1. Introduction and definitions

Let $\mathcal{A}$ be the class of functions $f$ of the form

$f(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}$

(1.1)

which are analytic in the open unit disk $\Delta = \{z\in\mathbb{C}:\left\vert z\right\vert < 1\}$ and normalized by the conditions $f(0) = 0$ and $f'(0) = 1$. An analytic function in a domain D is said to be univalent in D if it does not take the same value twice i.e, $f(z_1)\neq f(z_2)$ for all pairs of distinct points $z_1$ and $z_2$ in D.

The Koebe one-quarter theorem et al ^[3] ensures that the image of $\Delta$ under every univalent function $f\in\mathcal{A}$ contains the disk with the center at origin and of the radius $1/4$. Thus, every univalent function $f\in\mathcal{A}$ has an inverse $f^{-1}:f(\Delta)\rightarrow\Delta$, satisfying $f^{-1}(f(z)) = z, \; (z\in\Delta)$ and

${f}\left(f^{-1}(w)\right) = w\quad\left(|w|<r_0(f);\;r_0(f)\geq \frac{1}{4}\right).$

Moreover, it is easy to see that the inverse function has the series expansion of the form

$f^{-1}(w) = w-a_{2}w^{2}+(2{a_{2}}^{2}-a_{3})w^{3}-(5{a_{2}}^{3}-5a_{2}a_{3}+a_{4})w^{4}+... ;\ w \in \Delta,$

which implies that $f^{-1}$ is analytic. The derivative of $f^{-1}$ (see pp. 1038 ^[4]) is given by

$\frac{d}{dw} \left(f^{-1} \left(w\right)\right) = \frac{1}{f'\left(z\right)}.$

A function $f\in \mathcal{A}$ is said to be bi-univalent in $\Delta$ if both $f$ and $f^{-1}$ are univalent in $\Delta$. We denote the class of bi-univalent functions by $\sigma$.(see ^[2])

The function $f$ in class $\mathcal{A}$ is said to be starlike of order $\alpha$ where $0\le \alpha < 1$ in $\Delta$ if it satisfies the condition

$Re\{\frac{zf'(z)}{f(z)}\}>\alpha,$

where $z \in \Delta$. We denote the class of starlike functions of order $\alpha$ by $S^*(\alpha)$. The function $f$ of the form $(1)$ is said to be bi-starlike function of order $\alpha$ where $0 \le \alpha < 1$ if each of the following conditions are satisfied

$Re\{\frac{zf'(z)}{f(z)}\}>\alpha$

and

$Re\{\frac{wg'(w)}{g(w)}\}>\alpha,$

where $f \in \sigma$, $g = f^{-1}$ and $w = f(z)$. We denote the class of bi-starlike functions of order $\alpha$ by $S_{\sigma }^{*}(\alpha)$ (see ^[5]). If $f$ and $g$ are analytic functions in $\Delta$, we say that $f$ is subordinate to $g$, written as $f\prec g$, if there exists a Schwarz function $w$ analytic in $\Delta$, with $w(0) = 0$ and $\left\vert w(z)\right\vert < 1$ $(z\in\Delta)$, such that $f(z) = g\left(w(z)\right)$. In particular, when $g$ is univalent then the above definition reduces to $f(0) = 0$ and $f(\Delta)\subseteq g(\Delta)$.

The pre-Schwarzian derivative of $f$ is denoted by

${T_f}(z) = \frac{{f''(z)}}{{f'(z)}}$

and its norm is given by

$||{T_f}|| = \mathop {sup}\limits_{|z| < 1} (1 - |z{|^2})\left| {\frac{{f''(z)}}{{f'(z)}}} \right|.$

This norm have a significant meaning in the theory of Teichmuller spaces. For a univalent function $f$ it is well known that $||{T_f}|| < 6$. This is the best possible estimation.

Defining two subclasses for bi-univalent functions as follows

Definition 1.1. A function $f$ given by (1) is said to be in the class $S_{\sigma }^{*} \left[A, \, B\right],$ if the following conditions are satisfied

$\frac{zf'\left(z\right)}{f\left(z\right)} \prec \frac{1+Az}{1+Bz},$

$\frac{w\, g'\left(w\right)}{g\left(w\right)} \prec \frac{1+Aw}{1+Bw},$

where $f\in\sigma$, $g = f^{-1}$, $w = f(z)$, $w\in\Delta$ and $-1\le B < A\le1.$

Remark 1.1. If we take $A = (1-2\alpha)$ and $B = -1$ in the above Definition $1.1$ where $0\le\alpha < 1$, the class becomes $S_{\sigma }^{*} [(1-2\alpha), -1] \equiv S_{\sigma }^{*}(\alpha)$.

Remark 1.2. If we take $A = 1$ and $B = -1$ in the above Definition $1.1$, the class becomes $S_{\sigma }^{*} [1, -1] \equiv S_{\sigma }^{*}$.

Definition 1.2. A function $f$ given by (1) is said to be in the class $V_{\sigma }^{*} \left[A, \, B\right],$ if the following conditions are satisfied

$\left(\frac{z}{f\left(z\right)} \right)^{2} f'\left(z\right)\prec \frac{1+Az}{1+Bz},$

$\left(\frac{w\, }{g\left(w\right)} \right)^{2} g'\left(w\right)\prec \frac{1+Aw}{1+Bw},$

where $f\in\sigma$, $g = f^{-1}$, $w = f(z)$, $w\in\Delta$ and $-1\le B < A\le1.$

Remark 1.3. If we take $A = (1-2\alpha)$ and $B = -1$ in the above Definition $1.2$ where $0\le\alpha < 1$, the class becomes $V_{\sigma }^{*} [(1-2\alpha), -1] \equiv V_{\sigma }^{*}(\alpha)$.

Remark 1.4 If we take $A = 1$ and $B = -1$ in the above Definition $1.2$, the class becomes $V_{\sigma }^{*} [1, -1] \equiv V_{\sigma }^{*}$.

In this paper, we shall give the best norm estimation for the classes $S_\sigma^*[A, B]$ and $V_\sigma^*[A, B]$.

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2. Main result

Theorem 2.1. Let the function $f$ given by (1) be in the class $f\in S_{\sigma }^{*} \left[A, \, B\right]$, then

$\left\| T_{f} \right\| \le \min \left\{\frac{2\left(A-B\right)\left(A+2\right)}{\left(A+1\right)} , \frac{2\left(A-B\right)|A|}{\left(A+1\right)} \right\}.$

Proof. Since $f\in S_{\sigma }^{*} \left[A, \, B\right]$, let us assume that

$h\left(z\right) = \frac{zf'\left(z\right)}{f\left(z\right)} \prec \frac{1+Az}{1+Bz} = p(z).$

Using the definition of subordination, there exists a Schwarz function $\phi:\Delta\to\Delta$ with $\phi\left(0\right) = 0 \ $ and $|\phi \left(z\right)| < 1,$ such that

$h(z) = p\, o\, \phi(z) = \frac{1+A\phi \left(z\right)}{1+B\phi \left(z\right)}.$

Hence, $h(z)$ becomes

$h\left(z\right) = \frac{zf'\left(z\right)}{f\left(z\right)} = \frac{1+A\phi \left(z\right)}{1+B\phi \left(z\right)}.$

(2.1)

By logarithmic differentiation of $(3)$, we get

$\frac{1}{z} +\frac{f''\left(z\right)}{f'\left(z\right)} -\frac{f'\left(z\right)}{f\left(z\right)} = \frac{A}{\left(1+A\phi(z)\right)} -\frac{B}{\left(1+B\phi(z)\right)}.$

Above equation gives us the pre-Schwarzian derivative of $f$, i.e,

$T_{f}(z) = \frac{f''(z)}{f'(z)} = \frac{(A^{2}-AB)\phi(z)+2(A-B)}{(1+A\phi(z))(1+B\phi(z))},$

Setting $\phi \left(z\right) = id_{\Delta}$ (as $\phi$ belongs to the class of Schwarz functions and $\phi \left(z\right)\prec z\, \, on\, \, \Delta$) and rearranging the terms, we get

$\left(1-\left|z\right|^{2} \right)\left|T_{f} \left(z\right)\right| = \left(1-\left|z\right|^{2} \right)\left|\frac{\left(A^{2} -AB\right)z+2\left(A-B\right)}{\left(1+Az\right)\left(1+Bz\right)} \right|.$

Taking the supremum value both sides in the unit disc, the above equation becomes

$\mathop{\sup }\limits_{\left|z\right|<1} \left(1-\left|z\right|^{2} \right)\left|T_{f} \left(z\right)\right|\le \mathop{\sup }\limits_{\left|z\right|<1} \left(1-\left|z\right|^{2} \right)\left[\frac{\left(A^{2} -AB\right)\left|z\right|+2\left(A-B\right)}{\left(1+A\left|z\right|\right)\left(1+B\left|z\right|\right)} \right].$

As $-1\le B$, we get $\left(1-\left|z\right|\right)\le \left(1+B\left|z\right|\right)$, therefore the above inequality becomes

$\mathop{\sup }\limits_{\left|z\right|<1} \left(1-\left|z\right|^{2} \right)\left|T_{f} \left(z\right)\right|\le \mathop{\sup }\limits_{\left|z\right|<1} \left(1+\left|z\right|\right)\left[\frac{\left(A^{2} -AB\right)\left|z\right|+2\left(A-B\right)}{\left(1+A\left|z\right|\right)} \right].$

The above inequality gives us the norm of pre-Schwarzian derivative of $f$, denoted by $||T_{f}||$. To estimate the upper bound of $||T_{f}||$ in the unit disc $\Delta$, z must lead to $1$ and therefore

$\mathop{\lim }\limits_{z\to 1} \left(1+\left|z\right|\right)\left[\frac{\left(A^{2} -AB\right)\left|z\right|+2\left(A-B\right)}{\left(1+A\left|z\right|\right)} \right] = \frac{2\left(A-B\right)\left(A+2\right)}{\left(A+1\right)}.$

Finally we get

$\left\| T_{f} \right\| \le \frac{2\left(A-B\right)\left(A+2\right)}{\left(A+1\right)}.$

(2.2)

For the second part of the proof, let us assume that

$k(z) = \frac{w\, g'\left(w\right)}{g\left(w\right)} \prec \frac{1+Aw}{1+Bw} = p\left(w\right)$

where $z = f^{-1}(w) = g(w)$. By definition of subordination, there exists a Schwarz function $\phi :\Delta \to \Delta \ $ with $\phi \left(0\right) = 0 \ $ and $|\phi \left(z\right)| < 1,$ such that

$k(z) = p\, o\, \phi(z) = \frac{1+A\phi \left(z\right)}{1+B\phi \left(z\right)}.$

Since $f \in \sigma$, both $f$ and $f^{-1}$ are analytic and univalent in $\Delta$. The derivative of $f^{-1}$ is given by

$\frac{d(f^{-1}(w))}{w} = \frac{1}{f'(z)}.$

Therefore, $k(z)$ can be expressed as

$\frac{wg'(w)}{g(w)} = k\left(z\right) = \frac{f\left(z\right)}{zf'\left(z\right)} = \frac{1+A\phi \left(z\right)}{1+B\phi \left(z\right)}.$

(2.3)

Taking logarithmic differentiation of $(5)$, we get

$\frac{f'\left(z\right)}{f\left(z\right)} -\frac{1}{z} -\frac{f''\left(z\right)}{f'\left(z\right)} = \frac{A\phi '\left(z\right)}{\left(1+A\phi \left(z\right)\right)} -\frac{B\phi '\left(z\right)}{\left(1+B\phi \left(z\right)\right)}.$

Setting $\phi \left(z\right) = id_{\Delta}$ (as $\phi$ belongs to the class of Schwarz functions and $\phi \left(z\right)\prec z\, \, on\, \, \Delta$), we have

$\frac{f''\left(z\right)}{f'\left(z\right)} = \frac{Az\left(A-B\right)}{\left(1+Az\right)\left(1+Bz\right)}.$

Following the previous steps and using $(1-\left|z\right|)\le(1+B\left|z\right|),$ we get

$\mathop{\sup }\limits_{\left|z\right| < 1} \left(1-\left|z\right|^{2} \right)\left|T_{f} \left(z\right)\right|\le \mathop{\sup }\limits_{\left|z\right| < 1} \left(1+\left|z\right|\right)\left[\frac{|A|\left(A-B\right)\left|z\right|}{\left(1+A\left|z\right|\right)} \right].$

For upper bound of $||T_{f}||$, z must lead to 1, i.e,

$\mathop{\lim }\limits_{z\to 1} \left(1+\left|z\right|\right)\frac{|A|\left(A-B\right)\left|z\right|}{\left(1+A\left|z\right|\right)} = \frac{2\left(A-B\right)|A|}{\left(A+1\right)}.$

Therefore

$\left\| T_{f} \right\| \le \frac{2\left(A-B\right)|A|}{\left(A+1\right)}.$

(2.4)

Combining $(4)$ and $(6)$, the proof is complete.

Let $A = (1-2\alpha)$ and $B = -1$ in the above theorem where $0\le\alpha < 1$, the class becomes $S_{\sigma }^{*} [(1-2\alpha), -1] \equiv S_{\sigma }^{*}(\alpha)$.

Corollary 2.1. If $f\in S_{\sigma }^{*} (\alpha),$ then $|| T_{f}||\le min\{6-4\alpha, |2-4\alpha|\}$.

If $f$ is analytic and locally univalent in $\Delta$ such that $f\in S^{*} (\alpha),$ where $0 \le \alpha < 1$ then $|| T_{f}||\le ({6-4\alpha})$, which is due to Yamashita ^[1]. The above corollary generalizes the result for bi-univalent functions.

Let $A = 1$ and $B = -1$ in the above theorem, the class becomes $S_{\sigma }^{*} [1, -1] \equiv S_{\sigma }^{*}$.

Corollary 2.2. If $f \in S_{\sigma }^{*}$, then $\left\|T_{f}\right\| \le 6.$

The above corollary generalizes the norm estimation for bi-univalent functions.

Theorem 2.2. Let the function $f\left(z\right)$ given by (1) be in the class $V_{\sigma }^{*} \left[A, \, B\right]$, then

$\left\| T_{f} \right\| \le \min \left\{\frac{2\left(3+2A\right)\left(A-B\right)}{\left(A+1\right)} , \frac{2\left(A-B\right)\left(1+2A\right)}{\left(A+1\right)} \right\}.$

Proof. Since $f \in V_{\sigma }^{*} \left[A, \, B\right]$, let us assume that

$k(z) = \left(\frac{z}{f\left(z\right)} \right)^{2} f'\left(z\right) \prec \frac{1+Az}{1+Bz} = p(z).$

Therefore, there exists a Schwarz function $\phi :\Delta \to \Delta \ $ with $\phi \left(0\right) = 0 \ $ and $|\phi \left(z\right)| < 1$ such that

$k(z) = p\, o\, \phi(z) = \frac{1+A\phi \left(z\right)}{1+B\phi \left(z\right)}.$

Therefore, $k(z)$ can be expressed as

$k(z) = \left(\frac{z}{f\left(z\right)} \right)^{2} f'\left(z\right) = \frac{1+A\phi \left(z\right)}{1+B\phi \left(z\right)}.$

(2.5)

By logarithmic differentiation on $(7)$, we get

$\frac{2}{z} -\frac{2f'\left(z\right)}{f\left(z\right)} +\frac{f''\left(z\right)}{f'\left(z\right)} = \frac{A\phi '\left(z\right)}{\left(1+A\phi \left(z\right)\right)} -\frac{B\phi '\left(z\right)}{\left(1+B\phi \left(z\right)\right)},$

$\frac{f''\left(z\right)}{f'\left(z\right)} = \frac{2f'\left(z\right)}{f\left(z\right)} -\frac{2}{z} +\frac{A\phi '\left(z\right)}{\left(1+A\phi \left(z\right)\right)} -\frac{B\phi '\left(z\right)}{\left(1+B\phi \left(z\right)\right)}.$

Hence, the pre-Schwarzian derivative of $f$ becomes

$T_{f} \left(z\right) = \frac{f''\left(z\right)}{f'\left(z\right)} = \frac{2\left(1+A\phi \left(z\right)\right)}{z\left(1+B\phi \left(z\right)\right)} -\frac{2}{z} +\frac{A\phi '\left(z\right)}{\left(1+A\phi \left(z\right)\right)} -\frac{B\phi '\left(z\right)}{\left(1+B\phi \left(z\right)\right)}.$

Setting $\phi \left(z\right) = id_\Delta$, we get

$T_{f} \left(z\right) = \frac{f''\left(z\right)}{f'\left(z\right)} = \frac{B\left(A-B\right)+2Az\left(A-B\right)}{\left(1+Az\right)\left(1+Bz\right)} = \frac{\left(A-B\right)\left(3+2Az\right)}{\left(1+Az\right)\left(1+Bz\right)}.$

Therefore

${\mathop{\sup }\limits_{\left|z\right|<1} \left(1-\left|z\right|^{2} \right)\left|T_{f} \left(z\right)\right| \le \mathop{\sup }\limits_{\left|z\right|<1} \left(1+\left|z\right|\right)\left[\frac{\left(3+2A\left|z\right|\right)\left(A-B\right)}{\left(1+A\left|z\right|\right)} \right]}.$

Again, to estimate the upper bound of $||T_{f}||$, z must lead to 1 and hence we get

$\mathop{\lim }\limits_{z\to 1} \left(1+\left|z\right|\right)\frac{\left(A-B\right)\left(3+2A\left|z\right|\right)}{\left(1+A\left|z\right|\right)} = \frac{2\left(A-B\right)\left(3+2A\right)}{\left(A+1\right)}.$

Finally

$\left\| T_{f} \right\| \le \frac{2\left(A-B\right)\left(3+2A\right)}{\left(A+1\right)}.$

(2.6)

For the second part of the proof, its given $f \in V_{\sigma }^{*} \left[A, \, B\right]$ and therefore we assume

$k(z) = \left(\frac{w\, }{g\left(w\right)} \right)^{2} g'\left(w\right)\prec \frac{1+Aw}{1+Bw} = p(w)$

where $w = f(z)$, $g = f^{-1}$ and w $\in \Delta$. Since $f \in \sigma$ (as explained in the second part of previous theorem) we see,

$g' = \frac{d}{dw} \left(f^{-1} \left(w\right)\right) = \frac{1}{f'\left(z\right)}.$

Using above equation, $k(z)$ can be expressed as

$k(z) = \left(\frac{f\left(z\right)}{z} \right)^{2} \frac{1}{f'\left(z\right)} = \frac{1+A\phi \left(z\right)}{1+B\phi \left(z\right)}.$

By logarithmic differentiation of above equation, we get

$\frac{f''\left(z\right)}{f'\left(z\right)} = \frac{2\left(1+A\phi \left(z\right)\right)}{z\left(1+B\phi \left(z\right)\right)} -\frac{2}{z} -\frac{A\phi '\left(z\right)}{\left(1+A\phi \left(z\right)\right)} +\frac{B\phi '\left(z\right)}{\left(1+B\phi \left(z\right)\right)}.$

Setting $\phi \left(z\right) = id_\Delta$, the pre-Schwarzian derivative of $f$ becomes

$T_{f} \left(z\right) = \frac{f''\left(z\right)}{f'\left(z\right)} = \frac{\left(A-B\right)\left(1+2Az\right)}{\left(1+Az\right)\left(1+Bz\right)}$

Following the similar steps as in the first part of this theorem, we get

$||T_{f}|| \le \mathop{\sup }\limits_{\left|z\right|<1} \left(1+\left|z\right|\right)\frac{\left(A-B\right)\left(1+2A\left|z\right|\right)}{\left(1+A\left|z\right|\right)}.$

Finally,

$\left\| T_{f} \right\| \le \frac{2\left(A-B\right)\left(1+2A\right)}{\left(A+1\right)}.$

(2.7)

Combining $(8)$ and $(9)$, the proof is complete.

Let $A = (1-2\alpha)$ and $B = -1$ in the above theorem where $0\le\alpha < 1$, the class becomes $V_{\sigma }^{*} [(1-2\alpha), -1] \equiv V_{\sigma }^{*}(\alpha)$.

Corollary 2.3. If $f\in V_{\sigma }^{*} (\alpha),$ then $|| T_{f}||\le min\{10-8\alpha, 6-8\alpha\}.$

The above corollary deduces to the exact same norm estimation for analytic and bi-univalent functions in $\Delta$ which lies in a similar class denoted by $V_{\sigma}^{*}(\alpha)$ and is studied by Rahmatan ^[4].

Let $A = 1$ and $B = -1$ in the above theorem, the class becomes $V_{\sigma }^{*} [1, -1] \equiv V_{\sigma }^{*}$.

Corollary 2.4. If $f \in V_{\sigma }^{*}$, then $\left\|T_{f}\right\| \le 6.$

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Acknowledgements

The authors are thankful to the reviewers for their comments.

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Conflict of Interest

Authors declare that there is no conflict of interest.

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