Let $\mathcal{A}_p$ denote the class of functions $f$ of the form

$f(z)=z^p+\sum\limits_{k=1}^{\infty}a_{k+p}z^{k+p} (p\in\mathbb{N}=\{1, 2, \cdots\}), $

(1.1)

which are analytic in the open unit disk $\mathbb{U}=\{z:z\in\mathbb{C} \;\; {\rm and} \;\; |z|<1\}$. Also, let $\mathcal{A}_1=\mathcal{A}$.

Let $f, g\in\mathcal{A}_p$, where $f$ is given by (1.1) and $g$ is defined by

$g(z)={{z}^{p}}+\sum\limits_{k=1}^{\infty }{{{b}_{k+p}}{{z}^{k+p}}}.$

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

$(f*g)(z)=z^p+\sum\limits_{k=1}^{\infty}a_{k+p}b_{k+p}z^{k+p}=(g*f)(z).$

For two functions $f$ and $g$, analytic in $\mathbb{U}$, we say that the function $f$ is subordinate to $g$ in $\mathbb{U}$, if there exists a Schwarz function $\omega$, which is analytic in $\mathbb{U}$ with

$\omega(0)=0 \;\; {\rm {and}} \;\; |\omega(z)|<1 \;\; (z\in\mathbb{U}), $

such that

$f(z)=g(\omega(z)) \;\; (z\in\mathbb{U}).$

We denote this subordination by $f(z)\prec g(z)$. Furthermore, if the function $g$ is univalent in $\mathbb{U}$, then we have the following equivalence (see, for details, [3, 12]; see also [19, 20]):

$f(z)\prec g(z) (z\in\mathbb{U})\Longleftrightarrow f(0)=g(0) \;\; \mbox{and} \;\; f(\mathbb{U})\subset g(\mathbb{U}).$

A function $f\in\mathcal{A}$ is said to be the class $\mathcal{K}$ of convex functions in $\mathbb{U}$ if and only if

${\rm Re}\left(\frac{zf''(z)}{f'(z)}\right)>-1 (z\in\mathbb{U}).$

(1.2)

A function $f\in\mathcal{A}$ is said to be close-to-convex of order $\alpha (0\leq\alpha\leq1)$ in $\mathbb{U}$ if there exists a convex univalent function $h\in\mathcal{A}$ and a real $\beta$ such that

${\rm Re}\left(\frac{f'(z)}{e^{i\beta}h'(z)}\right)>\alpha (z\in\mathbb{U}).$

(1.3)

Janowski [11] introduced the class

$\mathcal{S}^*(a, b)=\left\{f\in\mathcal{A}: \frac{zf'(z)}{f(z)}\prec\frac{1+az}{1+bz} (z\in\mathbb{U};-1\leq b<a\leq1)\right\}.$

(1.4)

For $a=1, b=-1$, we have the class of starlike functions $\mathcal{S}^*=\mathcal{S}^*(1, -1)$.

For parameters $a_i\in\mathbb{C} (i=1, 2, \cdots, q)$ and $b_j\in\mathbb{C}\setminus\mathbb{Z}_0^- (\mathbb{Z}_0^-=0, -1, -2, \cdots; j=1, 2, \cdots, s)$, the generalized hypergeometric function ${_q F_s}(a_1, \cdots, a_q;b_1, \cdots, b_s;z)$ is defined by

$\begin{align} &_{q}{{F}_{s}}({{a}_{1}},\cdots ,{{a}_{q}};{{b}_{1}},\cdots ,{{b}_{s}};z)=\sum\limits_{k=0}^{\infty }{\frac{{{({{a}_{1}})}_{k}}\cdots {{({{a}_{q}})}_{k}}}{{{({{b}_{1}})}_{k}}\cdots {{({{b}_{s}})}_{k}}}}\frac{{{z}^{k}}}{k!} \\ &(q\le s+1;q,s\in {{\mathbb{N}}_{0}}=\mathbb{N}\cup \{0\};z\in \mathbb{U}), \\ \end{align}$

where $(\lambda)_k$ denotes the Pochhammer symbol defined, in terms of Gamma function, by

$(\lambda)_k=\frac{\Gamma(\lambda+k)}{\Gamma(\lambda)}=\left\{ \begin{array}{ll} 1 &(k=0; \lambda\in\mathbb{C}\setminus\{0\}), \\\\ \lambda(\lambda+1)\cdots(\lambda+k-1) &(k\in\mathbb{N}; \lambda\in\mathbb{C}).\\ \end{array} \right.$

Dziok and Srivastava in [7] (see also [8, 9]) considered a linear operator

$\mathcal{H}(a_1, \cdots, a_q;b_1, \cdots, b_s): \mathcal{A}_p\longrightarrow\mathcal{A}_p, $

defined by the Hadamard product

$\begin{array}{*{35}{l}} \mathcal{H}({{a}_{1}},\cdots ,{{a}_{q}};{{b}_{1}},\cdots ,{{b}_{s}})f(z)&=[{{z}^{p}}{{\cdot }_{q}}{{F}_{s}}({{a}_{1}},\cdots ,{{a}_{q}};{{b}_{1}},\cdots ,{{b}_{s}};z)]*f(z) \\ {}&={{z}^{p}}+\sum\limits_{k=1}^{\infty }{\frac{{{({{a}_{1}})}_{k}}\cdots {{({{a}_{q}})}_{k}}}{{{({{b}_{1}})}_{k}}\cdots {{({{b}_{s}})}_{k}}}}\frac{{{a}_{k+p}}}{k!}{{z}^{k+p}}, \\ \end{array}$

(1.5)

where $f\in\mathcal{A}_p$ is given by (1.1).

It follows from (1.5) that for all $j\in\{1, 2, \cdots, s\}$,

$b_j\mathcal{H}(b_j)f(z)=z\left[\mathcal{H}(b_j+1)f(z)\right]'+(b_j-p)\mathcal{H}(b_j+1)f(z),$

(1.6)

where, for convenice

$\mathcal{H}(b_j)f(z)=\mathcal{H}(a_1, \cdots, a_q;b_1, \cdots, b_j, \cdots, b_s)f(z)$

and

$\mathcal{H}(b_j+1)f(z)=\mathcal{H}(a_1, \cdots, a_q;b_1, \cdots, b_j+1, \cdots, b_s)f(z).$

The Dziok-Srivastava operator $\mathcal{H}(a_1, \cdots, a_q;b_1, \cdots, b_s)$ includes various linear operators, which were considered in earlier works, such as (for example) the linear operators introduced by Hohlov [10], Carlson and Shaffer [2], Ruschewyh [13] and Srivastava and Owa [18].

In particular, we mention here the Bernardi integral operator $\mathcal{J}_\nu: \mathcal{A}\longrightarrow\mathcal{A}$, defined by (see [1])

$\mathcal{J}_\nu[f(z)]=\frac{\nu+1}{z^\nu}\int_0^zt^{\nu-1}f(t)dt (\nu\in\mathbb{C}).$

(1.7)

Note that for $f(z)=z+a_2z^2+\cdots$, we have

$\mathcal{J}_\nu[f(z)]=\sum\limits_{k=1}^{\infty}\frac{\nu+1}{\nu+k}a_kz^k.$

(1.8)

Therefore the Bernardi operator and the Dziok-Srivastava operator are connected in the following way

$\mathcal{J}_\nu[f(z)]=\mathcal{H}(1+\nu, 1;\nu+2)f(z).$

Definition 1.1 Let us suppose

$-1\leq B\leq0, A\in\mathbb{C} \;\; {\rm {and}} \;\; |A|<1.$

(1.9)

We denote by $W_p(\mathcal{H}(b_j+1);A, B)$ the class of functions $f\in\mathcal{A}_p$ of form (1.1) which satisfy the following condition

$b_j\frac{\mathcal{H}(b_j)f(z)}{\mathcal{H}(b_j+1)f(z)}+p-b_j\prec\frac{1+Az}{1+Bz} (z\in\mathbb{U}).$

(1.10)

By using (1.6), condition (1.10) becomes

$\frac{z[z^{1-p}\mathcal{H}(b_j+1)f(z)]'}{z^{1-p}\mathcal{H}(b_j+1)f(z)} =\frac{z[\mathcal{H}(b_j+1)f(z)]'}{\mathcal{H}(b_j+1)f(z)}-p+1\prec\frac{1+Az}{1+Bz} (z\in\mathbb{U}).$

(1.11)

From (1.9), we see that

${\rm Re}\left(\frac{1+Az}{1+Bz}\right)>0 (z\in\mathbb{U}).$

(1.12)

Thus we have

$f\in W_p(\mathcal{H}(b_j+1);A, B)\Longrightarrow z^{1-p}\mathcal{H}(b_j+1)f(z)\in\mathcal{S}^*.$

Moreover, for $-1\leq B<A\leq1$, this means that $z^{1-p}\mathcal{H}(b_j+1)f(z)$ belongs to the class $\mathcal{S}^*(A, B)$ defined by (1.4).

Many interesting subclasses of analytic functions associated with the Dziok-Srivastava operator $\mathcal{H}(a_1, \cdots, a_q;b_1, \cdots, b_s)$ were investigated recently (for example) by Dziok [4, 5], Dziok and Sokol [6], Sokol [16, 17] and others. They obtained various properties and characterizations for these subclasses with respect to the parameters $a_i\in\mathbb{C} (i=1, 2, \cdots, q)$. However, in this paper, we aim to investigate some characterizations and inclusion relationships for the class $W_p(\mathcal{H}(b_j+1);A, B)$, which are in connection with the parameters $b_j\in\mathbb{C}\setminus\mathbb{Z}_0^- (\mathbb{Z}_0^-=0, -1, -2, \cdots; j=1, 2, \cdots, s)$.

First, we begin by proving the following two characterization theorems.

Theorem 2.1 If $f\in\mathcal{A}_p$ and $j\in\{1, 2, \cdots, s\}$, then

$z[z^{1-p}\mathcal{H}(b_j+1)f(z)]''=b_j[z^{1-p}\mathcal{H}(b_j)f(z)]'-b_j[z^{1-p}\mathcal{H}(b_j+1)f(z)]'.$

(2.1)

Proof From (1.6), we easily get

$z[\mathcal{H}(b_j+1)f(z)]'+(1-p)\mathcal{H}(b_j+1)f(z)=b_j\mathcal{H}(b_j)f(z)+(1-b_j)\mathcal{H}(b_j+1)f(z).$

(2.2)

Multiplying both sides of (2.2) by $z^{1-p}$, equality (2.2) becomes

$z[z^{1-p}\mathcal{H}(b_j+1)f(z)]' =b_j[z^{1-p}\mathcal{H}(b_j)f(z)]+(1-b_j)[z^{1-p}\mathcal{H}(b_j+1)f(z)].$

(2.3)

Then differentiating (2.3), we immediately obtain (2.1).

Theorem 2.2 If $f\in\mathcal{A}_p$ and $z^{1-p}\mathcal{H}(b_j+1)f(z)$ is convex univalent function, then $z^{1-p}\mathcal{H}(b_j)f(z)$ is close-to-convex of order ${\rm Re}\left(\frac{b_j-1}{|b_j|}\right)$ with respect to $z^{1-p}\mathcal{H}(b_j+1)f(z)$, where $j\in\{1, 2, \cdots, s\}$.

Proof From (2.1), we conclude that

$\frac{[z^{1-p}\mathcal{H}(b_j)f(z)]'}{[z^{1-p}\mathcal{H}(b_j+1)f(z)]'}=\frac{z[z^{1-p}\mathcal{H}(b_j+1)f(z)]''} {b_j[z^{1-p}\mathcal{H}(b_j+1)f(z)]'}+1.$

(2.4)

Hence, from (1.2) and (2.4), we have

$\begin{array}{*{35}{l}} \text{Re}\left\{ \frac{{{b}_{j}}}{|{{b}_{j}}|}\cdot \frac{{{[{{z}^{1-p}}\mathcal{H}({{b}_{j}})f(z)]}^{\prime }}}{{{[{{z}^{1-p}}\mathcal{H}({{b}_{j}}+1)f(z)]}^{\prime }}} \right\}&=\text{Re}\left\{ \frac{z}{|{{b}_{j}}|}\cdot \frac{{{[{{z}^{1-p}}\mathcal{H}({{b}_{j}}+1)f(z)]}^{\prime \prime }}}{{{[{{z}^{1-p}}\mathcal{H}({{b}_{j}}+1)f(z)]}^{\prime }}}+\frac{{{b}_{j}}}{|{{b}_{j}}|} \right\} \\ {}&>\text{Re}\left( \frac{{{b}_{j}}-1}{|{{b}_{j}}|} \right) \\ \end{array}$

and using (1.3), we obtain the asserted result.

In order to obtain inclusion properties, we first recall the following lemma.

Lemma 2.1 (see [12]) Let $\nu, A\in\mathbb{C}$ and $B\in[-1, 0]$ satisfy either

${\rm Re}[1+AB+\nu(1+B^2)]\geq|A+B+B(\nu+\overline{\nu})| \;\; {\rm for} \;\; B\in(-1, 0], $

(2.5)

or

$1+A>0, {\rm Re}[1-A+2\nu]\geq0 \;\; {\rm for} \;\; B=-1.$

(2.6)

If $f\in\mathcal{A}$ and $F(z)=\mathcal{J}_\nu[f(z)]$ is given by (1.7), then $F\in\mathcal{A}$ and

$\frac{zf'(z)}{f(z)}\prec\frac{1+Az}{1+Bz}\Longrightarrow\frac{zF'(z)}{F(z)}\prec\frac{1+Az}{1+Bz}.$

Theorem 2.3 If $f\in\mathcal{A}_p$ and $j\in\{1, 2, \cdots, s\}$, then

$z^{1-p}\mathcal{H}(b_j+1)f(z)=\mathcal{J}_{b_j-1}[z^{1-p}\mathcal{H}(b_j)f(z)], $

(2.7)

where $\mathcal{J}_{b_j-1}$ is the Bernardi operator (1.7) with $\nu=b_j-1$.

Proof From (1.5), we have

$\begin{array}{*{35}{l}} \mathcal{H}({{b}_{j}}+1)f(z)&={{z}^{p}}+\sum\limits_{k=1}^{\infty }{\frac{{{({{a}_{1}})}_{k}}\cdots {{({{a}_{q}})}_{k}}}{{{({{b}_{1}})}_{k}}\cdots {{({{b}_{j}}+1)}_{k}}\cdots {{({{b}_{s}})}_{k}}}}\frac{{{a}_{k+p}}}{k!}{{z}^{k+p}} \\ {}&={{z}^{p}}+\sum\limits_{k=1}^{\infty }{\frac{{{({{a}_{1}})}_{k}}\cdots {{({{a}_{q}})}_{k}}}{{{({{b}_{1}})}_{k}}\cdots {{({{b}_{j}})}_{k}}\left( \frac{{{b}_{j}}+k}{{{b}_{j}}} \right)\cdots {{({{b}_{s}})}_{k}}}}\frac{{{a}_{k+p}}}{k!}{{z}^{k+p}} \\ {}&=\mathcal{H}({{b}_{j}})f(z)*\left[ {{z}^{p}}+\sum\limits_{k=1}^{\infty }{\frac{{{b}_{j}}}{{{b}_{j}}+k}}{{z}^{k+p}} \right] \\ {}&={{z}^{p-1}}\left\{ {{z}^{1-p}}[\mathcal{H}({{b}_{j}})f(z)]*\left[ z+\sum\limits_{k=1}^{\infty }{\frac{{{b}_{j}}}{{{b}_{j}}+k}}{{z}^{k+1}} \right] \right\} \\ {}&={{z}^{p-1}}\left\{ {{z}^{1-p}}[\mathcal{H}({{b}_{j}})f(z)]*\left[ \sum\limits_{k=1}^{\infty }{\frac{({{b}_{j}}-1)+1}{({{b}_{j}}-1)+k}}{{z}^{k}} \right] \right\}. \\ \end{array}$

Hence, by (1.8) with $\nu=b_j-1$, we obtain

$\mathcal{H}(b_j+1)f(z)=z^{p-1}\mathcal{J}_{b_j-1}[z^{1-p}\mathcal{H}(b_j)f(z)], $

which implies that (2.7) holds.

Theorem 2.4 Let $m\in\mathbb{N}$ and $j\in\{1, 2, \cdots, s\}$. If $A\in\mathbb{C}$ and $B\in [-1,0]$ satisfy (2.5) or (2.6) with $\nu=b_j-1$, then

$W_p(\mathcal{H}(b_j);A, B)\subseteq W_p(\mathcal{H}(b_j+m);A, B).$

(2.8)

Proof Clearly, it is sufficient to prove (2.8) only for $m=1$. Let $f\in W_p(\mathcal{H}(b_j);A, B)$, then from (1.11) we have

$\frac{z[z^{1-p}\mathcal{H}(b_j)f(z)]'}{z^{1-p}\mathcal{H}(b_j)f(z)}\prec\frac{1+Az}{1+Bz} (z\in\mathbb{U}).$

(2.9)

By applying Lemma 2.1 and Theorem 2.3 to (2.9), we get

$\frac{z[z^{1-p}\mathcal{H}(b_j+1)f(z)]'}{z^{1-p}\mathcal{H}(b_j+1)f(z)}\prec\frac{1+Az}{1+Bz} (z\in\mathbb{U}), $

which means that $f\in W_p(\mathcal{H}(b_j+1);A, B)$.

It is natural to ask about the inclusion relation (2.8) when $m$ is not positive integer. Next, we will give a partial answer to the question by using a different method. We need the following lemma.

Lemma 2.2 (see [15]) Let $f\in\mathcal{K}$ and $g\in \mathcal{S}^*$. Then, for every analytic function $h$ in $\mathbb{U}$,

$\frac{(f*hg)(\mathbb{U})}{(f*g)(\mathbb{U})}\subset\overline{\rm co}[h(\mathbb{U})], $

where $\overline{co}[h(\mathbb{U})]$ denotes the closed convex hull of $h(\mathbb{U})$.

Theorem 2.5 If $f\in W_p(\mathcal{H}(b_j);A, B)$, $H(z)=z^{1-p}\mathcal{H}(b_j)f(z)\in\mathcal{S}^*$ and $G(z)=\sum\limits_{k=0}^{\infty}\frac{(b_j)_k}{(\widetilde{b_j})_k}z^{k+1}\in\mathcal{K}$, then $f\in W_p(\mathcal{H}(\widetilde{b_j});A, B)$ and $z^{1-p}\mathcal{H}(\widetilde{b_j})f(z)\in\mathcal{S}^*$.

Proof Let $f\in W_p(\mathcal{H}(b_j);A, B)$. Then by the definition of the class $W_p(\mathcal{H}(b_j);A, B)$, we have

$\frac{z[z^{1-p}\mathcal{H}(b_j)f(z)]'}{z^{1-p}\mathcal{H}(b_j)f(z)}=\frac{1+A\omega(z)}{1+B\omega(z)} =\phi[\omega(z)] (z\in\mathbb{U}),$

(2.10)

where $\phi$ is convex univalent mapping of $\mathbb{U}$ and $|\omega(z)|<1$ in $\mathbb{U}$ with $\omega(0)=0=\phi(0)-1$. Also, we have ${\rm Re}[\phi(z)]>0$ because of $H(z)\in\mathcal{S}^*$. Using (2.10) and the properties of convolution, we get

$\begin{array}{*{35}{l}} \frac{z{{[{{z}^{1-p}}\mathcal{H}(\widetilde{{{b}_{j}}})f(z)]}^{\prime }}}{{{z}^{1-p}}\mathcal{H}(\widetilde{{{b}_{j}}})f(z)}&=\frac{z{{\left[ \sum\limits_{k=0}^{\infty }{\frac{{{({{b}_{j}})}_{k}}}{{{(\widetilde{{{b}_{j}}})}_{k}}}}{{z}^{k+1}}*{{z}^{1-p}}\mathcal{H}({{b}_{j}})f(z) \right]}^{\prime }}}{\sum\limits_{k=0}^{\infty }{\frac{{{({{b}_{j}})}_{k}}}{{{(\widetilde{{{b}_{j}}})}_{k}}}}{{z}^{k+1}}*{{z}^{1-p}}\mathcal{H}({{b}_{j}})f(z)} \\ {}&=\frac{G(z)*z{H}'(z)}{G(z)*H(z)}=\frac{G(z)*\phi [\omega (z)]H(z)}{G(z)*H(z)}. \\ \end{array}$

(2.11)

Since $H(z)\in\mathcal{S}^*$, $G(z)\in\mathcal{K}$ and $\phi$ is convex univalent, then by applying Lemma 2.2 to (2.11), we conclude that (2.11) is subordinate to $\phi$ in $\mathbb{U}$. Thus, by (1.11), we obtain that $z^{1-p}\mathcal{H}(\widetilde{b_j})f(z)\in\mathcal{S}^*(A, B)\subseteq\mathcal{S}^*$ and so $f\in W_p(\mathcal{H}(\widetilde{b_j});A, B)$.

Lemma 3.1(see [14]) If either $0<a\leq c$ and $c\geq2$ when $a, c$ are real, or ${\rm Re}[a+c]\geq3$, ${\rm Re}[a]\leq {\rm Re}[c]$ and ${\rm Im}[a]={\rm Im}[c]$ when $a, c$ are complex, then the function

$f(z)=\sum\limits_{k=0}^{\infty}\frac{(a)_k}{(c)_k}z^{k+1} (z\in \mathbb{U})$

belongs to the class $\mathcal{K}$ of convex functions.

Corollary 3.1 If $b_j, \widetilde{b_j}$ are real such that $0<b_j\leq\widetilde{b_j}$ and $\widetilde{b_j}\geq2$ or $b_j, \widetilde{b_j}$ are complex $(b_j, \widetilde{b_j}\neq0, -1, -2, \cdots)$ such that ${\rm Re}[b_j+\widetilde{b_j}]\geq3$, ${\rm Re}[b_j]\leq {\rm Re}[\widetilde{b_j}]$ and ${\rm Im}[b_j]={\rm Im}[\widetilde{b_j}]$, then $W_p(\mathcal{H}(b_j);A, B)\subseteq W_p(\mathcal{H}(\widetilde{b_j});A, B)$.

Proof Since $A, B$ satisfy (1.12), so if $f\in W_p(\mathcal{H}(b_j);A, B)$, then $H(z)=z^{1-p}\mathcal{H}(b_j)f(z)\in\mathcal{S}^*$. By Lemma 3.1, the function

$G(z)=\sum\limits_{k=0}^{\infty}\frac{(b_j)_k}{(\widetilde{b_j})_k}z^{k+1} (z\in\mathbb{U})$

belongs to the class $\mathcal{K}$ of convex functions. Therefore, in view of Theorem 2.5, we obtain that $f\in W_p(\mathcal{H}(\widetilde{b_j});A, B)$.

Lemma 3.2 (see [12]) If $a, b, c$ are real and satisfy $-2\leq a<0$, $b\neq0$, $b\geq-1$ and $c>M(a, b)$, where

$M(a, b)=\max\{2+|a+b|, 1-ab\}, $

then the Gaussian hypergeometric function

${_2 F_1}(a, b, c;z)=\sum\limits_{k=0}^{\infty}\frac{(a)_k(b)_k}{(c)_k k!}z^{k}$

is convex in $\mathbb{U}$.

Corollary 3.2 Let $b_j\in(-1, 0)\cup(0, 1)$ and $j\in\{1, 2, \cdots, s\}$. If $\widetilde{b_j}>3+|b_j|$, then

$\sum\limits_{k=0}^{\infty}\frac{(b_j)_k}{(\widetilde{b_j})_k}z^{k+1}\in\mathcal{K}.$

Proof If we choose $b=1, a=b_j-1, c=\widetilde{b_j}-1$ in Lemma 3.2, then we obtain that

$F(z)=\sum\limits_{k=0}^{\infty}\frac{(b_j-1)_k}{(\widetilde{b_j}-1)_k}z^{k}$

is convex in $\mathbb{U}$ for $b_j\neq0, -1, -2, \cdots$; $-2\leq b_j-1<0$ and $\widetilde{b_j}-1>M(a, b)=2+|b_j|$. It is clear that $G(z)=\frac{\widetilde{b_j}-1}{b_j-1}[F(z)-1]\in\mathcal{K}$. After some calculations we have that

$G(z)=\sum\limits_{k=0}^{\infty}\frac{(b_j)_k}{(\widetilde{b_j})_k}z^{k+1}$

and this completes the proof.

Corollary 3.3 Let $b_j\in(-1, 0)\cup(0, 1)$ and $j\in\{1, 2, \cdots, s\}$. If $\widetilde{b_j}>3+|b_j|$, then

$W_p(\mathcal{H}(b_j);A, B)\subseteq W_p(\mathcal{H}(\widetilde{b_j});A, B).$

Proof The proof follows as the proof of Corollary 3.1 by using Corollary 3.2.

Corollary 3.4 Let $m\in\mathbb{N}$ and $j\in\{1, 2, \cdots, s\}$. If ${\rm Re}(b_j)>1$, then

$W_p(\mathcal{H}(b_j);A, B)\subseteq W_p(\mathcal{H}(b_j+m);A, B).$

Proof Obviously, it is sufficient to prove this corollary only for $m=1$. If $f\in W_p(\mathcal{H}(b_j);A, B)$, then $H(z)=z^{1-p}\mathcal{H}(b_j)f(z)\in\mathcal{S}^*(A, B)\subseteq\mathcal{S}^*$. Let us denote

$\frac{zH'(z)}{H(z)}=\frac{1+A\omega(z)}{1+B\omega(z)} =\phi[\omega(z)] (z\in\mathbb{U}), $

where $\phi$ is convex univalent and $|\omega(z)|<1$ in $\mathbb{U}$ with $\omega(0)=0=\phi(0)-1$ and ${\rm Re}[\phi(z)]>0$. If ${\rm Re}(b_j)>1$, then

$G(z)=\sum\limits_{k=1}^{\infty}\frac{(b_j-1)+1}{(b_j-1)+k}z^{k} (z\in\mathbb{U})$

belongs to the class $\mathcal{K}$ of convex functions (see [14]). Therefore, by (2.7), we have

$\begin{array}{*{35}{l}} \frac{z{{[{{z}^{1-p}}\mathcal{H}({{b}_{j}}+1)f(z)]}^{\prime }}}{{{z}^{1-p}}\mathcal{H}({{b}_{j}}+1)f(z)}&=\frac{{{[G(z)*zH(z)]}^{\prime }}}{G(z)*H(z)}=\frac{G(z)*z{H}'(z)}{G(z)*H(z)} \\ {}&=\frac{G(z)*\phi [\omega (z)]H(z)}{G(z)*H(z)}\in \overline{\text{co}}\phi (\mathbb{U}). \\ \end{array}$

Analogous to the proof of Theorem 2.5, we obtain that $f\in W_p(\mathcal{H}(b_j+1);A, B)$.