Properties of neighborhood for certain classes associated with complex order and m-q-p-valent functions with higher order

Abstract

In this paper, by using q-calculus (Jackson’s q-derivative) $D_{q,p}$ we defined new operator $D_{\lambda ,q,p}^{n}f^{(m)}(z)$. After that, we used this operator to introduce two new subclasses of multivalent analytic functions with complex order. Also, we obtained coefficients estimates and consequent inclusion relationships involving the $N_{j,\delta ,m}^{p,q}(f)$-neighborhood of these classes

Introduction

Let ${\mathcal {A}}_{j}(p)$ denote the class of functions in the form:

$$\begin{aligned} f(z)=z^{p}+\sum _{k=j+p}^{\infty }a_{k}z^{k}\; \;(j,p\in {{\mathbb {N}}} =\{1,2,\ldots \}), \end{aligned}$$

(1.1)

which are analytic and p-valent in the open is open unit disk${\mathbb {U}}= \{z:\left| z\right| <1\}.$We note that ${\mathcal {A}}_{1}(p)= {\mathcal {A}}(p)$(see [13, 30]) and ${\mathcal {A}}_{1}(1)={\mathcal {A}}.$ Also let T(p, j) denote the subclass of ${\mathcal {A}}_{j}(p)$which can express in the form:

$$\begin{aligned} f(z)=z^{p}-\sum _{k=j+p}^{\infty }a_{k}z^{k}\;\;(a_{k}\ge 0,\ j,p\in {\mathbb {N}} ). \end{aligned}$$

(1.2)

In recent years, the topic of q-calculus had attracted the attention of several researchers (see, for example, [2, 15, 16, 23, 34, 43,44,45]). Quantum calculus is the modern name for the investigation of calculus without limits. The quantum calculus or q-calculus began with Jackson in the early twentieth century, but this kind of calculus had already been worked out by Euler and Jacobi. In the general run, the q-calculus is used in various fields of Mathematics and Physics. Also, q-calculus appeared the connection between Mathematics and Physics. It had a lot of applications in different mathematical areas such as number theory, combinatorics, orthogonal polynomials, basic hyper-geometric functions and other sciences quantum theory, mechanics and the theory of relativity. Several convolutional and fractional calculus q-operators were defined by many researchers. The generalization of derivative and integral in q-calculus is known as q-analogue derivative and q-analogue integral. Recently, many authors used the q- analogue derivative and q-analogue integral to generalize many classes and many operators in Geometric Function Theory (see, for example, [14, 33, 40, 42]).

For a function $f(z)\in A(p)$ given by (1.1) (with $j=1)$ Jackson’s q-derivative (or q-difference) $D_{q,p}$ $(0<q<1)$is defined as follows:

$$\begin{aligned} D_{q,p}f(z)=\left\{ \begin{array}{ccc} \frac{f(z)-f(qz)}{(1-q)z} &{} &{} for\ z\ne 0, \\ &{} &{} \\ f^{\prime }(0)\ \ &{} &{} for\ z=0, \end{array} \underset{\ \ \ \ \ }{\overset{\ \ \ \ }{\underset{}{(f\in \mathcal {A(}p))}}} \right. \end{aligned}$$

(1.3)

provided that $f^{\prime }(0)$exists. From (1.1) (with $j=1$) and (1.3), we deduce that

$$\begin{aligned} D_{q,p}f(z)=\left[ p\right] _{q}z^{p-1}+\sum _{k=p+1}^{\infty }\left[ k\right] _{q}a_{k}z^{k-1}, \end{aligned}$$

(1.4)

such that q-integer number k $\left[ k\right] _{q}$ is defined by

$$\begin{aligned} \left[ k\right] _{q}=\frac{1-q^{k}}{1-q}=1+\sum \limits _{k=1}^{n-1}q^{k},\ 0<q<1,\ \left[ 0\right] _{q}=0. \end{aligned}$$

(1.5)

We observe that

$$\begin{aligned} \lim _{q\rightarrow 1^{-}}D_{q,p}f(z)=\lim _{q\rightarrow 1^{-}}\frac{ f(z)-f(qz)}{(1-q)z}=f^{\prime }(z), \end{aligned}$$

for a function fwhich is differentiable in a given subset of ${\mathbb {C}}$. For all $f(z)\in T(p,j),$we find ( see [25] )

$$\begin{aligned} f^{(m)}(z)&=\theta (p,m)z^{p-m}\nonumber \\&\quad -\sum \limits _{k=j+p}^{\infty }\theta (k,m)a_{k}z^{k-m}\ (p,j\in {\mathbb {N}},\ m\in {\mathbb {N}}_{0}={\mathbb {N}} \cup \{0\},\ p>m), \end{aligned}$$

(1.6)

where $\theta (p,m)$is defined by

$$\begin{aligned} \theta (p,m)=\frac{p!}{(p-m)!}=\left\{ \begin{array}{ll} 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ , &{} m=0, \\ p(p-1)\ldots (p-m+1), &{} m\ne 0. \end{array} \right. \ \end{aligned}$$

(1.7)

For $f\in T(p,j),$ we introduce the operator $D_{\lambda ,q,p}^{n}f^{(m)}:T(p,j)\rightarrow T(p,j)\ (\lambda \ge 0,\ n,m\in {\mathbb {N}} _{0},0<q<1,\ j,p\in {\mathbb {N}} ,\ p>m)$as follows:

$$\begin{aligned} D_{\lambda ,q,p}^{n}f^{(m)}(z)= & {} f^{(m)}(z),\\ D_{\lambda ,q,p}^{1}f^{(m)}(z)= & {} D_{\lambda ,q,p}\left( D_{\lambda ,q,p}^{0}f^{(m)}(z)\right) =(1-\lambda )f^{(m)}(z)+\frac{\lambda }{\left[ p-m \right] _{q}}zD_{q,p}(f^{(m)}(z)) \\= & {} \theta (p,m)z^{p\mathbf {-}m}-\sum \limits _{k=j+p}^{\infty }\theta (k,m) \left[ \tfrac{\left[ p-m\right] _{q}+\lambda \left( \left[ k-m\right] _{q}- \left[ p-m\right] _{q}\right) }{\left[ p-m\right] _{q}}\right] a_{k}z^{k-m},\\&D_{\lambda ,q,p}^{2}f^{(m)}(z) =D_{\lambda ,q,p}\left( D_{\lambda ,q,p}^{1}f^{(m)}(z)\right) =(1-\lambda )D_{\lambda ,q,p}^{1}f^{(m)}(z)+\frac{ \lambda }{\left[ p-m\right] _{q}}zD_{q,p}(D_{\lambda ,q,p}^{1}f^{(m)}(z)) \\= & {} \theta (p,m)z^{p\mathbf {-}m}-\sum \limits _{k=j+p}^{\infty }\theta (k,m) \left[ \tfrac{\left[ p-m\right] _{q}+\lambda \left( \left[ k-m\right] _{q}- \left[ p-m\right] _{q}\right) }{\left[ p-m\right] _{q}}\right] ^{2}a_{k}z^{k-m} \vdots \end{aligned}$$

$$\begin{aligned} D_{\lambda ,q,p}^{n}f^{(m)}(z)&=D_{\lambda ,q,p}\left( D_{\lambda ,q,p}^{n-1}f^{(m)}(z)\right) \nonumber \\&=(1-\lambda )D_{\lambda ,q,p}^{n-1}f^{(m)}(z)+ \frac{\lambda }{\left[ p-m\right] _{q}}zD_{q,p}(D_{\lambda ,q,p}^{n-1}f^{(m)}(z))(n\in {\mathbb {N}} ) \end{aligned}$$

(1.8)

From (1.2) and (1.8), we can obtain

$$\begin{aligned} =\theta (p,m)z^{p\mathbf {-}m}-\sum \limits _{k=j+p}^{\infty }\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}z^{k-m}, \end{aligned}$$

where

$$\begin{aligned} \Psi _{q,p}^{n,m}(k,\lambda )&=\left[ \frac{\left[ p-m\right] _{q}+\lambda \left( \left[ k-m\right] _{q}-\left[ p-m\right] _{q}\right) }{\left[ p-m \right] _{q}}\right] ^{n} \nonumber \\&\quad (\lambda \ge 0,\ m,n\in {\mathbb {N}} _{0},0<q<1,\ j,p\in {\mathbb {N}} ,p>m). \end{aligned}$$

(1.9)

We note that

• (1) $D_{\lambda ,q,p}^{n}f^{(0)}(z)=I_{q,p}^{n}(\lambda )f(z),$(Aouf and Madian [14], with $\varrho =0$]);

• (2) $\lim _{q\rightarrow 1^{-}}D_{1,q,p}^{n}f^{(m)}(z)=D_{p}^{n}f^{(m)}(z),$(Aouf [8, 9]);

• (3) $\lim _{q\rightarrow 1^{-}}D_{1,q,p}^{n}f^{(0)}(z)=D_{p}^{n}f(z)$(see [11, 18], Cătaş [24], with $l=0$] and [37]);

• (4) $\lim _{q\rightarrow 1^{-}}D_{1,q,1}^{n}f^{(0)}(z)=D^{n}f(z)$(see ([26, 27], with $l=0$));

• (5) $\lim _{q\rightarrow 1^{-}}D_{\lambda ,q,1}^{n}f^{(0)}(z)=D_{\lambda }^{n}f(z)$(see [1, 17, 21]);

• (6) $D_{1,q,1}^{n}f^{(0)}(z)=D_{q}^{n}f(z)$(see [32] ), $\lim _{q\rightarrow 1^{-}}D_{q}^{n}f(z)=D^{n}f(z)$(see Sălă gean [39] see also [10, 12]);

• (7) $\lim _{q\rightarrow 1^{-}}D_{1,q,p}^{n}f^{(m)}(z)=D_{p}^{n}f^{(m)}(z)$(see Aouf et al. [22]);

• (8) $\lim _{q\rightarrow 1^{-}}D_{\lambda ,q,p}^{n}f^{(m)}(z)=I_{\lambda ,p}^{n}f^{(m)}(z)$

  $$\begin{aligned} =\left\{ \begin{array}{c} f\in T(p,j):I_{\lambda ,p}^{n}f^{(m)}(z)=\theta (p,m)z^{p-m}-\sum \limits _{k=j+p}^{\infty }\left( \left[ \frac{p-m+\lambda \left( k-p\right) }{ p-m}\right] \right) ^{n}\theta (k,m)a_{k}z^{k-m}, \\ n,m\in {\mathbb {N}} _{0},\ j,p\in {\mathbb {N}} ,\lambda \ge 0,\ p>m \end{array} \right\} ; \end{aligned}$$

• (9) $D_{1,q,p}^{n}f^{(m)}(z)=I_{q,p}^{n}f^{(m)}(z)$

  $$\begin{aligned} =\left\{ \begin{array}{c} f\in T(p,j):I_{q,p}^{n}f^{(m)}(z)=\theta (p,m)z^{p-m}-\sum \limits _{k=j+p}^{\infty }\left( \frac{\left[ k-m\right] _{q}}{\left[ p-m \right] _{q}}\right) ^{n}\theta (k,m)a_{k}z^{k-m}, \\ n,m\in {\mathbb {N}} _{0},0<q<1,\ j,p\in {\mathbb {N}} ,p>m \end{array} \right\} ; \end{aligned}$$

• (10) $D_{1,q,p}^{n}f^{(0)}(z)=D_{q,p}^{n}f(z)$

  $$\begin{aligned} =\left\{ \begin{array}{c} f\in T(p,j):D_{q,p}^{n}f(z)=z^{p}-\sum \limits _{k=j+p}^{\infty }\left( \frac{ \left[ k\right] _{q}}{\left[ p\right] _{q}}\right) ^{n}a_{k}z^{k}, \\ n\in {\mathbb {N}} _{0},\ j,p\in {\mathbb {N}} ,\ 0<q<1 \end{array} \right\} . \end{aligned}$$

  Now by using $D_{\lambda ,q,p}^{n}f^{(m)}(z),$we defined the classes $F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$and $G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$in below definitions:

Definition 1

Assume$f(z)\in T(p,j),$then $f(z)\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$if it satisfies the following inequality:

$$\begin{aligned} \left| \frac{1}{b}\left[ \dfrac{(1-\sigma )zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{(1-\sigma )D_{\lambda ,q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right] \right| <\beta \end{aligned}$$

$$\begin{aligned}&(b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m). \end{aligned}$$

(1.10)

We observe that:

• (1) $\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=S_{j}(n,p,m,\sigma ,b,\beta )$ see Aouf et al. [22];

• (2) $F_{q,p}^{n,0}(j,\lambda ,\sigma ,b,\beta )=S_{q}^{n}(j,\lambda ,p,\sigma ,b,\beta )$see Aouf and madian [14], with $\varrho =0$];

• (3) $\lim _{q\rightarrow 1^{-}}F_{q,p}^{0,0}(j,\lambda ,\sigma ,b,\beta )=S_{j}(p,\sigma ,b,\beta )$ see Aouf and Mostafa [19], with $b_{k}=1$;

• (4) $\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )=F_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }+\sigma z^{2}(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime \prime }}{(1-\sigma )I_{\lambda ,p}^{n}f^{(m)}(z)+\sigma z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}-\left( p-m\right) \right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (5) $F_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=F_{q,p}^{n,m}(j,\sigma ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{(1-\sigma )zD_{q,p}(I_{q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(I_{q,p}^{n}f^{(m)}(z)))}{(1-\sigma )I_{q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(I_{q,p}^{n}f^{(m)}(z))}-\left[ p-m \right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (6) $\ F_{q,p}^{n,0}(j,1,\sigma ,b,\beta )=F_{q,p}^{n}(j,\sigma ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{(1-\sigma )zD_{q,p}(D_{q,p}^{n}f(z))+\sigma zD_{q,p}(zD_{q,p}(D_{q,p}^{n}f(z)))}{ (1-\sigma )D_{q,p}^{n}f(z)+\sigma zD_{q,p}(D_{q,p}^{n}f(z))}-\left[ p\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}$$

• (7) $\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=SF_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}{I_{\lambda ,p}^{n}f^{(m)}(z)}-\left( p-m\right) \right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (8) $F_{q,p}^{n,m}(j,1,0,b,\beta )=SF_{q,p}^{n,m}(j,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{ zD_{q,p}(I_{q,p}^{n}f^{(m)}(z))}{I_{q,p}^{n}f^{(m)}(z)}-\left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (9) $\ F_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=S_{q,p}^{n,m}(j,\lambda ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \dfrac{ zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}{D_{\lambda ,q,p}^{n}f^{(m)}(z)}- \left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},m,n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}},\ 0<q<1,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (10) $\lim _{q\rightarrow 1^{-}}F_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=KF_{p}^{n,m}(j,\lambda ,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ 1+\tfrac{z(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime \prime }}{(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime } }-\left( p-m\right) \right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (11) $F_{q,p}^{n,m}(j,1,1,b,\beta )=KF_{q,p}^{n,m}(j,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \tfrac{ D_{q,p}(zD_{q,p}(I_{q,p}^{n}f^{(m)}(z)))}{D_{q,p}(I_{q,p}^{n}f^{(m)}(z))}- \left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (12) $\ F_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=K_{q,p}^{n,m}(j,\lambda ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \dfrac{ D_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{D_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} . \end{aligned}$$

Definition 2

Assume $f(z)\in T(p,j),\;$if it satisfies (1.11), then$f(z)\in G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$

$$\begin{aligned} \left| \frac{1}{b}\left\{ (1-\sigma )\frac{D_{\lambda ,q,p}^{n}f^{(m)}(z)}{z^{p-m}}+\sigma \frac{D_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}{[p-m]_{q}z^{p-m-1}}-\theta (p,m)\right\} \right| <\beta \end{aligned}$$

$$\begin{aligned} (b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m). \end{aligned}$$

(1.11)

We note that:

• (1) $\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=K_{j}(n,p,m,\sigma ,b,\beta )\;$ see Aouf et al. [22];

• (2) $G_{q,p}^{n,0}(j,\lambda ,\sigma ,b,\beta )=K_{q}^{n}(j,\lambda ,p,\sigma ,b,\beta )\;$see Aouf and Madian [14], with $\varrho =0$];

• (3) $\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )=G_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ (1-\sigma )\frac{ I_{\lambda ,p}^{n}f^{(m)}(z)}{z^{p-m}}+\sigma \frac{(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}{(p-m)z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},m,n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (4) $\ G_{q,p}^{n,m}(j,1,\sigma ,b,\beta )=G_{q,p}^{n,m}(j,\sigma ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left\{ (1-\sigma )\frac{ I_{q,p}^{n}f^{(m)}(z)}{z^{p-m}}+\sigma \frac{D_{q,p}(I_{q,p}^{n}f^{(m)}(z))}{ [p-m]_{q}z^{p-m-1}}-\theta (p,m)\right\} \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},m,n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}},\ 0<q<1,0\le \sigma \le 1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (5) $\ G_{q,p}^{n,0}(j,1,\sigma ,b,\beta )=G_{q,p}^{n}(j,\sigma ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left\{ (1-\sigma )\frac{ D_{q,p}^{n}f(z)}{z^{p}}+\sigma \frac{D_{q,p}(D_{q,p}^{n}f(z))}{[p]_{q}z^{p-1} }-1\right\} \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}}^{*},n\in {\mathbb {N}}_{0},p,j\in {\mathbb {N}},\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}$$

• (6) $\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=L_{p}^{n,m}(j,\lambda ,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{(I_{\lambda ,p}^{n}f^{(m)}(z))^{\prime }}{(p-m)z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (7) $\ G_{q,p}^{n,m}(j,1,1,b,\beta )=M_{q,p}^{n,m}(j,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{ D_{q,p}(I_{q,p}^{n}f^{(m)}(z))}{[p-m]_{q}z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (8) $\ G_{q,p}^{n,0}(j,1,1,b,\beta )=G_{q,p}^{n}(j,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{ D_{q,p}(D_{q,p}^{n}f(z))}{[p]_{q}z^{p-1}}-1\right] \right|<\beta \right. \\&\left. b\in {\mathbb {C}} ^{*},n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0\le \sigma \le 1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}$$

• (9) $\lim _{q\rightarrow 1^{-}}G_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=O_{p}^{n,m}(j,\lambda ,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{I_{\lambda ,p}^{n}f^{(m)}(z)}{z^{p-m}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (10) $\ G_{q,p}^{n,m}(j,1,0,b,\beta )=R_{q,p}^{n,m}(j,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{ I_{q,p}^{n}f^{(m)}(z)}{z^{p-m}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (11) $\ G_{q,p}^{n,0}(j,1,0,b,\beta )=P_{q,p}^{n}(j,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{D_{q,p}^{n}f(z)}{ z^{p}}-1\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\ 0<q<1,\ 0<\beta \le 1\overset{}{}\right\} ; \end{aligned}$$

• (12) $G_{q,p}^{n,m}(j,\lambda ,1,b,\beta )=G_{q,p}^{n,m}(j,\lambda ,b,\beta )\;$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{D_{q,p}(D_{,\lambda ,q,p}^{n}f^{(m)}(z))}{[p-m]_{q}z^{p-m-1}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<q<1,\ 0<\beta \le 1,p>m\overset{}{}\right\} ; \end{aligned}$$

• (13) $G_{q,p}^{n,m}(j,\lambda ,0,b,\beta )=GL_{p}^{n,m}(j,\lambda ,b,\beta )$

  $$\begin{aligned}&\left\{ f\in T(p,j):\left| \frac{1}{b}\left[ \frac{I_{\lambda ,p}^{n}f^{(m)}(z)}{z^{p-m}}-\theta (p,m)\right] \right|<\beta ,\right. \\&\left. b\in {\mathbb {C}} ^{*},m,n\in {\mathbb {N}} _{0},p,j\in {\mathbb {N}} ,\lambda \ge 0,\ 0<\beta \le 1,p>m\overset{}{}\right\} . \end{aligned}$$

Now, as a results of Authors articles see ([3,4,5,6,7] [29, 31, 34, 35, 38]), we define the neighborhood $(j,\delta )$for $f\in T(p,j)$by

$$\begin{aligned} N_{j,\delta }^{p}(f)= & {} \left\{ g:g\in T(p,j),\ g(z)\right. \nonumber \\&\quad \left. =z^{p}-\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }k\vert a_{k}-b_{k\ }\vert \le \delta \right\} . \end{aligned}$$

(1.12)

In specially, if

$$\begin{aligned} h(z)=z^{p}\ (p\in {\mathbb {N}} ), \end{aligned}$$

(1.13)

we obtain

$$\begin{aligned} N_{j,\delta }^{p}(h)=\left\{ g:g\in T(p,j),\ g(z)=z^{p}-\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }k\left| b_{k\ }\right| \le \delta \right\} . \end{aligned}$$

(1.14)

Now, we define the $\left( q,j,\delta ,m\right) -$neighborhood for $f\in T(p,j)$ by

$$\begin{aligned} N_{j,\delta ,m}^{p,q}(f)= & {} \left\{ g:g\in T(p,j),\ g(z)=z^{p}\right. \nonumber \\&\quad \left. -\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}\left| a_{k}-b_{k\ }\right| \le \delta \right\} . \end{aligned}$$

(1.15)

In particular, if $h(z)\;$given by (1.12), we immediately have

$$\begin{aligned} N_{j,\delta ,m}^{p,q}(h)= & {} \left\{ g:g\in T(p,j),\ g(z)=z^{p}\right. \nonumber \\&\quad \left. -\sum \limits _{k=j+p}^{\infty }b_{k\ }z^{k}\ \text {and }\sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}\left| b_{k\ }\right| \le \delta \right\} . \end{aligned}$$

(1.16)

We note that

• (i) $\ N_{j,\delta ,0}^{p,q}(f)=N_{j,\delta }^{p,q}(f)\;$and$\ N_{j,\delta ,0}^{p,q}(h)=N_{j,\delta }^{p,q}(h)\;$(see [14, 20]);

• (ii) $\ \lim _{q\rightarrow 1^{-}}N_{j,\delta ,0}^{p,q}(f)=N_{j,\delta }^{p}(f)\;$and$\ \lim _{q\rightarrow 1^{-}}N_{j,\delta ,0}^{p,q}(h)=N_{j,\delta }^{p}(h)\;$(see [20] and Aouf et al. [22] ).

Preliminaries

On the other hand, we assume through the article that, $b\in {\mathbb {C}} ^{*},\ n,m\in {\mathbb {N}} _{0},\ p,j\in {\mathbb {N}} ,\ \lambda \ge 0,\ 0<q<1,\ 0\le \sigma \le 1$ ,$\ 0<\beta \le 1,p>m$ and$\ \Psi _{q,p}^{n,m}(k,\lambda )\;$is given by (1.9). To prove the main outcomes in the article we need Lemmas 1 and 2 below.

Lemma 1

Let$f\in T(p,j)$ is given by (1.2), then$f\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$

if and only if

$$\begin{aligned} \sum \limits _{k=j+p}^{\infty }{(}\left[ k-m\right] _{q}{+\beta }\left| b\right| {-}\left[ p-m\right] _{q}{ )[1+\sigma (}\left[ k-m\right] _{q}{-1)]\theta (k,m)\Psi }_{q,p}^{n,m} {(k,\lambda )a}_{k}\nonumber \\ {\le }{\beta }\left| b\right| \theta (p,m){ [1+\sigma (}\left[ p-m\right] _{q}{-1)].} \end{aligned}$$

(2.1)

Proof

If $f\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ).$ Then we have

$$\begin{aligned} \mathrm{Re}\left\{ \tfrac{(1-\sigma )zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{(1-\sigma )D_{\lambda ,q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right\} >-\beta \left| b\right| \ (z\in {\mathbb {U}} ), \end{aligned}$$

(2.2)

or, equivalently,

$$\begin{aligned} \mathrm{Re}\left\{ \tfrac{-\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}- \left[ p-m\right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }z^{k-p}}{[1+\sigma (\left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m \right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ z^{k-p}} \right\} >-\beta \left| b\right| . \end{aligned}$$

(2.3)

By setting $\left| {\small z}\right|$ $=r (0\le r<1)$in (2.3), the term in the denominator of the left hand side of (2.3) is positive for $0\le r<1$. Therefore, by Putting $r\longrightarrow 1$ through real values, (2.3) helps us to the desired assertion of Lemma 1.

Conversely, assume $\left| z\right| =1$and apply the hypothesis (2.1), from (2.3) we have

$$\begin{aligned}&\left| \dfrac{(1-\sigma )zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))+\sigma zD_{q,p}(zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z)))}{(1-\sigma )D_{\lambda ,q,p}^{n}f^{(m)}(z)+\sigma zD_{q,p}(D_{\lambda ,q,p}^{n}f^{(m)}(z))}-\left[ p-m\right] _{q}\right| \\&=\left| \frac{\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}- \left[ p-m\right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }z^{k-p}}{[1+\sigma (\left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m \right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ z^{k-p}} \right| \\&\le \frac{\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}-\left[ p-m \right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }\left| z\right| ^{k-p}}{[1+\sigma ( \left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ \left| z\right| ^{k-p}}\\&\le \tfrac{\sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}-\left[ p-m \right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k\ }}{[1+\sigma (\left[ p-m\right] _{q}-1)]\theta (p,m)-\sum \limits _{k=j+p}^{\infty }[1+\sigma (\left[ k-m\right] _{q}-1)]\theta (k,m)\Psi _{q,p}^{n,m}(k,\lambda )a_{k}\ }=\beta \left| b\right| . \end{aligned}$$

So, we have $f(z)\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$by applying the maximum modulus theorem, which completes the proof of Lemma 1. $\square$

Remark 1

Letting $q\rightarrow 1^{-}$ and $n=m=0$ in Lemma 1, we obtain the result obtained by Aouf and Mostafa [19], Lemma 1, with $b_{k}=1$].

The following lemma can be established similarly.

Lemma 2

Let $f\in T(p,j)$ is given by (1.2). Then$f\in G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$

if and only if

$$\begin{aligned} \sum \limits _{k=j+p}^{\infty }{[}\left[ p-m\right] _{q}{+\sigma (}\left[ k-m\right] _{q}{-}\left[ p-m\right] _{q}{)]\theta (k,m)\Psi }_{q,p}^{n,m}{(k,\lambda )a}_{k}{\le \beta } \left| b\right| \left[ p-m\right] _{q}{.} \end{aligned}$$

(2.4)

3- Inclusion results

In this part, we showed inclusion relations for each of the classes $F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$and $G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )\;$including $(q,j,\delta ,m)-\;$ neighborhood were defined by (1.15) and (1.16).

Theorem 1

Suppose$f\in T(p,j)\;$includes in $F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),\;$ then

$$\begin{aligned} F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ){\subset N}_{j,\eta ,m}^{p,q}{(h),} \end{aligned}$$

(3.1)

since h(z) is defined by (1.13) and $\eta$is given by

$$\begin{aligned} {\eta =}\tfrac{\left[ j+p-m\right] _{q}\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}{\ (}\left[ p-m\right] _{q}{>}\left| b\right| {).} \end{aligned}$$

(3.2)

Proof

Let $f\in F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$, then by using (2.1) of Lemma 1, we obtain

$$\begin{aligned}&(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )\sum \limits _{k=j+p}^{\infty }a_{k}\\&\le \sum \limits _{k=j+p}^{\infty }(\left[ k-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ k-m\right] _{q}-1)] {\theta (k,m)}\Psi _{q,p}^{n,m}(k,\lambda )a_{k} \end{aligned}$$

$$\begin{aligned} \le \beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}, \end{aligned}$$

(3.3)

which quickly gives

$$\begin{aligned} \sum \limits _{k=j+p}^{\infty }a_{k}\le \tfrac{\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}. \end{aligned}$$

(3.4)

Making use of (2.1) with (3.4), we obtain

$$\begin{aligned}&[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )\sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}a_{k}\le \\&\le \beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}+ \\&\quad (\left[ p-m\right] _{q}-\beta \left| b\right| )[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )\sum \limits _{k=j+p}^{\infty }a_{k}\\&\le \beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}+\frac{(\left[ p-m\right] _{q}-\beta \left| b\right| )\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})}\\&=\frac{\left[ j+p-m\right] _{q}\beta \left| b\right| [1+\sigma ( \left[ p-m\right] _{q}-1)]{\theta (p,m)}}{\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q}}. \end{aligned}$$

Hence

$$\begin{aligned} \sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}a_{k}&\le \tfrac{\left[ j+p-m\right] _{q}\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)] {\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}\nonumber \\&=\eta \ (\left[ p-m \right] _{q}>\left| b\right| ), \end{aligned}$$

(3.5)

by means of (1.14), we obtained (3.1) which asserted by Theorem 1. $\square$

Remark 2

Letting $q\rightarrow 1^{-}$ and $n=m=0\;$in Theorem 1, we obtain the result obtained by Aouf and Mostafa [19], Theorem 2, with $b_{k}=1$].

In a similar manner, we proved the following inclusion relationship by using (2.4) of Lemma 2 recompensed (2.1) of Lemma 1 on functions in $G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ).$

Theorem 2

Assume $f\in T(p,j)\;$includes in $G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$ then

$$\begin{aligned} G_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta ){\subset }N_{j,\delta ,m}^{p,q}{(h),} \end{aligned}$$

(3.6)

such that$h(z)\;$is defined by (1.13) and $\delta \;$is introduced by

$$\begin{aligned} {\delta =}\frac{\left[ j+p-m\right] _{q}\beta \left| b\right| \left[ p-m\right] _{q}}{[\left[ p-m\right] _{q}+\sigma (\left[ j+p-m \right] _{q}-\left[ p-m\right] _{q})]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}={.} \end{aligned}$$

(3.7)

4- Neighborhoods properties

In this section, we determine the neighborhood for each of the classes $F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$and $G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$. If there exists a function $\rho (z)\in F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ),$ satisfies (4.1), then $f(z)\in T(p,j)$is said to be in the class $F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$

$$\begin{aligned} \left| \frac{f(z)}{\rho (z)}-1\right|<\left[ p-m\right] _{q}-\gamma \ \ (z\in {\mathbb {U}};\ 0\le \gamma <\left[ p-m\right] _{q}). \end{aligned}$$

(4.1)

Analogously,if we find a function $\rho (z)\in G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )\;$which the inequality (4.1) achieve, then we can say for $f(z)\in T(p,j),\ f(z)\in G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ).$

Theorem 3

Let $f\in T(p,j)$includes in $F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$and

$$\begin{aligned} \gamma&=\left[ p-m\right] _{q}\nonumber \\&\quad -\tfrac{\eta (\left[ j+p-m \right] _{q}+\beta \vert b \vert -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}{\left[ j+p-m\right] _{q}\left\{ (\left[ j+p-m \right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )-\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]\right\} {\theta (p,m)}}{,} \end{aligned}$$

(4.2)

then

$$\begin{aligned} {N}_{j,\eta ,m}^{p,q}{(h)\subset }F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ){,} \end{aligned}$$

(4.3)

where

$$\begin{aligned} {\eta \le }\left[ p-m\right] _{q}\left[ j+p-m\right] _{q}\left\{ 1- \tfrac{\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)] {\theta (p,m)}}{(\left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{ \theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}\right\} {.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \end{aligned}$$

Proof

Assume $f\in N_{j,\eta ,m}^{p,q}(h).$ From (1.15) we find that

$$\begin{aligned} \sum \limits _{k=j+p}^{\infty }\left[ k-m\right] _{q}\left| a_{k}-b_{k}\right| \le \eta , \end{aligned}$$

(4.4)

which readily implies that

$$\begin{aligned} \sum \limits _{k=j+p}^{\infty }\left| a_{k}-b_{k}\right| \le \frac{ \eta }{\left[ j+p-m\right] _{q}}\ \ (p,\ j\in {\mathbb {N}}). \end{aligned}$$

(4.5)

Next, since $\rho (z)\in F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ),\;$by using (3.4), we have

$$\begin{aligned} \sum \limits _{k=j+p}^{\infty }b_{k}\le \tfrac{\beta \left| b\right| [1+\sigma (\left[ p-m\right] _{q}-1)]{\theta (p,m)}}{(\left[ j+p-m \right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}, \end{aligned}$$

(4.6)

so that

$$\begin{aligned} \left| \frac{f(z)}{\rho (z)}-1\right|&\le \frac{\sum \limits _{k=j+p}^{\infty }\left| a_{k}-b_{k}\right| }{ 1-\sum \limits _{k=j+p}^{\infty }b_{k}}\\&\le \tfrac{\eta (\left[ j+p-m\right] _{q}+\beta \left| b\right| - \left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]{\theta (j+p,m)}\Psi _{q,p}^{n,m}(j+p,\lambda )}{\left[ j+p-m\right] _{q}\left\{ ( \left[ j+p-m\right] _{q}+\beta \left| b\right| -\left[ p-m\right] _{q})[1+\sigma (\left[ j+p-m\right] _{q}-1)]\Psi _{q,p}^{n,m}(j+p,\lambda ) {\theta (j+p,m)}-\beta \left| b\right| [1+\sigma (\left[ p-m \right] _{q}-1)]{\theta (p,m)}\right\} }\\&=\left[ p-m\right] _{q}-\gamma , \end{aligned}$$

provided that $\gamma$ is given by (4.2) and by the above definition, $f\in F_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$, so the proof of Theorem 3 is finished. $\square$

The proof of Theorem 4 below is similar to the proof of Theorem 3, we omit the details involved.

Theorem 4

Let $f\in T(p,j)$includes in $G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )$and

$$\begin{aligned}&\gamma =\left[ p-m\right] _{q}\nonumber \\&\quad -\tfrac{\delta [\left[ p-m\right] _{q}+\sigma (\left[ j+p-m\right] _{q}-\left[ p-m\right] _{q})]\Psi _{q,p}^{n,m}(j+p,\lambda ){\theta (j+p,m)}}{\left[ j+p-m \right] _{q}\left\{ [\left[ p-m\right] _{q}+\sigma (\left[ j+p-m\right] _{q}- \left[ p-m\right] _{q})]\Psi _{q,p}^{n,m}(j+p,\lambda ){\theta (j+p,m) }-\beta \left[ p-m\right] _{q}\left| b\right| \right\} }{,} \end{aligned}$$

(4.7)

then

$$\begin{aligned} N_{j,\delta ,m}^{p,q}{(h)\subset }G_{q,p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ){,} \end{aligned}$$

(4.8)

where

$$\begin{aligned} {\delta \le }\left[ p-m\right] _{q}\left[ j+p-m\right] _{q}\left\{ 1- \tfrac{\beta \left[ p-m\right] _{q}\left| b\right| }{(\left[ p-m \right] _{q}+\sigma (\left[ j+p-m\right] _{q}-\left[ p-m\right] _{q})]\Psi _{q,p}^{n,m}(j+p,\lambda ){\theta (j+p,m)}}\right\} {.} \end{aligned}$$

(4.9)

Remarks

(1) Taking $m=0\;$in our outcomes, we obtain the outcomes obtained by Aouf and Madian [14], with $\varrho =0$]; (2) Taking $q\rightarrow 1^{-}\;$ and $\lambda =1\;$in our outcomes, we obtain the outcomes obtained by Aouf et al. [16]; (3) Taking $q\rightarrow 1^{-}\;$and $n=0\;$in our outcomes, we obtain the outcomes obtained by El- El-Ashwah et al. [28], with $b_{k}=1$ and $m=0$]; (4) Taking $q\rightarrow 1^{-}$ in Theorems 1,2,3 and 4, respectively, we obtain new outcomes for the classes $F_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),\ G_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),$ $F_{p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta )\;$and $G_{p}^{n,m(\gamma )}(j,\lambda ,\sigma ,b,\beta ),\;$ respectively;  (5) Taking (a) $\lambda =1$, (b) $\lambda =1\;$and $m=0,\;$(c) Letting $q\rightarrow 1^{-}$ and $\sigma =0,\;$(d)$\ \lambda =1\;$and $\sigma =0,\;$(E) $\sigma =0,\;$(F) Letting $q\rightarrow 1^{-}$ and $\sigma =1,\;$(G)$\ \lambda =1\;$and $\sigma =1,\;$(H) $\sigma =1,\;$in Lemma 1, Theorems 1 and 3, respectively, we get to new outcomes for the classes $F_{q,p}^{n,m}(j,\sigma ,b,\beta ),\ F_{q,p}^{n}(j,\sigma ,b,\beta ),SF_{p}^{n,m}(j,\lambda ,\sigma ,b,\beta ),SF_{q,p}^{n,m}(j,b,\beta ),S_{q,p}^{n,m}(j,\lambda ,b,\beta ),$ $KF_{p}^{n,m}(j,\lambda ,b,\beta ),\ KF_{q,p}^{n,m}(j,b,\beta )\;$and$\ K_{q,p}^{n,m}(j,\lambda ,b,\beta ),\;$ respectively;  (6) Taking (a) $\lambda =1$, (b) $\lambda =1\;$and $m=0,\;$(c) Letting $q\rightarrow 1^{-}$ and $\sigma =1,\;$(d)$\ \lambda =1\;$and $\sigma =1,\;$(E) $\lambda =\sigma =1\;$and $m=0,\;$(F) Letting $q\rightarrow 1^{-}$ and $\sigma =0,\;$(G)$\ \lambda =1\;$and $\sigma =0,\;$ (H) $\lambda =1\;$and $m=\sigma =0,$(I) $\sigma =1\;$(J) $\sigma =0\;$in Lemma 2, Theorems 2 and 4, respectively, we get to new outcomes for the classes $G_{q,p}^{n,m}(j,\sigma ,b,\beta ),\ G_{q,p}^{n}(j,\sigma ,b,\beta ),L_{p}^{n,m}(j,\lambda ,b,\beta ),M_{q,p}^{n,m}(j,b,\beta ),G_{q,p}^{n}(j,b,\beta ),O_{p}^{n,m}(j,\lambda ,b,\beta ),$

$R_{q,p}^{n,m}(j,b,\beta ),\ P_{q,p}^{n}(j,b,\beta ),G_{q,p}^{n,m}(j,\lambda ,b,\beta )\;$and $GL_{p}^{n,m}(j,\lambda ,b,\beta ),$respectively.

Conclusions

Throughout the paper, we defined new subclasses of complex order $F_{q,p}^{n,m}(j,\lambda ,\sigma ,b,\beta )$and $G_{q,p}^{n,m}(j, \lambda ,\sigma ,b,\beta )\;$by using $D_{\lambda ,q,p}^{n}f^{(m)}(z)\;$ operator. Also, we introduced coefficients estimates theorems and neighborhoods properties for this classes. This paper generalized many results for different authors. There was connection between q-analysis and (p,q)-analysis see Srivastava [41]. Srivastava [41], page 340] applied some obvious parametric, argument variations and considered $0<q<p\le 1,$ also translated the classical q-number $\left[ n\right] _{q}$ to $\left[ n\right] _{p,q}\;$as follows:

$$\begin{aligned} \left[ n\right] _{p,q}=\left\{ \begin{array}{ccc} \frac{p^{n}-q^{n}}{p-q} &{} if &{} n\in {\mathbb {N}} \\ 0 &{} if &{} n=0 \end{array} \right. =p^{n-1}\left[ n\right] _{\frac{q}{p}}. \end{aligned}$$