Applications of Fractional Derivatives for p-Valently α-Convex Functions of Order β

Abstract

Let $A_p$ denote the class of p-valently functions $f\left(z\right)$ in the open unit disc U with $f^{\left(j\right)}\left(0\right)=0(j=0,1,2,\dots ,p-1)$. In general, we write $f\left(z\right)\in A_p$ by $f\left(z\right)=z^p+{\sum }_{k=1}^{\infty }{a}_{p+k}{z}^{p+k}(p\in \mathbb{N}).$ For $f\left(z\right)\in A_p,p$-valently $\alpha$-convex functions of order $\beta$ in U are introduced. Some interesting properties for such functions can be seen. In this paper, we introduce fractional derivatives $D_z^{j+\lambda }f\left(z\right)$ for $f\left(z\right)\in A_p$ with $j=0,1,2,\dots ,p-1$ and $0\le \lambda <1$. Applying the fractional derivatives $D_z^{j+\lambda }f\left(z\right)$, we would like to generalize p-valently $\alpha$-convex functions of order $\beta$ in U. Some results for such functions $f\left(z\right)\in A_p$ are discussed with example functions. Further, a conjecture for our research in this paper is given with an example function.

1. Introduction

Let $A_p$ denote the class of functions $f\left(z\right)$ that are analytic and p-valent in the open unit disc $U=\{z\in C:|z|<1\}$ with $f^{\left(j\right)}\left(0\right)=0(j=0,1,2,\dots ,p-1)$ and $p\in \mathbb{N}=\{1,2,3,\dots \}$. In general, we write $f\left(z\right)\in A_p$ such that

$$f\left(z\right)=z^p+{\displaystyle \sum _{k=1}^{\infty }}{a}_{p+k}{z}^{p+k}.$$

(1)

If $f\left(z\right)\in A_p$ satisfies

$$Re\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>\beta (z\in U)$$

(2)

for $0\le \beta <p$, then $f\left(z\right)$ is said to be p-valently starlike of order $\beta$ in U. If we take $f\left(z\right)\in A_p$ given by

$$f\left(z\right)={\displaystyle \frac{z^p}{{(1-z)}^{2(p-\beta )}}}(0\le \beta <p),$$

(3)

then $f\left(z\right)$ satisfies

$$Re\left\{{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right\}=Re\left\{p-2(p-\beta )\left({\displaystyle \frac{z}{1-z}}\right)\right\}>\beta (z\in U).$$

(4)

Thus $f\left(z\right)$ is p-valently starlike of order $\beta$ in U.

Also, if $f\left(z\right)\in A_p$ satisfies

$$Re\left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)>\beta (z\in U)$$

(5)

for $0\le \beta <p$, then we say that $f\left(z\right)$ is p-valently convex of order $\beta$ in U. If we consider a function $f\left(z\right)\in A_p$ given by

$$f\left(z\right)=p{\int }_0^z{\displaystyle \frac{u^{p-1}}{{(1-u)}^{2(p-\beta )}}}du(0\le \beta <p),$$

(6)

then we see that

$$Re\left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)=Re\left\{p-2(p-\beta )\left({\displaystyle \frac{z}{1-z}}\right)\right\}>\beta (z\in U)$$

(7)

and $f\left(z\right)$ is p-valently convex of order $\beta$ in U.

In 1971, Jack [1] proved that if $f\left(z\right)\in A_1$ is convex of order $\beta$ in U, then $f\left(z\right)$ is starlike of order $\delta \left(\beta \right)$ in U, where

$$\delta \left(\beta \right)={\displaystyle \frac{2\beta -1+\sqrt{4{\beta }^2-4\beta +9}}{4}}(0\le \beta <1).$$

(8)

More recently, Guney and Owa [2] proved that if $f\left(z\right)\in A_p$ is p-valently convex of order $\beta$ in U, then $f\left(z\right)$ is p-valently starlike of order $\delta (p,\beta )$ in U, where

$$\delta (p,\beta )={\displaystyle \frac{2\beta -1+\sqrt{4{\beta }^2-4\beta +1+8p}}{4}}(0\le \beta <p).$$

(9)

With such p-valently starlike functions and p-valently convex functions, Dziok [3] has introduced p-valently $\alpha$-convex functions of order $\beta$ in U as follows.

If $f\left(z\right)\in A_p$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\right)>\beta (z\in U)$$

(10)

for some $\alpha \ge 0$ and $0\le \beta <p$, then we call the $f\left(z\right)$ p-valently $\alpha$-convex of order $\beta$ in U. If we consider $f\left(z\right)\in A_p$ given by

$$f\left(z\right)={\left\{{\displaystyle \frac{p}{\alpha }}{\int }_0^z{\displaystyle \frac{u^{\frac{p}{\alpha }-1}}{{(1+u)}^{\frac{2(p-\beta )}{\alpha }}}}du\right\}}^{\alpha }$$

(11)

with $\alpha \ge 0$ and $0\le \beta <p$, then $f\left(z\right)$ satisfies

$${\displaystyle \frac{1}{\alpha }}{\left(f\left(z\right)\right)}^{{\displaystyle \frac{1}{\alpha }}-1}{f}^{\prime }\left(z\right)=\left({\displaystyle \frac{p}{\alpha }}\right){\displaystyle \frac{z^{\frac{p}{\alpha }-1}}{{(1+z)}^{\frac{2(p-\beta )}{\alpha }}}}$$

(12)

and

$$\begin{array}{cc}\hfill & Re\left\{(1-\alpha ){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\right\}=p-2(p-\beta )Re\left({\displaystyle \frac{z}{1+z}}\right)\hfill \\ \hfill & >p-(p-\beta )=\beta (z\in U).\hfill \end{array}$$

(13)

Thus, $f\left(z\right)$ given by (11) is p-valently $\alpha$-convex of order $\beta$ in U.

The definition of p-valently $\alpha$-convex functions of order $\beta$ in U is given by Dziok [3]. But Dziok considered only the case $\beta =0$. We find the example function $f\left(z\right)$ given by (3) for p-valently $\alpha$-convex functions of order $\beta$ in U. If $p=1$ for such functions, we say that $f\left(z\right)$ is $\alpha$-convex function of order $\beta$ in U. But, Miller, Mocanu and Reade [4] only considered the case of $\beta =0$. There are numerous papers on starlike and convex functions in connection with multivalent functions. At various times, different mathematicians have obtained interesting results by working on various classes of $\alpha$-convex functions of order $\beta$ [5,6,7,8,9,10,11]. In this paper, p-valently $\alpha$-convex functions of order $\beta$ in U are used as in the proofs of our theorems.

In this paper, applying the Riemann–Liouville integral, we introduce the fractional derivative $D_z^{\lambda }f\left(z\right)$ of order $\lambda (0\le \lambda <1)$ of $f\left(z\right)\in A_p$ given by

$$\begin{array}{cc}\hfill D_z^{\lambda }f\left(z\right)& ={\displaystyle \frac{1}{\mathrm{\Gamma }(1-\lambda )}}{\displaystyle \frac{d}{dz}}{\int }_0^z{\displaystyle \frac{f\left(t\right)}{{(z-t)}^{\lambda }}}dt\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-\lambda )}}{z}^{p-\lambda }+{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{\mathrm{\Gamma }(p+k+1)}{\mathrm{\Gamma }(p+k+1-\lambda )}}{a}_{p+k}{z}^{p+k-\lambda }.\hfill \end{array}$$

(14)

Further, we define

$$D_z^{1+\lambda }f\left(z\right)={\displaystyle \frac{d}{dz}}\left(D_z^{\lambda }f\left(z\right)\right)={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p-\lambda )}}{z}^{p-1-\lambda }+{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{\mathrm{\Gamma }(p+k+1)}{\mathrm{\Gamma }(p+k-\lambda )}}{a}_{p+k}{z}^{p+k-1-\lambda }$$

(15)

and

$$\begin{array}{cc}\hfill D_z^{j+\lambda }f\left(z\right)& ={\displaystyle \frac{d}{dz}}\left(D_z^{j-1+\lambda }f\left(z\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p-j+1-\lambda )}}{z}^{p-j-\lambda }+{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{\mathrm{\Gamma }(p+k+1)}{\mathrm{\Gamma }(p+k-j+1-\lambda )}}{a}_{p+k}{z}^{p+k-j-\lambda },\hfill \end{array}$$

(16)

where $\mathrm{\Gamma }\left(x\right)$ is a Gamma function.

Previously, Hardy and Littlewood [12] studied fractional integrals. The fractional derivative of order $\lambda$ is defined as the derivative of the fractional integral of order $1-\lambda$. In 1957, Kuttner [13] considered fractional derivatives. Following this, we saw some research papers for fractional derivatives in [14,15,16]. Such fractional derivatives were defined by Owa [17] (or Owa and Srivastava [18]). With the above fractional derivatives, we consider $f\left(z\right)\in A_p$ satisfying

$$Re\left((1-\alpha ){\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{D}_z^{j+2+\lambda }f\left(z\right)}{D_z^{j+1+\lambda }f\left(z\right)}}\right)\right)>\beta (z\in U)$$

(17)

for some $\alpha \ge 0$ and $0\le \beta <p-j-\lambda$. If $\lambda =0$ and $j=0$, then (17) becomes (10). Therefore, the condition (9) is the generalization for p-valently $\alpha$-convex of order $\beta$ in U.

In 1997, Dziok [3] gave the following result.

Theorem 1 

(Dziok [3]). If $f\left(z\right)\in A_p$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\right)>0(z\in U)$$

(18)

for some $\alpha \ge 0$, then

$$Re\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>\beta (p,\alpha )(z\in U)$$

(19)

where

$$\beta (p,\alpha )=\left\{\begin{array}{cc}0;& (0\le \alpha <p)\\ {\displaystyle \frac{p\mathrm{\Gamma }\left(\frac{1}{2}+\frac{p}{\alpha }\right)}{\sqrt{\pi }\mathrm{\Gamma }\left(1+\frac{p}{\alpha }\right)}}& (\alpha \ge p)\end{array}\right.$$

(20)

Further, Miller, Mocanu and Reade [4,19] gave the following result.

Theorem 2 

(Miller, Mocanu, Reade [4,19]). If $f\left(z\right)\in A_1$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\right)>0(z\in U)$$

(21)

for some $\alpha \ge 0$, then

$$Re\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>\beta \left(\alpha \right)(z\in U),$$

(22)

where

$$\beta \left(\alpha \right)=\left\{\begin{array}{cc}0& (0\le \alpha <1)\\ {\displaystyle \frac{\mathrm{\Gamma }\left(\frac{1}{2}+\frac{1}{\alpha }\right)}{\sqrt{\pi }\mathrm{\Gamma }\left(1+\frac{1}{\alpha }\right)}}& (\alpha \ge 1)\end{array}\right.$$

(23)

The result is sharp for

$$f\left(z\right)={\left\{{\displaystyle \frac{1}{\alpha }}{\int }_0^z{\displaystyle \frac{u^{\frac{1}{\alpha }-1}}{{(1+u)}^{\frac{2}{\alpha }}}}du\right\}}^{\alpha }.$$

(24)

Recently, Guney and Owa [2] gave the following result.

Theorem 3 

(Guney, Owa [2]). If $f\left(z\right)\in A_p$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\right)>\beta (z\in U)$$

(25)

for some $\alpha \ge 0$ and $0\le \beta <p$, then

$$Re\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>\gamma (z\in U)$$

(26)

where

$$\gamma ={\displaystyle \frac{1}{4}}\left((2\beta -\alpha )+\sqrt{{(2\beta -\alpha )}^2+8\alpha p}\right).$$

(27)

2. Properties of Fractional Derivatives

To consider the properties of fractional derivatives, we need the following lemma due to Jack [1] (or due to Miller and Mocanu [20]).

Lemma 1 

(Jack [1], Miller, and Mocanu [20]). Let $w\left(z\right)$ be regular in U with $w\left(0\right)=0$. If $\left|w\right(z\left)\right|$ attains its maximum value on the circle $\left|z\right|=r$ at a point $z_0\in U$, then

$$z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right)$$

(28)

where $k\ge 1$.

With the above lemma, we prove

Theorem 4. 

If $f\left(z\right)\in A_p$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{D}_z^{j+2+\lambda }f\left(z\right)}{D_z^{j+1+\lambda }f\left(z\right)}}\right)\right)>\beta (z\in U)$$

(29)

for some $\alpha \ge 0$ and $0\le \beta <p-j-\lambda (j=1,2,\dots ,p-1;0\le \lambda <1)$ then

$$Re\left({\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right)>\gamma (z\in U),$$

(30)

where

$$\gamma ={\displaystyle \frac{1}{4}}\left((2\beta -\alpha )+\sqrt{{(2\beta -\alpha )}^2+8\alpha (p-j-\lambda )}\right).$$

(31)

Proof. 

From (16) for $D_z^{j+\lambda }f\left(z\right)$, we see that

$${\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}=p-j-\lambda$$

(32)

for $z=0$. We consider a function $w\left(z\right)$ given by

$${\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}=(p-j-\lambda ){\displaystyle \frac{1-tw\left(z\right)}{1-w\left(z\right)}}$$

(33)

with $t>0$ and $w\left(z\right)\ne 1$ in U. Then $w\left(z\right)$ is regular in U and $w\left(0\right)=0$. Therefore, if there exists a point $z_0\in U$ such that

$$\underset{\left|z\right|\le \left|z_0\right|}{max}\left|w\left(z\right)\right|=\left|w\left(z_0\right)\right|=1,$$

(34)

then Lemma 1 gives us

$$z_0{w}^{\prime }\left(z_0\right)=kw\left(z_0\right)(k\ge 1)$$

(35)

If $\left|w\right(z\left)\right|<1$ for all $z\in U$, then $w\left(z\right)$ satisfies

$$Re\left({\displaystyle \frac{1-tw\left(z\right)}{1-w\left(z\right)}}\right)>{\displaystyle \frac{1+t}{2}}(z\in U).$$

(36)

Thus, if we consider $\gamma$ given by

$$\gamma =(p-j-\lambda ){\displaystyle \frac{1+t}{2}}$$

(37)

then $f\left(z\right)$ satisfies (30). It follows from (33) that

$$\begin{array}{cc}\hfill & (1-\alpha ){\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{D}_z^{j+2+\lambda }f\left(z\right)}{D_z^{j+1+\lambda }f\left(z\right)}}\right)\hfill \\ \hfill & =(p-j-\lambda ){\displaystyle \frac{1-tw\left(z\right)}{1-w\left(z\right)}}+{\displaystyle \frac{\alpha z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}}-{\displaystyle \frac{\alpha tz{w}^{\prime }\left(z\right)}{1-tw\left(z\right)}}\hfill \end{array}$$

(38)

If there exists a point $z_0\in U$ which satisfies (34), then we obtain

$$\begin{array}{cc}\hfill & Re\left\{(1-\alpha ){\displaystyle \frac{z_0{D}_z^{j+1+\lambda }f\left(z_0\right)}{D_z^{j+\lambda }f\left(z_0\right)}}+\alpha \left(1+{\displaystyle \frac{z_0{D}_z^{j+2+\lambda }f\left(z_0\right)}{D_z^{j+1+\lambda }f\left(z_0\right)}}\right)\right\}\hfill \\ \hfill & =Re\left\{(p-j-\lambda ){\displaystyle \frac{1-tw\left(z_0\right)}{1-w\left(z_0\right)}}+{\displaystyle \frac{\alpha kw\left(z_0\right)}{1-w\left(z_0\right)}}-{\displaystyle \frac{\alpha ktw\left(z_0\right)}{1-tw\left(z_0\right)}}\right\}\hfill \end{array}$$

(39)

with $k\ge 1$. Letting $w\left(z_0\right)=e^{i\theta }(0<\theta <2\pi )$, we have

$$\begin{array}{cc}\hfill & Re\left({\displaystyle \frac{1-tw\left(z_0\right)}{1-w\left(z_0\right)}}\right)={\displaystyle \frac{1+t}{2}},\hfill \\ \hfill & Re\left({\displaystyle \frac{w\left(z_0\right)}{1-w\left(z_0\right)}}\right)=-{\displaystyle \frac{1}{2}},\hfill \end{array}$$

(40)

and

$$Re\left({\displaystyle \frac{tw\left(z_0\right)}{1-tw\left(z_0\right)}}\right)={\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1+tw\left(z_0\right)}{1-tw\left(z_0\right)}}-1\right)\ge {\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1-t}{1+t}}-1\right)\left(w\left(z_0\right)=-1\right)$$

(41)

Therefore, we obtain

$$\begin{array}{cc}\hfill & Re\left\{(1-\alpha ){\displaystyle \frac{z_0{D}_z^{j+1+\lambda }f\left(z_0\right)}{D_z^{j+\lambda }f}}\left(z\right)+\alpha \left(1+{\displaystyle \frac{z_0{D}_z^{j+2+\lambda }f\left(z_0\right)}{D_z^{j+1+\lambda }f\left(z_0\right)}}\right)\right\}\hfill \\ \hfill \le & \left\{(p-j-\lambda ){\displaystyle \frac{1+t}{2}}-{\displaystyle \frac{\alpha k}{2}}-{\displaystyle \frac{\alpha k}{2}}\left({\displaystyle \frac{1-t}{1+t}}-1\right)\right\}\hfill \\ \hfill \le & (p-j-\lambda ){\displaystyle \frac{1+t}{2}}-{\displaystyle \frac{\alpha }{2}}\left({\displaystyle \frac{1-t}{1+t}}\right)\hfill \end{array}$$

(42)

If we say that

$$\beta =(p-j-\lambda ){\displaystyle \frac{1+t}{2}}-{\displaystyle \frac{\alpha }{2}}\left({\displaystyle \frac{1-t}{1+t}}\right)$$

(43)

then $f\left(z\right)$ does not satisfy our condition (29). Thus, $w\left(z\right)$ should be $\left|w\right(z\left)\right|<1$ for all $z\in U$ if $f\left(z\right)$ satisfies (21). We note that

$$t={\displaystyle \frac{1}{2(p-j-\lambda )}}\left\{(2\beta -\alpha -2(p-j-\lambda ))+\sqrt{{(2\beta -\alpha )}^2+8\alpha (p-j-\lambda )}\right\}.$$

(44)

This gives us

$$\gamma =(p-j-\lambda ){\displaystyle \frac{1+t}{2}}={\displaystyle \frac{1}{4}}\left\{(2\beta -\alpha )+\sqrt{{(2\beta -\alpha )}^2+8\alpha (p-j-\lambda )}\right\}.$$

(45)

This completes the proof of the theorem.

• □

If we take $\alpha ={\displaystyle \frac{1}{2}}$ in Theorem 4, then we have

Corollary 1. 

If $f\left(z\right)\in A_p$ satisfies

$$Re\left({\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}+{\displaystyle \frac{z{D}_z^{j+2+\lambda }f\left(z\right)}{D_z^{j+1+\lambda }f\left(z\right)}}\right)>2\beta -1(z\in U)$$

(46)

for some $0\le \beta <p-j-\lambda$, then

$$Re\left({\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right)>\gamma (z\in U),$$

(47)

where

$$\gamma ={\displaystyle \frac{1}{8}}\left((4\beta -1)+\sqrt{{(4\beta -1)}^2+16(p-j-\lambda )}\right).$$

(48)

Letting $\lambda =0$ in Theorem 4, we see

Corollary 2. 

If $f\left(z\right)\in A_p$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{f}^{(j+1)}\left(z\right)}{f^{\left(j\right)}\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{f}^{(j+2)}\left(z\right)}{f^{(j+1)}\left(z\right)}}\right)\right)>\beta (z\in U)$$

(49)

for some $\alpha \ge 0$ and $0\le \beta <p-j$, then

$$Re\left({\displaystyle \frac{z{f}^{(j+1)}\left(z\right)}{f^{\left(j\right)}\left(z\right)}}\right)>\gamma (z\in U),$$

(50)

where

$$\gamma ={\displaystyle \frac{1}{4}}\left((2\beta -\alpha )+\sqrt{{(2\beta -\alpha )}^2+8\alpha (p-j)}\right).$$

(51)

Remark 1. 

If we take $j=0$ in Corollary 2, then we have Theorem 3 by Guney and Owa [2]. Letting $\alpha =1$ in Theorem 4, then we obtain

Corollary 3. 

If $f\left(z\right)\in A_p$ satisfies

$$Re\left(1+{\displaystyle \frac{z{D}_z^{j+2+\lambda }f\left(z\right)}{D_z^{j+1+\lambda }f\left(z\right)}}\right)>\beta (z\in U)$$

(52)

for some $0\le \beta <p-j-\lambda$, then

$$Re\left({\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right)>\gamma (z\in U),$$

(53)

where

$$\gamma ={\displaystyle \frac{1}{4}}\left((2\beta -1)+\sqrt{{(2\beta -1)}^2+8(p-j-\lambda )}\right).$$

(54)

Further, if we consider $p=1,j=0$ and $\lambda =0$ in Corollary 3, then we have the result by Jack [1].

Corollary 4. 

If $f\left(z\right)\in A_1$ satisfies

$$Re\left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)>\beta (z\in U).$$

(55)

for some $0\le \beta <1$, then

$$Re\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\right)>{\displaystyle \frac{1}{4}}\left((2\beta -1)+\sqrt{4{\beta }^2-4\beta +9}\right)$$

(56)

Next, we consider

Theorem 5. 

If $f\left(z\right)\in A_p$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{z^{p+1-j-\lambda }}}+\alpha \left(1+{\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right)\right)>\beta (z\in U)$$

(57)

for $\alpha \ge 0,\alpha \ne 1,0\le \beta <{\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}},j\in \mathbb{N}$ and $0\le \lambda <1$, then

$$Re\left({\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{z^{p+1-j-\lambda }}}\right)>\gamma (z\in U),$$

(58)

where

$$\gamma ={\displaystyle \frac{1}{4(1-\alpha )}}(2\beta -\alpha -2\alpha (p+1-j-\lambda )+\sqrt{M}),$$

(59)

and

$$\begin{array}{cc}\hfill M& ={(2\beta -\alpha )}^2+4{\alpha }^2(p+1-j-\lambda )(p+2-j-\lambda )\hfill \\ \hfill & +8\alpha (1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}-8\alpha \beta (p+1-j-\lambda ).\hfill \end{array}$$

(60)

Proof. 

We note that

$${\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{z^{p+1-j-\lambda }}}={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}$$

(61)

for $z=0$.

Noting that

$$Re\left({\displaystyle \frac{1-tw\left(z\right)}{1-w\left(z\right)}}\right)>{\displaystyle \frac{1+t}{2}}(z\in U)$$

(62)

if $\left|w\right(z\left)\right|<1(z\in U)$ from (28), we define a function $w\left(z\right)$ by

$${\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{z^{p+1-j-\lambda }}}={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1-tw\left(z\right)}{1-w\left(z\right)}}\right)$$

(63)

with $t>0$ and $w\left(z\right)\ne 1$ for $z\in U$, then $w\left(z\right)$ is analytic in U and $w\left(0\right)=0$. If $\left|w\right(z\left)\right|<1$ for all $z\in U$, then we have

$$\begin{array}{ccc}\hfill Re\left({\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{z^{p+1-j-\lambda }}}\right)& =& {\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}Re\left({\displaystyle \frac{1-tw\left(z\right)}{1-w\left(z\right)}}\right)\hfill \\ & >& {\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1+t}{2}}\right)(z\in U).\hfill \end{array}$$

(64)

Thus, we define $\gamma$ by

$$\gamma ={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1+t}{2}}\right).$$

(65)

Using (69), we obtain

$$\begin{array}{c}(1-\alpha ){\displaystyle \frac{z{D}_z^{j+\lambda }f\left(z\right)}{z^{p+1-j-\lambda }}}+\alpha \left(1+{\displaystyle \frac{z{D}_z^{j+1+\lambda }f\left(z\right)}{D_z^{j+\lambda }f\left(z\right)}}\right)=(1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1-tw\left(z\right)}{1-w\left(z\right)}}\right)\\ +\alpha (p+1-j-\lambda )+{\displaystyle \frac{\alpha z{w}^{\prime }\left(z\right)}{1-w\left(z\right)}}-{\displaystyle \frac{\alpha tz{w}^{\prime }\left(z\right)}{1-tw\left(z\right)}}\end{array}$$

(66)

If we suppose that there exists a point $z_0\in U$ such that

$$\underset{\left|z\right|\le z_0\mid }{max}\left|w\left(z\right)\right|=\left|w\left(z_0\right)\right|=1$$

(67)

then, by Lemma 1, we have

$$\begin{array}{ccc}& & Re\left\{(1-\alpha ){\displaystyle \frac{z_0{D}_z^{j+\lambda }f\left(z_0\right)}{z_0^{p+1-j-\lambda }}}+\alpha \left(1+{\displaystyle \frac{z_0{D}_z^{j+1+\lambda }f\left(z_0\right)}{D_z^{j+\lambda }f\left(z_0\right)}}\right)\right\}\hfill \\ & =& Re\left\{\begin{array}{c}(1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1-tw\left(z_0\right)}{1-w\left(z_0\right)}}\right)\\ +\alpha (p+1-j-\lambda )+{\displaystyle \frac{\alpha kw\left(z_0\right)}{1-w\left(z_0\right)}}-{\displaystyle \frac{\alpha ktw\left(z_0\right)}{1-tw\left(z_0\right)}}\end{array}\right\}.\hfill \end{array}$$

(68)

Letting $w\left(z_0\right)=e^{i\theta }(0<\theta <2\pi )$, we see that

$$\begin{array}{cc}\hfill & Re\left({\displaystyle \frac{1-tw\left(z_0\right)}{1-w\left(z_0\right)}}\right)={\displaystyle \frac{1+t}{2}},\hfill \\ \hfill & Re\left({\displaystyle \frac{w\left(z_0\right)}{1-w\left(z_0\right)}}\right)=-{\displaystyle \frac{1}{2}},\hfill \end{array}$$

(69)

and

$$Re\left({\displaystyle \frac{tw\left(z_0\right)}{1-tw\left(z_0\right)}}\right)={\displaystyle \frac{1}{2}}Re\left({\displaystyle \frac{1+tw\left(z_0\right)}{1-tw\left(z_0\right)}}-1\right)\ge {\displaystyle \frac{1}{2}}\left({\displaystyle \frac{1-t}{1+t}}-1\right)\left(w\left(z_0\right)=-1\right).$$

(70)

It follows from the above that

$$\begin{array}{cc}\hfill Re& \left\{(1-\alpha ){\displaystyle \frac{z_0{D}_z^{j+\lambda }f\left(z_0\right)}{z_0^{p+1-j-\lambda }}}+\alpha \left(1+{\displaystyle \frac{z_0{D}_z^{j+1+\lambda }f\left(z_0\right)}{D_z^{j+\lambda }f\left(z_0\right)}}\right)\right\}\hfill \\ \hfill & \le (1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1+t}{2}}\right)+\alpha (p+1-j-\lambda )-{\displaystyle \frac{\alpha k}{2}}\left({\displaystyle \frac{1-t}{1+t}}\right)\hfill \\ \hfill & \le (1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1+t}{2}}\right)+\alpha (p+1-j-\lambda )-{\displaystyle \frac{\alpha }{2}}\left({\displaystyle \frac{1-t}{1+t}}\right)\hfill \end{array}$$

(71)

Therefore, we consider $\beta$ such that

$$\beta =(1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1+t}{2}}\right)+\alpha (p+1-j-\lambda )-{\displaystyle \frac{\alpha }{2}}\left({\displaystyle \frac{1-t}{1+t}}\right).$$

(72)

Then $f\left(z\right)$ can not satisfy the condition (56). This means that there is no $z_0\in U$ such that for this $\beta$ and that $\left|w\right(z\left)\right|<1$ for all $z\in U$.

Equation (72) becomes

$$\begin{array}{c}(1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}{t}^2-\left(2\beta -\alpha -2(1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}-2\alpha (p+1-j-\lambda )\right)t\\ -\left(2\beta +\alpha -(1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}-2\alpha (p+1-j-\lambda )\right)\\ =0\end{array}$$

(73)

and

$$t={\displaystyle \frac{\mathrm{\Gamma }(p+1-j-\lambda )}{2(1-\alpha )\mathrm{\Gamma }(p+1)}}\left\{\left(2\beta -\alpha -2(1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}-2\alpha (p+1-j-\lambda )\right)+\sqrt{M}\right\},$$

where

$$\begin{array}{cc}\hfill & M={(2\beta -\alpha )}^2+4{\alpha }^2(p+1-j-\lambda )(p+2-j-\lambda )\hfill \\ \hfill & +8\alpha (1-\alpha ){\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}-8\alpha \beta (p+1-j-\lambda ).\hfill \end{array}$$

(74)

Thus, we have

$$\gamma ={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }(p+1-j-\lambda )}}\left({\displaystyle \frac{1+t}{2}}\right)={\displaystyle \frac{1}{4(1-\alpha )}}(2\beta -\alpha -2\alpha (p+1-j-\lambda )+\sqrt{M}).$$

(75)

• □

Considering $j=1$ and $\lambda =0$ in Theorem 5, we have

Corollary 5. 

If $f\left(z\right)\in A_p$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z{f}^{\prime }\left(z\right)}{z^p}}+\alpha \left(1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\right)\right)>\beta (z\in U)$$

(76)

for some $\alpha \ge 0,\alpha \ne 1,0\le \beta <p$, then

$$Re\left({\displaystyle \frac{z{f}^{\prime }\left(z\right)}{z^p}}\right)>\gamma (z\in U),$$

(77)

where

$$\gamma ={\displaystyle \frac{1}{4(1-\alpha )}}\left\{(2\beta -\alpha -2\alpha p)+\sqrt{{(2\beta -\alpha )}^2+4\alpha p(\alpha p+2-2\beta -\alpha )}\right\}.$$

(78)

3. Application for Dziok’s Theorem

In this section, we would like to consider the application of fractional derivatives for Theorem 1 by Dziok [3].

Theorem 6 

(Dziok [3]). Let $f\left(z\right)\in A_p,n\in \mathbb{N},m\in \mathbb{N}\cup \left\{0\right\}$, and $n>m$. If $f\left(z\right)$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}+\alpha \left(1+{\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+2}f\left(z^n\right)}{D_z^{\frac{m}{n}+1}f\left(z^n\right)}}\right)\right)>0(z\in U)$$

(79)

for some $\alpha \ge 0$, then

$$Re\left({\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}\right)>{\displaystyle \frac{\beta }{n}}(z\in U),$$

(80)

where

$$\beta =\left\{\begin{array}{cc}0;& (0\le \alpha <pn-m)\\ {\displaystyle \frac{(pn-m)\mathrm{\Gamma }\left(\frac{1}{2}+\frac{pn-m}{\alpha }\right)}{\sqrt{\pi }\mathrm{\Gamma }\left(1+\frac{pn-m}{\alpha }\right)}};& (\alpha \ge pn-m)\end{array}\right..$$

(81)

Proof. 

It follows that

$$D_z^{{\displaystyle \frac{m}{n}}}f\left(z\right)={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}}{z}^{p-{\displaystyle \frac{m}{n}}}+{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{\mathrm{\Gamma }(p+k+1)}{\mathrm{\Gamma }\left(p+k+1-\frac{m}{n}\right)}}{a}_{p+k}{z}^{p+k-{\displaystyle \frac{m}{n}}}$$

(82)

and that

$${\displaystyle \frac{\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}{\mathrm{\Gamma }(p+1)}}{D}_z^{{\displaystyle \frac{m}{n}}}f\left(z^n\right)=z^{pn-m}+{\displaystyle \sum _{k=1}^{\infty }}{\displaystyle \frac{\mathrm{\Gamma }(p+k+1)\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}{\mathrm{\Gamma }\left(p+k+1-\frac{m}{n}\right)\mathrm{\Gamma }(p+1)}}{a}_{p+k}{z}^{(p+k)n-m}.$$

(83)

Thus, we consider a function $g\left(z\right)$ by

$$g\left(z\right)={\displaystyle \frac{\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}{\mathrm{\Gamma }(p+1)}}{D}_z^{{\displaystyle \frac{m}{n}}}f\left(z^n\right),$$

(84)

then $g\left(z\right)\in A_{pn-m}$,

$${\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}={\displaystyle \frac{n{z}^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}$$

(85)

and

$$1+{\displaystyle \frac{z{g}^{″}\left(z\right)}{g^{\prime }\left(z\right)}}=n\left(1+{\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+2}f\left(z^n\right)}{D_z^{\frac{m}{n}+1}f\left(z^n\right)}}\right).$$

(86)

Therefore, we have

$$\begin{array}{cc}\hfill & \mathrm{Re}\left\{(1-\alpha ){\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}+\alpha \left(1+{\displaystyle \frac{z{g}^{″}\left(z\right)}{g^{\prime }\left(z\right)}}\right)\right\}\hfill \\ \hfill & =n\mathrm{Re}\left\{(1-\alpha ){\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}+\alpha \left(1+{\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+2}f\left(z^n\right)}{D_z^{\frac{m}{n}+1}f\left(z^n\right)}}\right)\right\}>0(z\in U)\hfill \end{array}$$

(87)

by the condition (79). Since $g\left(z\right)\in A_{pn-m}$, Theorem 1 gives us

$$Re\left({\displaystyle \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}}\right)=nRe\left\{{\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}\right\}>\beta (z\in U)$$

(88)

with

$$\beta =\left\{\begin{array}{cc}0;& (0\le \alpha <pn-m)\\ {\displaystyle \frac{(pn-m)\mathrm{\Gamma }\left(\frac{1}{2}+\frac{pn-m}{\alpha }\right)}{\sqrt{\pi }\mathrm{\Gamma }\left(1+\frac{pn-m}{\alpha }\right)}}& ;\end{array}(\alpha \ge pn-m).\right.$$

(89)

This completes the proof of the theorem. □

Remark 2. 

We consider a function $f\left(z\right)\in A_p$ given by

$$D_z^{{\displaystyle \frac{m}{n}}}f\left(z^n\right)={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}}{\left\{{\displaystyle \frac{pn-m}{\alpha }}{\int }_0^z{\displaystyle \frac{u^{\frac{pn-m}{\alpha }-1}}{{(1+u)}^{\frac{2(\beta -pn+m)}{\alpha }}}}du\right\}}^{\alpha }.$$

(90)

Then $f\left(z\right)$ satisfies

$$\begin{array}{c}{\left\{{\displaystyle \frac{\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}{\mathrm{\Gamma }(p+1)}}{D}_z^{{\displaystyle \frac{m}{n}}}f\left(z^n\right)\right\}}^{{\displaystyle \frac{1}{\alpha }}}={\displaystyle \frac{pn-m}{\alpha }}{\int }_0^z{\displaystyle \frac{u^{\frac{pn-m}{\alpha }-1}}{{(1+u)}^{\frac{2(\beta -pn+m)}{\alpha }}}}du\\ {\left({\displaystyle \frac{\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}{\mathrm{\Gamma }(p+1)}}\right)}^{{\displaystyle \frac{1}{\alpha }}}{\left(D_z^{{\displaystyle \frac{m}{n}}}f\left(z^n\right)\right)}^{{\displaystyle \frac{1}{\alpha }}-1}\left(D_z^{{\displaystyle \frac{m}{n}}+1}f\left(z^n\right)\right)n{z}^{n-1}=(pn-m){\displaystyle \frac{z^{\frac{pn-m}{\alpha }-1}}{{(1+z)}^{\frac{2(\beta -pn+m)}{\alpha }}}},\end{array}$$

(91)

and

$$\begin{array}{ccc}& & \left({\displaystyle \frac{1}{\alpha }}-1\right){\displaystyle \frac{n{z}^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}+{\displaystyle \frac{n{z}^n{D}_z^{\frac{m}{n}+2}f\left(z^n\right)}{D_z^{\frac{m}{n}+1}f\left(z^n\right)}}+(n-1)\hfill \\ & =& \left({\displaystyle \frac{pn-m}{\alpha }}-1\right)-{\displaystyle \frac{2(\beta -pn+m)}{\alpha }}\left({\displaystyle \frac{z}{z+1}}\right).\hfill \end{array}$$

(92)

From the above, we obtain

$$\begin{array}{ccc}& & n\left\{(1-\alpha ){\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}+\alpha \left(1+{\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+2}f\left(z^n\right)}{D_z^{\frac{m}{n}+1}f\left(z^n\right)}}\right)\right\}\hfill \\ & =& (pn-m)-2(\beta -pn+m)\left({\displaystyle \frac{z}{1+z}}\right)\hfill \end{array}$$

(93)

and

$$\begin{array}{cc}\hfill nRe\left\{(1-\alpha ){\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}\right.& \left.+\alpha \left(1+{\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+2}f\left(z^n\right)}{D_z^{\frac{m}{n}+1}f\left(z^n\right)}}\right)\right\}\hfill \\ \hfill & =(pn-m)-2(\beta -pn+m)Re\left({\displaystyle \frac{z}{1+z}}\right)\hfill \\ \hfill & >(pn-m)+(\beta -pn+m)=\beta (z\in U)\hfill \end{array}$$

(94)

Therefore, if $\beta =0$ in (90), then $f\left(z\right)$ satisfies (79). With the above remark, we give

Conjecture 1. 

Let $f\left(z\right)\in A_p,n\in \mathbb{N},m\in \mathbb{N}\cup \left\{0\right\}$, and $n>m$. If $f\left(z\right)$ satisfies

$$Re\left((1-\alpha ){\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}+\alpha \left(1+{\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+2}f\left(z^n\right)}{D_z^{\frac{m}{n}+1}f\left(z^n\right)}}\right)\right)>\beta (z\in U)$$

(95)

for some $\alpha \ge 0$ and $0\le \beta <pn-m$, then

$$Re\left({\displaystyle \frac{z^n{D}_z^{\frac{m}{n}+1}f\left(z^n\right)}{D_z^{\frac{m}{n}}f\left(z^n\right)}}\right)>{\displaystyle \frac{\gamma }{n}}(z\in U),$$

(96)

where

$$\gamma =\underset{0\le \theta <2\pi }{min}Re\left({\displaystyle \frac{e^{in\theta }{D}_z^{\frac{m}{n}+1}f\left(e^{in\theta }\right)}{D_z^{\frac{m}{n}}f\left(e^{in\theta }\right)}}\right)$$

(97)

and

$$D_z^{{\displaystyle \frac{m}{n}}}f\left(z^n\right)={\displaystyle \frac{\mathrm{\Gamma }(p+1)}{\mathrm{\Gamma }\left(p+1-\frac{m}{n}\right)}}{\left\{{\displaystyle \frac{pn-m}{\alpha }}{\int }_0^z{\displaystyle \frac{u^{\frac{pn-m}{\alpha }-1}}{{(1+u)}^{\frac{2(\beta -pn+m)}{\alpha }}}}du\right\}}^{\alpha }.$$

(98)

4. Discussion

In 1997, Dziok introduced p-valently $\alpha$-convex functions of order $\beta$ in U considering $\alpha$-convex functions $f\left(z\right)\in A_1$ in U. However, Dziok did not find starlikeness order for such functions. Dziok gave results only for the case of $\beta =0$ using the result by Miller, Mocanu and Reade. More recently, Guney and Owa found some results for starlikeness order for p-valently $\alpha$-convex functions of order $\beta$ in U. Considering the method by Guney and Owa, we discuss applications of p-valently $\alpha$-convex functions of order $\beta$ for fractional derivatives defined by Owa. As has been done in this paper, more discussion is needed for this problem because it is not easy to find the extremal function for p-valently $\alpha$-convex functions of order $\beta$ in U. For this reason, we give our conjecture. We hope we can solve this conjecture soon.