1 Introduction

Let \({\mathcal {A}}\) denote the class of analytic functions f defined in the open unit disc E with the normalizations \(f(0)=f'(0)-1=0\). Let \({\mathcal {A}}_{0}=\left\{ g: \, g(z)=f(z)/z, \, f\in {\mathcal {A}}\right\} \). Denote by S the subclass of \(\mathcal {A}\) consisting of univalent functions in E. A function \(f\in S\) is said to be starlike or convex, if f maps E conformally onto the domains which are respectively, starlike with respect to the origin or convex. The generalization of these two classes are given, respectively, by the following analytic characterizations:

$$\begin{aligned} S^{*}(\beta ) = \left\{ f\in {\mathcal {A}} : \text {Re}\left( \frac{zf'(z)}{f(z)}\right) >\beta , \quad 0\le \beta <1\right\} \end{aligned}$$

and

$$\begin{aligned} K(\beta ) = \left\{ f\in {\mathcal {A}} : \text {Re}\left( 1+\frac{zf''(z)}{f'(z)}\right) >\beta , \quad 0\le \beta <1\right\} . \end{aligned}$$

For \(\beta =0\), we usually set \(S^{*}(0)=S^{*}\) and \(K(0)=K\). A domain D in \(\mathbb {C}\) is close-to-convex if its complement in \(\mathbb {C}\) can be written as union of non-intersecting half lines. Let C denotes the class of close-to-convex normalized analytic functions in E.

For two functions \(f(z)=z+a_{2}z^{2}+a_{3}z^{3}+\cdots \) and \(g(z)=z+b_{2}z^{2}+b_{3}z^{3}+\cdots \) in \(\mathcal {A}\), their Hadamard product (or convolution) is the function \(f *g\) defined by

$$\begin{aligned} (f *g)(z)=z+ \sum _{n=2}^\infty a_nb_nz^n. \end{aligned}$$

For \(\mu >0\) and \(f\in {\mathcal {A}}\), we define the weighted integral transform

$$\begin{aligned} V_{\lambda , \mu }(f)(z)=\left( \int _{0}^{1}\lambda (t)\left( \frac{f(tz)}{t}\right) ^{\mu }dt\right) ^{\frac{1}{\mu }}, \end{aligned}$$
(1.1)

where powers are chosen so as to get the principal branch of \(V_{\lambda , \mu }(f)\) and \(\lambda \) is a non-negative real-valued integrable function satisfying the condition \(\displaystyle \int _{0}^{1}\lambda (t)dt=1\). One can note that \(V_{\lambda , \mu }(f)\) reduces to the linear integral transform \(V_{\lambda , 1}(f)\) for \(\mu =1\), which further contains some of the well-known operators such as Libera, Bernardi and Komatu as its special cases.

For \(\beta <1\), \(\alpha \ge 0\) and \(0\le \gamma \le 1\), let \({\mathcal {P}}_{\gamma }(\alpha ,\beta )\) denote the class of all those analytic functions f in \({\mathcal {A}}\) for which

$$\begin{aligned} \text {Re} \, \left\{ e^{i\eta }\left[ (1-\gamma )\left( \frac{f(z)}{z}\right) ^{\alpha }+\gamma \left( \frac{zf'(z)}{f(z)}\right) \left( \frac{f(z)}{z}\right) ^{\alpha }-\beta \right] \right\} >0, \end{aligned}$$
(1.2)

where \(z\in E\), \(\eta \in {\mathbb {R}}\) and again power is chosen to be the principal one. In 2002, for \(\eta =0\) in (1.2), Liu [14] gave a univalence criterion for functions in the class \({\mathcal {P}}_{\gamma }(\alpha ,\beta )\). For \(\alpha =\gamma =1\), Fournier and Ruscheweyh [12] and Ali and Singh [3] used the Duality Principle [19, 20] to prove starlikeness and convexity of the linear integral transform \(V_{\lambda , \alpha }(f)\), when f varies in the class \({\mathcal {P}}_{\gamma }(\alpha ,\beta )\). For \(\alpha =1\), Kim and Rønning [13] and Choi et al. [9] studied starlikeness and convexity of the linear transform \(V_{\lambda , \alpha }(f)\), \(f\in {\mathcal {P}}_{\gamma }(\alpha ,\beta )\). In 2008, Aghalary et al. [1] discussed the univalence of integral transform \(V_{\lambda , \alpha }(f)\) of the functions f in the class \({\mathcal {P}}_{\gamma }(\alpha ,\beta )\). Many researchers have worked on various generalizations of \({\mathcal {P}}_{\gamma }(\alpha ,\beta )\), for example, see [2, 4, 6, 7, 10, 16–18]. Recently, Ebadian et al. [11] studied the starlikeness of integral transform \(V_{\lambda , \alpha }(f)\) over the functions f in the class \({\mathcal {P}}_{\gamma }(\alpha ,\beta )\).

A function \(f\in {\mathcal {A}}\) is said to be in the class \({\mathcal {S}}_{\delta }(\nu )\) if

$$\begin{aligned} \text {Re} \, \left[ \left( 1+\frac{zf''(z)}{f'(z)}\right) +\left( \nu -1\right) \left( \frac{zf'(z)}{f(z)}\right) \right] >\delta , \end{aligned}$$
(1.3)

for \(\delta <\nu \le 1+\delta \) and \(0\le \delta <1\).

Remark 1.1

We list two interesting facts about the class \({\mathcal {S}}_{\delta }(\nu )\).

  1. (i)

    One can check easily that \(f\in {\mathcal {S}}_{\delta }(\nu )\) if and only if \(\displaystyle \frac{z^{2}f'(z)}{f(z)}\left( \frac{f(z)}{z}\right) ^{\nu }\) is starlike of order \(1+\delta -\nu \), where \(0\le 1+\delta -\nu <1\).

  2. (ii)

    For \(0<\alpha \le 1\), the class \(\displaystyle {\mathcal {S}}_{0}\left( \frac{1}{\alpha }\right) \) becomes a subclass of the class of \(\alpha \)-convex functions defined by Mocanu [15].

The main objective of the present paper is to solve the problem of finding the sharp estimate of the parameter \(\beta \) that ensures \(V_{\lambda , \alpha }(f)\) to be in the class \({\mathcal {S}}_{\delta }(\alpha )\) for \(f\in {\mathcal {P}}_{\gamma }(\alpha ,\beta )\).

2 Preliminaries

We shall need the duality theory of analytic functions to prove our result, so we include here some basic concepts and results from this theory. For a subset \({\mathcal {B}}\subset {\mathcal {A}}_{0}\) we define

$$\begin{aligned} {\mathcal {B}}^{*}=\left\{ g\in {\mathcal {A}}_{0}: \, (f*g)(z)\ne 0, \, z\in E, \, \text {for all} \, f\in {\mathcal {B}}\right\} . \end{aligned}$$

The set \({\mathcal {B}}^{*}\) is called the dual of \({\mathcal {B}}\). Further, the second dual of \({\mathcal {B}}\) is defined as \({\mathcal {B}}^{**}=({\mathcal {B}}^{*})^{*}\). The following result is fundamental in the theory.

Theorem A

([19], Corollary 1.1, Theorem 1.6) Let

$$\begin{aligned} {\mathcal {B}}=\left\{ \beta +(1-\beta )\left( \frac{1+xz}{1+yz}\right) : \, |x|=1, |y|=1\right\} , \, \, \beta \in {\mathbb {R}}, \, \beta \ne 1. \end{aligned}$$

We have

  1. (i)

    \(\displaystyle {\mathcal {B}}^{**}=\left\{ g\in {\mathcal {A}}_{0}: \, \exists \, {\phi }\in {\mathbb {R}} \, \text {such that} \, \text {Re} \, \left( e^{i\phi }(g(z)-\beta )\right) >0, \, \, z\in E\right\} \).

  2. (ii)

    If \(\Gamma _{1}\) and \(\Gamma _{2}\) are two continuous linear functionals on \(\mathcal {B}\) with \(0\not \in \Gamma _{2}\), then for every \(g\in {\mathcal {B}}^{**}\) we can find \({v}\in {\mathcal {B}}\) such that

    $$\begin{aligned} \frac{\Gamma _{1}(g)}{\Gamma _{2}(g)}=\frac{\Gamma _{1}(v)}{\Gamma _{2}(v)}. \end{aligned}$$

The basic reference to this theory is the book by Ruscheweyh [19] (see also [20]). We also need the following to state and prove our result.

Let \({\Lambda }:[0,1]\rightarrow \mathbb {R}\) integrable on [0,1] and positive on (0,1). Further, let

$$\begin{aligned} L_{\Lambda }(f):=\inf _{z\in {E}}\int _{0}^{1}\Lambda (t)\text {Re}\left( \frac{f(tz)}{tz}-\frac{1}{(1+t)^{2}}\right) dt, \quad f\in {C} \end{aligned}$$

and

$$\begin{aligned} L_{\Lambda }({C})=\inf _{f\in {{C}}}L_{\Lambda }(f). \end{aligned}$$

Fournier and Ruscheweyh [12] established the following:

Theorem B

  1. (i)

    If \(\displaystyle \frac{L_{\Lambda }({C})}{1-t^{2}}\) is a decreasing function on (0,1), then \(L_{\Lambda }({C})=0\).

  1. (ii)

    If \(\lambda :[0,1]\rightarrow \mathbb {R}\) is non-negative with \(\int _{0}^{1}\lambda (t)dt=1\), \(\Lambda (t)=\int _{t}^{1}\frac{\lambda (s)}{s}ds\) satisfies \(t\Lambda (t)\rightarrow 0\) for \(t\rightarrow 0^{+}\), then, for \(\beta <1\) given by

    $$\begin{aligned} \frac{\beta }{1-\beta }=-\int _{0}^{1}\lambda (t)\frac{1-t}{1+t}dt, \end{aligned}$$

    we have \(V_{\lambda , 1}({\mathcal {P}}_{1}(1,\beta ))\subset {S}\) and

    $$\begin{aligned} V_{\lambda ,1}({\mathcal {P}}_{1}(1,\beta ))\subset {S^{*}}\Leftrightarrow {L}_{\Lambda }({C})=0. \end{aligned}$$

    The value of \(\beta \) is sharp.

3 Main Results

For \(\alpha >0\) and \(\gamma >0\), define

$$\begin{aligned} \Lambda _{\gamma }(t)=\int _{t}^{1}\frac{\lambda (s)}{s^{\frac{\alpha }{\gamma }}}ds \end{aligned}$$
(3.1)

and

$$\begin{aligned} h(z):=h_{\delta }^{\alpha }(z)=\frac{\left( 1+\frac{\epsilon +1-2\alpha (1-\delta )}{2\alpha (1-\delta )}z\right) }{(1-z)^{2}}, \quad |\epsilon |=1, \end{aligned}$$
(3.2)

where \(0\le \delta <1\). Here for convenience, we write \(\Lambda _{1}(t)=\Lambda (t)\).

Theorem 3.1

Let \(\alpha \ge 1\), \(\gamma >0\) and \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\). Let \(\beta <1\) satisfies the condition

$$\begin{aligned} \frac{\beta -{1}/{2}}{1-\beta }=-\int _{0}^{1}{\lambda }(t)q(t)dt, \end{aligned}$$
(3.3)

where q(t) is the solution of the initial value problem

$$\begin{aligned}&\frac{d}{dt}\left( t^{{\alpha }/{\gamma }}q(t)\right) \nonumber \\&\quad =t^{\frac{\alpha }{\gamma }-1}\left[ \frac{2+(\alpha -2)(1+t)+(1-\alpha +\alpha \mu )t(1+t)-\alpha (1-\alpha +\alpha \mu )(1+t)^{2}}{\alpha \gamma (1-\mu )(1+t)^3}\right] ,\nonumber \\ \end{aligned}$$
(3.4)

with \(q(0)=1\). Assume that \(\displaystyle \lim _{t\rightarrow 0^{+}}t^{{\alpha }/{\gamma }}{\Lambda }_{\gamma }(t)=0\). Then, for \(f\in {\mathcal {P}}_{\gamma }(\alpha , \beta )\), integral operator \(V_{\lambda ,\alpha }(f)\) belongs to the class \({\mathcal {S}}_{\alpha \mu }(\alpha )\) if and only if

$$\begin{aligned}&\text {Re} \, \int _{0}^{1}{{\Lambda }_{\gamma }}(t) t^{\frac{\alpha }{\gamma }-1}\Bigl [\left( tzh'(tz)+{\alpha }{h(tz)}\right) \nonumber \\&\qquad -\Bigl (\frac{2+(\alpha -2)(1+t)+(1-\alpha +\alpha \mu )t(1+t) -\alpha (1-\alpha +\alpha \mu )(1+t)^{2}}{\alpha (1-\mu )(1+t)^3}\Bigr )\Bigr ]dt\nonumber \\&\quad \ge 0, \end{aligned}$$
(3.5)

where \(\Lambda _{\gamma }(t)\) and h(z) are defined by Eqs. (3.1) and (3.2), respectively. The value of \(\beta \) is sharp.

Proof

Define

$$\begin{aligned} H(z)=(1-\gamma )\left( \frac{f(z)}{z}\right) ^{\alpha }+\gamma \left( \frac{zf'(z)}{f(z)}\right) \left( \frac{f(z)}{z}\right) ^{\alpha }. \end{aligned}$$

Then the assumption \(f\in {\mathcal {P}}_{\gamma }(\alpha , \beta )\) means that \(\text {Re} \, \left\{ e^{i\eta }\left( \frac{H(z)-\beta }{1-\beta }\right) \right\} >0\) for some \(\eta \in {\mathbb {R}}\). Then, in the view of the Theorem A, we may restrict our attention to functions \(f\in {\mathcal {P}}_{\gamma }(\alpha , \beta )\) for which

$$\begin{aligned} H(z)=\beta +(1-\beta )\left( \frac{1+xz}{1+yz}\right) , \quad |x|=1, |y|=1. \end{aligned}$$
(3.6)

Thus,

$$\begin{aligned} H(z)=(1-\gamma )\left( \frac{f(z)}{z}\right) ^{\alpha }+\gamma \left( \frac{zf'(z)}{f(z)}\right) \left( \frac{f(z)}{z}\right) ^{\alpha }=\beta +(1-\beta )\left( \frac{1+xz}{1+yz}\right) . \end{aligned}$$

One can easily see that

$$\begin{aligned} G(z):=\left( \frac{f(z)}{z}\right) ^{\alpha }=\beta +\frac{\alpha (1-\beta )}{\gamma {z^{\frac{\alpha }{\gamma }}}}\int _{0}^{z}w^{\frac{\alpha }{\gamma }-1}\frac{1+xw}{1+yw}dw. \end{aligned}$$
(3.7)

Further

$$\begin{aligned} \left( \frac{zf'(z)}{f(z)}\right) \left( \frac{f(z)}{z}\right) ^{\alpha }=\frac{z}{\alpha }\frac{d}{dz}G(z)+G(z). \end{aligned}$$
(3.8)

Now, we let \(F(z)=V_{\lambda ,\alpha }(f)(z)\), where \(V_{\lambda ,\alpha }(f)\) is defined by (1.1). It follows from the part (i) of the Remark 1.1 that

$$\begin{aligned} F\in {\mathcal {S}}_{\alpha \mu }(\alpha ) \quad \text {iff} \quad \frac{z^{2}F'(z)}{F(z)}\left( \frac{F(z)}{z}\right) ^{\alpha }\in S^{*}(\delta ), \end{aligned}$$
(3.9)

where \(\delta =1+\alpha (\mu -1)\). Since \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\), therefore \(0\le \delta <1/2\). A well-known result from the theory of convolutions ([19], pp. 94) (also see [20]) states that

$$\begin{aligned} {\mathbb {F}}\in S^{*}(\delta ) \quad \Leftrightarrow \quad \frac{1}{z}({\mathbb {F}}*zh)(z)\ne 0, \quad z\in E, \end{aligned}$$

where \(h(z):=h_{\delta }^{\alpha }(z)\) is as defined in Eq. (3.2). Therefore

$$\begin{aligned} \frac{z^{2}F'(z)}{F(z)}\left( \frac{F(z)}{z}\right) ^{\alpha }\in S^{*}(\delta ) \end{aligned}$$

if and only if,

$$\begin{aligned} \left( \frac{zF'(z)}{F(z)}\left( \frac{F(z)}{z}\right) ^{\alpha }*{h}(z)\right) \ne 0. \end{aligned}$$

From the definition of F(z), one can see that

$$\begin{aligned} \frac{zF'(z)}{F(z)}\left( \frac{F(z)}{z}\right) ^{\alpha }= & {} \int _{0}^{1}{\lambda (t)}\frac{tzf'(tz)}{f(tz)}\left( \frac{f(tz)}{tz}\right) ^{\alpha }dt\\= & {} \int _{0}^{1}\frac{\lambda (t)}{1-tz}dt*\frac{zf'(z)}{f(z)}\left( \frac{f(z)}{z}\right) ^{\alpha }. \end{aligned}$$

In view of above equality, we see that \(F\in S^{*}(\delta )\) if and only if

$$\begin{aligned} 0\ne & {} \int _{0}^{1}\frac{\lambda (t)}{1-tz}dt*\frac{zf'(z)}{f(z)}\left( \frac{f(z)}{z}\right) ^{\alpha }*{{h}(z)}\\= & {} \int _{0}^{1}{\lambda (t)}h(tz)dt*\left( \frac{z}{\alpha }\frac{d}{dz}G(z)+G(z)\right) \quad \quad (\text {using (3.8)})\\= & {} \int _{0}^{1}{\lambda (t)}\left( \frac{1}{\alpha }tzh'(tz)+h(tz)\right) dt*{G(z)}\\= & {} \int _{0}^{1}{\lambda (t)}\left( \frac{1}{\alpha }tzh'(tz)+h(tz)\right) dt\\&*\left( \frac{\alpha (1-\beta )}{\gamma {z^{\frac{\alpha }{\gamma }}}}\int _{0}^{z}w^{\frac{\alpha }{\gamma }-1}\frac{1+xw}{1+yw}dw+\beta \right) \quad (\text {using (3.7)})\\= & {} (1-\beta )\left\{ \int _{0}^{1}{\lambda (t)}\left[ \frac{\alpha }{\gamma {z}^{{\alpha }/{\gamma }}}\int _{0}^{z}{w}^{\frac{\alpha }{\gamma }-1}\left( \frac{1}{\alpha }twh'(tw)+h(tw)\right) dw+\frac{\beta }{1-\beta }\right] dt\right\} \\&*\frac{1+xz}{1+yz}. \end{aligned}$$

Using another well-known result from the convolution theory [20, pp. 23], the last condition is equivalent to

$$\begin{aligned}&\text {Re} \, (1-\beta )\left\{ \int _{0}^{1}{\lambda (t)}\left[ \frac{\alpha }{\gamma {z}^{{\alpha }/{\gamma }}}\int _{0}^{z}{w}^{\frac{\alpha }{\gamma }-1}\left( \frac{1}{\alpha }twh'(tw)+h(tw)\right) dw+\frac{\beta }{1-\beta }\right] dt\right\} \ge {1}/{2}\\&{\text {or to}} \text { Re} \, (1-\beta )\left\{ \int _{0}^{1}{\lambda (t)}\left[ \frac{\alpha }{\gamma {z}^{{\alpha }/{\gamma }}}\int _{0}^{z}{w}^{\frac{\alpha }{\gamma }-1}\left( \frac{1}{\alpha }twh'(tw)+h(tw)\right) dw+\frac{\beta -1/2}{1-\beta }\right] dt\right\} \ge 0\\&{\text {or to}} \text { Re} \, \left\{ \int _{0}^{1}\frac{\lambda (t)}{{t}^{{\alpha }/{\gamma }}}\left[ \frac{\alpha {t}^{{\alpha }/{\gamma }}}{\gamma {z}^{{\alpha }/{\gamma }}}\int _{0}^{z}{w}^{\frac{\alpha }{\gamma }-1}\left( \frac{1}{\alpha }twh'(tw)+h(tw)\right) dw-{t}^{{\alpha }/{\gamma }}q(t)\right] dt\right\} \ge 0\\&\quad (\text {using (3.3)}). \end{aligned}$$

Since, \({{\Lambda }_{\gamma }}'(t)=-{\lambda (t)}/{t^{\frac{\alpha }{\gamma }}}\), therefore applying integration by parts, we have

$$\begin{aligned}&\text {Re} \, \int _{0}^{1}{{\Lambda }_{\gamma }}(t)t^{\frac{\alpha }{\gamma }-1}\Bigl [\frac{\alpha }{\gamma }\left( \frac{1}{\alpha }tzh'(tz)+{h(tz)}\right) \\&\quad -\Bigl (\frac{2+(\alpha -2)(1+t)+(1-\alpha +\alpha \mu )t(1+t)-\alpha (1-\alpha +\alpha \mu )(1+t)^{2}}{\alpha \gamma (1-\mu )(1+t)^3}\Bigr )\Bigr ]dt\ge 0 \end{aligned}$$

or equivalently,

$$\begin{aligned}&\text {Re} \, \int _{0}^{1}{{\Lambda }_{\gamma }}(t)t^{\frac{\alpha }{\gamma }-1}\Bigl [\left( tzh'(tz)+{\alpha }{h(tz)}\right) \\&\quad -\Bigl (\frac{2+(\alpha -2)(1+t)+(1-\alpha +\alpha \mu )t(1+t)-\alpha (1-\alpha +\alpha \mu )(1+t)^{2}}{\alpha (1-\mu )(1+t)^3}\Bigr )\Bigr ]dt\ge 0. \end{aligned}$$

Finally, to prove the sharpness, let \(f\in {\mathcal {P}}_{\gamma }(\alpha , \beta )\) be of the form for which

$$\begin{aligned}(1-\gamma )\left( \frac{f(z)}{z}\right) ^{\alpha }+\gamma \left( \frac{zf'(z)}{f(z)}\right) \left( \frac{f(z)}{z}\right) ^{\alpha }=\beta +(1-\beta )\frac{1+z}{1-z}.\end{aligned}$$

Using a series expansion we obtain that

$$\begin{aligned} \left( \frac{f(z)}{z}\right) ^{\alpha }=1+2(1-\beta )\sum _{n=1}^{\infty }\frac{1}{1+\frac{\gamma }{\alpha }n}z^{n}. \end{aligned}$$

Note that

$$\begin{aligned} \frac{zf'(z)}{f(z)}\left( \frac{f(z)}{z}\right) ^{\alpha }= & {} \frac{z}{\alpha }\frac{d}{dz}\left( \frac{f(z)}{z}\right) ^{\alpha }+\left( \frac{f(z)}{z}\right) ^{\alpha }\\= & {} 1+2(1-\beta )\sum _{n=1}^{\infty }\frac{1+\frac{n}{\alpha }}{1+\frac{\gamma }{\alpha }n}z^{n}. \end{aligned}$$

Further, for

$$\begin{aligned} F(z)=V_{\lambda ,\alpha }(f)(z):=\left( \int _{0}^{1}\lambda (t)\left( \frac{f(tz)}{t}\right) ^{\alpha }dt\right) ^{1/{\alpha }}, \end{aligned}$$

let

$$\begin{aligned} \mathcal {G}(z)=\frac{z^{2}F'(z)}{F(z)}\left( \frac{F(z)}{z}\right) ^{\alpha }. \end{aligned}$$

As seen before,

$$\begin{aligned} \mathcal {G}(z)= & {} \frac{z^{2}F'(z)}{F(z)}\left( \frac{F(z)}{z}\right) ^{\alpha }\nonumber \\= & {} z\left[ \int _{0}^{1}\frac{\lambda (t)}{1-tz}dt*\frac{zf'(z)}{f(z)}\left( \frac{f(z)}{z}\right) ^{\alpha }\right] \nonumber \\= & {} z+2(1-\beta )\sum _{n=1}^{\infty }\frac{1+\frac{n}{\alpha }}{1+\frac{\gamma }{\alpha }n}\tau _{n}z^{n+1}, \end{aligned}$$
(3.10)

where \(\tau _{n}=\int _{0}^{1}\lambda (t)t^{n}dt\).

From (3.4), it is a simple exercise to write q(t) in a series expansion as

$$\begin{aligned} q(t)=1+\sum _{n=1}^{\infty }(-1)^{n}\frac{\left( 1+\frac{n}{\alpha }\right) (n+\alpha -\alpha \mu )}{\alpha \left( 1+\frac{\gamma }{\alpha }n\right) (1-\mu )}t^{n}. \end{aligned}$$

Therefore, in view of (3.3), we have

$$\begin{aligned} \frac{\beta -\frac{1}{2}}{1-\beta }= & {} -\int _{0}^{1}\lambda (t)q(t)dt\\= & {} -\int _{0}^{1}\lambda (t)\left[ 1+\sum _{n=1}^{\infty }(-1)^{n}\frac{\left( 1+\frac{n}{\alpha }\right) (n+\alpha -\alpha \mu )}{\alpha \left( 1+\frac{\gamma }{\alpha }n\right) (1-\mu )}t^{n}\right] dt\\= & {} -1-\sum _{n=1}^{\infty }(-1)^{n}\frac{\left( 1+\frac{n}{\alpha }\right) (n+\alpha -\alpha \mu )}{\alpha \left( 1+\frac{\gamma }{\alpha }n\right) (1-\mu )}\int _{0}^{1}\lambda (t)t^{n}dt. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{1}{1-\beta }=-\frac{2}{1-\mu }\sum _{n=1}^{\infty }\frac{(-1)^{n}\tau _{n}\left( 1+\frac{n}{\alpha }\right) (n+\alpha -\alpha \mu )}{\alpha \left( 1+\frac{\gamma }{\alpha }n\right) }. \end{aligned}$$
(3.11)

Finally, we see that

$$\begin{aligned} \mathcal {G}'(z)= & {} 1+2(1-\beta )\sum _{n=1}^{\infty }\frac{\left( 1+\frac{n}{\alpha }\right) (n+1)}{1+\frac{\gamma }{\alpha }n}\tau _{n}z^{n}\\= & {} 1+2(1-\beta )\sum _{n=1}^{\infty }\frac{\left( 1+\frac{n}{\alpha }\right) (n+\alpha -\alpha \mu )}{1+\frac{\gamma }{\alpha }n}\tau _{n}z^{n}\\&+\left( 1-\alpha +\alpha \mu \right) 2(1-\beta )\sum _{n=1}^{\infty }\frac{\left( 1+\frac{n}{\alpha }\right) }{1+\frac{\gamma }{\alpha }n}\tau _{n}z^{n}. \end{aligned}$$

For \(z=-1\), we have

$$\begin{aligned} \mathcal {G}'(z)|_{z=-1}= & {} 1+2(1-\beta ) \left\{ \sum _{n=1}^{\infty }\frac{\left( 1+\frac{n}{\alpha }\right) (n+\alpha -\alpha \mu )}{1+\frac{\gamma }{\alpha }n}\tau _{n}(-1)^{n}\right. \\&\left. +\left( 1-\alpha +\alpha \mu \right) \sum _{n=1}^{\infty }\frac{\left( 1+\frac{n}{\alpha }\right) }{1+\frac{\gamma }{\alpha }n}\tau _{n}(-1)^{n}\right\} \\= & {} 1-\alpha (1-\mu )+\left( 1-\alpha +\alpha \mu \right) \left( -G(-1)-1\right) \quad \quad \\&(\text {using (3.10) and (3.11)})\\= & {} -\left( 1-\alpha +\alpha \mu \right) {G(-1)}. \end{aligned}$$

Thus \(z{\mathcal {G}}'(z)/{\mathcal {G}}(z)\) at \(z=-1\) equals \(\left( 1-\alpha +\alpha \mu \right) =\delta \). This implies that the result is sharp.\(\square \)

Remark 3.1

Theorem 3.1 yields several known results.

  1. (1)

    For \(\alpha =1\), Theorem 3.1 reduces to Theorem 2.3 in [5].

  2. (2)

    For \(\alpha =\gamma =1\), Theorem 3.1 reduces to Theorem 1 in [3].

4 Consequences of Theorem 3.1

For the function \({{\Lambda }_{\gamma }}(t)\), we define

$$\begin{aligned} \mathfrak {M}_{\gamma }(h):=&\,\text {Re} \, \int _{0}^{1}{{\Lambda }_{\gamma }}(t) t^{\frac{\alpha }{\gamma }-1}\Bigl [\left( tzh'(tz)+{\alpha }{h(tz)}\right) \\&-\Bigl (\frac{2+(\alpha -2)(1+t)+(1-\alpha +\alpha \mu )t(1+t)-\alpha (1-\alpha +\alpha \mu )(1+t)^{2}}{\alpha (1-\mu )(1+t)^3}\Bigr )\Bigr ]dt\\ \ge&\,0, \end{aligned}$$

where h is defined by the Eq. (3.2).

Theorem 4.1

Let \(\alpha \ge 1\) and \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\). Assume that \({{\Lambda }_{\gamma }}(t)\) is integrable on [0,1] and positive on (0,1) such that

$$\begin{aligned} \lim _{t\rightarrow 0^{+}}t^{{\alpha }/{\gamma }}{\Lambda }_{\gamma }(t)=0. \end{aligned}$$

Further suppose that

$$\begin{aligned} \frac{\left( -t{{\Lambda }_{\gamma }}'(t)+\left( \alpha -{\alpha }/{\gamma }\right) {{\Lambda }_{\gamma }}(t)\right) t^{\frac{1}{\gamma }(\alpha -1)}}{\left( \log \frac{1}{t}\right) ^{3-2\alpha (1-\mu )}}, \end{aligned}$$
(4.1)

is decreasing on (0,1) and \(\lambda (t)\) is decreasing in [0,1]. Then, for \(1/2\le \gamma \le 1\), one has \(\inf _{z\in {E}}\mathfrak {M}_{\gamma }(h)\ge 0\) for h defined by the equation (3.2).

Proof

As defined above

$$\begin{aligned} \mathfrak {M}_{\gamma }(h)= & {} \text {Re} \, \int _{0}^{1}{{\Lambda }_{\gamma }}(t)t^{\frac{\alpha }{\gamma }-1} \left[ \left( tzh'(tz)+{\alpha }{h(tz)}\right) \right. \\&\left. -\left( \frac{2+(\alpha -2)(1+t)+(1-\alpha +\alpha \mu )t(1+t)-\alpha (1-\alpha +\alpha \mu )(1+t)^{2}}{\alpha (1-\mu )(1+t)^3}\right) \right] dt\\= & {} \text {Re} \, \int _{0}^{1}{{\Lambda }_{\gamma }}(t)t^{\frac{\alpha }{\gamma } -1}\left[ (\alpha -1)\left( {h(tz)} -\frac{1-(1-\alpha +\alpha \mu )(1+t)}{\alpha (1-\mu )(1+t)^{2}}\right) \right. \\&\left. +\frac{d}{dt}t\left( {h(tz)}-\frac{1-(1-\alpha +\alpha \mu )(1+t)}{\alpha (1-\mu )(1+t)^{2}}\right) \right] dt\\= & {} \text {Re} \, \int _{0}^{1}{{\Lambda }_{\gamma }}(t)t^{\frac{\alpha }{\gamma }-\alpha }\frac{d}{dt}\left[ t^{\alpha }\left( {h(tz)}-\frac{1-(1-\alpha +\alpha \mu )(1+t)}{\alpha (1-\mu )(1+t)^{2}}\right) \right] dt. \end{aligned}$$

Integration by parts shows that

$$\begin{aligned} \mathfrak {M}_{\gamma }(h)= & {} \text {Re} \, \int _{0}^{1}\left( -t{{\Lambda }_{\gamma }}'(t)+\left( \alpha -\frac{\alpha }{\gamma }\right) {{\Lambda }_{\gamma }}(t)\right) t^{\frac{\alpha }{\gamma }-1}\\&\qquad \times \left( {h(tz)}-\frac{1-(1-\alpha +\alpha \mu )(1+t)}{\alpha (1-\mu )(1+t)^{2}}\right) dt. \end{aligned}$$

Note that

$$\begin{aligned} \frac{d}{dt}\left( -t{{\Lambda }_{\gamma }}'(t)+\left( \alpha -\frac{\alpha }{\gamma }\right) {{\Lambda }_{\gamma }}(t)\right) =(1-\alpha )t^{-{\alpha }/{\gamma }}{\lambda }(t)+t^{1-\frac{\alpha }{\gamma }}{\lambda }'(t)<0, \end{aligned}$$

as \(\alpha \ge 1\) and because of our hypotheses \(\lambda (t)\) is decreasing and non-negative on [0,1] so that \(\lambda (t)\ge \lambda (1)\ge 0\). By Eq. (4.2), above inequality implies that

$$\begin{aligned} -t{{\Lambda }_{\gamma }}'(t)+\left( \alpha -\frac{\alpha }{\gamma }\right) {{\Lambda }_{\gamma }}(t)>0, \quad t\in (0,1). \end{aligned}$$

Thus, the above condition along with Eq. (4.2) and Theorem B yield that \(\mathfrak {M}_{\gamma }(h)\ge 0\), which is equivalent to \(\inf _{z\in {E}}\mathfrak {M}_{\gamma }(h)\ge 0\). \(\square \)

For \(\alpha =1\) in Theorem 4.1 corresponds to the following result of Balasubramanian et al. [5]:

Corollary 4.1

Let \(1/2\le \gamma \le 1\) and \({{\Lambda }_{\gamma }}(t)\) be an integrable function on [0,1] and positive on (0,1) such that

$$\begin{aligned} \lim _{t\rightarrow 0^{+}}t^{{1}/{\gamma }}{\Lambda }_{\gamma }(t)=0. \end{aligned}$$

Further suppose that for \(0\le \mu \le 1/2\),

$$\begin{aligned} \frac{\left( -t{{\Lambda }_{\gamma }}'(t)+\left( 1-\frac{1}{\gamma }\right) {{\Lambda }_{\gamma }}(t)\right) }{\left( \log \frac{1}{t}\right) ^{1+2\mu }}, \end{aligned}$$
(4.2)

is decreasing on (0,1) and \(\lambda (t)\) is decreasing in [0,1]. Then, we have

$$\begin{aligned} \inf _{z\in {E}}\mathfrak {M}_{\gamma }(h)\ge 0 \end{aligned}$$

for h defined by the Eq. (3.2).

5 Applications

In 1955, Bazilevič [8] introduced the class of Bazilevič functions. Later in 1968, Thomas [22] defined the class \(B(\alpha )\) of Bazilevič function of type \(\alpha \) as follows:

$$\begin{aligned} B(\alpha )=\left\{ f\in {\mathcal {A}} : \text {Re}\left[ \frac{zf'(z)}{f(z)}\left( \frac{f(z)}{g(z)}\right) ^{\alpha }\right] >0\right\} , \end{aligned}$$

for some \(g\in {S}^{*}\). In [21], Singh considered the subclass \(B_{1}(\alpha )\) of \(B(\alpha )\) obtained by taking \(g(z)\equiv z\). Taking \(\gamma =1\) in \({\mathcal {P}}_{\gamma }(\alpha ,\beta )\), we obtain the following subclass:

$$\begin{aligned} {\mathcal {P}}(\alpha ,\beta ,\eta )=\left\{ f\in {\mathcal {A}} : \text {Re} \, \left\{ e^{i\eta }\left[ \left( \frac{zf'(z)}{f(z)}\right) \left( \frac{f(z)}{z}\right) ^{\alpha }-\beta \right] \right\} >0\right\} , \end{aligned}$$

where \(z\in E\), \(\eta \in {\mathbb {R}}\). It can be easily seen that the subclass \({\mathcal {P}}(\alpha ,0,0)\equiv B_{1}(\alpha )\), which is the subclass of Bazilevič functions and hence is the subclass of univalent functions S. Clearly, \({\mathcal {P}}(\alpha ,\beta ,\eta )\) is the larger class.

Writing \(\gamma =1\) in Theorems 3.1 and 4.1, and combining the results, we have the following:

Theorem 5.1

Assume that \(\alpha \ge 1\) and \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\). Let \(\beta <1\) satisfies the condition

$$\begin{aligned} \frac{\beta -{1}/{2}}{1-\beta }=-\int _{0}^{1}{\lambda }(t)q(t)dt, \end{aligned}$$

where q(t) is the solution of the initial value problem

$$\begin{aligned}&\frac{d}{dt}\left( t^{{\alpha }}q(t)\right) \nonumber \\&\quad =t^{{\alpha }-1}\left[ \frac{2+(\alpha -2)(1+t)+(1-\alpha +\alpha \mu )t(1+t)-\alpha (1-\alpha +\alpha \mu )(1+t)^{2}}{\alpha (1-\mu )(1+t)^3}\right] ,\nonumber \\ \end{aligned}$$
(5.1)

with \(q(0)=1\). Assume that \({\Lambda }(t)\) is integrable on [0,1] and positive on (0,1) such that \(\displaystyle \lim _{t\rightarrow 0^{+}}t^{{\alpha }}{\Lambda }(t)=0.\) If \(f\in {\mathcal {P}}(\alpha , \beta , \eta )\) and \({-t^{\alpha }{{\Lambda }}'(t)}/{\left( \log \frac{1}{t}\right) ^{3-2\alpha (1-\mu )}}\) is decreasing on (0,1), then integral operator \(V_{\lambda ,\alpha }(f)\) belongs to the class \({\mathcal {S}}_{\alpha \mu }(\alpha )\). The value of \(\beta \) is sharp.\(\square \)

To apply Theorem 5.1, it is sufficient to show that the function

$$\begin{aligned} g(t):=\frac{-t^{\alpha }{{\Lambda }}'(t)}{\left( \log \frac{1}{t}\right) ^{3-2\alpha (1-\mu )}} \end{aligned}$$

is decreasing in the interval (0,1). For \(\alpha \ge 1\) and \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\), it is obvious that

$$\begin{aligned} g(t)=\frac{\lambda (t)}{\left( \log \frac{1}{t}\right) ^{3-2\alpha (1-\mu )}} \end{aligned}$$

is decreasing in the interval (0, 1) provided

$$\begin{aligned} \left( \log \frac{1}{t}\right) {\lambda }'(t)+(3-2\alpha (1-\mu ))\frac{\lambda (t)}{t}\le 0. \end{aligned}$$
(5.2)

\(\square \)

In this section we shall look at some particular cases where we see how our results improve earlier results. Some well-known generalized integral operators are considered, and conditions are obtained under which these integral operators of Bazilevič functions belong to the class \({\mathcal {S}}_{\alpha \mu }(\alpha )\).

First, let \(\lambda \) be defined by

$$\begin{aligned} {\lambda }(t)=\frac{(1+a)^{p}}{\Gamma (p)}t^{a}\left( \log \frac{1}{t}\right) ^{p-1}, \ \ a>-1, \ \ p\ge 0. \end{aligned}$$

Then the integral operator

$$\begin{aligned} \mathcal {L}_{a}^{p,\alpha }[f](z)=\left( \frac{(1+a)^{p}}{\Gamma (p)}\int _{0}^{1}t^{a}\left( \log \frac{1}{t}\right) ^{p-1}\left( \frac{f(tz)}{t}\right) ^{\alpha }dt\right) ^{1/{\alpha }} \end{aligned}$$
(5.3)

is the generalized Komatu operator. Clearly, for \(\alpha =1\), generalized Komatu operator \(\mathcal {L}_{a}^{p,\alpha }\) reduces to Komatu operator \(\mathcal {L}_{a}^{p}\). For this particular case of \(\lambda \), the following result holds.

Theorem 5.2

Assume that \(\alpha \ge 1\), \(a>-1\), \(p\ge 1\) and \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\). Let \(\beta <1\) satisfy

$$\begin{aligned} \frac{\beta -{1}/{2}}{1-\beta }=-\frac{(1+a)^{p}}{\Gamma (p)}\int _{0}^{1}t^{a}\left( \log ({1}/{t})\right) ^{p-1}q(t)dt, \end{aligned}$$

where q is given by (5.1). If \(f\in {\mathcal {P}}(\alpha , \beta , \eta )\), then the generalized Komatu operator \(\mathcal {L}_{a}^{p,\alpha }[f]\) belongs to the class \({\mathcal {S}}_{\alpha \mu }(\alpha )\) if \(-1<a\le 0\) and \(p\ge 4-2\alpha (1-\mu )\). The value of \(\beta \) is sharp.

Proof

Set

$$\begin{aligned} {\lambda }(t)=\frac{(1+a)^{p}}{\Gamma (p)}t^{a}\left( \log \frac{1}{t}\right) ^{p-1}. \end{aligned}$$

Taking logarithmic derivative of \(\lambda (t)\), we have

$$\begin{aligned} \frac{{\lambda }'(t)}{\lambda (t)}=\frac{a}{t}-\frac{(p-1)}{t\log \frac{1}{t}}. \end{aligned}$$

Further, substituting values of \(\lambda (t)\) and \(\lambda '(t)\) in Eq. (5.2), we get

$$\begin{aligned} a\log \frac{1}{t}-p-2\alpha (1-\mu )+4\le 0. \end{aligned}$$

Since \(t<1\) implies \(\log \frac{1}{t}\ge 0\), thus condition (5.2) is satisfied whenever \(-1<a\le 0\) and \(p\ge 4-2\alpha (1-\mu )\). This completes the proof. \(\square \)

Taking \(\eta =0\) in Theorem 5.2, we obtain the following corollary:

Corollary 5.1

Assume that \(\alpha \ge 1\), \(a>-1\), \(p\ge 1\) and \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\). Let \(\beta <1\) satisfy

$$\begin{aligned} \frac{\beta -{1}/{2}}{1-\beta }=-\frac{(1+a)^{p}}{\Gamma (p)}\int _{0}^{1}t^{a}\left( \log ({1}/{t})\right) ^{p-1}q(t)dt, \end{aligned}$$

where q is given by (5.1). If f is a Bazilevič function in the class \({\mathcal {P}}(\alpha , \beta , 0)\), then the generalized Komatu operator \(\mathcal {L}_{a}^{p,\alpha }[f]\) belongs to the class \({\mathcal {S}}_{\alpha \mu }(\alpha )\) if \(-1<a\le 0\) and \(p\ge 4-2\alpha (1-\mu )\). The value of \(\beta \) is sharp.

Taking \(\alpha =1\) in Theorem 5.2 leads to an improvement of a particular case of Theorem 3.3 obtained by Balasubramanian et al. [5], which we state as a theorem.

Theorem 5.3

Assume that \(a>-1\), \(p\ge 1\) and \(0\le \mu \le \frac{1}{2}\). Let \(\beta <1\) satisfy

$$\begin{aligned} \frac{\beta -{1}/{2}}{1-\beta }=-\frac{(1+a)^{p}}{\Gamma (p)}\int _{0}^{1}t^{a}\left( \log ({1}/{t})\right) ^{p-1}q(t)dt, \end{aligned}$$

where q is given by

$$\begin{aligned} \frac{d}{dt}\left[ tq(t)\right] =\left[ \frac{1-\mu -(1+\mu )t}{(1-\mu )(1+t)^3}\right] , \end{aligned}$$

with \(q(0)=1\). If \(f\in {\mathcal {P}}(1, \beta , \eta )\), then the Komatu operator \(\mathcal {L}_{a}^{p}[f]\) is convex of order \(\mu \) if \(-1<a\le 0\) and \(p\ge 2(1+\mu )\). The value of \(\beta \) is sharp.

For another choice of \(\lambda \), let it now be given by

$$\begin{aligned} {\lambda }(t)=K{t}^{B-1}(1-t)^{C-A-B}{_2F_{1}(C-A, 1-A; C-A-B+1; 1-t)}, \end{aligned}$$

where \(A, B, C>0\) and K is a constant satisfying the normalization condition that \(\int _{0}^{1}{\lambda }(t)dt=1\). The integral transform \(V_{\lambda ,\alpha }\) in this case takes the form

$$\begin{aligned} H_{A,B,C,\alpha }[f](z)= & {} \left( K\int _{0}^{1}{t}^{B-1}(1-t)^{C-A-B} {}_{2}F_{1}(C-A, 1-A;\right. \\&\left. C-A-B+1; 1-t)\left( \frac{f(tz)}{t}\right) ^{\alpha }dt\right) ^{1/{\alpha }}, \end{aligned}$$

which is the generalized Hohlov operator. Clearly, for \(\alpha =1\), generalized Hohlov operator \(H_{A,B,C,\alpha }\) reduces to Hohlov operator \(H_{A,B,C}\). For this \(\lambda \), the following result holds.

Theorem 5.4

Assume that \(A,B,C>0\), \(\alpha \ge 1\) and \(1-\frac{1}{\alpha }\le \mu \le 1-\frac{1}{2\alpha }\). Let \(\beta <1\) satisfy

$$\begin{aligned} \frac{\beta -{1}/{2}}{1-\beta }= & {} -K\int _{0}^{1}{t}^{B-1} (1-t)^{C-A-B} {}_{2}F_{1}\\&(C-A, 1-A; C-A-B+1; 1-t)q(t)dt, \end{aligned}$$

where q is given by (5.1). If \(f\in {\mathcal {P}}(\alpha , \beta , \eta )\), then the generalized Hohlov operator \(H_{A,B,C,\alpha }[f]\) belongs to the class \({\mathcal {S}}_{\alpha \mu }(\alpha )\) if

$$\begin{aligned} 0<A\le 1, \quad 0<B\le 2-2\alpha (1-\mu ) \ \ \text {and} \ \ C\ge A+B+3-2\alpha (1-\mu ). \end{aligned}$$
(5.4)

The value of \(\beta \) is sharp.

Proof

Here

$$\begin{aligned} {\lambda }(t)=K{t}^{B-1}(1-t)^{C-A-B}{_2F_{1}(C-A, 1-A; C-A-B+1; 1-t)}. \end{aligned}$$

Taking logarithmic derivative of \(\lambda (t)\), we have

$$\begin{aligned} \frac{{\lambda }'(t)}{\lambda (t)}=\frac{B-1}{t}-\frac{C-A-B}{1-t}-\frac{\phi '(1-t)}{\phi (1-t)}, \end{aligned}$$

where \(\phi (1-t)={_2F_{1}(C-A, 1-A; C-A-B+1; 1-t)}\). One can see that under the given hypothesis the function \(\phi (1-t)\) and \(\phi '(1-t)\) are non-negative for \(t\in (0,1)\). The function \(\lambda \) satisfies condition (5.2) if

$$\begin{aligned}&t(1-t)\log \left( \frac{1}{t}\right) \frac{\phi '(1-t)}{\phi (1-t)}+(3-2\alpha (1-\mu )) \left( \log \left( \frac{1}{t}\right) +t-1\right) \\&+\,(2\alpha (1-\mu )-2-B)(1-t)\log \left( \frac{1}{t}\right) \ge 0. \end{aligned}$$

Since \(t<1\) implies \(\log \frac{1}{t}\ge 0\), thus the above inequality is satisfied whenever A, B and C satisfy condition (5.4). \(\square \)

Remark 5.1

For \(\alpha =1\), Theorem 5.4 improves the particular case of Theorem 3.2 obtained in [5] in the sense that the condition \(0<B\le \frac{1}{2}-\mu \) is now replaced by \(0<B\le 2\mu \).