1. Introduction

Let $A\left (p \right)$ denote the class of all functions

$f(z)=z^{p}+\sum_{n=1}^{\infty }a_{n+p}z^{n+p}, \qquad \qquad \left( p\in N=\{1, 2, 3.....\} \right) \label{a}$

(1.1)

which are analytic and p-valent in the open unit disk $E=\left\{ z:\left| z \right| < 1 \right\}$. For p=1, A(1)=A. Let f, $g\in A (p)$, where f is given by (1.1) and g is defined by

$g(z)={{z}^{p}}+\sum\limits_{n=1}^{\infty }{{{b}_{n+p}}}{{z}^{n+p}}, \ \ \ \ \ \ (z\in E).$

Then the Hadamard product (or convolution) $f*g$ of the functions f and g is defined by

$\left( f*g \right)(z)={{z}^{p}}+\sum\limits_{n=1}^{\infty }{{{a}_{n+p}}}{{b}_{n+p}}{{z}^{n+p}}=\left( g*f \right)(z).$

Let UCV and UST denote the usual classes of uniformly convex and uniformly starlike functions and are defined by

$\begin{align} & UCV=\left\{ f(z)\in A:Re\left( 1+\frac{z{{f}^{\prime \prime }}(z)}{{{f}^{\prime }}(z)} \right)>\left| \frac{z{{f}^{\prime \prime }}(z)}{{{f}^{\prime }}(z)} \right| \right\}, \ \ z\in E, \\ & UST=\left\{ f(z)\in A:Re\left( \frac{z{{f}^{\prime }}(z)}{f(z)} \right)>\left| \frac{z{{f}^{\prime }}(z)}{f(z)}-1 \right| \right\}, \ \ z\in E. \\ \end{align}$

hese classes were first introduced by Goodman ^[2,3] and further investigated by ^[14] and ^[6]. Kanas and Wiśniowska ^[4,5] introduced the conic domain Ω_k, k ≥ 0 as

${{\Omega }_{k}}=\left\{ u+iv:u>k\sqrt{{{\left( u-1 \right)}^{2}}+{{v}^{2}}} \right\}.$

For fixed k this domain represents the right half plane (k = 0), a parabola (k = 1), the right branch of hyperbola (0 < k < 1) and an ellipse (k > 1). For detail study about $\Omega _{k}$ and its generalizations, see ^[8,9,10]. The extremal functions for these conic regions are

${{p}_{k}}\left( z \right)=\left\{ \begin{align} & \frac{1+z}{1-z}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=0, \\ & 1+\frac{2}{{{\pi }^{2}}}{{\left( \log \frac{1+\sqrt{z}}{1-\sqrt{z}} \right)}^{2}}, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=1, \\ & \frac{1}{1-{{k}^{2}}}\cosh \left\{ \left( \frac{2}{\pi }\arccos k \right)\log \frac{1+\sqrt{z}}{1-\sqrt{z}} \right\}-\frac{{{k}^{2}}}{1-{{k}^{2}}}, \ \ \ \ \ \ \ \ \ \ \ \ 0<k<1, \\ & \frac{1}{{{k}^{2}}-1}\sin \left( \frac{\pi }{2K(\kappa )}\int_{0}^{\frac{u(z)}{\sqrt{\kappa }}}{}\frac{\text{d}t}{\sqrt{1-{{t}^{2}}}\sqrt{1-{{\kappa }^{2}}{{t}^{2}}}} \right)+\frac{{{k}^{2}}}{{{k}^{2}}-1}, \ \ k>1, \\ \end{align} \right.$

(1.2)

where

$u(z)=\frac{z-\sqrt{\kappa }}{1-\sqrt{\kappa }z}, \ \ z\in \mathbb{E}$

and $\kappa \in (0, 1)$ is chosen such that $k=\cosh \, (\pi K^{\prime }(\kappa)/(4K (\kappa)))$. Here $K (\kappa)$ is Legendre's complete elliptic integral of first kind and $K^{\prime }(\kappa)=K (\sqrt{1-\kappa ^{2}}$) and $K^{\prime }\left (t\right)$ is the complementary integral of $K\left (t\right)$.

Now we define the following:

Definition. Let $f\in A (p)$ given by (1.1) is said to belong to $k-UR_{p},$ $k\geq 0$ if it satisfies the following condition

$Re\left( \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!} \right)>k\left| \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!}-1 \right|, \ \ z\in E,$

where $f^{(p)}(z)$ is the $pth$ derivative of $f (z).$

Special Cases:

ⅰ) For $k=0,$ we have $0-UR_{p}=R_{p},$ introduced and studied by Noor et-al. ^[7].

ⅱ) For $k=0$, $p=1,$ we have $0-UR_{1}=R,$ introduced and studied by Singh et-al.^[15].

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2. Preliminary Results

Lemma 2.1. ^[12]. For $\alpha \leq 1$ and $\beta \leq 1$

$p(\alpha )\ast p(\beta )\subset p(\delta ), \qquad \delta =1-2(1-\alpha )(1-\beta ).$

The result is sharp.

Lemma 2.2. ^[1]. Let $\{d_{n}\}_{0}^{\infty }$ be a convex null sequence. Then the function

$q(z)=\frac{d_{0}}{2}+\sum_{n=1}^{\infty }d_{n}z^{n}$

is analytic in E and $Req (z)>0$ $\ \ z\in E$.

Lemma 2.3. ^[13]. For $0\leq \theta \leq \pi,$

$\frac{1}{2}+\sum_{n=1}^{m}\frac{\cos n\theta }{n+1}\geq 0.$

Lemma 2.4. ^[7]. If f and g belong to the class $R_{p}$ and

$h^{(p-1)}(z)=f^{(p-1)}(z)\ast g^{(p-1)}(z).$

Then h also belong to the class $R_{p}.$

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3. Main Result

Theorem 3.1. Let $f\in k-UR_{P}$ then

$Re\left( \frac{f^{(p)}(z)}{p!}\right) >\frac{k-1+2\log 2}{k+1}.$

Proof. Let $f\in k-UR_{p}$ then by definition, we have

$Re\left( \frac{f^{(p)}(z)+zf^{(p+1)}(z)}{p!}\right) >k\left \vert \frac{f^{(p)}(z)+zf^{(p+1)}(z)}{p!}-1\right \vert .$

After some simple computations, we have

$Re\left( \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!} \right)>\frac{k}{k+1},$

(3.1)

This can be written as

$Re\left( 1+\sum\limits_{n=1}^{\infty }{\frac{(p+n)!(n+1)}{n!}}{{a}_{n+p}}{{z}^{n}} \right)>\frac{k}{k+1},$

(3.2)

or

$Re\left( 1+\frac{1}{2}\sum\limits_{n=1}^{\infty }{\frac{(p+n)!(n+1)}{n!}}{{a}_{n+p}}{{z}^{n}} \right)>\frac{2k+1}{2k+2}.$

(3.3)

Consider the function

$h(z)=1+2\sum\limits_{n=1}^{\infty }{\frac{{{z}^{n}}}{n+1}}.$

(3.4)

Clearly h is analytic, $h (0)=1\,$ in E and

$Reh(z)=Re\left( 1-\frac{2}{z}[z+\log (1-z)] \right)>-1+2\log 2.$

(3.5)

From (3.3) and (3.4), we have

$\left( \frac{{{f}^{(p)}}(z)}{p!} \right)=\left( 1+\frac{1}{2}\sum\limits_{n=1}^{\infty }{\frac{(p+n)!(n+1)}{n!}}{{a}_{n+p}}{{z}^{n}} \right)*\left( 1+2\sum\limits_{n=1}^{\infty }{\frac{{{z}^{n}}}{n+1}} \right).$

(3.6)

Now using (3.3), (3.5) and Lemma 2.2 with $\alpha =\frac{2k+1}{2k+2}, \beta =-1+2\log 2$ and $\delta =\frac{k-1+2\log 2}{k+1},$ we have

$Re\left( \frac{{{f}^{(p)}}(z)}{p!} \right)>\frac{k-1+2\log 2}{k+1}.$

(3.7)

This completes the result.

For some spacial value of k and p we obtain the following known result.

Corollary 3.2. ^[7]. Let $f\in R_{p}$ then

$Re\left( \frac{f^{(p)}(z)}{p!}\right) >-1+2\log 2.$

Theorem 3.3. Let $f\in k-UR_{p}$ then

$Re\left( \frac{{{f}^{(p-1)}}(z)}{z} \right)>\frac{p!(2k+1)}{2k+2}.$

(3.8)

Proof. From (3.3), we have

$Re\left( 1+\frac{1}{2}\sum_{n=1}^{\infty }\frac{(p+n)!(n+1)}{n!}a_{n+p}z^{n}\right) >\frac{(2k+1)}{2k+2}.$

Now consider the convex null sequence $\{d_{n}\}_{0}^{\infty }$ define by $d_{0}=0,$ $d_{n}=\frac{2}{(n+1)^{2}},$ $n\geq 1,$ using Lemma 2.2, we have

$Re\left( \frac{1}{2}+\sum\limits_{n=1}^{\infty }{\frac{2}{{{(n+1)}^{2}}}}{{z}^{n}} \right)>0,$

or equivalently

$Re\left( 1+2\sum\limits_{n=1}^{\infty }{\frac{1}{{{(n+1)}^{2}}}}{{z}^{n}} \right)>\frac{1}{2}.$

(3.9)

From (3.3) and (3.9), we have

$\frac{{{f}^{(p-1)}}(z)}{p!z}=\left( 1+\frac{1}{2}\sum\limits_{n=1}^{\infty }{\frac{(p+n)!(n+1)}{n!}}{{a}_{n+p}}{{z}^{n}} \right)*\left( 1+2\sum\limits_{n=1}^{\infty }{\frac{1}{{{(n+1)}^{2}}}}{{z}^{n}} \right).$

(3.10)

From (3.10) and Lemma (2.1) with $\alpha =\frac{2k+1}{2k+2}$ and $\beta =\frac{1}{2},$ we have

$Re\left( \frac{{{f}^{(p-1)}}(z)}{z} \right)>\frac{p!(2k+1)}{2k+2}.$

(3.11)

Which is the required result.

Corollary 3.4. ^[7]. Let $f\in R_{p}$ then

$Re\left( \frac{{{f}^{(p-1)}}(z)}{z} \right)>\frac{p!}{2}, \ \ \ z\in E.$

Corollary 3.5. ^[15]. Let $f\in R$ then

$Re\left( \frac{f(z)}{z} \right)>\frac{1}{2}, \ \ \ z\in E.$

Theorem 3.6. Let $f\in k-UR_{p}$ then for every $n\geq 1$, the $n$th partial sum of f satisfies

$ReS_{n}^{(p)}(z, f)>\frac{p!k}{k+1}, \ \ \ z\in E.$

and hence $S_{n}(z, f)$ is $p-$valent in $E.$

Proof. From (3.2) and (3.4), we have

$\frac{s_{n}^{(p)}(z, f)}{p!}=\left( 1+\sum\limits_{n=1}^{\infty }{\frac{(p+n)!(n+1)}{p!n}}{{a}_{n+p}}{{z}^{n}} \right)*\left( 1+\sum\limits_{n=1}^{\infty }{\frac{{{z}^{n}}}{n+1}} \right).$

(3.12)

Putting $z=re^{i\theta },$ $0\leq r\leq 1,$ $0\leq \theta \leq \pi$ and the minimum principle for harmonic functions with Lemma 2.3, we have

$\begin{align} & Re\left( 1+\sum\limits_{n=1}^{k}{\frac{{{z}^{n}}}{n+1}} \right)=Re\left( 1+\sum\limits_{n=1}^{k}{\frac{{{r}^{n}}{{e}^{in\theta }}}{n+1}} \right), \ \text{0}\le \theta \le \pi \\ & =Re\left( 1+\sum\limits_{n=1}^{k}{\frac{{{r}^{n}}}{n+1}}(\cos n\theta +i\sin n\theta ) \right) \\ & =\left( 1+\sum\limits_{n=1}^{k}{\frac{{{r}^{n}}\cos n\theta }{n+1}} \right) \\ & =\left( 1+\sum\limits_{n=1}^{k}{\frac{{{r}^{n}}\cos n\theta }{n+1}} \right)\ge \frac{1}{2}. \\ \end{align}$

(3.13)

Using (3.2), (3.12), (3.13) and Lemma 2.1 with $\alpha =\frac{k}{k+1}$ and $\beta =\frac{1}{2},$ we have

$Re\left( s_{n}^{(p)}(z, f)\right) >\frac{p!k}{k+1}.$

(3.14)

This completes the proof. From the result given by ^[11], we see that $s_{n}(z, f)$ is p-valent in E for every n ≥ 1.

Corollary 3.7. ^[7]. Let $f\in R_{p}$, then for every n ≥ 1, the nth partial sum of f satisfies

$ReS_{n}^{(p)}(z, f)>0, \ \ \ \ z\in E$

and hence $s_{n}(z, f)$ is p-valent in E.

For k=1 we have the following corollary.

Corollary 3.8. ^[15]. Let $f\in 1-UR_{p}$, then for every n ≥ 1, the nth partial sum of f satisfies

$ReS{{'}_{n}}(z, f)>\frac{p!}{2}, \ \ \ \ \ \ z\in E$

and hence $s_{n}(z, f)$ is univalent in E.

Theorem 3.9. Let $f\in k-UR_{p},$ $g\in R_{p}$ and

$h^{(p-1)}(z)=f^{(p-1)}(z)\ast g^{(p-1)}(z).$

Then h belong to the class $k-UR_{p}.$

Proof. Since

$h^{(p-1)}(z)=f^{(p-1)}(z)\ast g^{(p-1)}(z).$

(3.15)

It follows that

$zh^{(p)}(z)=f^{(p)}(z)\ast g^{(p-1)}(z).$

(3.16)

After simple computations, (3.16) can be written as

$Re\left( \frac{{{h}^{(p)}}(z)+z{{h}^{(p+1)}}(z)}{p!} \right)=Re\left( \left( \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!} \right)*\left( \frac{{{g}^{(p-1)}}(z)}{zp!} \right) \right).$

(3.17)

From (3.17), (3.1), Corollary 3.4 and Lemma 2.1 with $\alpha =\frac{k}{k+1}$ and $\beta =\frac{1}{2},$ we get the required proof.

Corollary 3.10. ^[15]. If $f (z)=z+\sum_{n=2}^{\infty }a_{n}z^{n},$ and $g (z)=z+\sum_{n=2}^{\infty }b_{n}z^{n}$ belong to R then so does their Hadamard product

$h(z)=f(z)\ast g(z).$

Theorem 3.11. If f, $g\in R_{p}$, $h\in k-UR_{p}$ and

$\varphi ^{(p-1)}(z)=h^{(p-1)}(z)\ast f^{(p-1)}(z)\ast g^{(p-1)}(z).$

Then $\varphi \in k-UR_{p}.$

Proof. Suppose that

$m^{(p-1)}(z)=f^{(p-1)}(z)\ast g^{(p-1)}(z),$

(3.18)

and it is clear from Lemma 2.4 that, $m\in R_{p}$. From the hypothesis and (3.18), we have

$\varphi ^{(p-1)}(z)=h^{(p-1)}(z)\ast m^{(p-1)}(z).$

(3.19)

From (3.19) and Theorem 3.9, we get the required result.

Theorem 3.12. If $f_{1},$ $f_{2},$ $f_{3}, ...,$ $f_{n}$ belong to $R_{p}, h\in k-UR_{p}$ and

$g^{(p-1)}(z)=f_{1}^{(p-1)}(z)\ast f_{2}^{(p-1)}(z)\ast f_{3}^{(p-1)}(z)\ast ...\ast f_{n}^{(p-1)}(z)\ast h^{(p-1)}(z). \label{z}$

(3.20)

Then $g\in k-UR_{p}.$

Proof. For proving the above Theorem, we use the principle of mathematical induction. For n = 2, we have proved Theorem 3.11, thus (3.20) hold true for n = 2. Suppose that (3.20) hold true for $n=k;$ that is,

$g^{(p-1)}(z)=f_{1}^{(p-1)}(z)\ast f_{2}^{(p-1)}(z)\ast f_{3}^{(p-1)}(z)\ast ...\ast f_{k}^{(p-1)}(z)\ast h^{(p-1)}(z). \label{zz}$

(3.21)

Then $g\in k-UR_{p}.$

We have to prove that (3.20) hold true for $n=k+1,$ for this, consider

$g^{(p-1)}(z)=f_{1}^{(p-1)}(z)\ast f_{2}^{(p-1)}(z)\ast f_{3}^{(p-1)}(z)\ast ...\ast f_{k+1}^{(p-1)}(z)\ast h^{(p-1)}(z). \label{mn}$

(3.22)

Let

${{M}^{(p-1)}}=f_{1}^{(p-1)}*f_{2}^{(p-1)}*f_{3}^{(p-1)}*.........*f_{k}^{(p-1)}*{{h}^{(p-1)}}$

Then by hypothesis $M\in k-UR_{p}.$ Now (3.22) becomes

$g^{(p-1)}(z)=(M^{(p-1)}\ast f_{k+1}^{(p-1)})(z).$

(3.23)

Using Theorem 3.9, from (3.23), we have

$Re\left( \frac{g^{(p)}(z)+zg^{(p+1)}(z)}{p!}\right) >\frac{k}{k+1}.$

(3.24)

(3.24) now implies that $g\in k-UR_{p}.$ Therefore, the result is true for $n=k+1$ and hence by using mathematical induction, (3.20) holds true for all $n\geq 2.$ This completes the proof.

Theorem 3.13. If $f$, $g\in k-UR_{p}$ and

$h^{(p-1)}(z)=f^{(p-1)}(z)\ast g^{(p-1)}(z).$

Then h belong to the class $k-UR_{p}.$

Proof. Since

$h^{(p-1)}(z)=f^{(p-1)}(z)\ast g^{(p-1)}(z).$

(3.25)

Differentiation yields

$zh^{(p)}(z)=f^{(p)}(z)\ast g^{(p-1)}(z).$

(3.26)

After simplification, we have

$Re\left( \frac{{{h}^{(p)}}(z)+z{{h}^{(p+1)}}(z)}{p!} \right)=Re\left( \left( \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!} \right)*\left( \frac{{{g}^{(p-1)}}(z)}{zp!} \right) \right).$

(3.27)

From (3.27), (3.1), (3.11) and Lemma 2.1 with $\alpha =\frac{k}{k+1}$ and $\beta =\frac{2k+1}{2k+2},$ we have

$Re\left( \frac{h^{(p)}(z)+zh^{(p+1)}(z)}{p!}\right) >\frac{k}{k+1}.$

(3.28)

(3.28) implies that $h$ belong to $k-UR_{p}.$

Our next result give us a sufficient condition for the class $k-UR_{p}.$

Theorem 3.14. Let $f\in A (p)$ satisfies

$\sum\limits_{n=1}^{\infty }{\frac{(k-1)(n+1)(p+n)!}{p!n!}}\left| {{a}_{n+p}} \right|<1.$

(3.29)

Then $f\in k-UR_{p}.$

Proof. To prove the required result it is sufficient to show that

$k\left \vert \frac{f^{(p)}(z)+zf^{(p+1)}(z)}{p!}-1\right \vert -Re \left( \frac{f^{(p)}(z)+zf^{(p+1)}(z)}{p!}-1\right) <1$

(3.30)

Now

$\begin{align} & k\left| \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!}-1 \right|-Re\left( \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!}-1 \right) \\ & \le (k-1)\left| \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!}-1 \right| \\ & =(k-1)\left| \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)-p!}{p!} \right| \\ & =(k-1)\left| \sum\limits_{n=1}^{\infty }{\frac{(n+1)(p+n)!}{p!n!}}{{a}_{n+p}}{{z}^{n}} \right|. \\ \end{align}$

This can be written as

$\begin{align} & k\left| \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!}-1 \right|-Re\left( \frac{{{f}^{(p)}}(z)+z{{f}^{(p+1)}}(z)}{p!}-1 \right) \\ & \le (k-1)\left| \sum\limits_{n=1}^{\infty }{\frac{(n+1)(p+n)!}{p!n!}}{{a}_{n+p}} \right|\left| {{z}^{n}} \right| \\ \end{align}$

(3.31)

(3.13) is bounded above by 1 if (3.29) is satisfied. This completes the proof.

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Conflicts of Interest

All authors declare no conflicts of interest in this paper.

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