Some applications of second-order differential subordination for a class of analytic function defined by the lambda operator

1. Introduction

Let \( \mathbb{C} \) be complex plane and let \(\mathbb{ U } = \{z: z \in \mathbb{C} ~\text{and}~ |z| < 1\} = \mathbb{ U } \setminus \{0\}\) be an open unit disc in \( \mathbb{C} .\) Also let \(H(\mathbb{ U } )\) be a class of analytic functions in \(\mathbb{ U } .\) For \(n \in \mathbb{N}= \{1, 2, 3, \cdots , \} \) and \(a \in \mathbb{C} ,\) let \(H[a, n]\) be a subclass of \(H(\mathbb{ U } )\) formed by the functions of the form \[ f(z) = z+ a_n z^n + a_{n+1} z^{n+1} + \cdots \] with \(H_0 \equiv H[0, 1]\) and \(H \equiv H[1, 1].\) Suppose that \(A_n\) is a class of all analytic functions of the form

in the open unit disk \(\mathbb{ U } \) with \(A_1 = A.\) A function \(f \in H(\mathbb{ U } )\) is univalent if it is a one-to-one function in \(\mathbb{ U } .\) By \(S\), we denote a subclass of \(A\) formed by functions univalent in \(\mathbb{ U } .\) If a function \(f \in A\) maps \(\mathbb{ U } \) onto a convex domain and \(f\) is univalent, then \(f\) is called a convex function. By \[ K= \left\{ f\in A: \Re \left\{1+ \frac{zf''(z)}{f'(z)}\right\}> 0, \ \ z \in \mathbb{ U } \right\}, \] we denote a class of all convex functions defined in \(\mathbb{ U } \) and normalized by \(f(0) = 0\) and \(f'(0) = 1.\)

Let \(f\) and \(F\) be elements of \(H(\mathbb{ U } ).\) A function \(f\) is said to be subordinate to \(F\), if there exists a Schwartz function \(w\) analytic in \(\mathbb{ U } \) with \( w (0) = 0 ~~ \text{ and } ~~ |w(z) | < 1, \ \ z \in \mathbb{ U } , \) such that \(f(z) = F(w(z)).\) In this case, we write \( f(z) \prec F(z) ~~~ \text{ or } ~~ f \prec F.\) Furthermore, if the function \(F\) is univalent in \(\mathbb{ U } ,\) then we get the following equivalence [1,2]:

\[f(z) \prec F(z) \Leftrightarrow f(0) = F(0) ~~~ \text{ and}~~ f(\mathbb{ U } ) \prec F(\mathbb{ U } ). \] The method of differential subordinations (also known as the method of admissible functions) was first introduced by Miller and Mocanu in 1978 [3], and the development of the theory was originated in 1981 [4]. All details can be found in the book by Miller and Mocanu [2]. In recent years, numerous authors studied the properties of differential subordinations (see [5,6,7,8], etc.).

Let \( \Psi : \mathbb{C}^3 \times \mathbb{ U } \rightarrow \mathbb{C} \) and let \(h\) be univalent in \(\mathbb{ U } .\) If \(p\) is analytic in \(\mathbb{ U } \) and satisfies the second-order differential subordination:

then \(p\) is called the solution of differential subordination. The univalent function \(q\) is called a dominant of the solution of the differential subordination or, simply, a dominant if \( p \prec q\) for all \(p\) satisfying (2). The dominant \(q_1\) satisfying \(q_1 \prec q\) for all dominants \(q\) of (2) is called the best dominant of (2).

Let us recall lambda function [9] defined by:

\[ \lambda (z, s ) = \sum \limits _{k =2} ^{\infty} \frac{z^k }{ (2k +1) ^k }\] where \(z \in \mathbb{ U } , s \in \mathbb{C},\) when \(|z| < 1, \Re (s) > 1, \) when \(|z| = 1\) and let \(\lambda ^{(-1)} (z, s ) \) be defined such that \[ \lambda (z, s ) * \lambda ^{(-1)} (z, s ) = \frac{1}{(1- z) ^{ \mu +1}}, ~ \mu > -1 .\] We now define \(\left( z \lambda ^{(-1)} (z, s ) \right)\) as: \begin{align*} (z \lambda (z,s)) * \left( z \lambda ^{(-1)} (z, s ) \right) = \frac{z}{(1-z)^{\mu+1}} = z+ \sum \limits _{k =2} ^{\infty} \frac{ ( \mu +1)_{k -1}}{ (k -1) !} z^k , \mu > -1 \end{align*} and obtain the linear operator \( \mathcal{I}_{\mu}^{s} f(z)= \left( z \lambda ^{(-1)} (z, s ) \right) * f(z),\) where \(f \in A, z \in \mathbb{ U } \) and \( \left( z \lambda ^{(-1)} (z, s ) \right) = z+ \sum \limits _{k =2} ^{\infty} \frac{ ( \mu +1)_{k -1} (2k -1) ^s}{ (k -1) !} z^k .\) A simple computation gives us where where \(( \mu)_{k } \) is the Pochhammer symbol defined in terms of the Gamma function by: \[( \mu)_{k } = \frac{\Gamma ( \mu+k )}{\Gamma (\mu)}=\left\{ \begin{array}{ll} 1, & \hbox{ if \(k =0 \);} \\ \mu ( \mu +1) \cdots (\mu +k -1), & \hbox{ if \( k \in \mathbb{N} \)}. \end{array} \right . .\]

Definition 1. Let \(\mathfrak{ L} _{\mu , s } ( \varrho ) \) be a class of function \(f \in A\) satisfying the inequality \[ \mathfrak { \Re } \left( \mathcal{I}_ {\mu}^{s}f(z) \right) \geq \varrho ,\] where \( z\in \mathbb{ U }, \ \ 0 \leq \varrho < 1 \) and \(\mathcal{I}_ {\mu}^{s}f(z) \) is the Lambda operator.

Lemma 1. let \(h\) be a convex function with \(h(0)=a \) and let \(\gamma\in \mathbb { C^* }:= \mathbb { C } \setminus \{ 0 \} \) be a complex number with \( \mathfrak { \Re \{ \gamma \} }\geq 0. \) If \( p \in H[a, n] \) and

then \( p(z) \prec q(z) \prec h(z), \) where \( q(z) = \frac{\gamma}{nz^\frac{\gamma}{n}}\int \limits_{0}^{z}t^\frac{\gamma}{n-1}h(t)dt, ~~ z\in \mathbb{ U }.\) The function \(q\) is convex and is the best dominant for subordination (5).

Lemma 2. [10] Let \( \mathfrak { \Re \{ \mu \} }> 0, ~~n\in \mathbb { N } \) and \( w= \frac{n^2 + |\mu|^2-|n^2-\mu^2|}{4n \mathfrak { \Re \{ \mu \} }} . \) Also, let \(h\) be an analytic function in \(\mathbb { U } \) with \( h(0)=1.\) Suppose that \( \mathfrak { \Re }\left \{ 1+\frac{zh''(z)}{h'(z)} \right \} > -w. \) If \( p(z) = 1+p_nz^n+p_{n+1}z^{n+1}+ \cdots \) is analytic in \(\mathbb { U } \) and

then \( p(z) \prec q(z), \) where \(q\) is a solution of the differential equation \( q(z)+ \frac{n}{\mu}zq'(z) = h(z), ~~q(0)=1, \) given by \( q(z) = \frac{\mu}{nz^\frac{\mu}{n}}\int \limits_{0}^{z}t^\frac{\mu}{n-1}h(t) dt, ~~z\in \mathbb { U } . \) Moreover, \(q\) is the best dominant for the differential subordination (6).

Lemma 3. [11] Let \(r\) be a convex function in \(\mathbb { U } \) and let \( h(z)=r(z)+n\varrho zr'(z), ~~ z\in \mathbb { U } , \) where \(\varrho > 0 ~~ and ~~ n \in \mathbb { N }.\) If \( p(z)= r(0)+p_nz^n+p_{n+1}z^{n+1}+ \cdots , ~~z\in \mathbb { U } , \) is holomorphic in \( \mathbb { U } \) and \( p(z)+\varrho zp'(z)\prec h(z), ~~z\in \mathbb { U } , \) then \( p(z)\prec r(z) \) and this result is sharp.

In the present paper, we use the subordination results from [10] to prove our main results.

2. Main results

Theorem 1. The set \( \mathfrak{ L} _{\mu , s } ( \varrho ) \) is convex.

Proof. Let \( f_j(z)=z+\sum\limits_{k=2}^{\infty}a_{k,j}z^k, ~~ z\in \mathbb { U } , ~~ j= 1, \cdots , m \) be in the class \( \mathfrak{ L} _{\mu , s } ( \varrho ). \) Then, by Definition 1, we get

For any positive numbers \( \varsigma _1, \varsigma _2, \varsigma _3, \cdots ,\varsigma _m \) such that \( \sum\limits_{j=1}^{m}\varsigma _j=1 , \) it is necessary to show that the function \(h(z)=\sum\limits_{j=1}^{m}\varsigma _jf_j(z)\) is an element of \( \mathfrak{ L} _{\mu , s } ( \varrho ) \), i.e., Thus, we have If we differentiate (9) with respect to \(z\), then we obtain \[ (\mathcal{I}_ \mu ^s h(z))' = 1+\sum\limits_{k=2}^{\infty}kL(k,\mu,s) \left \{ \sum\limits_{j=1}^{m} \varsigma _ja_{k,j} \right \} z^{k-1}. \] Thus by using (8), we have \begin{align*} \mathfrak{ \Re } \left \{ (\mathcal{I}_ \mu ^s h(z))' \right \} = 1+\sum\limits_{j=1}^{m}\varsigma _j \mathfrak{ \Re } \left \{ \sum\limits_{k=2}^{\infty}kL(k,\mu,s)a_{k,j}z^{k-1} \right \} > 1+\sum\limits_{j=1}^{m}\varsigma _j(\varrho -1) = \varrho . \end{align*} Hence, inequality (7) is true and we arrive at the desired result.

Theorem 2. Let \(q\) be convex function in \(\mathbb {U} \) with \(q(0)=1\) and \( h(z)=q(z) + \frac{1}{\gamma+1}zq'(z), ~~ z\in \mathbb { U } , \) where \(\gamma\) is a complex number with \(\mathfrak{ \Re } \{ {\gamma} \} > -1\). If \(f\in \mathfrak{ L} _{\mu , s } ( \varrho ) \) and \( \aleph =\Upsilon_\gamma f,\) where

then implies that \( (\mathcal{I}_ \mu ^s \aleph (z))' \prec q(z) \) and this result is sharp.

Proof. In view of equality (10), we can write

Differentiating (12) with respect to \(z,\) we obtain \( (\gamma) \aleph (z)+z \aleph ~'(z)=(\gamma+1)f(z). \) Further, by applying the operator \(\mathcal{I}_ \mu ^s\) to the last equation, we get If we differentiate (13) with respect to \(z\), then we find By using the differential subordination given by (11) in equality (14), we obtain We define Hence, as a result of simple computations, we get \begin{align*} p(z) &= \left \{ z+ \sum\limits_{k=2}^{\infty}L(k,\mu,s)\frac{\gamma+1}{\gamma+k}a_kz^k \right \}' = 1+ p_1z+p_2z^2+ \cdots ,~~p\in H[1,1]. \end{align*} By using (16) in subordination (15), we obtain \begin{equation*} p(z)+\frac{1}{\gamma+1}zp'(z)\prec h(z)=q(z)+\frac{1}{\gamma+1}zq'(z),~~ z\in \mathbb{ U }. \end{equation*} If we use Lemma 2, then we write \( p(z)\prec q(z). \) Thus, we obtained the desired result and \(q\) is the best dominant.

Example 1. If we choose \( \gamma=i+1\) and \(q(z)= \frac{1+z}{1-z}, \) in Theorem 2, then we get \( h(z)=\frac{(i+2)-((i+2)z+2)z}{(i+2)(1-z)^2}. \) If \( f \in \mathfrak{ L} _{\mu , s } ( \varrho ) \) and \(\aleph\) is given as \( \aleph(z)= \Upsilon_if(z)=\frac{i+2}{z^{i+1}}\int\limits_0^zt^if(t)dt, \) then, by virtue of Theorem 2, we find \( (\mathcal{I}_ {\mu}^{s}f(z))'\prec h(z) = \frac{(i+2)-((i+2)z+2)z}{(i+2)(1-z)^2},\) implies \((\mathcal{I}_ {\mu}^{s}f(z))' \prec \frac{1+z}{1-z}.\)

Theorem 3. Let \( \mathfrak{ \Re }{ \left \{ \gamma \right \} } > -1 \) and \( w=\frac{1+|\gamma+1|^2-|\gamma^2+2\gamma|}{4\mathfrak{ \Re }{ \left \{ \gamma+1 \right \}}}. \) Suppose that \(h\) is an analytic function in \(\mathbb{ U } \) with \(h(0)=1\) and that \( \mathfrak { \Re }\left \{ 1+\frac{zh''(z)}{h'(z)} \right \} > -w. \) If \( f \in \mathfrak{ L} _{\mu , s } ( \varrho ) \) and \( \aleph = \Upsilon_\mu ^s f, \) where \(\aleph\) is defined by (10), then

implies that \( (\mathcal{I}_ \mu ^s \aleph(z))' \prec q(z), \) where \(q\) is the solution of the differential equation \( h(z) = q(z) + \frac{1}{\gamma+1}zq'(z), ~~q(0)=1, \) given by \( q(z) = \frac{\gamma+1}{z^{\gamma+1}}\int\limits_0^zt^\gamma f(t)dt. \) Moreover, \(q\) is the best dominant for subordination (17).

Proof. If we choose \( n=1 \) and \( \mu={\gamma+1} \) in Lemma 1, then the proof is obtained by means of the proof of Theorem 3.

Theorem 4. Let

be convex in \( \mathbb { U } \) with \( h(0)=1 .\) If \( f\in A \) and verifies the differential subordination \( (\mathcal{I}_ \mu ^s f(z))' \prec h(z), \) then \( (\mathcal{I}_ \mu ^s \aleph (z))' \prec q(z)=(2\varrho -1)+\frac{2(1-\varrho )(\gamma+1) \tau(\gamma)}{z^{\gamma+1}}, \) where \(\tau\) is given by the formula and \(\aleph\) is given by equation (10). The function \(q\) is convex and is the best dominant.

Proof. If \( h(z)=\frac{1+(2\varrho - 1)z}{1+z}, ~~ 0\leq \varrho < 1, \) then \( h\) is convex and, in view of Theorem 3, we can write \( (\mathcal{I}_ \mu ^s \aleph(z))' \prec q(z). \) Now, by using Lemma 1, we get \begin{align*} q(z) = \frac{\gamma+1}{z^{\gamma+1}}\int\limits_0^zt^\gamma h(t)dt = \frac{\gamma+1}{z^{\gamma+1}}\int\limits_0^zt^\gamma \left \{ \frac{1+(2\varrho -1)t}{1+t} \right \} dt = (2\varrho -1)+\frac{2(1-\varrho )(\gamma+1)}{z^{\gamma+1}}\tau(\gamma), \end{align*} where \(\tau\) is given by (19). Hence, we obtain \begin{equation*} (\mathcal{I}_ \mu ^s \aleph (z))' \prec q(z)=(2\varrho -1)+\frac{2(1-\varrho )(\gamma+1) \tau(\gamma)}{z^{\gamma+1}}. \end{equation*} The function \( q\) is convex. Moreover, it is the best dominant. Hence the theorem is proved.

Theorem 5. If \( 0 \leq \varrho < 1, 0 \leq \mu < 1, \delta \geq 0, \mathfrak{ \Re } \{ {\gamma} \} > -1, \) and \(\aleph= \Upsilon_\gamma f \) is defined by (10), then \( \Upsilon_\gamma( \mathfrak{ L} _{\mu , s } ( \varrho )) \subset \mathfrak{ L} _{\mu , s } ( \rho ), \) where

and \(\tau\) is given by (19).

Proof. Assume that \(h\) is given by equation (18), \(f \in \mathfrak{ L} _{\mu , s } ( \varrho ), \) and \( \aleph = \Upsilon_\gamma f \) is defined by (10). Then \(h\) is convex and, by Theorem 3, we deduce

where \(\tau\) is given by (19). Since \( q \) is convex, \( q( \mathbb{ U }) \) is symmetric about the real axis, and \( \mathfrak{ \Re } \{ {\gamma} \} > -1,\) we find \begin{align*} \mathfrak{ \Re } \left \{ (\mathcal{I}_ \mu ^s \aleph(z))' \right \} \geq \min \limits _{|z|=1} \mathfrak{ \Re } \{ {q(z)} \} = \mathfrak{ \Re } \{ q(1) \}= \rho(\gamma,\varrho ) = (2\varrho -1)+2(1-\varrho )(\gamma+1)(1-\varrho )\tau(\gamma). \end{align*} It follows from inequality (21) that \( \Upsilon_\gamma( \mathfrak{ L} _{\mu , s } ( \varrho )) \subset \mathfrak{ L} _{\mu , s } ( \rho ), \) where \(\rho\) is given by (20). Hence the theorem is proved.

Theorem 6. Let \(q\) be a convex function with \( q(0) = 1 \) and \(h\) be a function such that \( h(z)= q(z) + zq'(z), ~~ z\in \mathbb{ U }. \) If \( f \in A,\) then the subordination

implies that \( \frac{\mathcal{I}_ \mu ^s f(z)}{z} \prec q(z), \) and the result is sharp.

Proof. Let

Differentiating (23), we find \( (\mathcal{I}_ \mu ^s f(z))' = p(z)+zp'(z). \) We now compute \( p(z) \). This gives By using (24) in subordination (22), we find \( p(z)+zp'(z)\prec h(z)=q(z)+zq'(z). \) Hence, by applying Lemma 3, we conclude that \( p(z) \prec q(z) \) i.e., \( \frac{\mathcal{I}_ \mu ^s f(z)}{z} \prec q(z). \) This result is sharp and \(q\) is the best dominant. Hence the theorem is proved.

Example 2. If we take \( \mu = 0 \) and \( s = 1 \) in equality (4) and \( q(z) = \frac{1}{1-z} \) in Theorem 5, then \( h(z)=\frac{1}{(1-z)^2} \) and

Differentiating (25) with respect to \(z\), we get \begin{align*} (I_0^1f(z))' = 1+ \sum\limits_{k=2}^{\infty}\frac{(2k-1)}{(k-1)!}a_kz^{k-1} = 1+p_1z+p_2z^2+ \cdots , ~~ p \in H[1,1]. \end{align*} By using Theorem 5, we find \( (I_0^1f(z))'\prec h(z) = \frac{1}{(1-z)^2}. \) This yields \( \frac{I_0^1f(z)}{z} \prec q(z) = \frac{1}{1-z}. \)

Theorem 7. Let \( h(z)=\frac{1+(2\varrho -1)z}{1+z}, ~~ z\in \mathbb{ U } \) be convex in \( \mathbb { U } \) with \( h(0)=1\) and \( 0 \leq \varrho < 1.\) If \( f \in A \) satisfies the differential subordination

then \( \frac{\mathcal{I}_ \mu ^s f(z)}{z} \prec q(z)= (2\varrho -1)+\frac{2(1-\varrho )ln(1+z)}{z}. \) The function \(q\) is convex and, in addition, it is the best dominant.

Proof. Let

Differentiating (27), we find In view of (28), the differential subordination (26) becomes \( (\mathcal{I}_ \mu ^s f(z))' \prec h(z) = \frac{1+(2\varrho -1)z}{1+z},\) and by using Lemma 1, we deduce \( p(z)\prec q(z)=\frac{1}{z}\int h(t)dt = (2\varrho -1)+\frac{2(1-\varrho )ln(1+z)}{z}. \) Now, by virtue of relation (27) we obtained the desired result.

Corollary 1. If \( f \in \mathfrak{ L} _{\mu , s } ( \varrho ), \) then \( \mathfrak { \Re } \left( \frac{\mathcal{I}_ \mu ^s f(z)}{z} \right) > (2\varrho -1)+2(1-\varrho )ln(2). \)

Proof. If \( f \in \mathfrak { L} _{\mu , s } ( \varrho ), \) then it follows from Definition 1 that \( \mathfrak { \Re } \left \{ (\mathcal{I}_ \mu ^s f(z))' \right \} > \varrho , ~~ z \in \mathbb { U }, \) which is equivalent to \( (\mathcal{I}_ \mu ^s f(z))' \prec h(z) = \frac{1+(2\varrho -1)z}{1+z}. \) Now, by using Theorem 7, we obtain \begin{equation*} \frac{\mathcal{I}_ \mu ^s f(z)}{z} \prec q(z) = (2\varrho -1)+\frac{2(1-\varrho )ln(1+z)}{z}. \end{equation*} Since \(q\) is convex and \(q( \mathbb { U } ) \) is symmetric about the real axis, we conclude that \begin{equation*} \mathfrak { \Re } \left( \frac{\mathcal{I}_ \mu ^s f(z)}{z} \right) > \mathfrak { \Re } (q(1)) = (2\varrho -1)+2(1-\varrho )ln(2). \end{equation*}

Theorem 8. Let \(q\) be a convex function such that \(q(0)=1\) and \(h\) be the function given by the formula \( h(z)=q(z)+zq'(z), ~~z \in \mathbb { U }. \) If \( f \in A \) and verifies the differential subordination

then \( \frac{\mathcal{I}_ \mu ^s f(z)}{\mathcal{I}_ \mu ^s \aleph(z)} \prec q(z), ~~ z \in \mathbb { U }, \) and this result is sharp.

Proof. For function \( f \in A, \) given by Equation (1), we get \begin{equation*} \mathcal{I}_ \mu ^s \aleph (z) = z + \sum\limits_{k=2}^{\infty}L(k,\mu,s)\frac{\gamma+1}{k+\gamma}a_kb_kz^k, ~~z\in \mathbb { U }. \end{equation*} We now consider the function \begin{align*} p(z)=\frac{\mathcal{I}_ \mu ^s f(z)}{\mathcal{I}_ \mu ^s \aleph (z)} = \frac{z+\sum\limits_{k=2}^{\infty}L(k,\mu,s )a_kb_kz^k}{z+\sum\limits_{k=2}^{\infty}L(k,\mu,s)\frac{\gamma+1}{k+\gamma}a_kb_kz^k} = \frac{1+\sum\limits_{k=2}^{\infty}L(k,\mu,s)a_kb_kz^{k-1}}{1+\sum\limits_{k=2}^{\infty}L(k,\mu,s)\frac{\gamma+1}{k+\gamma}a_kb_kz^{k-1}}. \end{align*} In this case, we get \begin{equation*} (p(z))'=\frac{(\mathcal{I}_ \mu ^s f(z))'}{\mathcal{I}_ \mu ^s \aleph(z)} - p(z) \frac{(\mathcal{I}_ \mu ^s \aleph(z))'}{\mathcal{I}_ \mu ^s \aleph(z)}. \end{equation*} Then

By using relation (30) in inequality (29), we obtain \( p(z)+zp'(z)\prec h(z)=q(z)+zq'(z) \) and, by virtue of Lemma 3, \( p(z)\prec q(z), \) i.e., \( \frac{\mathcal{I}_ \mu ^s f(z)}{\mathcal{I}_ \mu ^s \aleph(z)} \prec q(z). \) Hence the theorem is proved.

Acknowledgments

The authors warmly thank the referees for the careful reading of the paper and their comments.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

The authors declare no conflict of interest.