Hermite-Hadamard type inequalities based on the Erdélyi-Kober fractional integrals

1.   Introduction

Let $\mathcal{A}$ be the class of functions $h$ of the form

where $h$ is analytic in $\mathbb{U} = \{z\in \mathbb{C}: |z| < 1\}$.

We denote $\mathcal{S}, \mathcal{S}^*$ and $\mathcal{K}$ the subclasses of $\mathcal{A}$ consisting of univalent, starlike and convex functions respectively (^[1,2]) and denote $\mathcal{P} = \{p:p(0) = 1, \mathrm{Re}p(z) > 0, z\in \mathbb{U}\}$.

An analytic function $s: \mathbb{U} = \{z:|z| < 1\}\rightarrow\mathbb{C}$ is subordinate to an analytic function $t: \mathbb{U}\rightarrow\mathbb{C}$, if there is a function $\nu$ satisfying $\nu(0) = 0\; \; \hbox{and}\; \; |\nu(z)| < 1\; \; (z\in\mathbb{U})$, such that $s(z) = t(\nu(z))(z\in\mathbb{U})$. Note that $s(z)\prec t(z)$. Especially, if $t$ is univalent in $\mathbb{U}$, then the following conclusion is true (see ^[1]):

In 1933, a classical Fekete-Szegö problem for $h(z) = z+\sum\limits_{n = 2}^\infty a_n z^n\in \mathcal{S}$ was introduced by Fekete and Szegö ^[3] as follows,

The result is sharp.

Using the subordination, the classes $\mathcal{S}^*(\phi)$ and $\mathcal{K}(\phi)$ of starlike and convex functions were defined by Ma and Minda ^[4] in 1994. The function $h(z)\in\mathcal{S}^*(\phi)$ iff $\frac{zh'(z)}{h(z)}\prec \phi(z)$ and $\frac{zh''(z)}{h'(z)}\prec \phi(z)$, where $h\in \mathcal{A}$ and $\phi\in \mathcal{P}$. Moreover, Fekete-Szegö problems of the classes were obtained by Ma and Minda ^[4]. The problem of Fekete-Szegö has always been a hot topic in geometry function theory. Many authors studied and obtained many results (see ^[5,6,7]).

Let $\phi(z) = \frac{1+Az}{1+Bz}$ and $-1\leq B < A\leq1$. The classes $\mathcal{S}^*(\phi)$ and $\mathcal{K}(\phi)$ reduce to $\mathcal{S}^*(\frac{1+Az}{1+Bz})$ and $\mathcal{K}(\frac{1+Az}{1+Bz})$, which are the classes of Janowski starlike and convex functions respectively (see ^[8]).

Without loss of generality, both $S^{\ast}(\frac{1+z}{1-z}) = S^{\ast}$ and $K(\frac{1+z}{1-z}) = K$ represent the well-known classes of starlike and convex function respectively.

In 2015, Mediratta et al. ^[9] introduced the family of exponential starlike functions $S^*(e^z)$, that is

or, equivalently

According to the properties of the exponential function $e^z$ and the subordination relationship, the class $S^*(e^z)$ maps the unit disc $\mathbb{U}$ onto a region, which is symmetric with respect to the real axis and 1.

In 1959, the class $\mathcal{S}_s^*$ of starlike functions with respect to symmetric points was introduced by Sakaguchi ^[10]. The function $h\in \mathcal{S}_s^*$ if and only if

In 1987, the classes $\mathcal{S}_c^*$ and $\mathcal{S}_{cs}^*$ of starlike functions with respect to conjugate points and symmetric conjugate points were introduced by El-Ashwa and Thomas ^[11] as follows,

For analytic functions $h(z)$ and $g(z)(z\in\mathbb{U})$. Let $S_H$ define the class of harmonic mappings with the following form (see ^[12,13])

where

In particular, $h$ is called the analytic part and $g$ is called the co-analytic part of $f$.

It is well known that the function $f = h+\overline{g}$ is locally univalent and sense preserving in $\mathbb{U}$ if and only if $|h'(z)| > |g'(z)|\; (z\in \mathbb{U})$(^[14]).

According to the above conclusion, the coefficient estimations, distortion theorems, integral expressions, Jacobi estimates and growth condition in geometric properties of covering theorem of the co-analytic part can be obtained by using the analytic part of harmonic functions. In the recent years, various subclasses of $S_{H}$ were researched by many authors as follows.

In 2007, the subclass of $S_{H}$ with $h\in \mathcal{K}$ was studied by Klimek and Michalski ^[15].

In 2014, the subclass of $S_{H}$ with $h\in \mathcal{S}$ was studied by Hotta and Michalski ^[16].

In 2015, the subclasses of $S_{H}$ with $h\in S^*(\frac{1+(1-2\beta)z}{1-z})$ and $h\in K(\frac{1+(1-2\beta)z}{1-z})$ were studied by Zhu and Huang ^[17].

In this paper, by using subordination relationship, we studied the subclasses of $S_H$ with $\frac{h(z)-\overline{h}(-\overline{z})}{2}\in S^*(e^z)$ and $\frac{h(z)-\overline{h}(-\overline{z})}{2}\in K(e^z)$.

Definition 1. Suppose $f = h+\overline{g}\in S_H$ of the form (1.3). Let $HS^{\ast, \alpha}_{sc}(e)$ denote the class of harmonic univalent exponential starlike functions with symmetric conjecture point consisting of $f$ with $h\in S_{sc}^{\ast}(e)$. That is, the function $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e)$ if and only if

Also, let $HK^{\alpha}_{sc}(e)$ denote the class of harmonic univalent convex exponential functions with symmetric conjecture point consisting of $f$ with $h\in K_{sc}(e)$, that is, $f = h+\overline{g}\in HK^{\alpha}_{sc}(e)$ if and only if

We know that $h(z)\in K_{sc}(e)\Longleftrightarrow zh'(z)\in S_{sc}^{\ast}(e)$.

2.   Preliminary preparation

In order to obtain our results, we need Lemmas as follows.

Lemma 1. (^[18]). Let $\omega(z) = c_{0}+c_{1}z+\ldots+c_{n}z^{n}+\ldots$ be analytic satisfying $|\omega(z)|\leq 1$ in $\mathbb{U}$. Then

and

Lemma 2. Let

we have

Lemma 3. If $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in S_{sc}^{\ast}(e)$, then

The estimate is sharp for $h$ given by

If $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in K_{sc}^{\ast}(e)$, then

The estimate is sharp for $h$ given by

Proof. Let $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in S_{sc}^{\ast}(e)$. According to Definition 1, we have

There exists a function $p(z) = 1+\sum\limits_{k = 1}^\infty p_kz^k$ satisfying

that is,

Using the results of Rogosinski ^[19], we have $|p_k|\leq 1$ for $k\geq 1$.

By means of comparing the coefficients of the both sides of (2.5), we get

and

Let $\phi(n) = 1+|a_3|+\cdots+|a_{2n-1}|$. It is easy to verify that

and

From (2.7), we have

According to (2.6), (2.7) and (2.8), we can obtain (2.3).

If $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in K_{sc}(e)$, then $zh'(z)\in S_{sc}^{\ast}(e)$. Using the results in (2.3), we can obtain (2.4) easily.

Lemma 4. Let $\mu \in \mathbb{C}$.

(1) If $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in S_{sc}^{\ast}(e),$ then

(2) If $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in K_{sc}(e),$ then

The estimates are sharp.

Proof. Let $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in S_{sc}^{\ast}(e)$. According to the subordination relationship and Definition 1, we get

that is,

where $\nu(z) = c_1z+c_2z^2+\cdots$ is an analytic function with $\nu(0) = 0$ and $|\nu(z)| < 1\; (z\in \mathbb{U})$.

By means of comparing the coefficients of two sides of (2.11), we get

Therefore, we have

Using the fact that (2.2) in Lemma 1, we obtain

The bound is sharp for $h$ given as follows,

Taking $h(z) = z+\sum\limits_{n = 2}^{\infty}a_{n}z^{n}\in K_{sc}(e)\Longleftrightarrow zh'(z) = z+\sum\limits_{n = 2}^{\infty}na_{n}z^{n}\in S_{sc}^{\ast}(e)$ into consideration, it is easy to obtain (2.10). The bound is sharp for $h$ given as follows,

Lemma 5. Suppose $h(z)\in \mathcal{A}$ and $|z| = r\in [0, 1)$.

(1) Let $h(z)\in S^*(e)$. Then

and

(2) Let $h(z)\in K(e)$. Then

and

Proof. Let $h(z)\in S^*(e)$ and $|z| = r\in [0, 1)$. According to the subordination relationship and Definition 1, there exists an analytic function $\nu(z) = c_1z+c_2z^2+\cdots$ satisfying $\nu(0) = 0$ and $|\nu(z)| < |z|$, such that

Thus, it is concluded that

that is,

Since

By (2.17) and (2.18), we get

From (2.16) and (2.19), we have

Similar to the previous proof. We let $h(z)\in K(e)$ and $|z| = r\in [0, 1)$, then

After simple calculation, we have

By (2.18) and (2.20), we get

Using the conclusion in ^[19], for $h(z)\in K(e)$, we have $\frac{2zh'(z)}{h(z)}-1\prec e^z$. According to the subordination relationship, we have

After simple calculation, we have

By (2.18) and (2.21), we get

Therefore, we complete the proof of Lemma 5.

Lemma 6. (^[20]). If $h(z)\in \mathcal{S}^*_{sc}(e)$, then $\frac{h(z)-\overline{h}(-\overline{z})}{2}\in \mathcal{S}^*(e)$.

Lemma 7. If $h(z)\in \mathcal{K}_{sc}(e)$, then $\frac{h(z)-\overline{h}(-\overline{z})}{2}\in \mathcal{K}(e)$.

Lemma 8. Suppose $h(z)\in \mathcal{A}$ and $|z| = r\in[0.1).$

(1) Let $h\in \mathcal{S}^*_{sc}(e)$. Then

where

(2) Let $h\in \mathcal{K}_{sc}(e)$. Then

where

Proof. Suppose $h(z)\in \mathcal{S}^*_{sc}(e)$. It is quite similar to the proof of Lemma 5, we have

According to Lemma 6 and (2.13) of Lemma 5, we have

By (2.24) and (2.25), we can obtain (2.22).

If $h(z)\in \mathcal{K}_{sc}(e)$, then

According to Lemmas 7 and (2.14) of Lemma 5, we have

By (2.26) and (2.27), we get

By (2.28), integrating along a radial line $\xi = te^{i\theta}$, we obtained immediately,

The verification for the remainder of (2.23) is given as follows. Let $H(z): = zh'(z)$, which is univalent. Suppose that $\xi_{1}\in \Gamma = H(\{z:|z| = r\})$ is the nearest point to the origin. By means of rotation, we suppose that $\xi_{1} > 0$ and $z_{1} = H^{-1}(\xi_{1})$. Let $\gamma = \{\xi:0\leq \xi\leq\xi_{1}\}$ and $L = H^{-1}(\gamma)$. If $\varsigma = H^{-1}(\xi)$, then $d\xi = H'(\varsigma)d\varsigma$. Hence

Thus the proof of Lemma 8 is completed.

3.   Main results

Next, the integral expressions for functions of the classes defined in Definition 1 are obtained.

Theorem 1. Let $\omega$ and $\nu$ be analytic in $\mathbb{U}$ with $|\omega(0)| = \alpha, \nu(0) = 0, |\omega(z)| < 1$ and $|\nu(z)| < 1$. If $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e)$. Then

where

Proof. Let $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e)$. Using Definition 1 and the subordination relationship, there exist analytic functions $\omega$ and $\nu$ satisfying $\omega(0) = b_1, \nu(0) = 0, |\omega(z)| < 1$ and $|\nu(z)| < 1\; (z\in \mathbb{U})$, such that

and

If we substitute $z$ by $-\overline{z}$ in (3.4), we obtain

It follows from (3.4) and (3.5) that

A routine computation for the equality (3.6) gives rise to the following equation,

Plugging (3.7) back into (3.3), we have

If the equality (3.8) is integrated from the both sides of it, then

Inserting (3.8) into (3.4), it is easy to show that

Thus, the proof of Theorem 1 is completed.

Taking Theorem 1 and $h\in K_{sc}(e)\Longleftrightarrow zh'(z)\in S^*_{sc}(e)$ into consideration, we get the following result.

Theorem 2. Let $\omega$ and $\nu$ be analytic in $\mathbb{U}$ satisfying $|\omega(0)| = \alpha, \nu(0) = 0, |\omega(z)| < 1$ and $|\nu(z)| < 1$. If $f\in HK^\alpha_{sc}(e)$, then

where $\varphi(\xi)$ is defined by (3.2).

In the following, the coefficient estimates of the class $HS^{\ast, \alpha}_{sc}(e)$ will be obtained.

Theorem 3. Let $h$ and $g$ be given by (1.3). If $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e)$, then

and

The estimates are sharp and the extremal function is

Specially, if $f\in HS^{\ast, 0}_{sc}(e),$ then

and

The estimates are sharp and the extremal function is

Proof. Let $h$ and $g$ be given by (1.3). Using the fact that $g' = \omega h'$ satisfying $\omega(z) = c_0+c_1z+c_2z^2+\cdots$ analytic in $\mathbb{U}$, we obtain

and

It is easy to show that

and

Since $g' = \omega h'$, it follows that $c_{0} = b_{1}$. By (2.1), it is obvious that $|c_k|\leq 1-\alpha^{2}$ for $k\in \mathbb{N}$. Therefore,

and

According to Lemma 3, (3.13) and (3.14), after the simple calculation, (3.9) and (3.10) can be obtained easily. We also obtain the extreme function.

By using the same methods in Theorem 3, the following results are obtained.

Theorem 4. Let $h$ and $g$ of the form (1.3). If $f = h+\overline{g}\in HK^{\alpha}_{sc}(e)$, then

and

For functions of the classes defined in the paper, Fekete-Szegö inequality of which are listed below.

Theorem 5. Let $f = h+\overline{g}$ with $h$ and $g$ given by (1.3) and $\mu \in \mathbb{C}$.

(1) If $f\in HS^{\ast, \alpha}_{sc}(e)$, then

and

(2) If $f\in HK^{\alpha}_{sc}(e)$, then

and

Proof. From the relation (3.11) and (3.12), we have

and

By (2.1), we have

and

According to Lemma 3 and Lemma 4, we can compete the proof of Theorem 5. The estimates above are sharp.

Paralleling the results of Zhu et al.^[17], the corresponding results for functions of the classes defined in the paper can be obtained. For example, the estimates of distortion, growth of $g$ and Jacobian of $f$ and so on.

Theorem 6. Let $|z| = r\in [0, 1).$

(1) If $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e)$, then

where $\phi_1(r)$ and $\phi_2(r)$ are given by (2.22).

Especially, let $\alpha = 0$, we have

(2) If $f = h+\overline{g}\in HK^{\alpha}_{sc}(e)$, then

where $\psi_1(r)$ and $\psi_2(r)$ are given by (2.23).

Especially, let $\alpha = 0$, we have

Proof. According to the relation $g' = \omega h'$, it is easy to see $\omega(z)$ satisfying $|\omega(0)| = |g'(0)| = |b_{1}| = \alpha$ such that (^[21]):

It is easy to show

A tedious calculation gives

Applying (3.17) and (2.22), we get (3.15). Similarly, applying (3.17) and (2.23), we get (3.16). Thus we complete the proof of Theorem 6.

Using the method analogous to that in proof of Lemma 8, we can obtain the following results.

Theorem 7. Let $|z| = r\in [0, 1).$

(1) If $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e),$ then

where $\phi_1(\xi)$ and $\phi_2(\xi)$ are given by (2.22).

Especially, let $\alpha = 0$, we have

(2) If $f = h+\overline{g}\in HK^{\alpha}_{sc}(e),$ then

where $\psi_1(\xi)$ and $\psi_2(\xi)$ are given by (2.23).

Especially, let $\alpha = 0$, we have

Next, the Jacobian estimates and growth estimates of $f$ are obtained.

Theorem 8. Let $|z| = r\in [0, 1).$

(1) If $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e)$, then

where $\phi_1(r)$ and $\phi_2(r)$ are given by (2.22).

(2) If $f = h+\overline{g}\in HK^{\alpha}_{sc}(e)$, then

where $\psi_1(r)$ and $\psi_2(r)$ are given by (2.23).

Proof. It is well known that Jacobian of $f = h+\overline{g}$ is

where $\omega$ satisfying $g' = \omega h'$ with $\omega(0) = 0$ and $|\omega(z)| < 1$ for $z\in \mathbb{U}$.

Let $f\in HS^{\ast, \alpha}_{sc}(e)$, plugging (3.17) and (2.22) back into (3.18), we get

and

Thus this completes the proof of (1). Plugging (3.17) and (2.23) back into (3.18), (2) of Theorem 8 can be proved by the same method as employed before.

Theorem 9. Let $|z| = r, \; 0\leq r < 1.$

(1) If $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(e)$, then

where $\phi_1(\xi)$ and $\phi_2(\xi)$ are given by (2.22).

(2) If $f = h+\overline{g}\in HK^{\alpha}_{sc}(e)$, then

where $\psi_1(\xi)$ and $\psi_2(\xi)$ are given by (2.23).

Proof. For any point $z = re^{i\theta}\in\mathbb{U},$ let $\mathbb{U}_{r} = \{z\in \mathbb{U}:|z| < r\}$ and denote

It is easy to see that $\mathbb{U}(0, d)\subseteq f(\mathbb{U}_{r})\subseteq f(\mathbb{U})$. Thus, there is $z_{r}\in\partial\mathbb{U}_{r}$ satisfying $d = |f(z_{r})|$. Let $L(t) = tf(z_{r})$ for $t\in[0, 1]$, then $\ell(t) = f^{-1}(L(t))$ is a well-defined Jordan arc. For $f = h+\overline{g}\in HS^{\ast, \alpha}_{sc}(\beta)$, by (2.22) and (3.17), we get

Using (2.22) and (3.17), the right side of (3.19) is obtained. The remainder of proofs is similar to that in (3.20) and so we omit.

According to (3.19) and (3.20), it follows that the covering theorems of $f$.

Theorem 10. Let $f = h+\overline{g}\in S_H$.

(1) If $f\in HS^{\ast, \alpha}_{sc}(e)$, then $\mathbb{U}_{R_1}\subset f(\mathbb{U})$, where

(2) If $f\in HK^{\alpha}_{sc}(e)$, then $\mathbb{U}_{R_2}\subset f(\mathbb{U})$, where

4.   Conclusion

In this paper, with the help of the analytic part $h$ satisfying certain conditions, we obtain the coefficients estimates of the co-analytic part $g$ and the geometric properties of harmonic functions. Applying the methods in the paper, the geometric properties of the co-analytic part and harmonic function with the analytic part satisfying other conditions can be obtained, which can enrich the research field of univalent harmonic mapping.

Acknowledgments

This work was supported by Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region(Grant No. NJZY20198; Grant No. NJZZ19209), Natural Science Foundation of Inner Mongolia Autonomous Region of China (Grant No. 2020MS01011; Grant No. 2019MS01023; Grant No. 2018MS01026) and the Program for Young Talents of Science and Technology in Universities of Inner Mongolia Autonomous Region Under Grant (Grant No. NJYT-18-A14).

Conflict of interest

The authors agree with the contents of the manuscript, and there is no conflict of interest among the authors.