Duality and Integral Transform of a Class of Analytic Functions

Abstract

For $\alpha , \gamma \ge 0$ and $\beta <1$, let ${\mathcal {W}}_{\beta }(\alpha , \gamma )$ denote the class of all normalized analytic functions $f$ in the open unit disk $E=\{z:|z|<1\}$ such that

$$\begin{aligned} \mathfrak {R}e^{i\phi }\left( (1-\alpha +2\gamma )\frac{f(z)}{z}+(\alpha -2\gamma )f'(z)+\gamma zf''(z)-\beta \right) >0, \, \, \, z\in E \end{aligned}$$

for some $\phi \in {\mathbb {R}}$. For $f\in {{\mathcal {W}}_{\beta }(\alpha , \gamma )}$, we consider the integral transform

$$\begin{aligned} V_{\lambda }(f)(z):=\int _{0}^{1}\lambda (t)\frac{f(tz)}{t}\mathrm{d}t, \end{aligned}$$

where $\lambda$ is a non-negative real-valued integrable function satisfying the condition $\int _{0}^{1}\lambda (t)\mathrm{d}t=1$. In a very recent paper, Ali et al. (J Math Anal Appl 385:808–822, 2012) discussed the starlikeness of the integral transform $V_{\lambda }(f)$ when $f\in {{\mathcal {W}}}_{\beta }(\alpha , \gamma )$. The aim of present paper is to find conditions on $\lambda (t)$ such that $V_{\lambda }(f)$ is starlike of order $\delta$ ($0\le \delta \le 1/2$) when $f\in {{\mathcal {W}}}_{\beta }(\alpha , \gamma )$. As applications, we study various choices of $\lambda (t)$, related to classical integral transforms.

1 Introduction

Let ${\mathcal {A}}$ denotes the class of analytic functions $f$ defined in the open unit disk $E=\{z : |z|<1 \}$ with the normalization $f(0)=f'(0)-1=0$ and ${\mathcal {A}}_{0}=\left\{ g: \, g(z)=f(z)/z, \, f\in {\mathcal {A}}\right\}$. Let $S$ be the subclass of $\mathcal {A}$ consisting of functions univalent in $E$. A function $f$ in $\mathcal {A}$ is said to be starlike of order $\beta$ if it satisfies

$$\begin{aligned} \mathfrak {R}\left( \frac{zf'(z)}{f(z)}\right) >\beta , \, \, \, z\in E, \end{aligned}$$

for some $\beta \, \, (0\le \beta <1)$. We denote by $S^{*}(\beta )$, the subclass of $S$ consisting of functions which are starlike of order $\beta$ in $E$. Set $S^{*}(0)=S^{*}$. For any 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 $\mathcal {A}$, the Hadamard product (or convolution) of $f$ and $g$ is the function $f *g$ defined by

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

For $f\in {\mathcal {A}}$, Fournier and Ruscheweyh [4] introduced the operator

$$\begin{aligned} V_{\lambda }(f)(z):=\int _{0}^{1}\lambda (t)\frac{f(tz)}{t}\mathrm{d}t, \end{aligned}$$

(1.1)

where $\lambda$ is a non-negative real-valued integrable function satisfying the condition $\int _{0}^{1}\lambda (t)\mathrm{d}t=1$. This operator contains some of the well-known operators (such as Alexander, Libera, Bernardi, and Komatu) as its special cases. This operator has been studied by a number of authors for various choices of $\lambda (t)$ [1, 3, 4, 6, 9, 12]. Fournier and Ruscheweyh [4] applied the Duality theory [10, 11] to prove the starlikeness of the linear integral transform $V_{\lambda }(f)$ over functions $f$ in the class

$$\begin{aligned} {\mathcal {P}}(\beta ):=\left\{ f\in {\mathcal {A}}:\exists \phi \in {{\mathbb {R}}} | \mathfrak {R}e^{i\phi }\left( f'(z)-\beta \right) >0, \, \, z\in E\right\} . \end{aligned}$$

In 2001, Kim and Rønning [5] investigated starlikeness properties of the integral transform (1.1) for functions $f$ in the class

$$\begin{aligned} {\mathcal {P}}_{\gamma }(\beta ):=\left\{ f\in {\mathcal {A}}:\exists \phi \in {{\mathbb {R}}} | \mathfrak {R}e^{i\phi }\left( (1-\gamma )\frac{f(z)}{z}+\gamma f'(z)-\beta \right) >0, \, \, z\in E\right\} . \end{aligned}$$

Recently in 2008, Ponnusamy and Rønning [9] discussed the same problem for the functions in the class

$$\begin{aligned} {\mathcal {R}}_{\gamma }(\beta ):=\left\{ f\in {\mathcal {A}}:\exists \phi \in {{\mathbb {R}}} | \mathfrak {R}e^{i\phi }\left( f'(z)+\gamma zf''(z)-\beta \right) >0, \, z\in E\right\} . \end{aligned}$$

It is evident that ${\mathcal {R}}_{\gamma }(\beta )$ is closely related to the class ${\mathcal {P}}_{\gamma }(\beta )$. Clearly, $f\in {\mathcal {R}}_{\gamma }(\beta )$ if and only if $zf'$ belongs to ${\mathcal {P}}_{\gamma }(\beta )$.

In a very recent paper, Ali et al. [1] discussed this problem for the functions $f$ in the class

$$\begin{aligned} \small {\mathcal {W}}_{\beta }(\alpha , \gamma )\!:=\!\left\{ f\in {\mathcal {A}}:\exists \phi \in {{\mathbb {R}}} | \mathfrak {R}e^{i\phi }\left( (1\!-\!\alpha \!+\!2\gamma )\frac{f(z)}{z}\!+\!(\alpha \!-\!2\gamma )f'(z)\!+\!\gamma zf''(z)\!-\!\beta \right) \!>\!0, \, z\!\in \! E\right\} . \end{aligned}$$

(1.2)

Note that ${{\mathcal {W}}_{\beta }(1, 0)}\equiv {\mathcal {P}}(\beta )$, ${{\mathcal {W}}_{\beta }(\alpha , 0)}\equiv {\mathcal {P}}_{\alpha }(\beta )$ and ${{\mathcal {W}}_{\beta }(1+2\gamma , \gamma )}\equiv {\mathcal {R}}_{\gamma }(\beta )$.

In Sect. 3 of the paper, we shall mainly tackle the problem: For given $0\le \delta \le 1/2$, to find conditions on $\beta$ such that $V_{\lambda }(f)\in S^{*}(\delta )$ whenever $f\in {{\mathcal {W}}_{\beta }(\alpha , \gamma )}$. In Sect. 4, we find easier criteria of starlikeness of the integral operator $V_{\lambda }(f)$. While in the last section of the paper, we discussed applications of results obtained for various choices of $\lambda (t)$.

To prove our result, we shall need the duality theory for convolutions, 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 basic reference to this theory is the book by Ruscheweyh [11] (see also [10]). We shall need the following fundamental result.

Theorem 1.1

(Duality Principle) Let

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

We have

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

• (2) 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

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

2 Preliminaries

We use the notations introduced in [1]. Let $\mu \ge 0$ and $\nu \ge 0$ satisfy

$$\begin{aligned} \mu +\nu =\alpha -\gamma \, \, \, \, \text {and} \, \, \, \, \mu \nu =\gamma . \end{aligned}$$

(2.1)

For $\gamma =0$, $\mu$ is also taken to be 0, in which case, $\nu =\alpha \ge 0$. Writing $\alpha =1+2\gamma$ in (2.1), we get $\mu +\nu =1+\gamma =1+\mu \nu$, or $(\mu -1)(1-\nu )=0$.

• (i) When $\gamma >0$, then writing $\mu =1$ gives $\nu =\gamma$.

• (ii) If $\gamma =0$, then $\mu =0$ and $\nu =\alpha =1$.

In the particular case $\alpha =1+2\gamma$, the values of $\mu$ and $\nu$ for $\gamma >0$ will be taken as $\mu =1$ and $\nu =\gamma$ respectively, while in the case when $\gamma =0$, we have $\mu =0$ and $\nu =1=\alpha$.

Define

$$\begin{aligned} {\phi }_{\mu ,\nu }(z)=1+\sum _{n=1}^{\infty }\frac{(n\nu +1)(n\mu +1)}{n+1}z^{n}, \end{aligned}$$

(2.2)

and

$$\begin{aligned} {\psi }_{\mu ,\nu }(z) = {\phi }_{\mu ,\nu }^{-1}(z)= & {} 1+\sum _{n=1}^{\infty }\frac{n+1}{(n\nu +1)(n\mu +1)}z^{n} \nonumber \\= & {} \int _{0}^{1}\int _{0}^{1}\frac{\mathrm{d}s\mathrm{d}t}{(1-t^{\nu }s^{\mu }z)^2}. \end{aligned}$$

(2.3)

Here ${\phi }_{\mu ,\nu }^{-1}$ denotes the convolution inverse of ${\phi }_{\mu ,\nu }$ such that ${\phi }_{\mu ,\nu }*{\phi }_{\mu ,\nu }^{-1}=z/(1-z)$. If we take $\gamma =0$, then $\mu =0$, $\nu =\alpha$ in (2.3), we have

$$\begin{aligned} {\psi }_{0,\alpha }(z)=1+\sum _{n=1}^{\infty }\frac{n+1}{n\alpha +1}z^{n}=\int _{0}^{1}\frac{\mathrm{d}t}{(1-t^{\alpha }z)^2}. \end{aligned}$$

If $\gamma >0$, then $\nu >0$, $\mu >0$, and making the change of variables $u=t^{\nu }$, $v=s^{\mu }$ results in

$$\begin{aligned} {\psi }_{\mu ,\nu }(z)=\frac{1}{\mu \nu }\int _{0}^{1}\int _{0}^{1}\frac{u^{1/{\nu }-1}v^{1/{\mu }-1}}{(1-uvz)^2}\mathrm{d}u\mathrm{d}v. \end{aligned}$$

Thus the function ${\psi }_{\mu ,\nu }$ can be written as

$$\begin{aligned} {\psi }_{\mu ,\nu }(z)=\left\{ \begin{array}{l@{\quad }l} \frac{1}{\mu \nu }\int _{0}^{1}\int _{0}^{1}\frac{u^{1/{\nu }-1}v^{1/{\mu }-1}}{(1-uvz)^2}\mathrm{d}u\mathrm{d}v, &{} \gamma >0; \\ \int _{0}^{1}\frac{\mathrm{d}t}{(1-t^{\alpha }z)^2}, &{} \gamma =0, \alpha >0. \end{array}\right. \end{aligned}$$

(2.4)

Further let $g$ be the solution of the initial-value problem

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}t^{1/{\nu }}(1+g(t))=\left\{ \begin{array}{l@{\quad }l} \frac{2}{\mu \nu }t^{1/{\nu }-1}\int _{0}^{1}\frac{1-\delta (1+st)}{(1-\delta )(1+st)^2}s^{1/{\mu }-1}\mathrm{d}s, &{}\quad \gamma >0; \\ \frac{2}{\alpha }\frac{{1-\delta (1+t)}}{(1-\delta )(1+t)^2}t^{1/{\alpha }-1}, &{} \quad \gamma =0, \alpha >0. \end{array}\right. \end{aligned}$$

(2.5)

satisfying $g(0)=1$, where $\delta \in [0,1/2]$. A simple calculation leads to the solution given by

$$\begin{aligned} g(t)=\frac{2}{\mu \nu }\int _{0}^{1}\int _{0}^{1}\frac{{1-\delta (1+swt)}}{(1-\delta )(1+swt)^2}s^{1/{\mu }-1}w^{{1/{\nu }-1}}\mathrm{d}s\mathrm{d}w-1. \end{aligned}$$

(2.6)

In particular

$$\begin{aligned} g_{\alpha }(t)=\frac{2}{\alpha }{t^{-1/{\alpha }}}\int _{0}^{t}{u^{{1/{\alpha }}-1}}\frac{{1-\delta (1+u)}}{(1-\delta )(1+u)^2}\mathrm{d}u-1, \, \, \gamma =0, \, \alpha >0. \end{aligned}$$

(2.7)

3 Main Results

Theorem 3.1

Let $\mu \ge 0$, $\nu \ge 0$ satisfy (2.1), and $\beta <1$ satisfy

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

(3.1)

where $g$ is the solution of the initial-value problem (2.5). Further let

$$\begin{aligned} \Lambda _{\nu }(t)= & {} \int _{t}^{1}\frac{\lambda (x)}{x^{1/{\nu }}}\mathrm{d}x, \, \, \, \, \, {\nu }>0, \end{aligned}$$

(3.2)

$$\begin{aligned} \Pi _{\mu ,\nu }(t)= & {} \left\{ \begin{array}{ll} \int _{t}^{1}\Lambda _{\nu }(x){x}^{1/{\nu }-1-1/{\mu }}\mathrm{d}x, &{}\quad \gamma >0;\\ \Lambda _{\alpha }(t), &{}\quad \gamma =0, (\mu =0, \, \nu =\alpha >0). \end{array}\right. \end{aligned}$$

(3.3)

and assume that $\displaystyle t^{1/{\nu }}\Lambda _{\nu }(t)\rightarrow 0$, and $\displaystyle t^{1/{\mu }}\Pi _{\mu ,\nu }(t)\rightarrow 0$ as $t\rightarrow 0^{+}$. Then for $\delta \in [0,1/2]$, we have $V_{\lambda }({{\mathcal {W}}}_{\beta }(\alpha , \gamma )) \subset S^{*}(\delta )$ if and only if ${\mathcal {L}}_{\Pi _{\mu ,\nu }}(h_{\delta })\ge 0$, where ${\mathcal {L}}_{\Pi _{\mu ,\nu }}(h_{\delta })$ and $h_{\delta }$ are defined by following equations:

$$\begin{aligned} {\mathcal {L}}_{\Pi _{\mu ,\nu }}(h_{\delta })=\left\{ \begin{array}{ll} \mathfrak {R}\int _{0}^{1}\Pi _{\mu ,\nu }(t)t^{1/{\mu }-1}\left( \frac{h(tz)}{tz}-\frac{1-\delta (1+t)}{(1-\delta )(1+t)^2}\right) \mathrm{d}t, &{} \quad \gamma >0; \\ \mathfrak {R}\int _{0}^{1}\Pi _{0,\alpha }(t)t^{1/{\alpha }-1}\left( \frac{h(tz)}{tz}-\frac{1-\delta (1+t)}{(1-\delta )(1+t)^2}\right) \mathrm{d}t, &{} \quad \gamma =0. \end{array}\right. \end{aligned}$$

(3.4)

and

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

(3.5)

respectively. This conclusion does not hold for any smaller values of $\beta$.

Proof

The case $\gamma =0 (\mu =0,\, \nu =\alpha$) corresponds to the Theorem 1.2 in [2], so we will prove the result only when $\gamma >0$.

Let

$$\begin{aligned} H(z)=(1-\alpha +2\gamma )\frac{f(z)}{z}+(\alpha -2\gamma )f'(z)+\gamma zf''(z). \end{aligned}$$

Since $\mu +\nu =\alpha -\gamma$ and $\mu \nu =\gamma$, then

$$\begin{aligned} H(z)= & {} (1+\gamma -(\alpha -\gamma ))\frac{f(z)}{z}+(\alpha -\gamma -\gamma )f'(z)+\gamma zf''(z)\\= & {} (1+\mu \nu -\mu -\nu )\frac{f(z)}{z}+(\mu +\nu -\mu \nu )f'(z)+\mu \nu zf''(z). \end{aligned}$$

Writing $f(z)=z+\sum _{n=2}^{\infty }a_{n}z^{n}$, we obtain from (2.2)

$$\begin{aligned} H(z)=1+\sum _{n=1}^{\infty }a_{n+1}{(n\nu +1)(n\mu +1)}z^{n}=f'(z)*\phi _{\mu ,\nu }(z), \end{aligned}$$

(3.6)

and (2.3) gives that

$$\begin{aligned} f'(z)=H(z)*\psi _{\mu ,\nu }(z). \end{aligned}$$

(3.7)

Now, let $f\in {{\mathcal {W}}_{\beta }(\alpha , \gamma )}$. Then, in the view of the Theorem 1.1, we may restrict our attention to functions $f\in {{\mathcal {W}}_{\beta }(\alpha , \gamma )}$ for which

$$\begin{aligned} H(z)= & {} (1-\alpha +2\gamma )\frac{f(z)}{z}+(\alpha -2\gamma )f'(z)+\gamma zf''(z)\\= & {} \beta +(1-\beta )\left( \frac{1+xz}{1+yz}\right) , \, \, \, |x|=|y|=1. \end{aligned}$$

Thus (3.7) gives

$$\begin{aligned} f'(z)=\left( (1-\beta )\frac{1+xz}{1+yz}+\beta \right) *\psi _{\mu ,\nu }(z), \end{aligned}$$

(3.8)

and therefore

$$\begin{aligned} \frac{f(z)}{z}=\frac{1}{z}\int _{0}^{z}\left( (1-\beta )\frac{1+xw}{1+yw}+\beta \right) \mathrm{d}w*\psi (z), \end{aligned}$$

(3.9)

here $\psi :=\psi _{\mu ,\nu }$. $\square$

Also, a well-known result from the theory of convolutions [11] (see also [10]) implies that

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

where

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

Hence $F\in S^{*}(\delta )$ if and only if

$$\begin{aligned} 0\ne & {} \frac{1}{z}(V_{\lambda }(f)(z)*h_{\delta }(z))=\frac{1}{z}\left[ \int _{0}^{1}\lambda (t)\frac{f(tz)}{t}\mathrm{d}t*h_{\delta }(z)\right] \\= & {} \int _{0}^{1}\frac{\lambda (t)}{1-tz}\mathrm{d}t*\frac{f(z)}{z}*\frac{h_{\delta }(z)}{z} \end{aligned}$$

Using (3.9), we have

$$\begin{aligned} 0\ne & {} \int _{0}^{1}\frac{\lambda (t)}{1-tz}\mathrm{d}t*\left[ \frac{1}{z}\int _{0}^{z}\left( (1-\beta )\frac{1+xw}{1+yw}+\beta \right) \mathrm{d}w*\psi (z)\right] *\frac{h_{\delta }(z)}{z}\\= & {} \int _{0}^{1}\frac{\lambda (t)}{1-tz}\mathrm{d}t*\frac{h_{\delta }(z)}{z}*\left[ \frac{1}{z}\int _{0}^{z}\left( (1-\beta )\frac{1+xw}{1+yw}+\beta \right) \mathrm{d}w\right] *\psi (z)\\= & {} \int _{0}^{1}\lambda (t)\frac{h_{\delta }(tz)}{tz}\mathrm{d}t*(1-\beta )\left[ \frac{1}{z}\int _{0}^{z}\left( \frac{1+xw}{1+yw}+\frac{\beta }{(1-\beta )}\right) \mathrm{d}w\right] *\psi (z)\\= & {} (1-\beta )\left[ \int _{0}^{1}\lambda (t)\frac{h_{\delta }(tz)}{tz}\mathrm{d}t+\frac{\beta }{(1-\beta )}\right] *\frac{1}{z}\int _{0}^{z}\frac{1+xw}{1+yw}\mathrm{d}w*\psi (z)\\= & {} (1-\beta )\left[ \int _{0}^{1}\lambda (t)\left( \frac{1}{z}\int _{0}^{z}\frac{h_{\delta }(tw)}{tw}\mathrm{d}w\right) \mathrm{d}t+\frac{\beta }{(1-\beta )}\right] *\frac{1+xz}{1+yz}*\psi (z). \end{aligned}$$

This holds if and only if [10], p. 23]

$$\begin{aligned}&\mathfrak {R}(1-\beta )\left[ \int _{0}^{1}\lambda (t)\left( \frac{1}{z}\int _{0}^{z}\frac{h_{\delta }(tw)}{tw}\mathrm{d}w\right) \mathrm{d}t+\frac{\beta }{(1-\beta )}\right] *\psi (z)\ge \frac{1}{2},\\\Leftrightarrow & {} \mathfrak {R}(1-\beta )\left[ \int _{0}^{1}\lambda (t)\left( \frac{1}{z}\int _{0}^{z}\frac{h_{\delta }(tw)}{tw}\mathrm{d}w\right) \mathrm{d}t+\frac{\beta }{(1-\beta )}-\frac{1}{2(1-\beta )}\right] *\psi (z)\ge 0,\\\Leftrightarrow & {} \mathfrak {R}\left[ \int _{0}^{1}\lambda (t)\left( \frac{1}{z}\int _{0}^{z}\frac{h_{\delta }(tw)}{tw}\mathrm{d}w\right) \mathrm{d}t+\frac{\beta }{(1-\beta )}-\frac{1}{2(1-\beta )}\right] *\psi (z)\ge 0,\\\Leftrightarrow & {} \mathfrak {R}\left[ \int _{0}^{1}\lambda (t)\left( \frac{1}{z}\int _{0}^{z}\frac{h_{\delta }(tw)}{tw}\mathrm{d}w\right) \mathrm{d}t-\frac{1}{2}+\frac{\beta }{2(1-\beta )}\right] *\psi (z)\ge 0,\\\Leftrightarrow & {} \mathfrak {R}\left[ \int _{0}^{1}\lambda (t)\left( \frac{1}{z}\int _{0}^{z}\frac{h_{\delta }(tw)}{tw}\mathrm{d}w-\frac{1+g(t)}{2}\right) \mathrm{d}t\right] *\psi (z)\ge 0, \, \, \, \, \text {(Using (3.1))}\\\Leftrightarrow & {} \mathfrak {R}\left[ \int _{0}^{1}\lambda (t)\left( \frac{h_{\delta }(tz)}{tz}-\frac{1+g(t)}{2}\right) \mathrm{d}t\right] *\frac{1}{z}\int _{0}^{z}\psi (w)\mathrm{d}w\ge 0,\\\Leftrightarrow & {} \mathfrak {R}\left[ \int _{0}^{1}\lambda (t)\left( \frac{h_{\delta }(tz)}{tz}\!-\!\frac{1\!+\!g(t)}{2}\right) \mathrm{d}t\right] *\sum _{n=0}^{\infty }\frac{z^n}{(n\nu \!+\!1)(n\mu \!+\!1)}\ge 0, \, \, \text { (Using (2.3))}\\\Leftrightarrow & {} \mathfrak {R}\int _{0}^{1}\lambda (t)\left( \sum _{n=0}^{\infty }\frac{z^n}{(n\nu +1)(n\mu +1)}*\frac{h_{\delta }(tz)}{tz}-\frac{1+g(t)}{2}\right) \mathrm{d}t \ge 0,\\\Leftrightarrow & {} \mathfrak {R}\int _{0}^{1}\lambda (t)\left( \int _{0}^{1}\int _{0}^{1}\frac{\mathrm{d}{\eta }\mathrm{d}{\zeta }}{(1-{\eta }^{\nu }{\zeta }^{\mu }z)}*\frac{h_{\delta }(tz)}{tz}-\frac{1+g(t)}{2}\right) \mathrm{d}t\ge 0,\\\Leftrightarrow & {} \mathfrak {R}\int _{0}^{1}\lambda (t)\left( \int _{0}^{1}\int _{0}^{1}\frac{h_{\delta }(tz{\eta }^{\nu }{\zeta }^{\mu })}{tz{\eta }^{\nu }{\zeta }^{\mu }}\mathrm{d}{\eta }\mathrm{d}{\zeta }-\frac{1+g(t)}{2}\right) \mathrm{d}t \ge 0, \end{aligned}$$

which can also be written as

$$\begin{aligned} \mathfrak {R}\int _{0}^{1}\lambda (t)\left( \int _{0}^{1}\int _{0}^{1}\frac{1}{\mu \nu }\frac{h_{\delta }(tzuv)}{tzuv}u^{1/{\nu }-1}v^{1/{\mu }-1}\mathrm{d}v\mathrm{d}u-\frac{1+g(t)}{2}\right) \mathrm{d}t\ge 0. \end{aligned}$$

Writing $w=tu$, we get

$$\begin{aligned} \mathfrak {R}\int _{0}^{1}\frac{\lambda (t)}{t^{1/{\nu }}}\left[ \int _{0}^{t}\int _{0}^{1}\frac{h_{\delta }(wzv)}{wzv}w^{1/{\nu }-1}v^{1/{\mu }-1}\mathrm{d}v\mathrm{d}w-{\mu \nu }{t^{1/{\nu }}}\frac{1+g(t)}{2}\right] \mathrm{d}t\ge 0. \end{aligned}$$

An integration by parts with respect to $t$ and (2.5) gives

$$\begin{aligned}&\mathfrak {R}\int _{0}^{1}\Lambda _{\nu }(t)\left[ \int _{0}^{1}\frac{h_{\delta }(tzv)}{tzv}t^{1/{\nu }-1}v^{1/{\mu }-1}\mathrm{d}v-t^{1/{\nu }-1}\int _{0}^{1}\frac{1-\delta (1+st)}{(1\!-\!\delta )(1\!+\!st)^2}s^{1/{\mu }\!-\!1}\mathrm{d}s\right] \mathrm{d}t\\&\quad \ge 0. \end{aligned}$$

Again writing $w=vt$ and $\eta =st$ reduces the above inequality to

$$\begin{aligned}&\mathfrak {R}\int _{0}^{1}\Lambda _{\nu }(t)t^{1/{\nu }-1/{\mu }\!-\!1}\left[ \int _{0}^{t}\frac{h_{\delta }(wz)}{wz}w^{1/{\mu }-1}\mathrm{d}w-\int _{0}^{t}\frac{1-\delta (1\!+\!\eta )}{(1\!-\!\delta )(1+\eta )^2}{\eta }^{1/{\mu }-1}\mathrm{d}\eta \right] \mathrm{d}t\\&\quad \ge 0, \end{aligned}$$

which after integration by parts with respect to $t$ yields

$$\begin{aligned} \mathfrak {R}\int _{0}^{1}\Pi _{\mu ,\nu }(t)t^{1/{\mu }-1}\left( \frac{h(tz)}{tz}-\frac{1-\delta (1+t)}{(1-\delta )(1+t)^2}\right) \mathrm{d}t\\&\quad \ge 0. \end{aligned}$$

Thus $F\in S^{*}(\delta )$ if and only if ${\mathcal {L}}_{\Pi _{\mu ,\nu }}(h_{\delta })\ge 0$.

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

$$\begin{aligned} (1-\alpha +2\gamma )\frac{f(z)}{z}+(\alpha -2\gamma )f'(z)+\gamma zf''(z)=\beta +(1-\beta )\frac{1+z}{1-z}. \end{aligned}$$

Using a series expansion we obtain that

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

Thus

$$\begin{aligned} F(z)=V_{\lambda }(f)(z)=\int _{0}^{1}\lambda (t)\frac{f(tz)}{t}\mathrm{d}t=z+2(1-\beta )\sum _{n=1}^{\infty }\frac{\tau _{n}}{(n\nu +1)(n\mu +1)}z^{n+1}, \end{aligned}$$

where $\tau _{n}=\int _{0}^{1}\lambda (t)t^{n}\mathrm{d}t$. From (2.6), it is a simple exercise to write $g(t)$ in a series expansion as

$$\begin{aligned} g(t)=1+\frac{2}{1-\delta }\sum _{n=1}^{\infty }\frac{(-1)^{n}(n+1-\delta )}{(n\nu +1)(n\mu +1)}t^{n}. \end{aligned}$$

(3.10)

Now, by (3.1) and (3.10), we have

$$\begin{aligned} \frac{\beta }{1-\beta }= & {} -\int _{0}^{1}\lambda (t)g(t)\mathrm{d}t\\= & {} -\int _{0}^{1}\lambda (t)\left[ 1+\frac{2}{1-\delta }\sum _{n=1}^{\infty }\frac{(-1)^{n}(n+1-\delta )}{(n\nu +1)(n\mu +1)}t^{n}\right] \mathrm{d}t\\= & {} -1-\frac{2}{1-\delta }\sum _{n=1}^{\infty }\frac{(-1)^{n}(n+1-\delta )}{(n\nu +1)(n\mu +1)}\int _{0}^{1}\lambda (t)t^{n}\mathrm{d}t. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{1}{1-\beta }=-\frac{2}{1-\delta }\sum _{n=1}^{ \infty }\frac{(-1)^{n}(n+1-\delta )\tau _{n}}{(n\nu +1)(n\mu +1)}. \end{aligned}$$

(3.11)

Finally, we see that

$$\begin{aligned} F'(z)=1+2(1-\beta )\sum _{n=1}^{\infty }\frac{(n+1)\tau _{n}}{(n\nu +1)(n\mu +1)}z^{n}. \end{aligned}$$

For $z=-1$, we have

$$\begin{aligned} F'(-1)= & {} 1+2(1-\beta )\sum _{n=1}^{\infty }\frac{(-1)^{n}(n+1)\tau _{n}}{(n\nu +1)(n\mu +1)}\\= & {} 1+2(1-\beta )\sum _{n=1}^{\infty }\frac{(-1)^{n}(n+1-\delta )\tau _{n}}{(n\nu +1)(n\mu +1)}+2(1-\beta )\sum _{n=1}^{\infty }\frac{(-1)^{n}\delta \tau _{n}}{(n\nu +1)(n\mu +1)}\\= & {} 1-(1-\delta )+\delta 2(1-\beta )\sum _{n=1}^{\infty }\frac{(-1)^{n}\tau _{n}}{(n\nu +1)(n\mu +1)}\\= & {} -\delta \left( -1+2(1-\beta )\sum _{n=1}^{\infty }\frac{(-1)^{n+1}\tau _{n}}{(n\nu +1)(n\mu +1)}\right) \\= & {} -\delta F(-1). \end{aligned}$$

Thus $zF'(z)/F(z)$ at $z=-1$ equals $\delta$. This implies that the result is sharp for the order of starlikeness.

4 Consequences of Theorem 3.1

Theorem 4.1

Let $0\le \delta \le 1/2$. Assume that both $\Pi _{\mu ,\nu }(t)$ and $\Lambda _{\nu }(t)$, as given in Theorem 3.1, are integrable on [0,1] and positive on (0,1). Further assume that $\mu \ge 1$, and

$$\begin{aligned} \frac{\Pi _{\mu ,\nu }(t)}{(1+t)(1-t)^{1+2\delta }} \, \, \text {is decreasing on (0, 1)}. \end{aligned}$$

(4.1)

If $\beta$ satisfies (3.1), then we have $V_{\lambda }({\mathcal {W}}_{\beta }(\alpha , \gamma ))\subset S^{*}(\delta )$, where $V_{\lambda }(f)$ is defined by (1.1).

Proof

For $\mu \ge 1$, the function $t^{{1/{\mu }}-1}$ is decreasing on (0,1). Thus the condition (4.1) along with [8], Theorem 2.3] gives

$$\begin{aligned} \mathfrak {R}\int _{0}^{1}\Pi _{\mu ,\nu }(t)t^{1/{\mu }-1}\left( \frac{h(tz)}{tz}-\frac{1-\delta (1+t)}{(1-\delta )(1+t)^2}\right) \mathrm{d}t\ge 0. \end{aligned}$$

The result now, follows from Theorem 3.1. $\square$

Below, we obtain the conditions to ensure starlikeness of $V_{\lambda }(f)$. As defined in Theorem 3.1, for $\gamma >0$,

$$\begin{aligned} \Pi _{\mu ,\nu }(t)=\int _{t}^{1}\Lambda _{\nu }(x){x}^{1/{\nu }-1-1/{\mu }}\mathrm{d}x, \, \, \, \text {and} \, \, \, \Lambda _{\nu }(t)=\int _{t}^{1}\frac{\lambda (x)}{x^{1/{\nu }}}\mathrm{d}x. \end{aligned}$$

In order to apply Theorem 4.1, we have to prove that the function

$$\begin{aligned} p(t)=\frac{\Pi _{\mu ,\nu }(t)}{(1+t)(1-t)^{1+2\delta }} \end{aligned}$$

is decreasing in (0,1). Since $p(t)>0$ and

$$\begin{aligned} \frac{p'(t)}{p(t)}=-\frac{\Lambda _{\nu }(t)}{t^{1-{1/{\mu }}-{1/{\nu }}}\Pi _{\mu , \nu }(t)}+\frac{2(t+\delta (1+t))}{1-t^{2}}, \end{aligned}$$

or equivalently,

$$\begin{aligned} \frac{p'(t)}{p(t)}=\frac{2(t+\delta (1+t))}{(1-t^{2})\Pi _{\mu ,\nu }(t)}\left\{ \Pi _{\mu ,\nu }(t)-\frac{(1-t^{2})\Lambda _{\nu }(t)t^{{1/{\nu }}-1-{1/{\mu }}}}{2(t+\delta (1+t))}\right\} , \end{aligned}$$

so it remains to show that $q(t)$ is increasing over (0,1), where

$$\begin{aligned} q(t):=\Pi _{\mu ,\nu }(t)-\frac{(1-t^{2})\Lambda _{\nu }(t)t^{{1/{\nu }}-1-{1/{\mu }}}}{2(t+\delta (1+t))}. \end{aligned}$$

Since $q(1)=0$, this will imply that $q(t)\le 0$, and thus $p(t)$ is decreasing on (0,1). Now

$$\begin{aligned} \displaystyle q'(t)= & {} -\frac{t^{{1/{\nu }}-1-{1/{\mu }}}(1+t)}{2(t+\delta (1+t))^{2}}\Bigg \{-\lambda (t)t^{-1/{\nu }}(1-t)(t+\delta (1+t))\nonumber \\&\left. +{\Lambda }_{\nu }(t)\left( \frac{(1\!-\!t)}{t}({1/{\nu }}\!-\!1\!-\!{1/{\mu }})(t\!+\!\delta (1\!+\!t))\!-\!(1\!-\!t\!-\!\delta (1+t))(1+2\delta )\right) \right\} .\nonumber \\ \end{aligned}$$

(4.2)

So, $q'(t)\ge 0$ for $t\in (0,1)$ is equivalent to the inequality $r(t)\le 0$, where $r(t)$ is equal to

$$\begin{aligned}&-\lambda (t)t^{-1/{\nu }}(1-t)(t+\delta (1+t))\\&\quad + \Lambda _{\nu }(t)\left( \frac{(1-t)}{t}({1/{\nu }}-1-{1/{\mu }})(t+\delta (1+t))-(1-t-\delta (1+t))(1+2\delta )\right) . \end{aligned}$$

By using the idea similar to the one used to prove Theorem 3.1 in [3], we can write

$$\begin{aligned} r(t)=-A(t)X(t)+\frac{Y(t)}{t}\int _{t}^{1}A(s)\mathrm{d}s, \end{aligned}$$

where,

$$\begin{aligned} A(t)= & {} \lambda (t)t^{-1/{\nu }}, \nonumber \\ X(t)= & {} (1-t)(t+\delta (1+t)), \nonumber \\ Y(t)= & {} X(t)({1/{\nu }}-1-{1/{\mu }})+Z(t), \nonumber \\ Z(t)= & {} -t(1-t-\delta (1+t))(1+2\delta ). \end{aligned}$$

(4.3)

Clearly, $A(t)>0$ and $X(t)>0$ for all $t\in (0,1).$

Case (i). If $Y(t)\le 0$ on (0,1), then $r(t)\le 0$ on (0,1) and thus the result follows.

Case (ii). When $Y(t)>0$. We may write

$$\begin{aligned} r(t)=\frac{Y(t)}{t}B(t), \, \, \text {where} \, \, B(t)=-A(t)\frac{tX(t)}{Y(t)}+\int _{t}^{1}A(s)\mathrm{d}s, \, \, \text {and} \, \, B(1)=0. \end{aligned}$$

Thus, to prove that $r(t)\le 0$, it is enough to prove that $B(t)$ is an increasing function of $t$. Now

$$\begin{aligned} B'(t)= & {} -A(t)\left[ \frac{A^{\prime }(t)}{A(t)}\frac{tX(t)}{Y(t)}+\left( \frac{tX}{Y}\right) ^{\prime }(t)+1\right] \\= & {} -t^{-1/{\nu }}\lambda (t)\left[ \left( \frac{t{\lambda }^{\prime }(t)}{\lambda (t)}-\frac{1}{\nu }\right) \frac{X(t)}{Y(t)}+\left( \frac{tX}{Y}\right) '(t)+1\right] . \end{aligned}$$

For $Y(t)>0$, $B'(t)\ge 0$ is equivalent to

$$\begin{aligned} \frac{t{\lambda }^{\prime }(t)}{\lambda (t)}\le \frac{1}{\nu }-\left[ 1+\left( \frac{tX}{Y}\right) ^{\prime }(t)\right] \frac{Y(t)}{X(t)}. \end{aligned}$$

(4.4)

Now, following three possibilities arise:

• (a). If $Y(t)>0$ throughout the interval (0,1), then (4.4) implies that $B'(t)\ge 0$ on (0,1). Thus, $B(t)$ is increasing in (0,1) which implies that, $B(t)\le B(1)=0$. Therefore, $r(t)\le 0$ on (0,1).

• (b). If $Y(t)>0$ on some interval $(0,t_{0})$ and $Y(t)\le 0$ on $[t_{0},1)$ for some $t_{0}\in (0,1)$, then (4.4) implies that $B'(t)\ge 0$ on $(0,t_{0})$. Thus, $B(t)$ is increasing in $(0,t_{0})$ which implies that, $B(t)\le B(t_{0})$ for any $t$ in $(0,t_{0})$. Since $B(t_{0}) \rightarrow -\infty$, this implies that $B(t)$ is negative. Therefore, $r(t)\le 0$ on $(0,t_{0})$. In view of Case (i), $r(t)\le 0$ whenever $Y(t)\le 0$. Thus, $r(t)\le 0$ on (0,1).

• (c). If $Y(t)\le 0$ on some interval $(0,t_{0}]$ and $Y(t)>0$ on $(t_{0},1)$ for some $t_{0}\in (0,1)$, then (4.4) implies that $B'(t)\ge 0$ on $(t_{0},1)$. Thus, $B(t)$ is increasing in $(t_{0},1)$ which implies that, $B(t)\le B(1)=0$ for any $t$ in $(t_{0},1)$. Therefore, $r(t)\le 0$ on $(t_{0},1)$. In view of Case (i), $r(t)\le 0$ whenever $Y(t)\le 0$ which implies that, $r(t)\le 0$ on (0,1).

Subcase (i). For $\delta =0$, $X(t)$ and $Y(t)$ reduces to the simple form

$$\begin{aligned} X(t)=t(1-t) \, \, \, \text {and} \, \, \, Y(t)=t(1-t)\left( \frac{1}{\nu }-2-\frac{1}{\mu }\right) . \end{aligned}$$

Clearly $Y(t)\le 0$ on (0,1) if $\displaystyle \frac{1}{\nu }-2-\frac{1}{\mu }\le 0$ or simply $\displaystyle \nu \ge {\mu }/{(2\mu +1)}$ and so $r(t)\le 0$ in this case. On the other hand, if $\displaystyle 0<\nu <{\mu }/{(2\mu +1)}$ on (0,1), then $Y(t)>0$ on (0,1) and thus (4.4) gives that

$$\begin{aligned} \frac{t{\lambda }'(t)}{\lambda (t)}\le 1+\frac{1}{\mu } \end{aligned}$$

on (0,1) and hence $r(t)\le 0$ throughout the interval (0,1).

In the case when $\gamma =0$, we have $\mu =0$, $\nu =\alpha >0$. Let

$$\begin{aligned} k(t):= {\Lambda }_{\alpha }(t)t^{{1/{\alpha }}-1}, \, \, \, \text {where} \, \, \, {\Lambda }_{\alpha }(t)=\int _{t}^{1}\frac{\lambda (x)}{x^{1/{\alpha }}}\mathrm{d}x. \end{aligned}$$

To apply Theorem 2.3 in [9] along with Theorem 3.1, the function $P(t)=\frac{k(t)}{(1+t)(1-t)^{1+2\delta }}$ must be shown decreasing on the interval (0,1). Since, $P(t)>0$ on (0,1) and

$$\begin{aligned} \frac{P'(t)}{P(t)}=\frac{2(t+\delta (1+t))}{(1-t^{2})k(t)}\left\{ \frac{(1-t^{2})k'(t)}{2(t+\delta (1+t))}+k(t)\right\} , \end{aligned}$$

thus, $P(t)$ is decreasing in (0,1) provided

$$\begin{aligned} Q(t):=k(t)+\frac{(1-t^{2})k'(t)}{2(t+\delta (1+t))}\le 0. \end{aligned}$$

Since, $Q(1)=0$, thus $Q(t)\le 0$ will certainly hold if $Q$ is increasing on (0, 1). Now $\displaystyle Q'(t)=\frac{(1+t)}{2(t+\delta (1+t))^{2}}\left\{ (1-t)(t+\delta (1+t))k''(t)+[2\delta (t+\delta (1+t))\right. \left. -(1-t)(1+\delta )]k'(t)\right\} ,$

where $\displaystyle (1-t)(t+\delta (1+t))k''(t)+[2\delta (t+\delta (1+t))-(1-t)(1+\delta )]k'(t)$ is equal to

$$\begin{aligned}&t^{{1/{\alpha }}-2}\left\{ t(1-t)(t+\delta (1+t)){\Lambda }_{\alpha }{''}(t) +\left[ 2\left( \frac{1}{\alpha }-1\right) (1-t)(t+\delta (1+t))\right. \right. \\&\quad +\,2t{\delta }(t+\delta (1+t))-t(1-t)(1+\delta )\Bigg ]\\&\quad \displaystyle {\Lambda }_{\alpha }{'}(t)+\left[ \left( \frac{1}{\alpha }-2\right) \frac{(1-t)}{t}(t+\delta (1+t))+2{\delta }(t+\delta (1+t))\right. \\&\quad \left. -(1-t)(1+\delta )\Bigg ] \left( \frac{1}{\alpha }-1\right) {\Lambda }_{\alpha }(t)\right\} . \end{aligned}$$

Thus, $Q'(t)\ge 0$, for $t\in (0,1)$, is equivalent to the inequality

$$\begin{aligned}&\left\{ t(1-t)(t+\delta (1+t)){\Lambda }_{\alpha }{''}(t) +\left[ 2\left( \frac{1}{\alpha }-1\right) (1-t)(t+\delta (1+t))\right. \right. \\&\quad +\,2t{\delta }(t+\delta (1+t))-t(1-t)(1+\delta )\Bigg ] {\Lambda }_{\alpha }{'}(t)\\&\quad \displaystyle +\left[ \left( \frac{1}{\alpha }-2\right) \frac{(1-t)}{t}(t+\delta (1+t)) +2{\delta }(t+\delta (1+t))\right. \\&\qquad \left. -(1-t)(1+\delta )\Bigg ] \left( \frac{1}{\alpha }-1\right) {\Lambda }_{\alpha }(t)\right\} \ge 0. \end{aligned}$$

The latter condition is equivalent to $\Delta (t)\ge 0$, where

$$\begin{aligned} \Delta (t)\equiv & {} \left\{ -t{\lambda }{'}(t)(1-t) (t+\delta (1+t))+{\lambda }(t)\left[ \left( 2-\frac{1}{\alpha }\right) (1-t) (t+\delta (1+t))\right. \right. \\&-2t\delta (t+\delta (1+t))+t(1-t)(1+\delta )\Bigg ]\\&+\left[ \left( \frac{1}{\alpha }-2\right) (1-t)(t+\delta (1+t))+ 2t\delta (t+\delta (1+t))\right. \\&\left. -t(1-t)(1+\delta )\Bigg ]\left( \frac{1}{\alpha }-1\right) t^{{1/{\alpha }-1}}{\Lambda }_{\alpha }(t)\right\} . \end{aligned}$$

A simple computation along with (4.3) shows that $\Delta$ can be rewritten as

$$\begin{aligned}&\small -tX(t){\lambda }'(t)+\left[ \left( 3-\frac{1}{\alpha }\right) X(t)-\left( X(t)+Z(t)\right) \right] {\lambda }(t)\nonumber \\&\quad +\left[ \left( \frac{1}{\alpha }-3\right) X(t)+\left( X(t)+Z(t)\right) \right] \left( \frac{1}{\alpha }-1\right) t^{{1/{\alpha }-1}}{\Lambda }_{\alpha }(t). \end{aligned}$$

(4.5)

Since ${\Lambda }_{\alpha }(t)\ge 0$ and setting

$$\begin{aligned} \left[ \left( \frac{1}{\alpha }-3\right) X(t)+\left( X(t)+Z(t)\right) \right] \left( \frac{1}{\alpha }-1\right) \ge 0, \end{aligned}$$

$\Delta \ge 0$ follows from

$$\begin{aligned} -tX(t){\lambda }'(t)+\left[ \left( 3-\frac{1}{\alpha }\right) X(t)-\left( X(t)+Z(t)\right) \right] {\lambda }(t)\ge 0. \end{aligned}$$

Since $X(t)$ is non-negative on (0,1), thus the inequality $\Delta \ge 0$ follows from

$$\begin{aligned} \frac{t{\lambda }{'}(t)}{{\lambda }(t)}\le & {} \left( 3-\frac{1}{\alpha }\right) \!-\!\frac{X(t)+Z(t)}{X(t)} \, \, \, \text {and} \, \, \, \left[ \left( \frac{1}{\alpha }-3\right) X(t)\!+\!\left( X(t)\!+\!Z(t)\right) \right] \left( \frac{1}{\alpha }\!-\!1\right) \nonumber \\\ge & {} 0. \end{aligned}$$

(4.6)

For $\delta =0$, (4.6) reduces to

$$\begin{aligned} \frac{t{\lambda }{'}(t)}{{\lambda }(t)}\!\le \!3-\frac{1}{\alpha } \, \, \text {for} \, \, \left( \frac{1}{\alpha }\!-\!1\right) \left( \frac{1}{\alpha }\!-\!3\right) \ge 0 \, \, \, \text {or equivalently for} \, \, \alpha \!\in \!(0,1/3]\cup [1,\infty ). \end{aligned}$$

These observations for $\delta =0$ lead to the following result by, Ali et al. [1], Theorem 4.2].

Corollary 4.1

Assume that both $\Pi _{\mu ,\nu }(t)$ and $\Lambda _{\nu }(t)$, as defined in Theorem 3.1 are integrable on [0,1], and positive on (0,1). Let $\lambda (t)$ be a normalized non-negative real-valued integrable function on [0,1]. Under the same conditions as stated in Theorem 3.1, if $\lambda$ satisfies

$$\begin{aligned} \frac{t{\lambda }'(t)}{\lambda (t)}\le \left\{ \begin{array}{ll} 1+\frac{1}{\mu }, &{}\quad \mu \ge 1 (\gamma >0); \\ 3-\frac{1}{\alpha }, &{}\quad \gamma =0, \alpha \in (0,1/3]\cup [1,\infty ), \end{array} \right. \end{aligned}$$

(4.7)

then $F(z)=V_{\lambda }(f)(z)\in S^{*}$. The conclusion does not hold for smaller values of $\beta$.

Subcase (ii). If $0<\delta \le 1/2$ with $\gamma >0$, then (4.4) can be rewritten as

$$\begin{aligned} \left( \frac{1}{\nu }-\frac{t{\lambda }'(t)}{\lambda (t)}\right) X(t)Y(t)\ge Y^{2}(t)+Y(t)(tX'(t)+X(t))-Y'(t)tX(t). \end{aligned}$$

Since $Y(t)=X(t)({1/{\nu }}-1-{1/{\mu }})+Z(t)$, so the above inequality is equivalent to

$$\begin{aligned}&\left( \frac{1}{\nu }-1-\frac{1}{\mu }\right) [X(t)+Z(t)]X(t)-\left( 1+\frac{1}{\mu }-\frac{t{\lambda }'(t)}{\lambda (t)}\right) \nonumber \\&\qquad \left[ \left( \frac{1}{\nu }-1-\frac{1}{\mu }\right) X(t)+Z(t)\right] \nonumber \\&\quad \le Z'(t)(tX(t))-Z(t)(tX(t))'-Z^{2}(t). \end{aligned}$$

(4.8)

Define $D(t)=t(1+\delta )-(1-\delta )$. Rewriting the expressions for $X(t)$ and $Z(t)$ in terms of $D(t)$, we get

$$\begin{aligned} X(t)=(1-t)(D(t)+1) \, \, \, \text {and} \, \, \, Z(t)=(1+2\delta )tD(t) \end{aligned}$$

and so a simple computation gives that

$$\begin{aligned} Z'(t)(tX(t))-Z(t)(tX)'(t)-Z^{2}(t)=2\delta (1+2\delta )t^{2}(1-D^{2}(t)). \end{aligned}$$

(4.9)

Since $D^{2}(t)\le 1$ for $t\in [0,1]$ thus (4.9) is non-negative in (0,1). Since $X(t)+Z(t)$ and $X(t)$ are non-negative on (0,1), so if $\displaystyle \left( {1}/{\nu }-1-{1}/{\mu }\right) \le 0$ or simply $\displaystyle \nu \ge {\mu }/{(\mu +1)}$, then the inequality (4.8) holds on the interval where $Y(t)>0$ and hence, $r(t)\le 0$ on (0,1).

While on the other hand, for $0<\delta \le 1/2$ with $\gamma =0$, from (4.6) we have

$$\begin{aligned} \frac{t{\lambda }{'}(t)}{{\lambda }(t)}\le & {} \left( 3\!-\!\frac{1}{\alpha }\right) \!-\!\frac{X(t)\!+\!Z(t)}{X(t)} \, \, \text {and} \, \, \left[ \left( \frac{1}{\alpha }\!-\!3\right) X(t)+\left( X(t)+Z(t)\right) \right] \left( \frac{1}{\alpha }-1\right) \\\ge & {} 0. \end{aligned}$$

Since $X(t)$ and $X(t)+Z(t)$ are non-negative on (0,1), thus equivalently,

$$\begin{aligned} \frac{t{\lambda }{'}(t)}{{\lambda }(t)}\le 3-\frac{1}{\alpha }, \, \, \, \text {for} \, \, \, \alpha \in (0,1/3]. \end{aligned}$$

Hence, for $0<\delta \le 1/2$ with $\gamma =0$, we have $\Delta \ge 0$ throughout the interval (0,1).

Thus, we see that above Corollary continues to hold for $\delta \in (0,1/2]$ but with some restrictions. More precisely, we have

Theorem 4.2

Let $\lambda (t)$ be a non-negative real-valued integrable function on [0,1]. Assume that both $\Pi _{\mu ,\nu }(t)$ and $\Lambda _{\nu }(t)$ are integrable on [0,1], and positive on (0,1). Let $\lambda$ satisfying the condition

$$\begin{aligned} \frac{t{\lambda }'(t)}{\lambda (t)}\le \left\{ \begin{array}{ll} 1+\frac{1}{\mu }, &{} \quad \mu \ge 1 (\gamma >0); \\ 3-\frac{1}{\alpha }, &{} \quad \gamma =0, \alpha \in (0,1/3]. \\ \end{array} \right. \end{aligned}$$

(4.10)

Let $f\in {{\mathcal {W}}_{\beta }(\alpha , \gamma )}$ with $\nu \ge {\mu }/{(\mu +1)}$, and $\beta <1$ with

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

(4.11)

where $g(t)$ is defined by (2.6) with $\delta \in (0,1/2]$.Then $F(z)=V_{\lambda }(f)(z)\in S^{*}(\delta )$. The conclusion does not hold for smaller values of $\beta$.

Remark 4.1

• 1. For $\alpha =1+2\gamma$ with $\gamma >0$ and $\mu =1$, Theorem 4.2 yields Theorem 3.1 in [3] with $0<\delta \le 1/2$.

• 2. With $\delta =0$, our Corollary 4.1 coincides with the Theorem 4.2 in [1].

5 Applications

In this section, we present a number of applications of Theorem 4.2 for various well-known integral operators. Let $(a)_{n}$ denote the Pochhammer symbol, defined in terms of the Gamma function, by

$$\begin{aligned} (a)_{n}:= \frac{\Gamma (a+n)}{\Gamma (a)}=\left\{ \begin{array}{ll} 1, &{}\quad n=0, \\ a(a+1)...(a+n-1), &{}\quad \hbox {n} \in \mathbb {N}. \end{array}\right. \end{aligned}$$

Define the Gaussian hypergeometric function by

$$\begin{aligned} _{2}F_{1}(a,b;c;z)=F(a,b;c;z)=\sum _{n=0}^{\infty }\frac{(a)_{n}(b)_{n}}{(c)_{n}(1)_{n}}z^{n}, \, \, \, |z|<1, \end{aligned}$$

where $a$, $b$ and $c$ are complex numbers with $c\ne 0,-1,-2,\ldots$. Note that the series $_{2}F_{1}$ converges absolutely for $z\in E$. Now let $\Phi$ be defined by $\Phi (1-t)=1+\sum _{n=1}^{\infty }b_{n}(1-t)^n$, $b_n\ge 0$ for $n\ge 1$, and

$$\begin{aligned} \lambda (t)=K{t}^{b-1}(1-t)^{c-a-b}{\Phi }(1-t), \end{aligned}$$

(5.1)

where $K$ is a constant chosen such that $\int _{0}^{1}\lambda (t)\mathrm{d}t=1$. The following result holds in this instance.

Theorem 5.1

Let $a$, $b$, $c$, $\alpha >0$, ${\nu }\ge {\mu }/(\mu +1)$ and $\beta <1$ satisfy

$$\begin{aligned} \frac{\beta }{1-\beta }=-K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)g(t)\mathrm{d}t, \end{aligned}$$

where $K$ is a constant such that $\displaystyle K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)\mathrm{d}t=1$ and $g$ is given by (2.6). Then for $\delta \in [0,1/2]$, we have $V_{\lambda }({{\mathcal {W}}}_{\beta }(\alpha , \gamma )) \subset S^{*}(\delta )$ provided the following condition hold

$$\begin{aligned} c\ge a+b \, \, \, \text {and} \, \, \, b\le \left\{ \begin{array}{ll} 2+\frac{1}{\mu }, &{}\quad \gamma >0 (\mu \ge 1); \\ 4-\frac{1}{\alpha }, &{} \quad \gamma >0, \alpha \in (1/4,1/3],\\ \end{array}\right. \end{aligned}$$

(5.2)

where

$$\begin{aligned} V_{\lambda }(f)(z)=K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)\frac{f(tz)}{t}\mathrm{d}t. \end{aligned}$$

The value of $\beta$ is sharp.

Proof

Using (5.1), we have

$$\begin{aligned} \frac{t{\lambda }'(t)}{{\lambda }(t)}=(b-1)-\frac{(c-a-b)t}{1-t}-\frac{t{\Phi }'(1-t)}{{\Phi }(1-t)}. \end{aligned}$$

The condition (4.10) is satisfied when

$$\begin{aligned} (b-1)-\frac{(c-a-b)t}{1-t}-\frac{t{\Phi }'(1-t)}{{\Phi }(1-t)}\le \left\{ \begin{array}{ll} 1+\frac{1}{\mu }, &{}\quad \mu \ge 1 (\gamma >0); \\ 3-\frac{1}{\alpha }, &{}\quad \gamma =0, \alpha \in (0,1/3]. \\ \end{array} \right. \end{aligned}$$

Since $\Phi (1-t)=1+\sum _{n=1}^{\infty }b_{n}(1-t)^n$, $b_n\ge 0$ for $n\ge 1$, so the functions $\Phi (1-t)$ and ${\Phi }'(1-t)$ are non-negative in (0,1). Therefore, a simple computation of $(b-1)-\frac{(c-a-b)t}{1-t}$ with $c-a-b\ge 0$, implies that the condition (4.10) is satisfied whenever $b$ satisfies (5.2). Hence the result follows by applying Theorem 4.2. $\square$

Writing $\gamma =0$, $\alpha >0$ in Theorem 5.1 leads to the following corollary:

Corollary 5.1

Let $a$, $b$, $c$, $\alpha >0,$ and $\beta <1$ satisfy

$$\begin{aligned} \frac{\beta }{1-\beta }=-K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)g_{\alpha }(t)\mathrm{d}t, \end{aligned}$$

where $K$ is a constant such that $\displaystyle K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)\mathrm{d}t=1$ and $g_{\alpha }$ is given by (2.7). If $f\in {\mathcal {W}}_{\beta }(\alpha , 0)\equiv {\mathcal {P}}_{\alpha }(\beta )$, then the function

$$\begin{aligned} V_{\lambda }(f)(z)=K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)\frac{f(tz)}{t}\mathrm{d}t \end{aligned}$$

belongs to $S^{*}(\delta )$ with $\delta \in (0,1/2]$ whenever $a$, $b$, $c$ are related by $c\ge a+b$ and $b\le 4-\frac{1}{\alpha }$, $\alpha \in (1/4,1/3]$, for all $t\in (0,1)$. The value of $\beta$ is sharp.

Writing $\alpha =1+2\gamma$, $\gamma >0$ and $\mu =1$ in Theorem 5.1 gives the following corollary, which is an improvement of the Theorem 4.3 in [3]:

Corollary 5.2

Let $a$, $b$, $c>0$, $\gamma \ge 1/2$ and $\beta <1$ satisfy

$$\begin{aligned} \frac{\beta }{1-\beta }=-K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)g_{\gamma }(t)\mathrm{d}t, \end{aligned}$$

where $K$ is constant such that $\displaystyle K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)\mathrm{d}t=1$ and $g_{\gamma }$ is given by (2.7). If $f\in {\mathcal {W}}_{\beta }(1+2\gamma , \gamma )$, then the function

$$\begin{aligned} V_{\lambda }(f)(z)=K\int _{0}^{1}t^{b-1}(1-t)^{c-a-b}{\Phi }(1-t)\frac{f(tz)}{t}\mathrm{d}t \end{aligned}$$

belongs to $S^{*}(\delta )$ with $\delta \in (0,1/2]$ whenever $a$, $b$, $c$ are related by $c\ge a+b$ and $0<b\le 3$, for all $t\in (0,1)$ and $\gamma >1/2$. The value of $\beta$ is sharp.

The following special case of Theorem 5.1 corresponds to Bernardi operator, which we state as a theorem.

Theorem 5.2

Let $c>-1$, $\nu \ge {\mu }/(\mu +1)$ and $\beta <1$ satisfy

$$\begin{aligned} \frac{\beta }{1-\beta }=-(c+1)\int _{0}^{1}t^{c}g(t)\mathrm{d}t, \end{aligned}$$

where $g$ in given by (2.6). If $f\in {\mathcal {W}}_{\beta }(\alpha , \gamma )$, then the Bernardi Transform

$$\begin{aligned} V_{\lambda }(f)(z)=(1+c)\int _{0}^{1}t^{c-1}f(tz)\mathrm{d}t \end{aligned}$$

belongs to $S^{*}(\delta )$ with $\delta \in (0,1/2]$ if

$$\begin{aligned} c\le \left\{ \begin{array}{ll} 1+\frac{1}{\mu }, &{}\quad \mu \ge 1 (\gamma >0); \\ 3-\frac{1}{\alpha }, &{}\quad \gamma =0, \alpha \in (1/4,1/3]. \\ \end{array}\right. \end{aligned}$$

The value of $\beta$ is sharp.

Taking $\gamma =0$, $\alpha >0$ Theorem 5.2 reduces to the following corollary:

Corollary 5.3

Let $-1<c\le 3-1/{\alpha }$, $\alpha \in (1/4,1/3]$ and $\beta <1$ satisfy

$$\begin{aligned} \frac{\beta }{1-\beta }=-(c+1)\int _{0}^{1}t^{c}g_{\alpha }(t)\mathrm{d}t, \end{aligned}$$

where $g_{\alpha }$ is given by (2.7). If $f\in {\mathcal {W}}_{\beta }(\alpha , 0)\equiv {\mathcal {P}}_{\alpha }(\beta )$, then the function

$$\begin{aligned} V_{\lambda }(f)(z)=(1+c)\int _{0}^{1}t^{c-1}f(tz)\mathrm{d}t \end{aligned}$$

belongs to $S^{*}(\delta )$ with $\delta \in (0,1/2]$. The value of $\beta$ is sharp.

Remark 5.1

For $\alpha =1+2\gamma$, $\gamma >0$ and $\mu =1$ in Theorem 5.2 yields Corollary 4.1 in [3].

To prove the next theorem, we define

$$\begin{aligned} {\lambda }(t)=\left\{ \begin{array}{ll} (a+1)(b+1)\frac{t^{a}(1-t^{b-a})}{b-a}, &{}\quad b\ne a; \\ (a+1)^{2}t^{a}\log (1/t), &{} \quad b=a, \end{array}\right. \end{aligned}$$

(5.3)

where $b>-1$ and $a>-1$.

Theorem 5.3

Let $b>-1$, $a>-1$, ${\nu }\ge {\mu }/(\mu +1)$ and $\alpha >0$. Let $\beta <1$ satisfy

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

where $g$ is given by (2.6) and $\lambda (t)$ is defined by (5.3). If $f\in {\mathcal {W}}_{\beta }(\alpha , \gamma )$, then the convolution operator

$$\begin{aligned} G_{f}(a,b;z)=\left\{ \begin{array}{ll} \frac{(a+1)(b+1)}{b-a}\int _{0}^{1}{t^{a-1}(1-t^{b-a})}f(tz)\mathrm{d}t, &{} b\ne a; \\ {(a+1)^2}\int _{0}^{1}{t^{a-1}}\log (1/t)f(tz)\mathrm{d}t, &{} b=a. \end{array}\right. \end{aligned}$$

belongs to $S^{*}(\delta )$ with $\delta \in (0,1/2]$ if

$$\begin{aligned} a\le \left\{ \begin{array}{ll} 1+\frac{1}{\mu }, &{} \quad \mu \ge 1 (\gamma >0); \\ 3-\frac{1}{\alpha }, &{} \quad \gamma =0, \alpha \in (1/4,1/3]. \\ \end{array} \right. \end{aligned}$$

(5.4)

The value of $\beta$ is sharp.

Proof

We omitted the proof as it follows on the same lines as discussed in Theorem 5.3 [1].

Remark 5.2

• 1. For $\alpha =1+2\gamma$, $\gamma >0$ and $\mu =1$ in Theorem 5.3 yields Theorem 4.1 in [3].

• 2. For $\gamma =0$, Theorem 5.3 gives a result similar to Theorem 2.1 [2].

Now, we define

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

In this case, $V_{\lambda }$ reduces to the Komatu operator

$$\begin{aligned} V_{\lambda }(f)(z)=\frac{(1+a)^{p}}{\Gamma (p)}\int _{0}^{1}\left( \log \left( \frac{1}{t}\right) \right) ^{p-1}t^{a-1}f(tz)\mathrm{d}t, \, \, \, a>-1, \, \, p\ge 0. \end{aligned}$$

For $p = 1$ Komatu operator gives the Bernardi integral operator. For this $\lambda$, the following result holds.

Theorem 5.4

Let $-1<a$, $\alpha >0$, $p\ge 1$, ${\nu }\ge {\mu }/(\mu +1)$ and $\beta <1$ satisfy

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

where $g$ is given by (2.6). If $f\in {\mathcal {W}}_{\beta }(\alpha , \gamma )$, then the function

$$\begin{aligned} {\Phi }_{p}(a;z)*f(z)=\frac{(1+a)^{p}}{\Gamma (p)}\int _{0}^{1}\left( \log \left( \frac{1}{t}\right) \right) ^{p-1}t^{a-1}f(tz)\mathrm{d}t \end{aligned}$$

belongs to $S^{*}(\delta )$ with $\delta \in (0,1/2]$ if

$$\begin{aligned} a\le \left\{ \begin{array}{ll} 1+\frac{1}{\mu }, &{} \quad \mu \ge 1 (\gamma >0); \\ 3-\frac{1}{\alpha }, &{} \quad \gamma =0, \alpha \in (1/4,1/3]. \\ \end{array}\right. \end{aligned}$$

(5.5)

The value of $\beta$ is sharp.

Proof

Since

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

therefore, using the fact that $\log (1/t)>0$ for $t\in (0,1)$, and $p\ge 1$, condition (4.10) is satisfied whenever $a$ satisfies (5.5).

Remark 5.3

Setting $\alpha =1+2\gamma$, $\gamma >0$ and $\mu =1$ in Theorem 5.4, we get Theorem 4.2 in [3].