New Results about Radius of Convexity and Uniform Convexity of Bessel Functions

Abstract

We determine in this paper new results about the radius of uniform convexity of two kinds of normalization of the Bessel function $J_{\nu }$ in the case $\nu \in (-2,-1)$, and provide an alternative proof regarding the radius of convexity of order alpha. We then compare results regarding the convexity and uniform convexity of the considered functions and determine interesting connections between them.

1. Introduction

Let $U\left(r\right)=\{z\in \mathbb{C}:|z|<r\}$ be the disk, centered at zero, of radius $r,$ where $r>0.$

We denote by $U\left(r\right)=U(0,r).$

We say that a function f of the form

$$f\left(z\right)=z+a_2{z}^2+\dots$$

(1)

is convex on $U\left(r\right)$ if and only if $f\left(U\right(r\left)\right)$ is a convex domain in the set $\mathbb{C}$ and the function f is univalent.

We know that the function f is convex on $U\left(r\right)$ if and only if

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)>0,z\in U\left(r\right).$$

We say that f is a convex function of order $\alpha$ on $U\left(r\right)$ if

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

The radius of convexity of order $\alpha$ for f is defined by the equality

$$\begin{array}{c}\hfill r_f^c\left(\alpha \right)=sup\left\{r\in (0,\infty ):Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)>\alpha ,z\in U\left(r\right)\right\}.\end{array}$$

(2)

We say that f is uniformly convex in the disk $U\left(r\right)$ if the function f has the form in (1), it is a convex function, and it has the property that the arc $f\left(\gamma \right)$ is convex for every circular arc $\gamma$ contained in the disk $U\left(r\right)$ with center $\zeta$, also in $U\left(r\right)$. The function f is uniformly convex in the disk $U\left(r\right)$ if and only if

$$\begin{array}{c}\hfill Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)>\left|\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right|,z\in U\left(r\right).\end{array}$$

We know that the radius of uniform convexity is defined by

$$\begin{array}{c}\hfill r_f^{uc}\left(\alpha \right)=sup\left\{r\in (0,\infty ):Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)>\left|\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right|,z\in U\left(r\right)\right\}.\end{array}$$

(3)

The Bessel function of the first kind is defined by

$$J_{\nu }\left(z\right)=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n!\mathsf{\Gamma }(n+\nu +1)}{\left(z/2\right)}^{2n+\nu }.$$

Consider the following normalized forms:

$$\begin{array}{c}\hfill g_{\nu }\left(z\right)=2^{\nu }\mathsf{\Gamma }(1+\nu )z^{1-\nu }{J}_{\nu }\left(z\right)=z-\frac{1}{4(\nu +1)}{z}^3+\dots ,\end{array}$$

(4)

and

$$\begin{array}{c}\hfill h_{\nu }\left(z\right)=2^{\nu }\mathsf{\Gamma }(1+\nu )z^{1-\nu /2}{J}_{\nu }\left(z^{\frac{1}{2}}\right)=z-\frac{1}{4(\nu +1)}{z}^2+\dots ,\end{array}$$

(5)

where $\nu$ is a real number and $-2<\nu <-1$, and $g_{\nu }$ and $h_{\nu }$ are entire functions.

This article can be considered a continuation of previous papers [1,2] which dealt with geometric properties of Bessel functions.

For more details about the geometric properties of Bessel functions, interested readers are referred to the following papers: [1,3,4,5,6,7,8,9,10,11,12,13].

The aim of this work is to determine the radius of convexity of order $\alpha$, $r_f^c\left(\alpha \right)$ for $f=g_{\nu }$ and $f=h_{\nu }$ and the radius of uniform convexity $r_f^{uc}\left(\alpha \right)$ for the case $\nu \in (-2,-1)$ and to derive an interesting connection between the convexity and uniform convexity.

In the next section, we provide several results which are necessary later in this work.

2. Preliminaries

Lemma 1

([14], p. 483, Hurwitz). If $\nu \in (-2,-1),$ then $J_{\nu }\left(z\right)$ has exactly two purely imaginary conjugate complex zeros, and all the other zeros are real.

The zeros $z^{-\nu }{J}_{\nu }\left(z\right)$ are taken to be $\pm j_{\nu ,n}$, where $n\in {\mathbb{N}}^{*}=\{1,2,3,\dots \}.$ We may suppose, without restricting the generality, that $j_{\nu ,1}=ia,a>0,$ and $0<a<j_{\nu ,2}<j_{\nu ,3}<\cdots <j_{\nu ,n}<\cdots .$

Lemma 2

([14], p. 502). The following equality holds

$$\sum _{n=1}^{\infty }\frac{1}{j_{\nu ,n}^2}=\frac{1}{4(\nu +1)}.$$

(6)

Lemma 3

([8]). In the notations of Lemma 2, we have

$$\frac{z{g}_{\nu }^{\prime }\left(z\right)}{g_{\nu }\left(z\right)}=1-2\sum _{n=1}^{\infty }\frac{z^2}{j_{\nu ,n}^2-z^2},$$

(7)

and

$$\frac{z{h}_{\nu }^{\prime }\left(z\right)}{h_{\nu }\left(z\right)}=1-\sum _{n=1}^{\infty }\frac{z}{j_{\nu ,n}^2-z}.$$

(8)

The series are uniformly convergent on every compact subset of $\mathbb{C}\backslash \{\pm j_{\nu ,n}:n\in {\mathbb{N}}^{*}\}.$

Lemma 4

([9]). If $v\in \mathbb{C}$, $\delta \in \mathbb{R}$, and $\delta >\rho \ge \left|v\right|$, then

$$\begin{array}{c}\hfill \left|\frac{v}{\delta -v}\right|\le \frac{\rho }{\delta -\rho }\mathrm{and}\left|\frac{v}{{(\delta -v)}^2}\right|\le \frac{\rho }{{(\delta -\rho )}^2}.\end{array}$$

Proof. 

The following implications hold

$$|\delta -v|\ge \delta -\rho \Rightarrow \frac{1}{|\delta -v|}\le \frac{1}{\delta -\rho }\Rightarrow \left|\frac{1}{{(\delta -v)}^2}\right|\le \frac{1}{{(\delta -\rho )}^2}.$$

If the last two inequalities are multiplied by the inequality $\left|v\right|\le \rho ,$ we obtain the desired results. □

Lemma 5.

If $v\in \mathbb{C}$, $\delta ,\gamma \in \mathbb{R},$$\gamma \ge \delta >\rho \ge \left|v\right|$, then

$$\left|\frac{v^2}{(\delta +v)(\gamma -v)}\right|\le \frac{{\rho }^2}{(\delta -\rho )(\gamma +\rho )}.$$

(9)

Proof. 

We can prove the second inequality of the following equivalence:

$$\left|\frac{1}{(\delta +v)(\gamma -v)}\right|\le \frac{1}{(\delta -\rho )(\gamma +\rho )}\iff (\delta -\rho )(\gamma +\rho )\le \left|(\delta +v)(\gamma -v)\right|,$$

(10)

where $\gamma \ge \delta >\rho \ge \left|v\right|$.

We prove the inequality (10) in two steps.

Let $v=x+iy$; then, it is obvious that

$$\left|(\delta +v)(\gamma -v)\right|=\sqrt{[{(\delta +x)}^2+y^2]\left[{(+y^2+\gamma -x)}^2\right]}\ge \left|(\gamma -x)(\delta +x)\right|,$$

(11)

where $\gamma \ge \delta >\rho \ge \sqrt{x^2+y^2}$.

On the other hand, a simple calculation results in

$$(\delta +x)(\gamma -x)\ge (\delta -\rho )(\gamma +\rho ),x\in [-\rho ,\rho ].$$

(12)

It is easily seen that (11) and (12) imply the second inequality of (10). Finally, multiplying the inequality ${\rho }^2\ge {\left|v\right|}^2$ by the first inequality of (10), we obtain (9) and the proof is complete. □

Lemma 6.

If $v\in \mathbb{C}$, $\delta ,\gamma \in \mathbb{R}$, and $\gamma \ge \delta >\rho \ge \left|v\right|$, then

$$\left|\frac{2v^2[2\gamma \delta +(\gamma -\delta )v]}{{(\gamma -v)}^2{(\delta +v)}^2}\right|\le \frac{2r^2[2\gamma \delta -(\gamma -\delta )\rho ]}{{(\gamma +\rho )}^2{(\delta -\rho )}^2}.$$

(13)

Proof. 

The inequality obviously holds provided that $\gamma =\delta$ (see (10)), thus, we have to prove it in the case that $\gamma >\delta .$

We can then prove the following inequality:

$$\left|\frac{2\gamma \delta +(\gamma -\delta )v}{(\delta +v)(\gamma -v)}\right|\le \frac{2\gamma \delta -(\gamma -\delta )\rho }{(\delta -\rho )(\gamma +\rho )},\gamma \ge \delta >\rho \ge \left|v\right|.$$

(14)

We define $z=x+iy$ and define the mapping

$$\varphi :[-\rho ,\rho ]\to \mathbb{R},\varphi \left(y\right)=\frac{{(\omega +x)}^2+y^2}{[{(\delta +x)}^2+y^2][{(\gamma -x)}^2+y^2]},\omega =\frac{2\gamma \delta }{\gamma -\delta }.$$

Then, we have

$${\varphi }^{\prime }\left(y\right)=2y\frac{[{(\delta +x)}^2+y^2][{(\gamma -x)}^2+y^2]-[{(\delta +x)}^2+{(\gamma -x)}^2+2y^2][{(\omega +x)}^2+y^2]}{{[{(\delta +x)}^2+y^2]}^2{[{(\gamma -x)}^2+y^2]}^2}.$$

As ${\varphi }^{\prime }\left(y\right)<0,y\in (0,\rho )$ and ${\varphi }^{\prime }\left(y\right)>0,y\in (-\rho ,0)$, it follows that

$$\varphi \left(y\right)\le \varphi \left(0\right)=\frac{{(\omega +x)}^2}{\left[{(\delta +x)}^2\right]\left[{(\gamma -x)}^2\right]},y\in [-\rho ,\rho ].$$

(15)

We can determine the maximum of the function

$$\phi :[-\rho ,\rho ]\to \mathbb{R},\phi \left(x\right)=\frac{\omega +x}{(\delta +x)(\gamma -x)}.$$

We have

$${\phi }^{\prime }\left(x\right)=\frac{x^2+2\omega x-\gamma \delta }{{(\delta +x)}^2{(\gamma -x)}^2}$$

The derivative ${\phi }^{\prime }\left(x\right)=0$ has one positive root, $x_1=\sqrt{{\omega }^2+\gamma \delta }-\omega$, and one negative root, $x_2=-\sqrt{{\omega }^2+\gamma \delta }-\omega .$ As $x_2<-r$ and $x_1\in (-\rho ,\rho ),$ it follows the inequality

$$\begin{array}{c}\hfill \frac{\omega +x}{(\delta +x)(\gamma -x)}=\phi \left(x\right)\le max\{\phi (-\rho ),\phi \left(\rho \right)\}=\phi (-\rho )=\frac{\omega -\rho }{(\delta -\rho )(\gamma +\rho )}\end{array}$$

(16)

for every $x\in [-\rho ,\rho ].$ From (15) and (16), we have (14). Finally, multiplying the inequalities (14), $|v^2|\le {\rho }^2$ and the first inequality of (10), we infer (13). □

Lemma 7.

If the functions $g_{\nu }$ and $h_{\nu }$ are defined by (4) and (5), respectively, then

$$\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}=z\frac{z{J}_{\nu +2}\left(z\right)-3J_{\nu +1}\left(z\right)}{J_{\nu }\left(z\right)-z{J}_{\nu +1}\left(z\right)}.$$

(17)

$$\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}=\frac{z{J}_{\nu +2}\left(z^{\frac{1}{2}}\right)-4z^{\frac{1}{2}}{J}_{\nu +1}\left(z^{\frac{1}{2}}\right)}{4J_{\nu }\left(z^{\frac{1}{2}}\right)-2z^{\frac{1}{2}}{J}_{\nu +1}\left(z^{\frac{1}{2}}\right)}.$$

(18)

Proof. 

We differentiate the equality (4), and at the second time we differentiate it logarithmically. After multiplying by z, we obtain the following equality:

$$\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}=\frac{z^2{J}_{\nu }^{″}\left(z\right)+2z(1-\nu )J_{\nu }^{\prime }\left(z\right)+\nu (\nu -1)J_{\nu }\left(z\right)}{z{J}_{\nu }^{\prime }\left(z\right)+(1-\nu )J_{\nu }\left(z\right)}.$$

The function $J_{\nu }$ is a solution of the Bessel differential equation; thus, we can replace the function $z^2{J}_{\nu }^{″}$ using the equality $z^2{J}_{\nu }^{″}\left(z\right)=({\nu }^2-z^2)J_{\nu }\left(z\right)-z{J}_{\nu }^{\prime }\left(z\right)$, and it follows that

$$\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}=\frac{z(1-2\nu )J_{\nu }^{\prime }\left(z\right)+(2{\nu }^2-\nu -z^2)J_{\nu }\left(z\right)}{z{J}_{\nu }^{\prime }\left(z\right)+(1-\nu )J_{\nu }\left(z\right)}.$$

In the second step, we use the following well-known equality: $z{J}_{\nu }^{\prime }\left(z\right)=\nu J_{\nu }\left(z\right)-z{J}_{\nu +1}\left(z\right)$, and infer

$$\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}=\frac{z(2\nu -1)J_{\nu +1}\left(z\right)-z^2{J}_{\nu }\left(z\right)}{J_{\nu }\left(z\right)-z{J}_{\nu +1}\left(z\right)}.$$

Finally, we replace $z{J}_{\nu }\left(z\right)$ in the numerator by $z{J}_{\nu }\left(z\right)=2(\nu +1)J_{\nu +1}\left(z\right)-z{J}_{\nu +2}\left(z\right),$ and obtain (17).

We differentiate equality (5) twice, similarly to the case of the function $g_{\nu },$ and obtain

$$\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}=\frac{\nu (\nu -2)J_{\nu }\left(z^{\frac{1}{2}}\right)+(3-2\nu )z^{\frac{1}{2}}{J}_{\nu }^{\prime }\left(z^{\frac{1}{2}}\right)+z{J}_{\nu }^{″}\left(z^{\frac{1}{2}}\right)}{2(2-\nu )J_{\nu }\left(z^{\frac{1}{2}}\right)+2z^{\frac{1}{2}}{J}_{\nu }^{\prime }\left(z^{\frac{1}{2}}\right)}.$$

We use the equality $z{J}_{\nu }^{″}\left(z^{\frac{1}{2}}\right)=({\nu }^2-z)J_{\nu }\left(z^{\frac{1}{2}}\right)-z^{\frac{1}{2}}{J}_{\nu }^{\prime }\left(z^{\frac{1}{2}}\right),$ and obtain

$$\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}=\frac{(2{\nu }^2-2\nu -z)J_{\nu }\left(z^{\frac{1}{2}}\right)+(2-2\nu )z^{\frac{1}{2}}{J}_{\nu }^{\prime }\left(z^{\frac{1}{2}}\right)}{2(2-\nu )J_{\nu }\left(z^{\frac{1}{2}}\right)+2z^{\frac{1}{2}}{J}_{\nu }^{\prime }\left(z^{\frac{1}{2}}\right)}.$$

Now, using the equality $z^{\frac{1}{2}}{J}_{\nu }^{\prime }\left(z^{\frac{1}{2}}\right)=\nu J_{\nu }\left(z^{\frac{1}{2}}\right)-z^{\frac{1}{2}}{J}_{\nu +1}\left(z^{\frac{1}{2}}\right),$ we infer

$$\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}=\frac{(2\nu -2)z^{\frac{1}{2}}{J}_{\nu +1}\left(z^{\frac{1}{2}}\right)-z{J}_{\nu }\left(z^{\frac{1}{2}}\right)}{4J_{\nu }\left(z^{\frac{1}{2}}\right)-2z^{\frac{1}{2}}{J}_{\nu +1}\left(z^{\frac{1}{2}}\right)},$$

and combining this with the equality $z^{\frac{1}{2}}{J}_{\nu }\left(z^{\frac{1}{2}}\right)=2(\nu +1)J_{\nu +1}\left(z^{\frac{1}{2}}\right)-z^{\frac{1}{2}}{J}_{\nu +2}\left(z^{\frac{1}{2}}\right)$, (18) follows. □

3. Main Results

Theorem 1.

If $\alpha \in [0,1)$ and $\nu \in (-2,-1)$, then the radius of convexity of order α for the mapping $g_{\nu }$ is $r_{\nu }^c\left(\alpha \right)=r_1,$ where $r_1$ is the unique root of the equation

$$1+r\frac{I_{\nu +2}\left(r\right)+3I_{\nu +1}\left(r\right)}{I_{\nu +1}\left(r\right)+r{I}_{\nu }\left(r\right)}=\alpha$$

(19)

in the interval $(0,a).$

Proof. 

According to the proof of Theorem 1 [2], the equalities

$$\frac{z{g}_{\nu }^{\prime }\left(z\right)}{g_{\nu }\left(z\right)}=1-2\sum _{n=1}^{\infty }\frac{z^2}{j_{\nu ,n}^2-z^2},\sum _{n=1}^{\infty }\frac{1}{j_{\nu ,n}^2}=\frac{1}{4(\nu +1)}.$$

imply

$$\frac{z{g}_{\nu }^{\prime }\left(z\right)}{g_{\nu }\left(z\right)}=1-\frac{a^2}{2(1+\nu )}\frac{z^2}{a^2+z^2}-2\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}.$$

The logarithmic differentation of this equality leads to

$$\begin{array}{c}\hfill 1+\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}=\\ \hfill 1-\frac{a^2}{2(1+\nu )}\frac{z^2}{a^2+z^2}-2\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}-\\ \hfill \frac{\frac{a^2}{1+\nu }\frac{a^2{z}^2}{{(a^2+z^2)}^2}+2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{2z^4[2a^2{j}_{\nu ,n}^2+z^2(j_{\nu ,n}^2-a^2)]}{{(a^2+z^2)}^2{(j_{\nu ,n}^2-z^2)}^2}}{1-\frac{a^2}{2(1+\nu )}\frac{z^2}{a^2+z^2}-2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}}.\end{array}$$

(20)

It is proven in Theorem 1 [2] that the radius of starlikeness, $r_{g_{\nu }}^{*}$, for the function $g_{\nu }$ is the smallest root of the equation

$$\begin{array}{c}\hfill 1+\frac{a^2}{2(1+\nu )}\frac{r^2}{a^2-r^2}\\ \hfill -2\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{r^4}{(a^2-r^2)(j_{\nu ,n}^2+r^2)}=ir\frac{g_{\nu }^{\prime }\left(ir\right)}{g_{\nu }\left(ir\right)}=0,\end{array}$$

in the interval $(0,a).$ Thus, we have

$$0<r_{g_{\nu }}^{*}<a<j_{\nu ,2}<j_{\nu ,3}<\cdots <j_{\nu ,n}<\cdots .$$

(21)

Taking into account that $\nu +1<0,$ the equality (20) implies the following inequality:

$$\begin{array}{c}\hfill Re\left(1+\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right)\ge \\ \hfill 1+\frac{a^2}{2(1+\nu )}\left|\frac{z^2}{a^2+z^2}\right|-2\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\left|\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}\right|-\\ \hfill \frac{-\frac{a^2}{1+\nu }\left|\frac{a^2{z}^2}{{(a^2+z^2)}^2}\right|+2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\left|\frac{2z^4[2a^2{j}_{\nu ,n}^2+z^2(j_{\nu ,n}^2-a^2)]}{{(a^2+z^2)}^2{(j_{\nu ,n}^2-z^2)}^2}\right|}{1+\frac{a^2}{2(1+\nu )}\left|\frac{z^2}{a^2+z^2}\right|-2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\left|\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}\right|}\end{array}$$

(22)

for every $z\in U\left(r_{\nu }^{*}\right).$

Using $\delta =a^2,$ $\rho =r^2$ and $v=z^2$ in Lemma 4, we obtain

$$\begin{array}{c}\hfill \frac{a^2}{2(1+\nu )}\left|\frac{z^2}{a^2+z^2}\right|\ge \frac{a^2}{2(1+\nu )}\frac{r^2}{a^2-r^2}\mathrm{and}\frac{a^2}{2(1+\nu )}\left|\frac{z^2}{{(a^2+z^2)}^2}\right|\ge \\ \hfill \frac{a^2}{{(a^2-r^2)}^2}\frac{r^2}{2(1+\nu )}.\end{array}$$

(23)

In a similar manner, Lemma 5 and Lemma 6 imply that

$$\begin{array}{c}\hfill \left|\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}\right|\le \frac{r^4}{(a^2-r^2)(j_{\nu ,n}^2+r^2)}\\ \hfill \left|\frac{2z^4[2a^2{j}_{\nu ,n}^2+z^2(j_{\nu ,n}^2-a^2)]}{{(a^2+z^2)}^2{(j_{\nu ,n}^2-z^2)}^2}\right|\le \frac{2r^4[2a^2{j}_{\nu ,n}^2-r^2(j_{\nu ,n}^2-a^2)]}{{(a^2-r^2)}^2{(j_{\nu ,n}^2+r^2)}^2}.\end{array}$$

(24)

Now, inequalities (22)–(24) imply the following inequality:

$$\begin{array}{c}\hfill Re\left(1+\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right)\ge \\ \hfill 1+\frac{a^2}{2(1+\nu )}\frac{r^2}{a^2-r^2}-2\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{r^4}{(a^2-r^2)(j_{\nu ,n}^2+r^2)}-\\ \hfill \frac{-\frac{a^2}{1+\nu }\frac{a^2{r}^2}{{(a^2-r^2)}^2}+2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{2r^4[2a^2{j}_{\nu ,n}^2-r^2(j_{\nu ,n}^2-a^2)]}{{(a^2-r^2)}^2{(j_{\nu ,n}^2+r^2)}^2}}{1+\frac{a^2}{2(1+\nu )}\frac{z^2}{a^2-r^2}-2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{r^4}{(a^2-r^2)(j_{\nu ,n}^2+r^2)}}=1+\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)}=\mathsf{\Phi }\left(r\right),\end{array}$$

(25)

provided that $a>r_{g_{\nu }}^{*}>\left|z\right|,$ where $r_{g_{\nu }}^{*}$ verifies the inequalities (21).

The following equalities hold: $\mathsf{\Phi }\left(0\right)=1$ and ${lim}_{r\nearrow r_{g_{\nu }}^{*}}\mathsf{\Phi }\left(r\right)=-\infty .$ Consequently, equation $1+\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)}=\alpha$ has a real root in the interval $(0,r_{g_{\nu }}^{*}).$ The smallest positive real root of the equation $1+\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)}=\alpha$ is denoted by $r_{g_{\nu }}^c\left(\alpha \right),$ and this root is the radius of convexity of order $\alpha$ of the function $g_{\nu }.$ The first equality of Lemma 7 and the equality $J_{\nu }\left(iz\right)=i^{\nu }{I}_{\nu }\left(z\right)$ imply that the equation $1+\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)}=\alpha$ is equivalent to (19).

□

We determine the radius of uniform convexity of the mapping $g_{\nu }$ in the next theorem.

Theorem 2.

If $\nu \in (-2,-1),$ then the radius of uniform convexity for the mapping $g_{\nu }$ is $r_{\nu }^{*}\left(\alpha \right)=r_2,$ where $r_2$ is the smallest positive root of the equation

$$\frac{1}{2}+r\frac{I_{\nu +2}\left(r\right)+3I_{\nu +1}\left(r\right)}{I_{\nu +1}\left(r\right)+r{I}_{\nu }\left(r\right)}=0$$

(26)

in the interval $(0,r_{\nu }^{*}).$

Proof. 

Equality (20) implies the following inequality:

$$\begin{array}{c}\hfill \left|\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right|\le \\ \hfill -\frac{a^2}{2(1+\nu )}\left|\frac{z^2}{a^2+z^2}\right|+2\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\left|\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}\right|+\\ \hfill \frac{-\frac{a^2}{1+\nu }\left|\frac{a^2{z}^2}{{(a^2+z^2)}^2}\right|+2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\left|\frac{2z^4[2a^2{j}_{\nu ,n}^2+z^2(j_{\nu ,n}^2-a^2)]}{{(a^2+z^2)}^2{(j_{\nu ,n}^2-z^2)}^2}\right|}{1+\frac{a^2}{2(1+\nu )}\left|\frac{z^2}{a^2+z^2}\right|-2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\left|\frac{z^4}{(a^2+z^2)(j_{\nu ,n}^2-z^2)}\right|}.\end{array}$$

(27)

We can again use inequalities (22) and (23), and in combination with (27), we have

$$\begin{array}{c}\hfill \left|\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right|\le -\frac{a^2}{2(1+\nu )}\frac{r^2}{a^2-r^2}+2\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{r^4}{(a^2-r^2)(j_{\nu ,n}^2+r^2)}+\\ \hfill \frac{-\frac{a^2}{1+\nu }\frac{a^2{r}^2}{{(a^2-r^2)}^2}+2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{2r^4[2a^2{j}_{\nu ,n}^2-r^2(j_{\nu ,n}^2-a^2)]}{{(a^2-r^2)}^2{(j_{\nu ,n}^2+r^2)}^2}}{1+\frac{a^2}{2(1+\nu )}\frac{z^2}{a^2-r^2}-2{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}\frac{r^4}{(a^2-r^2)(j_{\nu ,n}^2+r^2)}}=-\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)}.\end{array}$$

Inequalities (25) and (27) imply

$$\begin{array}{c}\hfill Re\left(1+\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right)-\left|\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right|\ge 1+2\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)},z\in U\left(r_{\nu }^{*}\right).\end{array}$$

(28)

The smallest positive root of the equation $1+2\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)}=0$ in the interval $(0,r_{\nu }^{*})$ is denoted by $r_{\nu }^{uc}.$ According to (28), the value $r_{\nu }^{uc}$ is the biggest with the property that

$$\begin{array}{c}\hfill Re\left(1+\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right)-\left|\frac{z{g}_{\nu }^{″}\left(z\right)}{g_{\nu }^{\prime }\left(z\right)}\right|>0,z\in U\left(r_{\nu }^{uc}\right).\end{array}$$

Lemma 7 and the equality $J_{\nu }\left(iz\right)=i^{\nu }{I}_{\nu }\left(z\right)$ imply that the equation $1+2\frac{ir{g}_{\nu }^{″}\left(ir\right)}{g_{\nu }^{\prime }\left(ir\right)}=0$ is equivalent to (26), completing the proof. □

Theorems 1 and 2 imply the following result.

Corollary 1.

The mapping $g_{\nu }$ is uniformly convex in the disk $U\left(r\right)$ if and only if it is convex of order $\frac{1}{2}.$

Theorem 3.

If $\alpha \in [0,1)$ and $\nu \in (-2,-1),$ then the radius of convexity of order α for the mapping $h_{\nu }$ is $r_{h_{\nu }}^c\left(\alpha \right)=r_3,$ where $r_3$ is the smallest real root of the equation

$$1+\frac{r{I}_{\nu +2}\left(r^{\frac{1}{2}}\right)+4r^{\frac{1}{2}}{I}_{\nu +1}\left(r^{\frac{1}{2}}\right)}{4I_{\nu }\left(r^{\frac{1}{2}}\right)+2r^{\frac{1}{2}}{I}_{\nu +1}\left(r^{\frac{1}{2}}\right)}=\alpha$$

(29)

in the interval $(0,r_{h_{\nu }}^{*}).$

Proof. 

According to the proof of Theorem 2 [2], the equalities

$$\frac{z{h}_{\nu }^{\prime }\left(z\right)}{h_{\nu }\left(z\right)}=1-\sum _{n=1}^{\infty }\frac{z}{j_{\nu ,n}^2-z},\sum _{n=1}^{\infty }\frac{1}{j_{\nu ,n}^2}=\frac{1}{4(\nu +1)}$$

imply

$$\frac{z{h}_{\nu }^{\prime }\left(z\right)}{h_{\nu }\left(z\right)}=1-\frac{a^2}{4(\nu +1)}·\frac{z}{a^2+z}-\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{z^2}{\left(a^2+z\right)\left(j_{\nu ,n}^2-z\right)},$$

where $z\in U(0,r).$

The logarithmic differentiation of the equality leads to

$$1+\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}=1-\frac{a^2}{4(\nu +1)}·\frac{z}{a^2+z}-\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{z^2}{\left(a^2+z\right)\left(j_{\nu ,n}^2-z\right)}-$$

$$-\frac{\frac{a^2}{4(1+\nu )}·\frac{a^{2}z}{{\left(a^2+z\right)}^2}+{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{z^2\left[2a^2{j}_{\nu ,n}^2+z\left(j_{\nu ,n}^2-a^2\right)\right]}{{\left(j_{\nu ,n}^2-z\right)}^2{\left(a^2+z\right)}^2}}{1-\frac{a^2}{4a^2+z}·\frac{z}{(\nu +1)}-{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{z^2}{\left(j_{\nu ,n}^2-z\right)\left(a^2+z\right)}}.$$

(30)

It is proven in [2] that the radius of starlikeness, $r_{h_{\nu }}^{*}$, for function $h_{\nu }$ is the smallest root of the equation

$$\frac{-r{h}_{\nu }^{\prime }(-r)}{h_{\nu }(-r)}=0,r\in \left(0,a^2\right),z\in U(0,r).$$

However,

$$\frac{-r{h}_{\nu }^{\prime }(-r)}{h_{\nu }(-r)}=1+\frac{a^2}{4(\nu +1)}·\frac{r}{a^2-r}-$$

$$-\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{r^2}{\left(a^2-r\right)\left(j_{\nu ,n}^2+r\right)}=0,r\in \left(0,a^2\right).$$

Taking into the account that $\nu +1<0,$ we obtain from relation (30)

$$Re\left(1+\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right)\ge \frac{a^2}{4(\nu +1)}·\left|\frac{z}{a^2+z}\right|-\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\left|\frac{z^2}{\left(a^2+z\right)\left(j_{\nu ,n}^2-z\right)}\right|-$$

$$-\frac{\frac{-a^2}{4(\nu +1)}·\left|\frac{a^{2}z}{{\left(a^2+z\right)}^2}\right|+{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\left|\frac{z^2\left[2a^2{j}_{\nu ,n}^2+z\left(j_{\nu ,n}^2-a^2\right)\right]}{{\left(a^2+z\right)}^2{\left(j_{\nu ,n}^2-z\right)}^2}\right|}{1+\frac{a^2}{4(\nu +1)}·\left|\frac{z}{a^2+z}\right|-{\sum }_n^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\left|\frac{z^2}{\left(a^2+z\right)\left(j_{\nu ,n}^2-z\right)}\right|}$$

(31)

and $z\in U(0,r),r\in \left(0,r_{h_{\nu }}^{*}\right)$. We obtain from Lemmas 4 and 5 the following inequality:

$$Re\left(1+\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right)\ge 1+\frac{a^2}{4(\nu +1)}·\frac{r}{a^2-r}-\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{r^2}{\left(a^2-r\right)\left(a^2+r\right)}-$$

$$-\frac{-\frac{a^2}{4(\nu +1)}·\frac{a^{2}r}{{\left(a^2-r\right)}^2}+{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{r^2\left[2a^2{j}_{\nu ,n}^2-r\left(j_{\nu ,n}^2-a^2\right)\right]}{{\left(a^2-r\right)}^2{\left(j_{\nu ,n}^2+r\right)}^2}}{1+\frac{a^2}{4(\nu +1)}·\frac{r}{a^2-r}-{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{r^2}{\left(a^2-r\right)\left(j_{\nu ,n}^2-r\right)}}=$$

(32)

$$=1-\frac{r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)}=\psi \left(r\right),a>r_{h_{\nu }}^{*}>\left|z\right|,$$

similarly to the proof of Theorem 1. The mapping

$$\psi :\left(0,r_{h_{\nu }}^{*}\right)\to \mathbb{R},\psi \left(r\right)=1+\frac{-r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)},$$

is strictly decreasing, and $a>r_{h_{\nu }}^{*}>\left|z\right|.$

We then have ${\displaystyle \underset{r\nearrow r_{h_{\nu }}^{*}}{lim}}\psi \left(r\right)=-\infty ,$ $\psi \left(0\right)=1,$ and the equation

$$1+\frac{-r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)}=\alpha$$

has at least one real root in the interval $\left(0,r_{h_{\nu }}^{*}\right).$

The smallest positive real root of the equation $1-\frac{r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)}=\alpha$ is denoted by $r_{h_{\nu }}^c\left(\alpha \right),$ and this root is the radius of convexity of order $\alpha$ of the function $h_{\nu }.$ The second equality of Lemma 7 and the equality $J_{\nu }\left(iz\right)=i^{\nu }{I}_{\nu }\left(z\right)$ imply that the equation $1-\frac{r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)}=\alpha$ is equivalent to (29). □

Theorem 4.

If $\alpha \in [0,1)$ and $\nu \in (-2,-1),$ then the radius of uniform convexity of $h_{\nu }$ is $r_{h_{\nu }}^{*}\left(\alpha \right)=r_4,$ where $r_4$ is the smallest positive root of the equation

$$\frac{r{I}_{\nu +2}\left(r^{\frac{1}{2}}\right)+4r^{\frac{1}{2}}{I}_{\nu +1}\left(r^{\frac{1}{2}}\right)}{4I_{\nu }\left(r^{\frac{1}{2}}\right)+2r^{\frac{1}{2}}{I}_{\nu +1}\left(r^{\frac{1}{2}}\right)}=\frac{1}{2}$$

(33)

in the interval $(0,r_{h_{\nu }}^{*}).$

Proof. 

Equality (30) implies the following inequality:

$$\left|\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right|\le -\frac{a^2}{4(\nu +1)}·\left|\frac{z}{a^2+z}\right|+\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\left|\frac{z^2}{\left(a^2+z\right)\left(j_{\nu ,n}^2-z\right)}\right|+$$

$$+\frac{\frac{-a^2}{4(\nu +1)}·\left|\frac{a^{2}z}{{\left(a^2+z\right)}^2}\right|+{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\left|\frac{z^2\left[2a^2{j}_{\nu ,n}^2+z\left(j_{\nu ,n}^2-a^2\right)\right]}{{\left(a^2+z\right)}^2{\left(j_{\nu ,n}^2-z\right)}^2}\right|}{1+\frac{a^2}{4(\nu +1)}·\left|\frac{z}{a^2+z}\right|-{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\left|\frac{z^2}{\left(a^2+z\right)\left(j_{\nu ,n}^2-z\right)}\right|}.$$

(34)

We obtain the following from the relation (31), Lemma 4, and the relation (34):

$$\left|\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right|\le -\frac{a^2}{4(\nu +1)}·\frac{r}{a^2-r}+\sum _{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{r^2}{\left(a^2-r\right)\left(j_{\nu ,n}^2+r\right)}+$$

$$+\frac{\frac{a^2}{4(\nu +1)}·\frac{a^{2}r}{{\left(a^2-r\right)}^2}+{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{r^2\left[2a^2{j}_{\nu ,n}^2-r\left(j_{\nu ,n}^2-a^2\right)\right]}{{\left(a^2-r\right)}^2{\left(j_{\nu ,n}^2+r\right)}^2}}{1-\frac{a^2}{4(a^2-r)}·\frac{r}{\nu +1}-{\sum }_{n=2}^{\infty }\frac{a^2+j_{\nu ,n}^2}{j_{\nu ,n}^2}·\frac{r^2}{\left(a^2-r\right)\left(j_{\nu ,n}^2+r\right)}}=\frac{r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)},\left|z\right|\le r<a^2.$$

Inequalities (32) and (34) imply

$$\begin{array}{c}\hfill Re\left(1+\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right)-\left|\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right|\ge 1-\frac{2r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)},z\in U\left(r_{h_{\nu }}^{*}\right).\end{array}$$

(35)

The smallest positive root of the equation $1-\frac{2r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)}=0$ in the interval $(0,r_{h_{\nu }}^{*})$ is denoted by $r_{h_{\nu }}^{uc}.$

According to (35), the value $r_{h_{\nu }}^{uc}$ is the biggest with the property that

$$Re\left(1+\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right)-\left|\frac{z{h}_{\nu }^{″}\left(z\right)}{h_{\nu }^{\prime }\left(z\right)}\right|>0,z\in U\left(r_{h_{\nu }}^{uc}\right).$$

The equation $1-\frac{2r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)}=0$ is equivalent to (33), completing the proof. Lemma 7 and the equality $J_{\nu }\left(iz\right)=i^{\nu }{I}_{\nu }\left(z\right)$ imply that the equation $1-\frac{2r{h}_{\nu }^{″}(-r)}{h_{\nu }^{\prime }(-r)}=0$ is equivalent to (33). □

From Theorems 3 and 4, we obtain the following corollary.

Corollary 2.

The function $h_{\nu }$ is uniformly convex in the disk $U\left(r\right)$ if and only if it is convex of order $\frac{1}{2}.$