An extension of Petrović’s inequality for \(h-\)convex (\(h-\)concave) functions in plane

1. Introduction

Let \(h:[c,d]\to \mathbb{R}\) be a non-negative function and \((0,1)\subseteq [c,d]\). A function \(f :[a,b] \to \mathbb{R}\) is said to be an \(h-\)convex, if \(f\) is non-negative for all \(x,y \in [a,b]\) and \(\alpha\in (0,1),\) one has If above inequality is reversed, then \(f\) is said to be \(h\)-concave. The \(h-\)convex function was introduced by Varo\v{s}anec in [1]. This function generalized convex function and many other generalization of convex function like \(s-\)convex function, Godunova-Levin function, \(s-\)Godunova-Levin function and \(P-\)function given in [1, 2, 3].

Remark 1. Particular value of \(h\) in inequality (1) gives us the following results:

1. \(h(\alpha)=\alpha\) gives the convex functions.
2. \(h(\alpha)=1\) gives the \(P-\)functions.
3. \(h(\alpha)=\alpha^s\) and \(\alpha \in (0,1)\) gives the \(s-\)convex functions of second sense.
4. \(h(\alpha)=\frac{1}{\alpha}\) and \(\alpha \in (0,1)\) gives the Godunova-Levin functions.
5. \(h(\alpha)=\frac{1}{\alpha^s}\) and \(\alpha \in (0,1)\) gives the \(s-\)Godunova-Levin functions of second sense.
In case of \(h-\)concavity, following results are valid:
6. \(h(\alpha)=1\) gives the reverse \(P-\)functions.
7. \(h(\alpha)=\frac{1}{\alpha}\) gives the reverse Godunova-Levin functions.
8. \(h(\alpha)=\frac{1}{\alpha^s}\) gives the reverse \(s-\)Godunova-Levin functions of second sense.

In [4], Dragomir gave the definition of convex functions on coordinates. Following his idea, the \(h-\)convex on coordinates was introduced by Alomari et al. in [5].

Definition 1. Let \(\Delta=[a_1,b_1]\times[a_2,b_2]\subseteq\mathbb{R}^2\) and \(f:\Delta\to \mathbb{R}\) be a mapping. Define partial mappings

and Also let interval \([c,d]\) contains \((0,1)\) and \(h:[c,d] \to \mathbb{R}\) be a positive function. A mapping \(f:\Delta \to \mathbb{R}\) is said to be \(h-\)convex (\(h-\)concave) on \(\Delta,\) if the partial mappings defined in (2) and (3) are \(h-\)convex (\(h-\)concave) on \([a,b]\) and \([c,d]\) respectively for all \(y\in [c,d]\) and \(x\in [a,b].\)

Remark 2. From above definition, one can deduce the definitions of those particular cases on coordinates.

In [6](also see [7, p. 154]), Petrović proved the following result, which is known as Petrović's inequality in the literature.

Theorem 2. Suppose that \(\left( {{x_1},...,{x_n}} \right)\) and \(({p_1},...,{p_n})\) be non-negative n-tuples such that \(\sum_{k=1}^{n}p_kx_k\geq x_i\)\;\; for \(i=1,...,n\) and \(\sum_{k=1}^{n}p_kx_k\in[0,a]\). If \(f\) is a convex function on \([0,a]\), then the inequality

is valid.

A function \(h:[c,d] \to \mathbb{R}\) is said to be a submultiplicative function if for all \(x,y\in [c,d].\) If the above inequality is reversed, then \(h\) is said to be supermultiplicative function. If equality holds in the above inequality, then \(h\) is said to be multiplicative function. By considering \(h\) to be supermultiplicative along with other condition, in the following theorem generalization of Petrović's inequality was proved by Rehman et al. in [8].

Theorem 3. Let \((x_1,...,x_n)\) be non-negative n-tuples and \((p_1,...,p_n)\) be positive n-tuples such that

Also let \(h:[0,\infty) \to \mathbb{R^+}\) be a supermultiplicative function such that If \(f:[0,\infty) \to \mathbb{R}\) be an \(h-\)convex function on \([0,\infty)\), then The following reverse version of above theorem was also proved in [8].

Theorem 4. Let \((x_1,...,x_n)\) be non-negative n-tuples and \((p_1,...,p_n)\) be positive n-tuples and the conditions given in (6) are valid. Also let \(h:[0,a] \to \mathbb{R^+}\) be a submultiplicative function such that

If \(f:[0,a] \to \mathbb{R}\) be an \(h-\)concave function on \([0,a]\), then reverse of (8) is valid.

In recent years, \(h-\)Convex functions are considered in literature by many researchers and mathematicians, for example, see [1, 3, 5, 9] and references there in. Many authors worked on Petrović's inequality by giving results related to it, for example see [6, 7 10] and it has been generalized for \(m-\)convex functions by Bakula et al. in [11]. In [12], Petrović's inequality was generalized on coordinates by using the definition of convex functions on coordinates. In this paper, Petrović's inequality is generalized for \(h-\)convex functions on coordinates, when \(h\) is supermultiplicative function. When \(h\) is submultiplicative, Petrović's inequality is generalized for \(h-\)concave functions on coordinates.

2. Main results

The following theorem consist the result for generalized Petrović's inequality for \(h-\)convex functions on coordinates.

Theorem 5. Let \((x_1,...,x_n)\) and \((y_1,...,y_n)\) be non-negative n-tuples, \((p_1,...,p_n)\) and \((q_1,...,q_n)\) be positive n-tuples such that

and Also let \(h:[0,\infty) \to \mathbb{R^+}\) be a supermultiplicative function such that (7) is valid. If \(f:[0,a]\times [0,b] \to \mathbb{R}\) be an \(h-\)convex function on coordinates, then where \(x_i>c_1, y_j>c_2.\)

Proof. Let \(f_x:[0,a]\to\mathbb{R}\) and \(f_y:[0,b]\to\mathbb{R}\) be mappings such that \(f_x(v)=f(x,v)\) and \(f_y(u)=f(u,y).\) Since \(f\) is coordinated \(h-\)convex on \([0,a]\times [0,b]\), therefore \(f_y\) is \(h-\)convex on \([0,b]\), so by Theorem 3, one has \begin{align*} \sum\limits_{j=1}^{n}p_jf_y(x_j)\leq\frac{\sum\limits_{j=1}^{n}p_jh(x_j-c_1)}{h\left(\sum\limits_{k=1}^{n}p_kx_k-c_1 \right) }f_y\left(\sum\limits_{k=1}^{n}p_kx_k\right) +\left( \sum\limits_{j=1}^{n}p_j-\frac{\sum\limits_{j=1}^{n}p_jh(x_j-c_1)}{h\left(\sum\limits_{k=1}^{n}p_kx_k-c_1 \right) }\right)f_y(c_1). \end{align*} This is equivalent to \begin{align*} \sum\limits_{j=1}^{n}p_jf(x_j,y)\leq\frac{\sum\limits_{j=1}^{n}p_jh(x_j-c_1)}{h\left(\sum\limits_{k=1}^{n}p_kx_k -c_1\right) }f\left(\sum\limits_{k=1}^{n}p_kx_k,y\right)+\left(\sum\limits_{j=1}^{n}p_j-\frac{\sum\limits_{j=1}^{n}p_jh(x_j-c_1)}{h\left(\sum\limits_{k=1}^{n}p_kx_k -c_1\right) }\right)f(c_1,y), \end{align*} by setting \(y=y_j,\) we get \begin{align*} \sum\limits_{j=1}^{n}p_jf(x_j,y_j)\leq\frac{\sum\limits_{j=1}^{n}p_jh(x_j-c_1)}{h\left(\sum\limits_{k=1}^{n}p_kx_k -c_1\right) }f\left(\sum\limits_{k=1}^{n}p_kx_k,y_j\right) +\left(\sum\limits_{j=1}^{n}p_j-\frac{\sum\limits_{j=1}^{n}p_jh(x_j-c_1)}{h\left(\sum\limits_{k=1}^{n}p_kx_k -c_1\right) }\right)f(c_1,y_j). \end{align*} Multiplying above inequality by \(p_j\) and taking sum for \(j=1,...,n,\) one has

Now again by Theorem 4, one has
\( \sum\limits_{j=1}^{n}q_j f\left(\sum\limits_{k=1}^{n}p_kx_k,y_j\right)\leq\frac{\sum\limits_{j=1}^{n}q_jh(y_j-c_2)}{h\left(\sum\limits_{k=1}^{n}q_ky_k-c_2\right)}f\left(\sum\limits_{k=1}^{n}p_kx_k,\sum\limits_{j=1}^{n}q_jy_j\right)+\left(\sum\limits_{j=1}^{n}q_j-\frac{\sum\limits_{j=1}^{n}q_jh(y_j-c_2)}{h\left(\sum\limits_{k=1}^{n}q_ky_k-c_2\right)}\right)f\left(\sum\limits_{k=1}^{n}p_kx_k,c_2\right) \)\\ and \begin{align*} \sum\limits_{j=1}^{n}q_jf\left(c_1,y_j\right)\leq\frac{\sum\limits_{j=1}^{n}q_jh(y_j-c_2)}{h\left(\sum\limits_{k=1}^{n}q_ky_k -c_2\right) }f\left(c_1,\sum\limits_{j=1}^{n}q_jy_j\right) +\left(\sum\limits_{j=1}^{n}q_j-\frac{\sum\limits_{j=1}^{n}q_jh(y_j-c_2)}{h\left(\sum\limits_{k=1}^{n}q_ky_k-c_2 \right)}\right)f(c_1,c_2). \end{align*} Putting these values in inequality (13), we get the required result.

In the following theorem, we give the Petrović's inequality for \(h-\)convex functions on coordinates.

Theorem 6. Let the conditions given in Theorem 5 are valid. % Also let \(h:[0,\infty)^2 \to \mathbb{R^+}\) be a supermultiplicative function. If \(f:[0,a]\times [0,b] \to \mathbb{R}\) be an \(h-\)convex function on coordinates, then

Proof. If we take \(c_1=0=c_2\) in Theorem 5, we get the required result.

In the following corollary, we give the Petrović's inequality for convex functions on coordinates which is given in [12].

Theorem 7. Let the conditions given in Theorem 5 are valid. If \(f:[0,a]\times [0,b] \to \mathbb{R}\) be a convex function on coordinates, then

Proof. If we take \(h(x)=x\) for all \(x\in [0,\infty)\), then it satisfied the condition imposed on \(h\) given in Theorem 5. Hence using this value of \(h\) in above theorem gives the required result.

One can see that the condition on function \(h\) given in (7) restrict us to give Petrović's type inequalities for particular cases of \(h-\)convex functions given in Remark 1. If we consider reverse inequality in (7), then it covers some of particular cases but for \(h-\)concave function. In the following theorem, reverse of (12) has been concluded. The notable thing is the requirements of submultiplicity and reverse of (7) for function \(h\) along with \(h\)-concavity of the function \(f\).

Theorem 8.\label{231} Let \((x_1,...,x_n)\) and \((y_1,...,y_n)\) be non-negative n-tuples, \((p_1,...,p_n)\) and \((q_1,...,q_n)\) be positive \(n-\)tuples such that (10) and (11) are valid. Also let \(h:[0,\infty) \to \mathbb{R^+}\) be a submultiplicative function such that (9) is valid. If \(f:[0,a]\times [0,b] \to \mathbb{R}\) be an \(h-\)concave function on coordinates, then the reverse of inequality (12) holds.

Proof. By using Theorem 4 and following the steps of Theorem 5, one can deduce the required results.

In the following theorem, we give the Petrović's inequality for \(h-\)concave functions on coordinates.

Theorem 9. Let the conditions given in Theorem 8 are valid. Also let \(h:[0,\infty) \to \mathbb{R^+}\) be a submultiplicative function. If \(f:[0,a]\times [0,b] \to \mathbb{R}\) be an \(h-\)concave function on coordinates, then the reverse of inequality (14) is valid.

Proof. If we take \(c_1=0=c_2\) in Theorem 8, we get the required result.

In the following theorem, we give the Petrović's inequality for concave functions on coordinates.

Theorem 10. Let the conditions given in Theorem 8 are valid. If \(f:[0,a]\times [0,b] \to \mathbb{R}\) be a concave function on coordinates, then then the reverse of inequality (15) is valid.

Proof. If we take \(h(x)=x\) and \(c_1=0=c_2\) in Theorem 8, we get the required result.

Theorem 10. Let \((x_1,...,x_n)\) and \((y_1,...,y_n)\) be non-negative n-tuples, \((p_1,...,p_n)\) and \((q_1,...,q_n)\) be positive n-tuples such that (10) and (11) are valid. If \(f:[0,a]\times [0,b] \to \mathbb{R}\) is reverse \(P-\)function on coordinates, then

Remark 3. Consider \(h(x)=\frac{1}{x}\), then \( h(\alpha)+h(1-\alpha)=\frac{1}{\alpha}+\frac{1}{1-\alpha} >1 \text{ for all } \alpha \in (0,1). \) Using above value of \(h\) in Theorem 8 gives Petrović type inequality for reverse Godunova-Levin functions on coordinates.

Remark 4. Let us consider \(H(h)=h(\alpha)+h(1-\alpha)-1, \alpha \in (0,1),\) we take \(g_1(\alpha):=H(\alpha^s)=\alpha^s+(1-\alpha)^s-1 ,\text{ where } s\in(0,1).\) In [8], it has been shown that \(g_1\) is positive by considering different values of \(\alpha\) and \(s\) in interval \((0,1)\), therefore \(h(\alpha)=\alpha^s\) for \(\alpha, s \in (0,1)\) satisfied the conditions of Theorem 8, but it doesn't satisfies the conditions of Theorem 5. Hence the above value of \(h\) in Theorem 8 leads us to the Petrović type inequalities for reverse of \(s-\)Godunova-Levin on coordinates.

Remark 5. Let us consider \(g_2(\alpha):=H\left(\frac{1}{\alpha^s}\right)=\frac{1}{\alpha^s}+\frac{1}{(1-\alpha)^s}-1,\text{ where } s\in(0,1).\) This function is also discussed in [8] and it has been shown that \(g_2\) is positive for different values of \(\alpha\) and \(s\) in \((0,1)\). Thus it satisfied the conditions of Theorem 8, but it doesn't satisfy the conditions of Theorem 5. Hence the above value of \(h\) in Theorem 8 leads us to the Petrović type inequalities for \(s-\)concave function on coordinates.

Acknowledgments

The authors are very grateful to the editor and reviewers for their careful and meticulous reading of the paper. The research work 3rd author is supported by Higher Education Commission of Pakistan under NRPU 2017-18, Project No. 7962.

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Competing Interests

The author(s) do not have any competing interests in the manuscript.