Results of Certain Subclasses of Univalent Function Related to Bessel Functions

Abstract

In this article, we established the necessary and sufficient conditions as well as the inclusion relations for a few subclasses of univalent functions associated with Bessel functions. Furthermore, we investigated an integral operator linked to Bessel functions and elaborated on several mapping properties. The study includes various theorems, corollaries and the consequences derived from the main results.

1. Introduction and Preliminaries

Define $\mathcal{A}$ as the collection of functions $\hslash :\mathbb{U}⟶\mathbb{C}$ that fulfill the normalization requirements $\hslash \left(0\right)=0$ and ${\hslash }^{\prime }\left(0\right)=1$. These functions are analytic in the region $\mathbb{U}=\left\{z:\left|z\right|<1\right\}$ and are represented by the following equation:

$$\hslash \left(z\right)=z+\sum _{s=2}^{\infty }{d}_s{z}^s.$$

(1)

Additionally, we define $\mathcal{S}$ as a subset of $\mathcal{A},$ where the functions within $\mathcal{S}$ are univalent in $\mathbb{U}.$ Let ${\mathcal{S}}_1$ represent the subfamily of $\mathcal{S}$ that includes functions ℏ for which both ℏ and its derivative ${\hslash }^{\prime }$ are univalent in $\mathbb{U}.$ A function $\hslash \left(z\right)$ expressed in the form (1) is classified as belonging to ${\mathcal{S}}_m$ if both ℏ and its first m derivatives are univalent in $\mathbb{U}.$ Let ${\mathcal{S}}^{\star }\left(\delta \right)$ and $\mathcal{C}\left(\delta \right)$ represent the subclasses of $\mathcal{S}$ that include functions which are starlike of order $\delta$ and convex of order $\delta ,$ where $0\le \delta <1.$ The analytic characterizations of these two classes are provided as follows:

$${\mathcal{S}}^{\star }\left(\delta \right)=\left\{\hslash \in \mathcal{S}:\Re \left({\displaystyle \frac{z{\hslash }^{\prime }\left(z\right)}{\hslash \left(z\right)}}\right)>\delta ,0\le \delta <1\right\}$$

and

$$\mathcal{C}\left(\delta \right)=\left\{\hslash \in \mathcal{S}:\Re \left(1+{\displaystyle \frac{z{\hslash }^{″}\left(z\right)}{{\hslash }^{\prime }\left(z\right)}}\right)>\delta ,0\le \delta <1\right\}.$$

Further, $\mathcal{C}\equiv \mathcal{C}\left(0\right)$ and ${\mathcal{S}}^{\star }\equiv \mathcal{S}\left(0\right)$, the well-known standard class of convex functions. It is an established fact that

$$\hslash \in \mathcal{C}\left(\delta \right)⟺z{\hslash }^{\prime }\in {\mathcal{S}}^{*}\left(\delta \right).$$

A function $\hslash \in \mathcal{A}$ is considered to be in the $\mathcal{UCV}$ class of uniformly convex functions in the unit disk $\mathbb{U}$ if it qualifies as a normalized convex function in $\mathbb{U}.$ Furthermore, it possesses the characteristic that for every circular arc $\Gamma$ situated in $\mathbb{U},$ with center $\xi$ also in $\mathbb{U},$ the image curve $\hslash (\Gamma )$ remains a convex arc. The concept of uniformly convex functions, designated as $\mathcal{UCV},$ was introduced by Goodman [1]. Rønning [2] showed that a function $\hslash \left(z\right)$ represented in the form (1) is in $\mathcal{UCV}$ if and only if

$$\Re \left\{1+(z-\xi ){\displaystyle \frac{{\hslash }^{″}\left(z\right)}{{\hslash }^{\prime }\left(z\right)}}\right\}\ge 0,(z,\xi )\in \mathbb{U}\times \mathbb{U}.$$

He also introduced the concept of uniformly starlike functions and the analytic criteria is as follows: $\hslash \in \mathcal{UST}$ if and only if

$$\Re \left\{{\displaystyle \frac{\hslash \left(z\right)-\hslash \left(\xi \right)}{(z-\xi ){\hslash }^{\prime }\left(z\right)}}\right\}\ge 0,(z,\xi )\in \mathbb{U}\times \mathbb{U}.$$

In addition, we identify two important subclasses of $\mathcal{S},$ known as $\alpha -\mathcal{UCV}$ and $\alpha -\mathcal{ST},$ which include functions that are $\alpha$-uniformly convex and $\alpha$- starlike in $\mathbb{U},$ respectively. The analytic representations of these two classes are provided by

$$\alpha -\mathcal{UCV}:=\left\{\hslash \in \mathcal{S}:\Re \left(1+{\displaystyle \frac{z{\hslash }^{″}\left(z\right)}{{\hslash }^{\prime }\left(z\right)}}\right)>\alpha \left|{\displaystyle \frac{z{\hslash }^{″}\left(z\right)}{{\hslash }^{\prime }\left(z\right)}}\right|(z\in \mathbb{U};0\le \alpha <\infty )\right\}$$

and

$$\alpha -\mathcal{ST}:=\left\{\hslash \in \mathcal{S}:\Re \left({\displaystyle \frac{z{\hslash }^{\prime }\left(z\right)}{\hslash \left(z\right)}}\right)>\alpha \left|{\displaystyle \frac{z{\hslash }^{\prime }\left(z\right)}{\hslash \left(z\right)}}-1\right|(z\in \mathbb{U};0\le \alpha <\infty )\right\}.$$

The $\alpha -\mathcal{UCV}$ class was defined by Kanas and Wiśniowska [3], where they explored its geometric definition and its links to conic domains. The $\alpha -\mathcal{ST}$ class was studied in [4], showing its connection to the $\alpha -\mathcal{UCV}$ class via the well-known Alexander relation that relates the standard classes of convex and starlike functions. For more comprehensive developments regarding the $\alpha -\mathcal{UCV}$ and $\alpha -\mathcal{ST}$ classes, see the contributions of Kanas and Srivastava [5]. In this case, when $\alpha =1,$ we can conclude that

$$1-\mathcal{UCV}\equiv \mathcal{UCV},1-\mathcal{ST}=\mathcal{SP},$$

(2)

where $\mathcal{UCV}$ and $\mathcal{SP}$ represent well-known classes of uniformly convex functions and parabolic starlike functions within the disk $\mathbb{U}$ (for further information, we refer to sources [1,6,7,8,9,10,11,12,13,14,15,16,17]). Notably, Srivastava and Mishra [18] conducted a comprehensive and cohesive examination of the $\mathcal{UCV}$ and $\mathcal{SP}$ classes by employing a specific fractional calculus operator.

Let $\mathcal{T}$ denotes the class of all functions given by

$$\hslash \left(z\right)=z-\sum _{s=2}^{\infty }{d}_s{z}^s,a_s\ge 0,$$

(3)

normalized by the conditions $\hslash \left(0\right)={\hslash }^{\prime }\left(0\right)-1=0$ which are analytic and univalent in $\mathbb{U}.$ We denote by ${\mathcal{T}}^{*}\left(\delta \right)$ and $\mathcal{K}\left(\delta \right)$ the subclasses of $\mathcal{T}$ that are, respectively, starlike of order $\delta$ and convex of order $\delta .$ Silverman [19] investigated functions in the classes ${\mathcal{T}}^{*}\left(\delta \right)=\mathcal{T}\cap {\mathcal{S}}^{*}\left(\delta \right)$ and $\mathcal{K}\left(\delta \right)=\mathcal{T}\cap \mathcal{C}\left(\delta \right).$ Furthermore, let ${\mathcal{T}}_1$ be the subfamily of $\mathcal{T}$ consisting of functions ℏ for which ℏ and ${\hslash }^{\prime }$ are univalent in $\mathbb{U}$. It is clear that the second coefficient of a function in ${\mathcal{T}}_1$ cannot vanish. Therefore, class ${\mathcal{T}}_1$ is non-empty as the function ${\displaystyle z-\frac{z^2}{2}}$ belongs to the class ${\mathcal{T}}_1.$ A function $\hslash \left(z\right)$ given in the form (3) is said to be in ${\mathcal{T}}_M$ if ℏ and its first M derivatives are univalent in $\mathbb{U}$. If $\hslash \in {\mathcal{T}}_M$, then ℏ is said to be in ${\mathcal{T}}_{\infty }$.

In 1995, for $f\in \mathcal{A}$, Dixit and Pal [20] defined the class ${\mathcal{R}}^{\lambda }(E,F)$ as follows: For $\lambda \in \mathbb{C}\setminus \left\{0\right\}$ and $-1\le F<E\le 1,$ a function $\hslash \in {\mathcal{R}}^{\lambda }(E,F)$ if it satisfies the following condition

$$\left|{\displaystyle \frac{{\hslash }^{\prime }\left(z\right)-1}{(E-F)\lambda -F[{\hslash }^{\prime }\left(z\right)-1]}}\right|<1,z\in \mathbb{U}.$$

1.1. Definition

Definition 1.

A function $\hslash \left(z\right)$ expressed in the form (1) is classified as belonging to ${\mathcal{S}}_1$ if both ℏ and its first derivatives are univalent in $\mathbb{U}.$ Similarly we can define that a function $\hslash \left(z\right)$ expressed in the form (1) is classified as belonging to ${\mathcal{S}}_2$ if $\hslash ,$ ${\hslash }^{\prime }$ and ${\hslash }^{″}$ are univalent in $\mathbb{U}.$

Definition 2.

A function $\hslash \left(z\right)$ expressed in the form (3) is classified as belonging to ${\mathcal{T}}_1$ if both ℏ and its first derivatives are univalent in $\mathbb{U}.$ Similarly we can define that a function $\hslash \left(z\right)$ expressed in the form (3) is classified as belonging to ${\mathcal{T}}_2$ if $\hslash ,$ ${\hslash }^{\prime }$ and ${\hslash }^{″}$ are univalent in $\mathbb{U}.$

Remark 1.

There exists an interesting relation between the classes ${\mathcal{S}}_1,$ ${\mathcal{S}}_2,$ ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$, as shown here:

$${\mathcal{T}}_1=\mathcal{T}\cap {\mathcal{S}}_{1}and{\mathcal{T}}_2=\mathcal{T}\cap {\mathcal{S}}_2.$$

1.2. Bessel Function

The specific solution to the second-order linear homogeneous differential equation

$$z^2{w}^{″}\left(z\right)+az{w}^{\prime }\left(z\right)+\left[b{z}^2-q^2+(1-a)q\right]w\left(z\right)=0,(b\in \mathbb{C}anda,q\in \mathbb{R})$$

known as the generalized Bessel function of the first kind and order $q.$ The solution is represented as $w_{q,a,b}\left(z\right)$ and can be expressed through the infinite sum formulation given by

$$w\left(z\right)=w_{q,a,b}\left(z\right)=\sum _{s=0}^{\infty }{\displaystyle \frac{{(-b)}^s}{s!\Gamma \left(q+s+\frac{a+1}{2}\right)}}{\left({\displaystyle \frac{z}{2}}\right)}^{2s+q}z\in \mathbb{U}.$$

(4)

In this context, $\Gamma$ refers to the Euler gamma function. Baricz [21] investigated the generalized Bessel function of the first kind. In 2010, Baricz and Frasin [22] proved the univalence of specific integral operators associated with generalized Bessel functions. Furthermore, Baricz in partnership with Ponnusamy [23] identified criteria for the starlikeness and convexity of these functions (see [24] also). In 2008, Baricz [25] introduced and studied Bessel functions. When setting $a=b=1$ in (4), the Bessel functions are defined as follows

$${\mathcal{J}}_q\left(z\right)=\sum _{s=0}^{\infty }{\displaystyle \frac{{(-1)}^s}{s!\Gamma (q+s+1)}}{\left({\displaystyle \frac{z}{2}}\right)}^{2s+q},z\in \mathbb{U}.$$

For $a=-b=1$ in (4), we obtained the modified Bessel functions which are defined [25] by

$${\mathcal{I}}_q\left(z\right)=\sum _{s=0}^{\infty }{\displaystyle \frac{1}{s!\Gamma (q+s+1)}}{\left({\displaystyle \frac{z}{2}}\right)}^{2s+q},z\in \mathbb{U}.$$

The spherical Bessel functions [25] are derived for the case where $a-1=b=1$ in Equation (4), as defined by

$${\mathcal{K}}_q\left(z\right)=\sqrt{{\displaystyle \frac{2}{z}}}{\mathcal{J}}_{q+{\displaystyle \frac{1}{2}}}\left(z\right)=\sum _{s=0}^{\infty }{\displaystyle \frac{{(-1)}^s}{s!\Gamma \left(q+s+\frac{3}{2}\right)}}{\left({\displaystyle \frac{z}{2}}\right)}^{2s+q},z\in \mathbb{U}.$$

Under particular conditions, Bessel functions of the first kind may be reduced to elementary functions like sine and cosine, and modified Bessel functions of the first kind can be simplified to hyperbolic sine and cosine. According to the series expansion $w_{q,a,b},$ we have $w_{q,a,b}\notin \mathcal{S}$; therefore, we consider the following transformation:

$$v_{q,a,b}\left(z\right)=2^q\Gamma \left(q+{\displaystyle \frac{a+1}{2}}\right)z^{-q/2}{w}_{q,a,b}\left(\sqrt{z}\right).$$

By virtue of (4) following with a simple computation, it is easy to see that the function $w_{q,a,b}\left(\sqrt{z}\right)$ is a single-valued function for $q\in \mathbb{R}.$ Through the application of the Pochhammer symbol, defined in terms of the Euler gamma function, we can express the function $v_{q,a,b}\left(z\right)$ with the following series representation

$$v_q\left(z\right)=v_{q,a,b}\left(z\right)=\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{\left(k\right)}_s\left(s\right)!}}{z}^s,$$

(5)

for $k=q+{\displaystyle \frac{a+1}{2}}\notin {\mathbb{Z}}_0^{-}=\left\{0,-1,-2,\cdots \right\}$. For $a=-b=1$ in (5), then the function $v_{q,a,b}\left(z\right)\equiv {\ell }_q\left(z\right)$ have the following series representation

$${\ell }_q\left(z\right)=v_{q,1,-1}\left(z\right)=\sum _{s=0}^{\infty }{\displaystyle \frac{z^s}{4^s{\left(m\right)}_s\left(s\right)!}}{z}^s$$

for some $m=q+1\notin {\mathbb{Z}}_0^{-}.$ For $z=1$, we have

$$v_q\left(1\right)=\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{\left(k\right)}_s\left(s\right)!}}.$$

(6)

Further, we have the following three derivatives in their initial form

$$v_q^{\prime }\left(1\right)={\displaystyle \frac{\left(\frac{-b}{4}\right)}{\left(k\right)}}{v}_{q+1}\left(1\right),$$

(7)

$$v_q^{″}\left(1\right)={\displaystyle \frac{{\left(\frac{-b}{4}\right)}^2}{{\left(k\right)}_2}}{v}_{q+2}\left(1\right),$$

(8)

$$v_q^{\prime \prime \prime }\left(1\right)={\displaystyle \frac{{\left(\frac{-b}{4}\right)}^3}{{\left(k\right)}_3}}{v}_{q+3}\left(1\right),$$

(9)

$$z{v}_q\left(z\right)=z+\sum _{s=2}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{z}^s$$

(10)

and

$${\mathcal{U}}_1\left(z\right)=z(2-v_q\left(z\right))=z-\sum _{s=2}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{z}^s.$$

(11)

For any function $\hslash \in \mathcal{A}$ given in the form (1) and $g\in \mathcal{A}$ given by $g\left(z\right)=z+{\sum }_{s=2}^{\infty }{b}_s{z}^s$, we define the Hadamard product (or Convolution) of ℏ and g by

$$\left(\hslash \ast g\right)\left(z\right)=z+\sum _{s=2}^{\infty }{d}_s{b}_s{z}^s,z\in \mathbb{U}.$$

We now introduce the linear operators ${\mathcal{H}}_1,{\mathcal{J}}_1:\mathcal{A}⟶\mathcal{A}$ by

$${\mathcal{H}}_1\left(z\right)=z{v}_q\left(z\right)\ast \hslash \left(z\right)$$

and

$${\mathcal{J}}_1\left(z\right)={\mathcal{U}}_1\left(z\right)\ast \hslash \left(z\right),$$

Consequently, for any function $\hslash \left(z\right)$ presented in (1), we can state that

$${\mathcal{H}}_1\left(z\right)=z+\sum _{s=2}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s{z}^s$$

and

$${\mathcal{J}}_1\left(z\right)=z-\sum _{s=2}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s{z}^s.$$

If we take $a=1$ and $b=-1,$ then the operators ${\mathcal{H}}_1$ and ${\mathcal{J}}_1$ reduce to the following operators ${\mathcal{H}}_1^{\ast }$ and ${\mathcal{J}}_1^{\ast }.$ Consequently, for any function $\hslash \left(z\right)$ presented in (1), we can state that

$${\mathcal{H}}_1^{\ast }\left(z\right)=z+\sum _{s=2}^{\infty }{\displaystyle \frac{d_s{z}^s}{4^{s-1}{\left(m\right)}_{s-1}{\left(1\right)}_{s-1}}}$$

and

$${\mathcal{J}}_1^{\ast }\left(z\right)=z-\sum _{s=2}^{\infty }{\displaystyle \frac{d_s{z}^s}{4^{s-1}{\left(m\right)}_{s-1}{\left(1\right)}_{s-1}}},$$

where $m=q+1\notin {\mathbb{Z}}_0^{-}.$

To outline our key results, it is essential to utilize the following lemmas.

Lemma 1

([26], Corollary 2). Let a function $\hslash \left(z\right)\in \mathcal{S}$ be given in the form (1), such that $d_2\ne 0$. Then, $\hslash \in {\mathcal{S}}_1$ if

$$\sum _{s=3}^{\infty }s(s-1)|d_s|\le 2|d_2|.$$

Lemma 2

([26], Theorem 2). Let a function $\hslash \left(z\right)$ be given in the form (3), such that $d_2>0$. Then, $\hslash \in {\mathcal{T}}_1$ if

$$\sum _{s=3}^{\infty }s(s-1)d_s\le 2d_2.$$

Further, the condition given in Lemma 2 is also sufficient for $0<d_2\le {\displaystyle \frac{1}{3}}$.

Lemma 3

([26], Theorem 6). If a function $\hslash \left(z\right)$ given in the form (3), with ${\prod }_{s=2}^{M+1}{d}_s\ne 0,$ then $\hslash \in {\mathcal{S}}_M$ if

$$\sum _{s=r+2}^{\infty }(s-r)(s-r+1)\cdots s|d_s|\le (r+1)!|d_{r+1}|,$$

for $r=1,2,\cdots ,M$.

Lemma 4

([26], Theorem 6). If a function $\hslash \left(z\right)$ given in the form (3), with ${\prod }_{s=2}^{M+1}{d}_s\ne 0,$ then $\hslash \in {\mathcal{T}}_M$ if

$$\sum _{s=r+2}^{\infty }(s-r)(s-r+1)\cdots s{d}_s\le (r+1)!d_{r+1},$$

for $r=1,2,\cdots ,M$.

Lemma 5

([20]). If a function $\hslash \in {\mathcal{R}}^{\lambda }(E,F)$ is expressed in the form given by (1), then it follows that

$$|d_s|\le {\displaystyle \frac{(E-F)\left|\lambda \right|}{s}},s\ge 2.$$

(12)

The result is sharp.

Lemma 6

([6]). If $\hslash \in \mathcal{S}$, $\hslash \left(z\right)=z+{\sum }_{s=2}^{\infty }{d}_s{z}^{s}z\in \mathbb{U}$, then

$$|d_s|\le s\mathrm{for}s=2,3\cdots .$$

Equality $|d_s|=s$ for a given $s\ge 2$ holds if and only if ℏ is a rotation of the Koebe function.

In this article, the authors established the necessary and sufficient conditions as well as the inclusion relations for a few subclasses of univalent functions associated with Bessel functions. Furthermore, we investigate an integral operator linked to Bessel functions and elaborate on several mapping properties. The study includes various theorems, corollaries and the consequences derived from the main results.

2. Main Result

Theorem 1.

If $b<0$ and $k>0$, then a necessary condition for $z{v}_q\left(z\right)$ to belong to ${\mathcal{S}}_1$ is that inequality

$$v_q^{″}\left(1\right)+2v_q^{\prime }\left(1\right)\le -{\displaystyle \frac{b}{k}}$$

(13)

holds. If $0\le -b\le {\displaystyle \frac{2k}{3}}$, then condition (13) is necessary and also sufficient for ${\mathcal{U}}_1\left(z\right)$ to be in ${\mathcal{T}}_1$.

Proof. 

In order to verify that $z{v}_q\left(z\right)$ is an element of ${\mathcal{S}}_1$ as stated in Lemma 1, it suffices to show that the inequality below is satisfied:

$$\sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\le -{\displaystyle \frac{b}{2k}}.$$

(14)

The left-hand side can now be represented in the following manner:

$$\begin{array}{cc}\hfill \sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}& =\sum _{s=3}^{\infty }s{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}\hfill \\ \hfill & =\sum _{s=3}^{\infty }(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}+2\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}\hfill \\ \hfill & ={\left({\displaystyle \frac{-b}{4}}\right)}^2\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-3}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}+2\left({\displaystyle \frac{-b}{4}}\right)\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-2}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}.\hfill \end{array}$$

Nothing that ${\left(a\right)}_s=a{(a+1)}_{s-1},$

$$\begin{array}{cc}\hfill \sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}& ={\left({\displaystyle \frac{-b}{4}}\right)}^2{\displaystyle \frac{1}{k(k+1)}}\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-3}}{{(k+2)}_{s-3}{\left(1\right)}_{s-3}}}\hfill \\ \hfill & +2\left({\displaystyle \frac{-b}{4k}}\right)\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-2}}{{(k+1)}_{s-2}{\left(1\right)}_{s-2}}}\hfill \\ \hfill & ={\left({\displaystyle \frac{-b}{4}}\right)}^2{\displaystyle \frac{1}{k(k+1)}}{v}_{q+2}\left(1\right)+2{\displaystyle \frac{-b}{4k}}\left[v_{q+1}\left(1\right)-1\right]\hfill \end{array}$$

$$\sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}=v_q^{″}\left(1\right)+2v_q^{\prime }\left(1\right)+{\displaystyle \frac{2b}{4k}}.$$

(15)

The upper bound of the last expression is ${\displaystyle \frac{-b}{2k}}$, which occurs if and only if the inequality in (13) is met.

By virtue of Lemma 2, condition (13) is both necessary and sufficient for ${\mathcal{U}}_1\left(z\right)$ to be in ${\mathcal{T}}_1$. □

For $a=1$ and $b=-1$, we can improve the assertion of Theorem 1 as follows.

Corollary 1.

If $m>0$, then a necessary and sufficient condition for $z{\ell }_q\left(z\right)$ to belong to ${\mathcal{S}}_1$ is that inequality

$${\displaystyle \frac{v_{q+2}\left(1\right)}{16(m+1)}}+{\displaystyle \frac{v_{q+1}\left(1\right)}{2}}\le 1,$$

(16)

holds. This condition (16) is also sufficient for $z(2-{\ell }_q\left(z\right))$ to be in ${\mathcal{T}}_1.$

Theorem 2.

If $b<0$ and $k>0$, then a necessary condition for $z{v}_q\left(z\right)$ to belong to ${\mathcal{S}}_2$ is that inequality:

$$v_q^{\prime \prime \prime }\left(1\right)+3v_q^{″}\left(1\right)\le {\displaystyle \frac{3b^2}{8{\left(k\right)}_2}},$$

(17)

holds. This condition (17) is also necessary for ${\mathcal{U}}_1\left(z\right)$ to be in ${\mathcal{T}}_2.$

Proof. 

In order to verify that $z{v}_q\left(z\right)$ is an element of ${\mathcal{S}}_2$ as stated in Lemma 3, it suffices to show that the inequality below is satisfied:

$$\sum _{s=4}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\le 3{\displaystyle \frac{{\left(-\frac{b}{4}\right)}^2}{{\left(k\right)}_2}}.$$

(18)

The left-hand side can now be represented in the following manner:

$$\begin{array}{cc}\hfill \sum _{s=4}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}& =\sum _{s=4}^{\infty }s{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}\hfill \\ & =\sum _{s=4}^{\infty }(s-3+3){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}\hfill \\ & =\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-4}}}+3\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}\hfill \\ & ={\left({\displaystyle \frac{-b}{4}}\right)}^3\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-4}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-4}}}+3{\left({\displaystyle \frac{-b}{4}}\right)}^2\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-3}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}.\hfill \end{array}$$

Noting that ${\left(a\right)}_s=a{(a+1)}_{s-1}$

$$\begin{array}{cc}\hfill \sum _{s=4}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}& ={\left({\displaystyle \frac{-b}{4}}\right)}^3{\displaystyle \frac{1}{{\left(k\right)}_3}}\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-4}}{{(k+3)}_{s-4}{\left(1\right)}_{s-4}}}\hfill \\ & +3{\left({\displaystyle \frac{-b}{4}}\right)}^2{\displaystyle \frac{1}{{\left(k\right)}_2}}\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-3}}{{(k+2)}_{s-3}{\left(1\right)}_{s-3}}}\hfill \\ & ={\left({\displaystyle \frac{-b}{4}}\right)}^3{\displaystyle \frac{1}{{\left(k\right)}_3}}{v}_{q+3}\left(1\right)+3{\left({\displaystyle \frac{-b}{4}}\right)}^2{\displaystyle \frac{1}{{\left(k\right)}_2}}\left[v_{q+2}\left(1\right)-1\right]\hfill \end{array}$$

$$\sum _{s=4}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}=v_q^{\prime \prime \prime }\left(1\right)+3v_q^{″}\left(1\right)-{\displaystyle \frac{3{\left(-\frac{b}{4}\right)}^2}{{\left(k\right)}_2}}.$$

(19)

The upper bound of the last expression is bounded above by ${\displaystyle 3\frac{{\left(-\frac{b}{4}\right)}^2}{{\left(k\right)}_2}}$ if the inequality in (17) is met.

By Lemma 4, the condition (17) is also sufficient for ${\mathcal{U}}_1\left(z\right)$ to be in ${\mathcal{T}}_2.$ □

Let $a=1$ and $b=-1$; we can improve the assertion of Theorem 2 as follows.

Corollary 2.

If $m>0$ (with $m\ne 0,-1,-2,\dots$), then a sufficient condition for $z{\ell }_m\left(z\right)$ to belong to ${\mathcal{S}}_2$ is that inequality

$${\displaystyle \frac{v_{q+3}\left(1\right)}{4(m+2)}}+3v_q\left(1\right)\le 4,$$

(20)

holds. This condition (20) is also sufficient for $z(2-{\ell }_q\left(z\right))$ to be in ${\mathcal{T}}_2.$

Theorem 3.

Let $f\in \mathcal{S}$. If $k>0$, then a sufficient condition for the Bessel operator ${\mathcal{H}}_1\left(z\right)$ to belong to ${\mathcal{S}}_1$ is that the coefficient inequality

$$v_q^{\prime \prime \prime }\left(1\right)+5v_q^{″}\left(1\right)+4v_q^{\prime }\left(1\right)\le 2\left(-{\displaystyle \frac{b}{k}}\right),$$

(21)

must hold. This condition (21) is also sufficient for ${\mathcal{J}}_1\left(z\right)$ to be in ${\mathcal{T}}_1.$

Proof. 

In order to verify that ${\mathcal{H}}_1\left(z\right)$ is an element of ${\mathcal{S}}_1$ as stated in Lemma 1, it suffices to show that the inequality below is satisfied: consider

$$\sum _{s=3}^{\infty }s(s-1)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|\le -{\displaystyle \frac{b}{k}}.$$

(22)

The left-hand side can now be represented in the following manner:

$$\begin{array}{cc}\hfill \sum _{s=3}^{\infty }s(s-1)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|& =\sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\left|d_s\right|\hfill \\ \hfill & \le \sum _{s=3}^{\infty }{s}^2(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\mathrm{by}\mathrm{using}\mathrm{Lemma}6,\hfill \\ \hfill & =\sum _{s=3}^{\infty }{\left[s-2+2\right]}^2{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}.\hfill \end{array}$$

$$\sum _{s=3}^{\infty }{\left[(s-2)+2\right]}^2{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}=\sum _{s=3}^{\infty }(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}+4\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}+4\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}$$

$$\begin{array}{c}\hfill =\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-4}}}+\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}+4\sum _{s=1}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+1}}{{\left(k\right)}_{s+1}{\left(1\right)}_s}}+4\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+2}}{{\left(k\right)}_{s+2}{\left(1\right)}_s}}\end{array}$$

$$\begin{array}{c}\hfill =\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+3}}{{\left(k\right)}_{s+3}{\left(1\right)}_s}}+5\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+2}}{{\left(k\right)}_{s+2}{\left(1\right)}_s}}+4\sum _{s=1}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+1}}{{\left(k\right)}_{s+1}{\left(1\right)}_s}}\\ \hfill ={\displaystyle \frac{{\left(\frac{-b}{4}\right)}^3}{{\left(k\right)}_3}}\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{(k+3)}_s{\left(1\right)}_s}}+5{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^2}{{\left(k\right)}_2}}\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{(k+2)}_s{\left(1\right)}_s}}+4{\displaystyle \frac{\left(\frac{-b}{4}\right)}{\left(k\right)}}\left(\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{(k+1)}_s{\left(1\right)}_s}}-1\right)\end{array}$$

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{{\left(\frac{-b}{4}\right)}^3}{{\left(k\right)}_3}}{v}_{q+3}\left(1\right)+5{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^2}{{\left(k\right)}_2}}{v}_{q+2}\left(1\right)+4{\displaystyle \frac{\left(\frac{-b}{4}\right)}{\left(k\right)}}{v}_{q+1}\left(1\right)-4{\displaystyle \frac{\left(\frac{-b}{4}\right)}{\left(k\right)}}\hfill \\ \hfill & =v_q^{\prime \prime \prime }\left(1\right)+5v_q^{″}\left(1\right)+4v_q^{\prime }\left(1\right)-{\displaystyle \frac{b}{k}}.\hfill \end{array}$$

(23)

The upper bound of the last expression is ${\displaystyle \frac{b}{k}}$, which occurs if and only if the inequality in (21) is met. □

Let $a=1$ and $b=-1$; we can improve the assertion of Theorem 3 as follows.

Corollary 3.

If $m>0$, then a sufficient condition for ${\mathcal{H}}_1^{\ast }\left(z\right)\in {\mathcal{S}}_1$ is that the inequality

$${\displaystyle \frac{1}{64(m+1)(m+2)}}{v}_{q+3}\left(1\right)+{\displaystyle \frac{5}{16(m+1)}}{v}_{q+2}\left(1\right)+v_{q+1}\left(1\right)\le 2$$

(24)

holds well. This condition (24) is also sufficient for ${\mathcal{J}}_1^{\ast }\left(z\right)$ to be in ${\mathcal{T}}_1.$

Theorem 4.

Let $k>0$ and $f\in \mathcal{S}$. A sufficient condition for the Bessel operator ${\mathcal{H}}_1\left(z\right)\in {\mathcal{S}}_2$ is that

$$v_q^{iv}\left(1\right)+7v_q^{\prime \prime \prime }\left(1\right)+9v_p^{″}\left(1\right)\le {\displaystyle \frac{9b^2}{8{\left(k\right)}_2}}$$

(25)

must hold. This condition (25) is sufficient for ${\mathcal{J}}_1$ to be in ${\mathcal{T}}_2.$

Proof. 

In order to verify that ${\mathcal{H}}_1\left(z\right)$ is an element of ${\mathcal{S}}_2$ as stated in Lemma 3, it suffices to show that the inequality below is satisfied:

$$\sum _{s=4}^{\infty }s(s-1)(s-2)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|\le {\displaystyle \frac{9b^2}{16{\left(k\right)}_2}}.$$

(26)

The left-hand side can now be represented in the following manner:

$$\begin{array}{cc}\hfill \sum _{s=4}^{\infty }s(s-1)(s-2)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{n-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|& =\sum _{s=4}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\left|d_s\right|\hfill \\ \hfill & \le \sum _{s=4}^{\infty }{s}^2(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\mathrm{by}\mathrm{using}\mathrm{Lemma}6,\hfill \\ \hfill & =\sum _{s=4}^{\infty }{\left[s-3+3\right]}^2{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}\hfill \end{array}$$

$$\sum _{s=4}^{\infty }{\left[(s-3)+3\right]}^2{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}=\sum _{s=4}^{\infty }(s-3){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-4}}}+9\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}+6\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-4}}}$$

$$\begin{array}{c}\hfill =\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-5}}}+\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-4}}}+9\sum _{s=1}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+2}}{{\left(k\right)}_{s+2}{\left(1\right)}_s}}+6\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+3}}{{\left(k\right)}_{s+3}{\left(1\right)}_s}}\end{array}$$

$$\begin{array}{c}\hfill =\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+4}}{{\left(k\right)}_{s+4}{\left(1\right)}_s}}+7\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+3}}{{\left(k\right)}_{s+3}{\left(1\right)}_s}}+9\sum _{s=1}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+2}}{{\left(k\right)}_{s+2}{\left(1\right)}_s}}\\ \hfill ={\displaystyle \frac{{\left(\frac{-b}{4}\right)}^4}{{\left(k\right)}_4}}\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{(k+4)}_s{\left(1\right)}_s}}+7{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^3}{{\left(k\right)}_3}}\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{(k+3)}_s{\left(1\right)}_s}}+9{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^2}{{\left(k\right)}_2}}\left(\sum _{s=1}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{(k+2)}_s{\left(1\right)}_s}}-1\right)\end{array}$$

$$\begin{array}{cc}\hfill & ={\displaystyle \frac{{\left(\frac{-b}{4}\right)}^4}{{\left(k\right)}_4}}{v}_{q+4}\left(1\right)+7{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^3}{{\left(k\right)}_3}}{v}_{q+3}\left(1\right)+9{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^2}{{\left(k\right)}_2}}{v}_{q+2}\left(1\right)-9{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^2}{{\left(k\right)}_2}}\hfill \\ \hfill & =v_q^{iv}\left(1\right)+7v_q^{\prime \prime \prime }\left(1\right)+9v_q^{″}\left(1\right)-9{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^2}{{\left(k\right)}_2}}.\hfill \end{array}$$

(27)

The upper bound of the last expression is ${\displaystyle \frac{9b^2}{16{\left(k\right)}_2}}$ which occurs if and only if the inequality in (25) is met. □

For $a=1$ and $b=-1$, we can improve the assertion of Theorem 4 as follows.

Corollary 4.

Let $m>0$ and $f\in \mathcal{S}$. A sufficient condition for the Bessel operator ${\mathcal{J}}_1^{\ast }\left(z\right)\in {\mathcal{S}}_2$ is that the following inequality

$${\displaystyle \frac{1}{16(m+2)(m+3)}}{v}_{q+4}\left(1\right)+{\displaystyle \frac{1}{4(m+2)}}+9v_{q+2}\left(1\right)\le 18$$

(28)

must hold well. This condition (28) is also sufficient for ${\mathcal{J}}_1^{\ast }\left(z\right)$ to be in ${\mathcal{T}}_2.$

Theorem 5.

Let $k>0$ and $f\in {\mathcal{R}}^{\lambda }(E,F)$. A sufficient condition for the Bessel operator ${\mathcal{H}}_1\left(z\right)\in {\mathcal{S}}_1$ is that the Bessel’s inequality

$$v_q^{\prime }\left(1\right)\le 2$$

(29)

is satisfied.

Proof. 

To prove that ${\mathcal{H}}_1\left(z\right)$ belongs to ${\mathcal{S}}_1$, it is sufficient to show that the condition

$$\sum _{s=3}^{\infty }s(s-1)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|\le {\displaystyle \frac{(E-F)\left|\lambda \right|(-b/4)}{\left(k\right)}}$$

(30)

holds. The left-hand side can now be represented in the following manner:

$$\begin{array}{cc}\hfill \sum _{s=3}^{\infty }s(s-1)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|& =\sum _{s=3}^{\infty }s(s-1)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\right|\left|d_s\right|\hfill \\ \hfill & \le \sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{\displaystyle \frac{(E-F)\left|\lambda \right|}{s}}\mathrm{by}\mathrm{using}\mathrm{Lemma}5\hfill \\ \hfill & =(E-F)\left|\lambda \right|\sum _{s=1}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+1}}{{\left(k\right)}_{s+1}{\left(1\right)}_s}}\hfill \end{array}$$

Noting that ${\left(a\right)}_s=a{(a+1)}_{s-1}$

$$\begin{array}{cc}\hfill \sum _{s=3}^{\infty }s(s-1)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|& ={\displaystyle \frac{(E-F)\left|\lambda \right|(-b)}{4\left(k\right)}}\left(\sum _{s=0}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^s}{{(k+1)}_s{\left(1\right)}_s}}-1\right)\hfill \\ \hfill & ={\displaystyle \frac{(E-F)\left|\lambda \right|(-b)}{4\left(k\right)}}{v}_{q+1}\left(1\right)-{\displaystyle \frac{(E-F)\left|\lambda \right|b}{4\left(k\right)}}\hfill \end{array}$$

$$={\displaystyle \frac{(E-F)\left|\lambda \right|(-b)}{4\left(k\right)}}{v}_q^{\prime }\left(1\right)-{\displaystyle \frac{(E-F)\left|\lambda \right|(-b)}{4\left(k\right)}}.$$

(31)

The upper bound of the last expression is ${\displaystyle \frac{(E-F)\left|\lambda \right|(b/4)}{k}}$ if the inequality in (29) is met. □

For $a=1$ and $b=-1$, we can improve the assertion of Theorem 5 as follows.

Corollary 5.

Let $m>0$. Then, a sufficient condition for ${\mathcal{H}}_1^{\ast }\left(z\right)\in {\mathcal{R}}^{\lambda }(E,F)$ is that the inequality

$$v_{q+1}\left(1\right)\le 2$$

holds well.

Theorem 6.

Let $k>0$ and $f\in {\mathcal{R}}^{\lambda }(E,F)$. A sufficient condition for the Bessel operator ${\mathcal{H}}_1\left(z\right)\in {\mathcal{S}}_2$ is that the Bessel’s inequality

$$v_q^{\prime }\left(1\right)\le {\displaystyle \frac{3}{32}}{\displaystyle \frac{b^2}{{\left(k\right)}_2}}$$

(32)

is satisfied.

Proof. 

By Lemma 3, to prove that ${\mathcal{H}}_1\left(z\right)$ belongs to ${\mathcal{S}}_2$, it is sufficient to show that the condition

$$\sum _{s=3}^{\infty }s(s-1)(s-2)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|\le 3!{\displaystyle \frac{{(E-F)\left|\lambda \right|(-b/4)}^2}{4{\left(k\right)}_2}}$$

(33)

holds. By using Lemma 5, the left-hand side can now be represented in the following manner:

$$\begin{array}{cc}\hfill \sum _{s=3}^{\infty }s(s-1)(s-2)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|& =\sum _{s=3}^{\infty }s(s-1)(s-2)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}\right|\left|d_s\right|\hfill \\ \hfill & \le \sum _{s=3}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{\displaystyle \frac{(E-F)\left|\lambda \right|}{s}}\hfill \\ \hfill & =(E-F)\left|\lambda \right|\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}\hfill \\ \hfill \mathrm{Noting}\mathrm{that}{\left(a\right)}_s=a{(a+1)}_{s-1}\\ \hfill \sum _{s=3}^{\infty }s(s-1)\left|{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-1}}}{d}_s\right|& \le {\displaystyle \frac{(E-F)\left|\lambda \right|b^2}{16{\left(k\right)}_2}}\left(\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-3}}{{(k+2)}_{s-3}{\left(1\right)}_{s-3}}}\right)\hfill \\ \hfill & =(E-F)\left|\lambda \right|v_q^{″}.\hfill \end{array}$$

The upper bound of the last expression is $3!{\displaystyle \frac{{(E-F)\left|\lambda \right|(-b/4)}^2}{4{\left(k\right)}_2}}$ if the inequality in (32) is met. □

An Integral Operator: The next theorem yields similar conclusions in relation to a specific integral operator $\mathcal{G}(k,b,z)$ and is defined as follows:

$$\mathcal{G}(k,b,z)={\int }_0^z\left(2-v_q\left(t\right)\right)dt.$$

(34)

In terms of the power series expansion, we can express $\mathcal{G}(k,b,z)$ as

$$\mathcal{G}(k,b,z)=z-\sum _{s=2}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_s}}{z}^s.$$

(35)

Theorem 7.

If $b<0$ and $k>0$, then, a necessary condition for $\mathcal{G}(k,b,z)$ to belong to ${\mathcal{T}}_1$ is that the inequality

$$v_q^{\prime }\left(1\right)\le -{\displaystyle \frac{b}{2k}}$$

(36)

holds well. The condition (36) is also sufficient if $0\le -b\le {\displaystyle \frac{4k}{3}}$.

Proof. 

In order to verify that $\mathcal{G}(k,b,z)$ is an element of ${\mathcal{T}}_1$ as stated in Lemma 2, it suffices to show that the inequality below is satisfied:

$$\sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_s}}\le {\displaystyle \frac{\left(-\frac{b}{4}\right)}{k}}$$

(37)

$$\begin{array}{cc}\hfill \sum _{s=3}^{\infty }s(s-1){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_s}}& =\sum _{s=3}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-2}}}\hfill \\ & ={\displaystyle \frac{1}{k}}\sum _{s=1}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s+1}}{{(k+1)}_{s+1}{\left(1\right)}_s}}\hfill \\ & ={\displaystyle \frac{-b}{4k}}\left[v_{q+1}\left(1\right)-1\right]\hfill \\ & =v_q^{\prime }\left(1\right)-{\displaystyle \frac{\left(-\frac{b}{4}\right)}{k}}.\hfill \end{array}$$

The upper bound of the last expression is ${\displaystyle \frac{-b}{4k}}$ which occurs if the inequality in (36) holds.

By virtue of Lemma 2, this condition (36) is also sufficient for $\mathcal{G}(k,b,z)$ to be in ${\mathcal{T}}_1$. □

For $a=1$ and $b=-1$, we can improve the assertion of Theorem 7 as follows.

Corollary 6.

If $m>0$ (with $m\ne 0,-1,-2,\dots$), then a necessary and sufficient condition for $\mathcal{G}(m,b,z)$ to belong to ${\mathcal{T}}_1$ is that inequality

$$v_{q+1}\left(1\right)\le 2$$

(38)

Theorem 8.

Let $b<0$ and $k>0$. A necessary condition for the integral operator $\mathcal{G}(k,b,z)$ to belong to ${\mathcal{T}}_2$ is that the inequality

$$v_q^{″}\left(1\right)\le {\displaystyle \frac{b^2}{8{\left(k\right)}_2}}$$

(39)

holds good.

Proof. 

In order to verify that $\mathcal{G}(k,b,z)$ is an element of ${\mathcal{T}}_2$ as stated in Lemma 2, it suffices to show that the inequality below is satisfied:

$$\sum _{s=4}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_s}}\le {\displaystyle \frac{{\left(-\frac{b}{4}\right)}^2}{{\left(k\right)}_2}}$$

(40)

$$\begin{array}{cc}\hfill \sum _{s=4}^{\infty }s(s-1)(s-2){\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_s}}& =\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{{\left(k\right)}_{s-1}{\left(1\right)}_{s-3}}}\hfill \\ & =\sum _{s=4}^{\infty }{\displaystyle \frac{{\left(\frac{-b}{4}\right)}^{s-1}}{k(k+1){(k+2)}_{s-3}{\left(1\right)}_{s-3}}}\hfill \\ & ={\displaystyle \frac{{\left(-\frac{b}{4}\right)}^2}{{\left(k\right)}_2}}\left[v_{q+2}\left(1\right)-1\right]\hfill \\ & =v_q^{″}\left(1\right)-{\displaystyle \frac{{\left(-\frac{b}{4}\right)}^2}{{\left(k\right)}_2}}\hfill \end{array}$$

The upper bound of the last expression is ${\displaystyle \frac{{\left(-\frac{b}{4}\right)}^2}{{\left(k\right)}_2}}$ which occurs if and only if the inequality in (39) is met. □

For $a=1$ and $b=-1$, we can improve the assertion of Theorem 8 as follows.

Corollary 7.

Let $m>0$. Then, a necessary condition for $\mathcal{G}(m,b,z)$ to belong to ${\mathcal{T}}_2$ is that inequality

$$v_{q+2}\left(1\right)\le 2$$

(41)

holds well.

3. Conclusions

We have established the necessary conditions for Bessel functions and a Bessel operator to belong to the classes ${\mathcal{S}}_1,$${\mathcal{T}}_1$ and ${\mathcal{S}}_2$. Notably, we have also determined the mapping properties of the integral operator $\mathcal{G}(k,b,z).$ The findings presented in this paper possess a universal nature and are anticipated to have various applications within the theory of special functions. Furthermore, it is possible to derive new results for the subclasses discussed in this paper in conjunction with specific probability distribution series and Mittag-Leffler functions. The sharp upper bound for the second coefficients of the functions whose derivatives are all univalent is found. Interestingly, there are no such functions in the classes ${\mathcal{S}}_1,$ ${\mathcal{S}}_M,$ ${\mathcal{T}}_1,$ and ${\mathcal{T}}_M$, whose second coefficient attains this sharp upper bound. The method of analysis for finding such sharp functions is still eluding.