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On \(m\)-convex functions of the second type

M. Sertbaş1
1Karadeniz Technical University, Faculty of Sciences, Department of Mathematics, 61080, Trabzon, Türkiye
Copyright © M. Sertbaş. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this study, we give a new \(m\)-convex function that is called an \(m\)-convex of the second type and its some properties. Moreover, some integral inequalities are examined for each \(m\)-convex function of the second type.

Keywords: m-convex function, integral inequalities

1. Introduction and preliminaries

The concept of a convex function is equivalent to the set of all points forming the convex function’s graph being convex [1]. Thus, problems involving convex functions are associated with convex sets. As a result of a thorough understanding of convex geometry, many mathematicians are better able to address problems from a geometric perspective.

Convex functions can occasionally be found in proof methods given in various mathematics topics. For example, Kittaneh used a simple property of convex functions to provide a better lower bound on the numerical radius of bounded linear operators. Consequently, the theory of convex functions remains a popular topic. Various classes of convex functions are studied by many mathematicians. One of these is the class of m-convex functions given by Toader, whose definition is given below [2].

A function \(f : [0,b] \to \mathbb{R}\) is called an \(m\)-convex function, where \(m \in (0,1]\) and \(b>0\), if for all \(x,y \in [0,b]\) and \(t \in [0,1]\) we have \[f\left(tx + m(1-t)y\right) \leq tf(x) + m(1-t)f(y).\]

If \(m=1\), then it is a convex function. If \(f(tx) \leq tf(x)\), then it is a star-shaped function.

The following relations are true for a non negative continuous function which vanishes at the origin [3] Convex functions \(\subset\) Star – shaped functions and also \(m\)-Convex functions \(\subset\) Star – shaped. Integral inequalities of \(m\)-convex functions have also been studied by some mathematicians [47]. Furthermore, inequalities satisfied by functions defined on linear operators can also be found in the literature [8,9].

On the other hand, Takahashi introduced the concept of convexity at a metric distance where the linear vector space conditions are not satisfied [10]. Fixed point theorems on these convex abstract metric spaces have become a subject of considerable study [1114]. Moreover, the definition of \(m\)-convexity in \(b\)-metric spaces is defined by taking inspiration from \(m\)-convex functions in [15]. However, this definition corresponds to a very special case of convex metric spaces [16].

In this article, we adapt the definition given by Sertbaş et al. to functions [16]. First, we give some properties of these functions. Then, we calculate some basic integral inequalities.

2. \(m\)-convex functions of the second type

Definition 1. \(f : [0,b] \to \mathbb{R}\), \(b>0\) is called to be an \(m\)– convex of the second type for some \(m \in (0,1]\) if \[f\left(tx + (1-t)my\right) \leq tf(x) + (1-mt)f(y),\] for any \(x,y \in [0,b]\) and \(t \in [0,1]\).

It is obvious that a non negative \(m\)-convex function and non negative continuous convex function which vanishes at the origin are an \(m\)-convex of the second type.

Theorem 1. If a function \(f : [0,b] \to \mathbb{R}\), \(b>0\) is a second type \(m\)-convex with \(m \in (0,1)\), then the following statements are satisfied;

i) \(f\) is a non negative function,

ii) The sequence \(\{f(m^n x)\}\), \(n \in \mathbb{N}\) converges for all \(x \in [0,b]\) and \[\lim_{n\to+\infty} f(m^n x) \leq \min\{f(0),f(x)\},\] is satisfied.

iii) \(f\) is continuous at zero if and only if \(f(0) \leq f(x)\) for each \(x \in [0,b]\).

Proof. In this case, for any \(x,y \in [0,b]\) \[f(y)=f(y+0mx) \leq f(y)+(1-m)f(x),\] and then \(f(x)\geq 0\) for each \(x\in[0,b]\). On the other hand, for any \(x\in[0,b]\) \[f(mx)=f(0+(1-0)mx) \leq f(x),\] is true. Therefore, \[0 \leq f(m^n x) \leq f(m^{n-1}x) \leq \cdots \leq f(mx) \leq f(x),\] is holds for \(n \in \mathbb{N}\). It means that the sequence \(\{f(m^n x)\}\) is monotone deceasing and bounded and so it converges. Also, for \(n \in \mathbb{N}\) \[f(m^n x)=f(m^n x+(1-m^n)0) \leq m^n f(x)+(1-m^{n+1})f(0),\] is correct. From these results \[\lim_{n\to+\infty} f(m^n x) \leq \min\{f(0),f(x)\},\] is obtained. Moreover, if \(f\) is continuous at zero, then from the last limit the inequality \(f(0)\leq f(x)\) is get for each \(x\in[0,b]\).

On the contrary, suppose that the inequality \(f(0)\leq f(x)\) is true for each \(x\in[0,b]\). In this case, for any \(x\in[0,b]\) there exist an element \(t\in[0,1]\) such that \(x=tb\) and \[ 0\leq f(x)-f(0)=f(tb)-f(0)\leq tf(b)+(1-mt)f(0)-f(0)=t[f(b)-mf(0)],\tag{1}\] is correct and so \(f\) continuous at zero. ◻

Corollary 1. If a function \(f : [0,b] \to \mathbb{R}\), \(b>0\) is an \(m\)-convex of the second type with \(m\in(0,1)\) and continuous at zero, then \[0 \leq \underline{\lim}_{x\to0}\frac{f(x)-f(0)}{x} \leq \overline{\lim}_{x\to0}\frac{f(x)-f(0)}{x} \leq \frac{f(b)-mf(0)}{b}.\]

Also, if \(f(x)\) is vanish at origin and continuous on \([0,b]\), then it is differentiable at \(x=0\).

Proof. According to previous theorem and the inequality (1) the first assertion can be proved.

If \(f(0)=0\), then it is a star shaped function. Therefore, it is obtained the second assertion by using Theorem 6 in [3]. ◻

Theorem 2. Assume that a function \(f : [0,b] \to \mathbb{R}\), \(b>0\) is an \(m\)-convex of the second type with \(m\in(0,1)\), then it is bounded on \([0,b]\) and following statements are hold

i) \(f\) is a bounded function on \([0,b]\),

ii) For any \(x\in[0,b]\), \(\overline{\lim}_{y\to x} f(my) \leq f(x)\).

Proof. For any \(x\in[0,b]\) there exist a unique element \(t\in[0,1]\) such that \(x=tb\) and \[0 \leq f(x)=f(tb)\leq tf(b)+(1-mt)f(0) \leq f(b)+f(0),\] this means that \(f\) is a bounded function on \([0,b]\).

On the other hand, \(\delta\) is a positive or negative number and small enough such that \(x+n\delta\in[0,b]\), \(n\in\mathbb{N}\). In this case, \[f(m(x+\delta))=f\left(\frac{m(x+n\delta)}{n}+\left(1-\frac{1}{n}\right)mx\right)\leq \frac{1}{n}f(m(x+n\delta))+ \left(1-\frac{m}{n}\right)f(x).\]

If \(\delta\to0\), then it must be \(n\to+\infty\). Since \(f\) is a bounded function, \[\overline{\lim}_{y\to x}f(my)\leq f(x),\] is get. This idea is in [17]. ◻

3. Some inequalities

Theorem 3. Let \(f : [0,b] \to \mathbb{R}\), \(b>0\) be an \(m\)-convex of the second type and Lebesgue integrable on \([ma,c]\), \(0\leq a<c\leq b\). Then \[f\left(m\frac{a+c}{2}\right)\leq \frac{3-m}{2(c-a)}\int_a^c f(x)\,dx \leq (3-m)\left(\frac{c-ma}{c-a}\right)\frac{f(c)+(2-m)f(a)}{4}.\]

Proof. Because \(f\) is an \(m\)-convex of the second type a, for all \(x,y\in[a,b]\) we have \[f\left(m\frac{x+y}{2}\right) \leq \frac{f(x)+(2-m)f(y)}{2}.\]

If \(x=ta+(1-t)c\) and \(y=tc+(1-t)a\) are chosen, then we have \[f\left(m\frac{a+c}{2}\right)\leq \frac{(f(ta+(1-t)c)+(2-m)f(tc+(1-t)a))}{2}.\]

We obtain by integrating the last inequality \[f\left(m\frac{a+c}{2}\right)\leq \frac{3-m}{2(c-a)}\int_a^c f(x)\,dx.\]

On the other hand, \[\int_a^c f(y)\,dy \leq \int_{ma}^c f(y)\,dy \leq (c-ma)\frac{f(c)+(2-m)f(a)}{2}.\]

This is completed the proof of theorem. ◻

Theorem 4. Suppose that \(f : [0,b] \to \mathbb{R}\), \(b>0\) is a function and \(f’\) is an \(m\)-convex of the second type on \([0,mb]\), then for \(0\leq a<b\) \[\frac{f(mb)+(2-m)f(ma)}{3-m}-\frac{1}{(b-a)}\int_a^b f(mx)\,dx \leq \frac{m(b-a)}{3-m}\frac{(2-m) f'(b)+f'(a)}{4}.\]

Proof. Because \(f’\) is bounded on \([0,mb]\) and the points of discontinuity of a derivative function are only of the second type, \(f’\) must be continuous on \([0,mb]\). In this case, we use the idea given in [18], and so \[\int_0^1 (1-(3-m)t)f'(tma+(1-t)mb)\,dt = \frac{(2-m)f(ma)+f(mb)}{m(b-a)}- \frac{3-m}{m(b-a)^2}\int_a^b f(mx)\,dx,\] is get. Because \(f’\) is an \(m\)-convex of the second type on \([0,mb]\), \[\begin{aligned} \int_0^1 &(1-(3-m)t)f'(tma+(1-t)mb)\,dt \leq \int_0^1 (1-2t)f'(tma+(1-t)mb) \\ &\leq \int_0^1 |1-2t|f'(tma+(1-t)mb)\,dt \leq \int_0^1 |1-2t|\bigl(tf'(a)+(1-mt)f'(b)\bigr)\,dt \\ &=\frac{(2-m) f'(b)+f'(a)}{4}. \end{aligned}\]

From this inequality, \[\frac{f(mb)+(2-m)f(ma)}{3-m}-\frac{1}{(b-a)}\int_a^b f(mx)\,dx \leq \frac{m(b-a)}{3-m}\frac{(2-m) f'(b)+f'(a)}{4}.\] ◻

Corollary 2. Under the previous theorem, the following inequality is obtained \[f(ma)-\frac{1}{(b-a)}\int_a^b f(mx)\,dx < (b-a)\frac{(2-m) f'(b)+f'(a)}{8},\]

Proof. Since \(f’\) is a non negative continuous function on \([0,mb]\), \(f(x)=f(0)+\int_0^x f'(\tau)d\tau\) is a monotone increasing function. Also, \(\frac{m}{3-m}<\frac{1}{2}\) is true for \(m<1\). In this case, the desired result from the previous theorem is obtained. ◻

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