 1,$ and the maximum is attained only by the rotations of the function $l(z)=z/(1-z).$ Here ${{\mathcal C}}$ denotes the class of convex functions i.e $f\in{{\mathcal S}}$ such that $f({{\mathbb D}})$ is convex. This conjecture was recently settled in @OPW13. In fact the conjecture has been solved for a wider class of functions (the class of starlike functions of order $\beta$, $0\le \beta<1$), which includes the class $\mathcal{C}$; see also Corollary \[cor3\]. In @PW13, the Yamashita conjecture problem for the class of $\phi$-spirallike functions of order $\beta ~(0\leq \beta <1)$ and $\phi \in (-\pi /2,\pi /2)$ have also been settled (see @Lib67 for the definition of $\phi$-spirallike function). Recent work in this direction can also be found in @OPW14. There are several other classes of analytic univalent functions having interesting geometric properties for which solution of the Yamashita conjecture problem would be of interesting to readers working in this field.

Our objective in this paper is to give a partial solution to a problem posed in @PW13 by considering the following subfamily of the family of starlike functions introduced by Padmanabhan @Pad68.

\[def1\] A function $f\in{{\mathcal A}}$ is said to be in ${{\mathcal S}}(\alpha)$, $0<\alpha \le 1$, if $$\left|\left(\frac{zf'(z)}{f(z)}-1\right){\Big/} \left(\frac{zf'(z)}{f(z)}+1\right) \right|<\alpha,
~~\mbox{ equivalently, }~~
\frac{zf'(z)}{f(z)} \prec \frac{1+\alpha z}{1-\alpha z},$$ for all $z\in {{\mathbb D}}$.

It is evident to see that ${{\mathcal S}}(\alpha)\subset {{\mathcal S}}t(\beta).$ One can also verify that $k_{\beta}(z):=z/(1-z)^{2(1-\beta)}$ belongs to ${{\mathcal S}}t(\beta),$ whereas, the function $k_{\alpha}(z):=z/(1-\alpha z)^{2}\in {{\mathcal S}}(\alpha).$ In this paper, we prove the Yamashita conjecture for functions in a more general family than ${{\mathcal S}}(\alpha)$. In particular, the conjecture will also follow for functions in ${{\mathcal S}}(\alpha)$. The generalization is now defined below.

\[def2\] A function $f\in{{\mathcal A}}$ is said to belong to the class ${{\mathcal S}}(\alpha, \beta)$, $0<\alpha\leq 1$, $0\leq \beta<1$, if $$\left|\left(\frac{zf'(z)}{f(z)}-1\right){\Big/} \left(\frac{zf'(z)}{f(z)}+1-2\beta\right) \right|
<\alpha,$$ i.e., $$\label{eq1}
\frac{zf'(z)}{f(z)} \prec \frac{1+(1-2\beta)\alpha z}{1-\alpha z},$$ for all $z\in {{\mathbb D}}$.

A general form of this definition is earlier introduced by Aouf (see @Aou88 [Definition 2]). Definition of ${{\mathcal S}}(\alpha, \beta)$ says that the domain values of $\displaystyle {zf'(z)}/{f(z)}$ lie in the disk of radius $ \displaystyle {2\alpha(1-\beta)}/({1-\alpha^2})$ centred at $({1+\alpha^2(1-2\beta)})/({1-\alpha^2})$. We see that if $\beta =0,$ then Definition \[def2\] turns into Definition \[def1\]. The function $$\label{eq2}
 k_{\alpha,\beta}(z):=z(1-\alpha z)^{-2(1-\beta)},\quad  z\in {{\mathbb D}},\,  0<\alpha\leq 1, \, 0\leq \beta<1,$$ belongs to the family ${{\mathcal S}}(\alpha, \beta)$ and in this context it plays the role of extremal function for ${{\mathcal S}}(\alpha, \beta)$. Also, one notes that $${{\mathcal S}}(\alpha, \beta)\subset {{\mathcal S}}(\alpha)
\subset {{\mathcal S}}t(\beta) \subset {{\mathcal S}}t;\, 
{{\mathcal S}}(1, \beta)={{\mathcal S}}t(\beta),\, {{\mathcal S}}(\alpha, 0)={{\mathcal S}}(\alpha) ~\mbox{and}~ {{\mathcal S}}(1)={{\mathcal S}}t.$$ Consequently, $$\label{eq3}
 k_{1,\beta}(z)=k_{\beta}(z),\, k_{\alpha,0}(z)=k_{\alpha}(z),\, k_{1,0}(z)=k_{1}(z)=k(z).$$ The images of several sub-disks ${{\mathbb D}}_r$, $0<r\le 1$, under the functions $k_{\alpha,\beta}(z)$ for different values of $\alpha$ and $\beta$ have been described in Figure 1.

\[b\]<span>0.5</span> ![Images of the disks ${{\mathbb D}}$ and ${{\mathbb D}}_{0.8}$ under $k_{1/2,0}$ and $k_{4/5,1/3}$.](f1.eps "fig:"){width="6.7cm"}

The image domain $k_{1/2,0}({{\mathbb D}})$

\[b\]<span>0.45</span> ![Images of the disks ${{\mathbb D}}$ and ${{\mathbb D}}_{0.8}$ under $k_{1/2,0}$ and $k_{4/5,1/3}$.](f2.eps "fig:"){width="6.7cm"}

The image domain $k_{4/5,1/3}({{\mathbb D}}_{0.8})$

In this article, our main aim is to examine the maximum area problem for the functions of type $(z/f)$ so-called the Yamashita conjecture problem, when $f\in {{\mathcal S}}(\alpha,\beta)$. By looking into the behavior of $z/k_{\alpha,\beta}(z)$ (see Section 3), we expect the following theorem whose proof is given in Section 3.

\[thm1\]<span>**(Main Theorem)**</span> Let $0<\alpha\leq 1$ and $0\leq \beta <1$. If $f(z)\in {{\mathcal S}}(\alpha,\beta)$ and $z/f(z)$ is a non-vanishing analytic function in ${{\mathbb D}}$, then we have the maximal area $$\begin{aligned}
 \displaystyle \max_{f\in {{\mathcal S}}(\alpha,\beta)} \Delta \left(\rho,\frac{z}{f}\right)
 & = 4\pi{\alpha}^2(1-\beta)^2{\rho}^2 {}_2F_1(2\beta-1,2\beta-1;2;{\alpha}^2{\rho}^2),\quad |z|<\rho\\
 & =:A_{\alpha,\beta}(\rho)
 \end{aligned}$$ for all $\rho, \, 0<\rho\leq 1$. The maximum is attained only by the rotations of the function $ k_{\alpha,\beta}(z)$ defined by $(\ref{eq2}).$

This generalizes the main results which are discussed in @OPW13 and @Yam90.

In Section 2, we prepare some basic results and use them to prove our main theorem in Section 3.

**Preparatory Results**
=======================

If $f\in {{\mathcal S}}$ then $z/f(z)$ is non-vanishing in ${{\mathbb D}}$. Hence, it can be described as Taylor's series of the form $$\label{eq-1-sec.2}
\frac{z}{f(z)}=1+\sum_{n=1}^{\infty}b_nz^n, \, z\in {{\mathbb D}}.$$

We first derive a coefficient estimate in series form for a function $f$ of the form (\[eq-1-sec.2\]) when $f\in {{\mathcal S}}(\alpha,\beta).$

\[lem1\] Let $0<\alpha\leq 1$, $0\leq \beta <1$, and $f(z)\in {{\mathcal S}}(\alpha,\beta)$. If $g(z)$ is a non-vanishing analytic function in ${{\mathbb D}}$ of the form $(\ref{eq-1-sec.2})$, then it necessarily satisfies the coefficient inequality $$\sum_{k=1}^{\infty} \left(k^2-(k-2(1-\beta))^2 \alpha^2 \right)|b_k|^2
\leq 4(1-\beta)^2 \alpha^2.$$

Let $g(z)={z}/{f(z)}$ be of the form (\[eq-1-sec.2\]). Note that the logarithmic derivative gives $$\frac{zg'(z)}{g(z)}=1-\frac{zf'(z)}{f(z)}.$$ Since $f(z)\in {{\mathcal S}}(\alpha,\beta)$, the subordination relation (\[eq1\]) says that there exists an analytic function $ w:{{\mathbb D}}\rightarrow \overline{{{\mathbb D}}}$ with $w(0)=1$ such that $$\frac{zf'(z)}{f(z)}=\frac{1+\alpha(1-2\beta)zw(z)}{1-\alpha zw(z)}, \, z\in {{\mathbb D}},$$ and hence $$\frac{g'(z)}{g(z)}=\frac{-2\alpha(1-\beta)w(z)}{1-\alpha zw(z)}.$$ Writing this in series form, we get $$\sum_{k=1}^{\infty}kb_kz^{k-1}=\alpha\left(-2(1-\beta)+\sum_{k=1}^{\infty}(k-2(1-\beta))b_kz^k\right)w(z).$$ After a minor re-arrangement, we obtain $$\begin{aligned}
\sum_{k=1}^{n}kb_kz^{k-1}+\sum_{k=n+1}^{\infty}kb_kz^{k-1} 
& =\alpha\left(-2(1-\beta)+\sum_{k=1}^{n-1}(k-2(1-\beta))b_kz^k\right)w(z) \\
& + \alpha\left(\sum_{k=n}^{\infty}(k-2(1-\beta))b_kz^k\right)w(z). \end{aligned}$$ By Clunie's method @Clu59 (see also @CK60 [@Rob70; @Rog43]), we obtain $$\sum_{k=1}^{n}k^2|b_k|^2{\rho}^{2k-2}\leq {\alpha}^2\left(4(1-\beta)^2
+\sum_{k=1}^{n-1}\left(k-2(1-\beta)\right)^2|b_k|^2{\rho}^{2k}\right),$$ equivalently, $$\label{lem-eq1}
\sum_{k=1}^{n} k^2 |b_k|^2{\rho}^{2k-2}
 -\alpha^2 \sum_{k=1}^{n-1} (k-2(1-\beta))^2  |b_k|^2{\rho}^{2k}
 \leq 4(1-\beta)^2 \alpha^2.$$ If we take $\rho=1$ and allow $n\rightarrow \infty$, then we obtain the desired inequality $$\sum_{k=1}^{\infty} \left(k^2-(k-2(1-\beta))^2 \alpha^2 \right)|b_k|^2
\leq 4(1-\beta)^2 \alpha^2.$$ The proof of our lemma is now complete.

We remark that the special choices $\alpha=1$ and $\beta=0$ turned Lemma \[lem1\] into the well-known Area Theorem for $f\in {{\mathcal S}}$ (see for instance @Good83 [Theorem 11, p. 193 of Vol-2]).

We now prepare a lemma using a new technique introduced in @OPW13 and this lemma plays an important role to prove our main theorem in this paper.

\[lem2\] Let $0<\alpha\leq 1,\, 0\leq \beta<1$, and $f(z)\in {{\mathcal S}}(\alpha,\beta)$. For $|z|<\rho\le 1$ suppose that $$\frac{z}{f(z)}= 1+ \sum_{k=1}^{\infty}b_kz^k ~\mbox{and}~\ (1+\alpha z)^{2-2\beta}
=1+ \sum_{k=1}^{\infty}{(-1)}^kc_kz^k.$$ Then the relation $$\label{eq5}
\sum_{k=1}^N k|b_k|^2 {\rho}^{2k} \leq \sum_{k=1}^{N}k|c_k|^2 {\rho}^{2k}$$ is valid for all $N\in \mathbb{N}$.

We divide our proof into three steps.

<span>**Step-I: Clunie's method.**</span>

Rewrite (\[lem-eq1\]) in the following form: $$\sum_{k=1}^{n-1} \left(k^2-(k-2(1-\beta))^2 \alpha^2{\rho}^2 \right)|b_k|^2{\rho}^{2k-2}+n^2|b_n|^2{\rho}^{2n-2}
\leq 4(1-\beta)^2 \alpha^2.$$ Multiply by $\rho^2$ on both sides we obtain $$\label{eq6}
\sum_{k=1}^{n-1} \left(k^2-(k-2(1-\beta))^2 \alpha^2{\rho}^2 \right)|b_k|^2{\rho}^{2k}+n^2|b_n|^2{\rho}^{2n}
\leq 4(1-\beta)^2\alpha^2 \rho^2.$$ The function $b(z)=(1+\alpha z)^{2-2\beta}$ clearly shows that the equality in $(\ref{eq6})$ attains with $b_k=(-1)^kc_k.$

<span>**Step-II: Cramer's Rule.**</span>

We consider the inequalities (\[eq6\]) for $n = 1,\ldots,N$, and multiply the $n$-th coefficient by a factor $\lambda_{n,N}$ for each $n$. These factors are chosen in such a way that the addition of the left sides of the modified inequalities is equivalent to the left side of (\[eq5\]). The calculation of the factors $\lambda_{n,N}$ leads to the following system of linear equations: $$\label{eq7}
k=k^2\lambda_{k,N}+\sum_{n=k+1}^{N} {\lambda}_{n,N} \left(k^2-(k-2(1-\beta))^2 \alpha^2{\rho}^2 \right),
\, k = 1, \cdots , N.$$ Since the matrix of the system (\[eq7\]) is an upper triangular matrix with positive integers as diagonal elements, the solution of the system is uniquely determined. Cramer's rule allows us to write the solution of the system (\[eq7\]) in the form $$\lambda_{n,N}=\frac{((n-1)\ !)^2}{(N\ ! )^2} 
\mbox{Det$~A_{n,N}$},$$ where $A_{n,N}$ is the $(N-n+1)\times(N-n+1)$ matrix constructed as follows: $$A_{n,N}=\left[\begin{array}{cccc}
n & n^2-(n-2(1-\beta))^2\alpha^2 \rho^2 & \cdots & n^2-(n-2(1-\beta))^2\alpha^2 \rho^2\\
n+1 & (n+1)^2 & \cdots & (n+1)^2 -(n+1-2(1-\beta))^2\alpha^2 \rho^2\\
\vdots & \vdots & \vdots & \vdots \\
N & 0 & \cdots & N^2\\
\end{array} \right]$$ If we replace $\alpha^2 \rho^2$ by $r^2$ and $2(1-\beta)$ by $\gamma$ in (\[eq7\]), then the equation is equivalent to @OPW13 [(8)]. The rest of the proof now similarly follows as explained in @OPW13.

We now establish a preliminary result concerning necessary and sufficient conditions for a function to be in ${{\mathcal S}}(\alpha,\beta).$

\[lem3\] Let $0<\alpha\leq 1 $ and $0\leq \beta <1$. Then $f(z)\in {{\mathcal S}}(\alpha,\beta)$ if and only if $ F$ defined by $F(z)=z\left({f(z)}/{z}\right)^{\frac{1}{1-\beta}} \in {{\mathcal S}}(\alpha), \,z\in {{\mathbb D}}.$

Let $F(z)=z\left({f(z)}/{z}\right)^{\frac{1}{1-\beta}}.$ Taking logarithm derivative on both sides and simplify, we get $$z\frac{F'(z)}{F(z)}=1+\frac{1}{1-\beta} \left(z\frac{f'(z)}{f(z)}-1\right).$$ By componendo and dividendo rule, we have $$\label{eq4}
 \left |\left(z\frac{F'(z)}{F(z)}-1\right){\Big /} \left(z\frac{F'(z)}{F(z)}+1\right) \right|
 =\left |\left(z\frac{f '(z)}{f(z)}-1\right){\Big /} \left(z\frac{f'(z)}{f(z)}+1-2\beta\right) \right|.$$ By Definition \[def2\] we get $f(z)\in {{\mathcal S}}(\alpha,\beta)$ if and only if $F(z)\in {{\mathcal S}}(\alpha)$.

**Proof of the Main Theorem**
=============================

As an initial observation, from (\[eq2\]) we see that $$\frac{z}{k_{\alpha,\beta}(z)} =(1-\alpha z)^{2(1-\beta)}
= 1+\sum_{n=1}^\infty c_nz^n$$ where $\displaystyle c_n= \frac{(\zeta)_n}{(1)_n} {\alpha}^n$ and $\zeta =-2(1-\beta).$

Hence, we apply the area formula for the function $\displaystyle z/k_{\alpha,\beta}(z)$ and obtain $$\begin{aligned}
\pi^{-1}\Delta\left(\rho,\frac{z}{k_{\alpha,\beta}}\right) & =
\sum_{n=1}^{\infty}n|c_n|^2 {\rho}^{2n}, \quad |z|<\rho \\
 & = \sum_{n=1}^{\infty}n \frac{(\zeta)_n(\zeta)_n}{(1)_n(1)_n}{\alpha}^{2n}{\rho}^{2n} \\
& = {\zeta}^2{\alpha}^2{\rho}^2\sum_{n=0}^{\infty} 
\frac{(\zeta+1 )_n(\zeta+1)_n}{(2)_n(1)_n}{\alpha}^{2n}{\rho}^{2n} \\ 
& =  4{\alpha}^2(1-\beta)^2{\rho}^2 {}_2F_1(2\beta-1,2\beta-1;2;{\alpha}^2{\rho}^2)\\
 & = {\pi}^{-1}A_{\alpha,\beta}(\rho).\end{aligned}$$ At this point let us write $A_{\alpha,\beta}(\rho)$, $0<\rho\leq 1$, in the following form: $$A_{\alpha,\beta}(\rho)=4\pi {\alpha}^{2}(1-\beta)^2 {\rho}^{2}
\sum_{n=0}^{\infty}\frac{(2\beta-1)_{n}^2}{(1)_{n}(2)_n}{\alpha}^{2n}{\rho}^{2n}.$$ Because the series on the right hand side has positive coefficients, $A_{\alpha,\beta}(\rho)$ is a non-decreasing and convex function of the real variable $\rho.$ Thus, $A_{\alpha,\beta}(\rho)\leq A_{\alpha,\beta}(1)$, i.e. $$A_{\alpha,\beta}(\rho)\leq 4\pi {\alpha}^{2}(1-\beta)^2
\sum_{n=0}^{\infty}\frac{(2\beta-1)_{n}^2}{(1)_{n}(2)_n}{\alpha}^{2n}.$$ In order to look for the Dirichlet-finite function $g_{\alpha,\beta}(z)=z/k_{\alpha,\beta}(z)$, we now comprise Table 1 for the values of $A_{\alpha,\beta}(1)$ for various choices of $\alpha$ and $\beta$, and add pictures of the images of the unit disk under the extremal functions $g_{\alpha,\beta}(z)$ for the corresponding values of $\alpha$ and $\beta$ (see Figures 2 to 3).

   **Values of $\beta$**   **Values of $\alpha$**   **Approximate Values of $A_{\alpha,\beta}(1)$**  
  ----------------------- ------------------------ ------------------------------------------------- --
            2/3                     1/4                                0.0875754                     
                                    2/3                                0.638452                      
                                    5/6                                 1.01889                      
            4/5                     1/4                                0.0317791                     
                                    2/3                                0.245872                      
                                    5/6                                0.415385                      

Table 1

\[b\]<span>0.45</span> ![Images of the disks ${{\mathbb D}}$ and ${{\mathbb D}}_{0.9}$ under $g_{2/3,2/3}$ and $g_{5/6,2/3}$.](f3.eps "fig:"){width="6.5cm"}

The image domain $g_{2/3,2/3}({{\mathbb D}})$

\[b\]<span>0.45</span> ![Images of the disks ${{\mathbb D}}$ and ${{\mathbb D}}_{0.9}$ under $g_{2/3,2/3}$ and $g_{5/6,2/3}$.](f4.eps "fig:"){width="6.45cm"}

The image domain $g_{5/6,2/3}({{\mathbb D}}_{0.9})$

\[b\]<span>0.45</span> ![Images of the disks ${{\mathbb D}}_{0.8}$ and ${{\mathbb D}}_{0.7}$ under $g_{1/4,4/5}$ and $g_{5/6,4/5}$.](f5.eps "fig:"){width="6.5cm"}

The image domain $g_{1/4,4/5}({{\mathbb D}}_{0.8})$

\[b\]<span>0.45</span> ![Images of the disks ${{\mathbb D}}_{0.8}$ and ${{\mathbb D}}_{0.7}$ under $g_{1/4,4/5}$ and $g_{5/6,4/5}$.](f6.eps "fig:"){width="7.2cm"}

The image domain $g_{5/6,4/5}({{\mathbb D}}_{0.7})$

It is now time for us to prove the main theorem.

For $z\in {{\mathbb D}},$ we know by Lemma \[lem3\] that $$f(z)\in {{\mathcal S}}(\alpha,\beta) \Longleftrightarrow 
F(z)=z\left( \frac{f(z)}{z}\right)^{\frac{1}{1-\beta}} \in {{\mathcal S}}(\alpha).$$ Further $F(z) \in {{\mathcal S}}(\alpha)$ gives $$\frac{z}{F(z)}=\left(\frac{z}{f(z)}\right)^{\frac{1}{1-\beta}} \prec (1-\alpha z)^2
=\left(\frac{z}{k_{\alpha,\beta}(z)}\right)^{\frac{1}{1-\beta}},$$ i.e. $$\frac{z}{f(z)} \prec (1-\alpha z)^{2(1-\beta)}= \frac{z}{k_{\alpha,\beta}(z)}.$$ If $$\frac{z}{f(z)}= 1+ \sum_{n=1}^{\infty}b_nz^n ~\mbox{and}~\ \frac{z}{k(z)}
=1+ \sum_{n=1}^{\infty}c_nz^n,\quad |z|<\rho,$$ then the extension of Rogosinski's result observed by Goluzin @Dur83 [Theorem 6.3, p.193] yields $$\sum_{n=1}^{\infty}n|b_n|^2 {\rho}^{2n} \leq \sum_{n=1}^{\infty}n|c_n|^2 {\rho}^{2n}.$$

That is $$\Delta \left(\rho,\frac{z}{f}\right) \leq \Delta \left(\rho,\frac{z}{k_{\alpha,\beta}}\right)
= \pi {\zeta}^2{\alpha}^2{\rho}^2 {}_2F_1(\zeta +1,\zeta+1;2;{\alpha}^2{\rho}^2),\quad \zeta+1=2\beta-1$$ whenever the sequence $\{n{\rho}^{2n}\}$ is decreasing function of $\rho,\, 0<\rho\leq \frac{1}{\sqrt{2}}$. Thus, the theorem is obviously true for $0<\rho\leq \frac{1}{\sqrt{2}}$. On other hand, in order to present a proof to include the case $ \rho>\frac{1}{\sqrt{2}}$, it suffices to prove $$\sum_{n=1}^N n|b_n|^2 {\rho}^{2n} \leq \sum_{n=1}^{N}n|c_n|^2 {\rho}^{2n},\,
N\in \mathbb{N},\, \rho\in (0,1).$$ This follows from Lemma \[lem2\] and hence the proof of Theorem \[thm1\] is complete.

If we choose ${\beta=0}$ in Theorem \[thm1\], then we get the following Yamashita conjecture problem solved for functions in the Padmanabhan class ${{\mathcal S}}(\alpha)$:

\[thm2\] Let for $0<\alpha \leq 1,\, f\in {{\mathcal S}}(\alpha)$ and $z/f(z)$ be a non-vanishing analytic function in ${{\mathbb D}}$. Then we have $$\displaystyle \max_{f\in {{\mathcal S}}(\alpha)}\Delta \left(\rho,\frac{z}{f}\right)=2\pi {\alpha}^2\rho^2(2+{\alpha}^2\rho^2)
     =: A_{\alpha}(\rho)$$ for all $\rho,\, 0<\rho\leq 1$. The maximum is attained only by the rotation of the function $k_{\alpha}(z)$ defined by $(\ref{eq3}).$

We now collect the values of $A_{\alpha}(1)$ in the form of a table (see Table 2) for several values of $\alpha$ and affix geometrical pictures of the images of the unit disk under the extremal functions $g_{\alpha}(z)=z/k_{\alpha}(z)=(1-\alpha z)^{2}$ for the corresponding $\alpha$ values (see Figure 4).

\[b\]<span>0.45</span> ![Images of the disks ${{\mathbb D}}$ and ${{\mathbb D}}_{0.9}$ under $g_{2/3}$ and $g_{8/9}$.](f7.eps "fig:"){width="6.5cm"}

The image domain $g_{2/3}({{\mathbb D}})$

\[b\]<span>0.45</span> ![Images of the disks ${{\mathbb D}}$ and ${{\mathbb D}}_{0.9}$ under $g_{2/3}$ and $g_{8/9}$.](f8.eps "fig:"){width="6.7cm"}

The image domain $g_{8/9}({{\mathbb D}}_{0.9})$

   **Values of $\alpha$**   **Approximate Values of $A_{\alpha}(1)$**     
  ------------------------ ------------------------------------------- -- --
            1/4                             0.809942                      
            2/3                              6.82616                      
            5/6                              11.7567                      
            8/9                              13.8515                      

Table 2

If ${\alpha=1}$ and $\beta=1/2$, then as a consequence of Theorem \[thm1\] we get

\[cor2\]@OPW13 [Theorem 2] We have $$\displaystyle \max_{f\in {{\mathcal S}}t(1/2)}~\Delta \left(\rho,\frac{z}{f}\right)
    =\pi \rho^2 
 \quad \mbox{for $0<\rho\leq 1$},$$ where the maximum is attained only by the rotation of the Koebe function $k(z)$ defined by $(\ref{eq3}).$

Moreover, if we choose ${\alpha=1}$ in Theorem \[thm1\], we get

\[cor3\]@OPW13 [Theorem 3] Let $f\in {{\mathcal S}}t(\beta)$ for some $0\leq \beta <1$. Then we have $$\displaystyle \max_{f\in {{\mathcal S}}t(\beta)}\Delta \left(\rho,\frac{z}{f}\right)
    =4\pi (1-\beta)^2\rho^2 {}_2F_1(2\beta-1,2\beta-1;2;\rho^2) 
 \quad \mbox{for $0<\rho\leq 1$},$$ where the maximum is attained only by the rotation of the function $k_{\beta}(z)$ defined by $(\ref{eq3}).$

**Concluding Remark**
=====================

For $-1\leq B<A \le 1,$ the Janowski class ${{\mathcal S}}^*(A,B)$ is defined by the subordination relation $${{\mathcal S}}^*(A,B):=\left \{f\in {{\mathcal A}}: \frac{zf'(z)}{f(z)}\prec \frac{1+Az}{1+Bz},\quad z\in {{\mathbb D}}\right\}.$$ The class ${{\mathcal S}}^*(A,B)$ is introduced in @Jan73 and studied by number of researchers in this field. It is evident that ${{\mathcal S}}^*(A,B)\subset {{\mathcal S}}t$. In @PW13, it has been reported that Yamashita's conjecture is an open problem to prove for convex functions of order $\beta$ and more generally, for functions in the class ${{\mathcal S}}^*(A,B)$ and also for the class of functions $f$ for which $zf'(z)\in {{\mathcal S}}^*(A,B)$. In particular, the choices $A=(1-2\beta)\alpha$ and $B=-\alpha$ turn the class ${{\mathcal S}}^*(A,B)$ into the class ${{\mathcal S}}(\alpha,\beta)$. Therefore, a partial solution to the above open problem has been solved in this paper.

<span>**Acknowledgements.**</span> This research was started in the mid 2013 when the authors were visited ISI Chennai Centre. We would like to thank Professors R. Pavaratham and S. Ponnusamy for useful discussion in this topic. The second author also acknowledges the support of the National Board for Higher Mathematics, Department of Atomic Energy, India (grant no. 2/39(20)/2010-R$\&$D-II). The authors thank one of the referees for useful comments on the paper.

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[^1]: ${}^*$ The corresponding author

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