1 Introduction
Convexity is an important concept in many branches of mathematics, pure and applied. In particular, many important integral inequalities are based on a convexity assumption of a certain function, such as Jensen’s inequality, the Hermite-Hadamard inequality, the Hardy-Littlewood-Pólya majoration inequality, Petrović’s inequality, Popoviciui’s convex function inequality, and many others. For more details as regards inequalities via convex functions, we refer the reader to the monograph [1]. However, for many encountered problems, convexity is not satisfied. This leads to the necessity to extend this concept.
In the last 60 years, great attention has been focused on the generalization of the notion of convexity. Let us cite some references in this direction. In [2], Definetti introduced the class of quasi-convex functions. In [3], Mangasarian introduced the notion of pseudo-convex functions. Polyak [4] defined the concept of strongly convex functions. The class of ε-convex functions was introduced by Hyers and Ulam [5]. In [6], Varosanec introduced the notion of h-convexity that includes the class of s-convex functions (see Hudzik [7]). For other work in this direction, we refer the reader to [8‐12] and the references therein.
Advertisement
In this paper, we present a new concept of convexity that depends on a certain function satisfying some axioms. This new notion generalizes different types of convexity, including ε-convex functions, α-convex functions, h-convex functions, and many others. Moreover, some integral inequalities are established via this new notion of convexity. As particular cases, we retrieve several existing inequalities from the literature.
2 An implicit convexity concept
We denote by \(\mathcal{F}\) the family of mappings \(F: \mathbb{R}\times \mathbb{R}\times\mathbb{R}\times(0,1)\to\mathbb{R}\) satisfying the following axioms:
(A1)
If \(u_{i}\in L^{1}(0,1)\), \(i=1,2,3\), then for every \(\lambda\in (0,1)\), we have
$$\int_{0}^{1} F \bigl(u_{1}(t),u_{2}(t),u_{3}(t), \lambda \bigr)\,dt= F \biggl( \int_{0}^{1} u_{1}(t)\, dt, \int_{0}^{1} u_{2}(t)\,dt, \int_{0}^{1} u_{3}(t)\,dt,\lambda \biggr). $$
(A2)
For every \(u\in L^{1}(0,1)\), \(w\in L^{\infty}(0,1)\), and \((z_{1},z_{2})\in\mathbb{R}^{2}\), we have where \(T_{F,w}:\mathbb{R}\times\mathbb{R}\times\mathbb{R}\to\mathbb {R}\) is a function that depends on \((F,w)\), and it is nondecreasing with respect to the first variable.
$$\int_{0}^{1} F \bigl(w(t)u(t),w(t)z_{1},w(t)z_{2},t \bigr)\,dt=T_{F,w} \biggl( \int_{0}^{1} w(t)u(t)\,dt,z_{1},z_{2} \biggr), $$
(A3)
For any \((w,u_{1},u_{2},u_{3})\in\mathbb{R}^{4}\), \(u_{4}\in(0,1)\), we have where \(L_{w}\in\mathbb{R}\) is a constant that depends only on w.
$$wF(u_{1},u_{2},u_{3},u_{4})=F(wu_{1},wu_{2},wu_{3},u_{4})+L_{w}, $$
We introduce the new concept of convexity as follows.
Definition 2.1
Let \(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be a given function. We say that f is a convex function with respect to some \(F\in\mathcal{F}\) (or F-convex function) iff
$$ F \bigl(f \bigl(tx+(1-t)y \bigr),f(x),f(y),t \bigr)\leq0,\quad (x,y,t)\in [a,b]\times[a,b]\times(0,1). $$
(2.1)
Advertisement
The following property follows immediately from (2.1).
Proposition 2.2
Let
\(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be an
F-convex function, for some
\(F\in\mathcal{F}\). Then
$$F \bigl(f(x),f(x),f(x),t \bigr)\leq0, \quad (x,t)\in[a,b]\times(0,1). $$
Proof
Taking \(x=y\) in (2.1), the desired inequality follows. □
Now, we give some examples of F-convex functions.
Example 2.3
Let \(\varepsilon\geq0\), and let \(f: [a,b]\to\mathbb{R}\), \((a,b)\in \mathbb{R}^{2}\), \(a< b\), be an ε-convex function, that is (see [5]), Define the function \(F: \mathbb{R}\times\mathbb{R}\times\mathbb {R}\times[0,1]\to\mathbb{R}\) by Let \(u_{i}\in L^{1}(0,1)\), \(i=1,2,3\), and let \(\lambda\in[0,1]\). We have Therefore, the function F satisfies axiom (A1). Now, let \(u\in L^{1}(0,1)\), \(w\in L^{\infty}(0,1)\), and \((z_{1},z_{2})\in \mathbb{R}^{2}\). We have where \(T_{F,w}:\mathbb{R}\times\mathbb{R}\times\mathbb{R}\to\mathbb {R}\) is defined by Then the function F satisfies axiom (A2). Now, let \((w,u_{1},u_{2},u_{3})\in\mathbb{R}^{4}\) and \(u_{4}\in(0,1)\). We have Therefore, axiom (A3) is satisfied with Thus we proved that \(F\in\mathcal{F}\). On the other hand, since f is ε-convex, for all \((x,y,t)\in[a,b]\times[a,b]\times(0,1)\), we have As a consequence, f is an F-convex function.
$$f \bigl(tx+(1-t)y \bigr)\leq tf(x)+(1-t)f(y)+\varepsilon, \quad (x,y,t)\in [a,b] \times[a,b]\times[0,1]. $$
$$ F(u_{1},u_{2},u_{3},u_{4})=u_{1}-u_{4}u_{2}-(1-u_{4})u_{3}- \varepsilon,\quad (u_{1},u_{2},u_{3},u_{4}) \in \mathbb{R}\times\mathbb{R}\times\mathbb{R}\times[0,1]. $$
(2.2)
$$\begin{aligned} \int_{0}^{1} F \bigl(u_{1}(t),u_{2}(t),u_{3}(t), \lambda \bigr)\,dt =& \int_{0}^{1} \bigl(u_{1}(t)-\lambda u_{2}(t)-(1-\lambda)u_{3}(t)-\varepsilon \bigr)\,dt \\ =& \int_{0}^{1} u_{1}(t)\,dt-\lambda \int_{0}^{1} u_{2}(t)\,dt-(1-\lambda) \int_{0}^{1} u_{3}(t)\,dt-\varepsilon \\ =& F \biggl( \int_{0}^{1} u_{1}(t)\,dt, \int_{0}^{1} u_{2}(t)\,dt, \int_{0}^{1} u_{3}(t)\,dt,\lambda \biggr). \end{aligned}$$
$$\begin{aligned} & \int_{0}^{1} F \bigl(w(t)u(t),w(t)z_{1},w(t)z_{2},t \bigr)\,dt \\ &\quad= \int_{0}^{1} \bigl(w(t)u(t)-tw(t)z_{1}-(1-t)w(t)z_{2}- \varepsilon \bigr)\,dt \\ &\quad= \int_{0}^{1} w(t)u(t)\,dt-z_{1} \int_{0}^{1} tw(t)\,dt-z_{2} \int_{0}^{1}(1-t)w(t)\,dt-\varepsilon \\ &\quad=T_{F,w} \biggl( \int_{0}^{1} w(t)u(t)\,dt,z_{1},z_{2} \biggr), \end{aligned}$$
$$\begin{aligned} &T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} tw(t)\,dt \biggr)u_{2}- \biggl( \int _{0}^{1}(1-t)w(t)\,dt \biggr)u_{3}- \varepsilon, \\ &\quad (u_{1},u_{2},u_{3})\in\mathbb{R} \times\mathbb{R}\times\mathbb{R}. \end{aligned}$$
(2.3)
$$\begin{aligned} wF(u_{1},u_{2},u_{3},u_{4}) =&w \bigl(u_{1}-u_{4}u_{2}-(1-u_{4})u_{3}- \varepsilon \bigr) \\ =&wu_{1}-u_{4}(wu_{2})-(1-u_{4}) (wu_{3})-w\varepsilon \\ =& \bigl(wu_{1}-u_{4}(wu_{2})-(1-u_{4}) (wu_{3})-\varepsilon \bigr)+ (1-w)\varepsilon \\ =&F(wu_{1},wu_{2},wu_{3},u_{4})+(1-w) \varepsilon. \end{aligned}$$
$$ L_{w}=(1-w)\varepsilon. $$
(2.4)
$$\begin{aligned} F \bigl(f \bigl(tx+(1-t)y \bigr),f(x),f(y),t \bigr) =&f \bigl(tx+(1-t)y \bigr)-tf(x)-(1-t)f(y)-\varepsilon \leq 0. \end{aligned}$$
Remark 2.4
Taking \(\varepsilon=0\) in the above example, we observe that any convex function \(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), is an F-convex function with respect to the function \(F: \mathbb{R}\times \mathbb{R}\times\mathbb{R}\times(0,1)\to\mathbb{R}\) defined by
$$F(u_{1},u_{2},u_{3},u_{4})=u_{1}-u_{4}u_{2}-(1-u_{4})u_{3}, \quad(u_{1},u_{2},u_{3},u_{4})\in \mathbb{R}\times\mathbb{R}\times\mathbb{R}\times(0,1). $$
Example 2.5
Let \(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be an α-convex function, \(0<\alpha\leq1\), that is, Define the function \(F: \mathbb{R}\times\mathbb{R}\times\mathbb {R}\times[0,1]\to\mathbb{R}\) by Let \(u_{i}\in L^{1}(0,1)\), \(i=1,2,3\), and let \(\lambda\in[0,1]\). We have Therefore, the function F satisfies axiom (A1). Now, let \(u\in L^{1}(0,1)\), \(w\in L^{\infty}(0,1)\), and \((z_{1},z_{2})\in\mathbb{R}^{2}\). We have where \(T_{F,w}:\mathbb{R}\times\mathbb{R}\times\mathbb{R}\to\mathbb {R}\) is defined by Then the function F satisfies axiom (A2). Now, let \((w,u_{1},u_{2},u_{3})\in\mathbb{R}^{4}\) and \(u_{4}\in(0,1)\). We have Therefore, axiom (A3) is satisfied with Thus we proved that \(F\in\mathcal{F}\). On the other hand, since f is α-convex, for all \((x,y,t)\in [a,b]\times[a,b]\times(0,1)\), we have As a consequence, f is an F-convex function.
$$f \bigl(tx+(1-t)y \bigr)\leq t^{\alpha}f(x)+ \bigl(1-t^{\alpha}\bigr)f(y),\quad (x,y,t)\in [a,b]\times[a,b]\times[0,1]. $$
$$ F(u_{1},u_{2},u_{3},u_{4})=u_{1}-u_{4}^{\alpha}u_{2}- \bigl(1-u_{4}^{\alpha}\bigr)u_{3}, \quad (u_{1},u_{2},u_{3},u_{4})\in \mathbb{R}\times\mathbb{R}\times\mathbb{R}\times[0,1]. $$
(2.5)
$$\begin{aligned} \int_{0}^{1} F \bigl(u_{1}(t),u_{2}(t),u_{3}(t), \lambda \bigr)\,dt =& \int_{0}^{1} \bigl(u_{1}(t)- \lambda^{\alpha}u_{2}(t)- \bigl(1-\lambda^{\alpha}\bigr)u_{3}(t) \bigr)\,dt \\ =& \int_{0}^{1} u_{1}(t)\,dt- \lambda^{\alpha}\int_{0}^{1} u_{2}(t)\,dt- \bigl(1-\lambda ^{\alpha}\bigr) \int_{0}^{1} u_{3}(t)\,dt \\ =& F \biggl( \int_{0}^{1} u_{1}(t)\,dt, \int_{0}^{1} u_{2}(t)\,dt, \int_{0}^{1} u_{3}(t)\,dt,\lambda \biggr). \end{aligned}$$
$$\begin{aligned} & \int_{0}^{1} F \bigl(w(t)u(t),w(t)z_{1},w(t)z_{2},t \bigr)\,dt \\ &\quad= \int_{0}^{1} \bigl(w(t)u(t)-t^{\alpha}w(t)z_{1}- \bigl(1-t^{\alpha}\bigr)w(t)z_{2} \bigr) \,dt \\ &\quad= \int_{0}^{1} w(t)u(t)\,dt-z_{1} \int_{0}^{1} t^{\alpha}w(t) \,dt-z_{2} \int _{0}^{1} \bigl(1-t^{\alpha}\bigr)w(t) \,dt \\ &\quad=T_{F,w} \biggl( \int_{0}^{1} w(t)u(t)\,dt,z_{1},z_{2} \biggr), \end{aligned}$$
$$\begin{aligned} &T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} t^{\alpha}w(t)\,dt \biggr)u_{2}- \biggl( \int_{0}^{1} \bigl(1-t^{\alpha}\bigr)w(t)\,dt \biggr)u_{3}, \\ &\quad(u_{1},u_{2},u_{3})\in \mathbb{R}\times\mathbb{R}\times\mathbb{R}. \end{aligned}$$
(2.6)
$$\begin{aligned}[b] wF(u_{1},u_{2},u_{3},u_{4})&=w \bigl(u_{1}-u_{4}^{\alpha}u_{2}- \bigl(1-u_{4}^{\alpha}\bigr)u_{3} \bigr) =wu_{1}-u_{4}^{\alpha}(wu_{2})- \bigl(1-u_{4}^{\alpha}\bigr) (wu_{3}) \\ &=F(wu_{1},wu_{2},wu_{3},u_{4}). \end{aligned} $$
$$ L_{w}=0. $$
(2.7)
$$\begin{aligned} F \bigl(f \bigl(tx+(1-t)y \bigr),f(x),f(y),t \bigr) =&f \bigl(tx+(1-t)y \bigr)-t^{\alpha}f(x)- \bigl(1-t^{\alpha}\bigr)f(y)\\ \leq& 0. \end{aligned}$$
Example 2.6
Let \(h: J\to[0,\infty)\) be a given function which is not identical to 0, where J is an interval in \(\mathbb{R}\) such that \((0,1)\subseteq J\). Let \(f: [a,b]\to[0,\infty)\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be a h-convex function, that is (see [6]), We suppose that \(h\in L^{1}(0,1)\). Define the function \(F: \mathbb {R}\times\mathbb{R}\times\mathbb{R}\times(0,1)\to\mathbb{R}\) by Let \(u_{i}\in L^{1}(0,1)\), \(i=1,2,3\), and let \(\lambda\in[0,1]\). We have Therefore, the function F satisfies axiom (A1). Now, let \(u\in L^{1}(0,1)\), \(w\in L^{\infty}(0,1)\), and \((z_{1},z_{2})\in\mathbb{R}^{2}\). We have where \(T_{F,w}:\mathbb{R}\times\mathbb{R}\times\mathbb{R}\to\mathbb {R}\) is defined by Then the function F satisfies axiom (A2). Now, let \((w,u_{1},u_{2},u_{3})\in\mathbb{R}^{4}\) and \(u_{4}\in(0,1)\). We have Therefore, axiom (A3) is satisfied with Thus we proved that \(F\in\mathcal{F}\). On the other hand, since f is h-convex, for all \((x,y,t)\in [a,b]\times[a,b]\times(0,1)\), we have As a consequence, f is an F-convex function.
$$f \bigl(tx+(1-t)y \bigr)\leq h(t)f(x)+h(1-t)f(y),\quad (x,y,t)\in[a,b]\times [a,b] \times(0,1). $$
$$\begin{aligned} &F(u_{1},u_{2},u_{3},u_{4})=u_{1}-h(u_{4}) u_{2}-h(1-u_{4})u_{3}, \\ &\quad (u_{1},u_{2},u_{3},u_{4}) \in\mathbb{R}\times\mathbb{R}\times\mathbb{R}\times(0,1). \end{aligned}$$
(2.8)
$$\begin{aligned} \int_{0}^{1} F \bigl(u_{1}(t),u_{2}(t),u_{3}(t), \lambda \bigr)\,dt =& \int_{0}^{1} \bigl(u_{1}(t)-h(\lambda) u_{2}(t)-h(1-\lambda)u_{3}(t) \bigr)\,dt \\ =& \int_{0}^{1} u_{1}(t)\,dt-h(\lambda) \int_{0}^{1} u_{2}(t)\,dt-h(1-\lambda) \int _{0}^{1} u_{3}(t)\,dt \\ =& F \biggl( \int_{0}^{1} u_{1}(t)\,dt, \int_{0}^{1} u_{2}(t)\,dt, \int_{0}^{1} u_{3}(t)\,dt,\lambda \biggr). \end{aligned}$$
$$\begin{aligned} & \int_{0}^{1} F \bigl(w(t)u(t),w(t)z_{1},w(t)z_{2},t \bigr)\,dt \\ &\quad= \int_{0}^{1} \bigl(w(t)u(t)-h(t) w(t)z_{1}-h(1-t)w(t)z_{2} \bigr)\,dt \\ &\quad= \int_{0}^{1} w(t)u(t)\,dt-z_{1} \int_{0}^{1} h(t) w(t)\,dt-z_{2} \int _{0}^{1}h(1-t)w(t)\,dt \\ &\quad=T_{F,w} \biggl( \int_{0}^{1} w(t)u(t)\,dt,z_{1},z_{2} \biggr), \end{aligned}$$
$$ \begin{aligned}[b] &T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} h(t) w(t)\,dt \biggr)u_{2}- \biggl( \int_{0}^{1} h(1-t) w(t)\,dt \biggr)u_{3},\\ & \quad (u_{1},u_{2},u_{3})\in\mathbb{R}\times \mathbb{R}\times\mathbb{R}. \end{aligned} $$
(2.9)
$$\begin{aligned} wF(u_{1},u_{2},u_{3},u_{4}) =&w \bigl(u_{1}-h(u_{4}) u_{2}-h(1-u_{4})u_{3} \bigr) \\ =&wu_{1}-h(u_{4}) (wu_{2})-h(1-u_{4}) (wu_{3}) \\ =&F(wu_{1},wu_{2},wu_{3},u_{4}). \end{aligned}$$
$$ L_{w}=0. $$
(2.10)
$$\begin{aligned} F \bigl(f \bigl(tx+(1-t)y \bigr),f(x),f(y),t \bigr) =&f \bigl(tx+(1-t)y \bigr)-h(t) f(x)-h(1-t)f(y) \\ \leq& 0. \end{aligned}$$
3 Integral inequalities involving F-convex functions
Some integral inequalities via F-convex functions are presented in this section.
Theorem 3.1
Let
\(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be an
F-convex function, for some
\(F\in\mathcal{F}\). Suppose that
\(f\in L^{1}(a,b)\). Then
$$\begin{aligned} &F \biggl(f \biggl(\frac{a+b}{2} \biggr),\frac{1}{b-a} \int_{a}^{b} f(x)\,dx,\frac{1}{b-a} \int_{a}^{b} f(x)\,dx,\frac{1}{2} \biggr) \leq0, \end{aligned}$$
(3.1)
$$\begin{aligned} &T_{F,1} \biggl(\frac{1}{b-a} \int_{a}^{b} f(x)\,dx,f(a),f(b) \biggr)\leq0. \end{aligned}$$
(3.2)
Proof
Since f is an F-convex function, for every \(u,v\in[a,b]\), we have Taking we obtain Using axiom (A1), we get On the other hand, we have Therefore, which proves (3.1).
$$F \biggl(f \biggl(\frac{1}{2}u+ \biggl(1-\frac{1}{2} \biggr)v \biggr),f(u),f(v),\frac{1}{2} \biggr)\leq0. $$
$$u=ta+(1-t)b, \qquad v=tb+(1-t)a,\quad t\in[0,1], $$
$$F \biggl(f \biggl(\frac{a+b}{2} \biggr), f \bigl(ta+(1-t)b \bigr),f \bigl(tb+(1-t)a \bigr),\frac {1}{2} \biggr)\leq0. $$
$$F \biggl( \int_{0}^{1} f \biggl(\frac{a+b}{2} \biggr)\,dt, \int_{0}^{1} f \bigl(ta+(1-t)b \bigr)\,dt, \int_{0}^{1} f \bigl(tb+(1-t)a \bigr)\,dt, \frac{1}{2} \biggr)\leq0. $$
$$\int_{0}^{1} f \bigl(ta+(1-t)b \bigr)\,dt= \int_{0}^{1} f \bigl(tb+(1-t)a \bigr)\,dt= \frac{1}{b-a} \int _{a}^{b} f(x)\,dx. $$
$$F \biggl( \int_{0}^{1} f \biggl(\frac{a+b}{2} \biggr)\,dt, \frac{1}{b-a} \int_{a}^{b} f(x)\,dx, \frac{1}{b-a} \int_{a}^{b} f(x)\,dx,\frac{1}{2} \biggr) \leq0, $$
Again, since f is an F-convex function, for every \(t\in(0,1)\), we have Using axiom (A2) with \(w\equiv1\), and integrating over \((0,1)\) with respect to the variable t, we obtain that is, which proves (3.2). □
$$F \bigl(f \bigl(ta+(1-t)b \bigr),f(a),f(b),t \bigr)\leq0. $$
$$T_{F,1} \biggl( \int_{0}^{1} f \bigl(ta+(1-t)b \bigr)\,dt,f(a),f(b) \biggr)\leq0, $$
$$T_{F,1} \biggl(\frac{1}{b-a} \int_{a}^{b} f(x)\,dx,f(a),f(b) \biggr)\leq0, $$
Remark 3.2
The following lemma will be useful later (see [13]).
Lemma 3.3
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\). Then
$$\frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx=\frac{b-a}{2} \int_{0}^{1} (1-2t)f' \bigl(ta+(1-t)b \bigr)\,dt. $$
We have the following result.
Theorem 3.4
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\). Suppose that
Then
(i)
\(|f'|\)
is
F-convex on
\([a,b]\), for some
\(F\in\mathcal{F}\).
(ii)
The function
\(t\in(0,1)\mapsto L_{w(t)}\)
belongs to
\(L^{1}(0,1)\), where
\(w(t)=|1-2t|\).
$$ T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr)+ \int_{0}^{1} L_{w(t)}\,dt\leq0. $$
(3.3)
Proof
Since \(|f'|\) is F-convex, we have Multiplying this inequality by \(w(t)\) and using axiom (A3), we get Integration over \((0,1)\) with respect to the variable t and using axiom (A2), we obtain On the other hand, from Lemma 3.3, we have Since \(T_{F,w}\) is nondecreasing with respect to the first variable, we deduce that which proves (3.3). □
$$F \bigl(\bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr|,t \bigr)\leq0,\quad t\in(0,1). $$
$$F \bigl(w(t)\bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|,w(t)\bigl|f'(a)\bigr|,w(t)\bigl|f'(b)\bigr|,t \bigr)+L_{w(t)}\leq0,\quad t\in(0,1). $$
$$T_{F,w} \biggl( \int_{0}^{1}w(t)\bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|\,dt,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr) + \int_{0}^{1} L_{w(t)}\,dt\leq0. $$
$$\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\, dx \biggr\vert \leq \int_{0}^{1} w(t)\bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|\,dt. $$
$$T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}- \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr)+ \int_{0}^{1} L_{w(t)}\,dt\leq0, $$
Another similar result is given by the following theorem.
Theorem 3.5
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\), and let
\(p>1\). Suppose that
\(|f'|^{p/(p-1)}\)
is
F-convex on
\([a,b]\), for some
\(F\in\mathcal{F}\), and
\(|f'|\in L^{p/(p-1)}(a,b)\). Then
where
$$ T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr)\leq0, $$
(3.4)
$$A(p,f)= \biggl(\frac{2}{b-a} \biggr)^{\frac{p}{p-1}}(p+1)^{\frac {1}{p-1}} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert ^{\frac{p}{p-1}}. $$
Proof
Since \(|f'|^{\frac{p}{p-1}}\) is F-convex, we have Using axiom (A2) with \(w\equiv1\), and integrating over \((0,1)\) with respect to the variable t, we get On the other hand, using Lemma 3.3 and Hölder’s inequality, we obtain that is, Since \(T_{F,1}\) is nondecreasing with respect to the first variable, we obtain which proves (3.4). □
$$F \bigl(\bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|^{\frac{p}{p-1}}, \bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}},t \bigr)\leq0, \quad t\in(0,1). $$
$$T_{F,1} \biggl( \int_{0}^{1} \bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|^{\frac{p}{p-1}}\,dt,\bigl|f'(a)\bigr|^{\frac {p}{p-1}},\bigl|f'(b)\bigr|^{\frac{p}{p-1}} \biggr)\leq0. $$
$$\begin{aligned} & \biggl(\frac{2}{b-a} \biggr)^{\frac{p}{p-1}}(p+1)^{\frac{1}{p-1}} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert ^{\frac {p}{p-1}} \\ &\quad\leq \int_{0}^{1} \bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|^{\frac{p}{p-1}}\,dt, \end{aligned}$$
$$A(p,f)\leq \int_{0}^{1} \bigl|f' \bigl(ta+(1-t)b \bigr)\bigr|^{\frac{p}{p-1}}\,dt. $$
$$T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr)\leq0, $$
4 Particular cases
As consequences of the presented theorems, we obtain in this section some integral inequalities for different (and independent) kinds of convexity.
4.1 The case of ε-convexity
We have the following Hermite-Hadamard inequalities for ε-convex functions.
Corollary 4.1
Let
\(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be an
ε-convex function, \(\varepsilon\geq0\). Suppose that
\(f\in L^{1}(a,b)\). Then
$$f \biggl(\frac{a+b}{2} \biggr)-\varepsilon\leq\frac{1}{b-a} \int_{a}^{b} f(x)\, dx\leq\frac{f(a)+f(b)}{2}+ \varepsilon. $$
Proof
From Example 2.3, we know that an ε-convex function is an F-convex. Using (2.2) and (2.3) with \(w\equiv1\), we have and So, applying Theorem 3.1, we obtain the desired result. □
$$F(u_{1},u_{2},u_{3},u_{4})=u_{1}-u_{4}u_{2}-(1-u_{4})u_{3}- \varepsilon,\quad (u_{1},u_{2},u_{3},u_{4}) \in \mathbb{R}\times\mathbb{R}\times\mathbb{R}\times[0,1], $$
$$T_{F,1}(u_{1},u_{2},u_{3})=u_{1}- \frac{u_{2}+u_{3}}{2}-\varepsilon,\quad (u_{1},u_{2},u_{3}) \in\mathbb{R}\times\mathbb{R}\times\mathbb{R}. $$
Taking \(\varepsilon=0\) in Corollary 4.1, we obtain the following standard Hermite-Hadamard inequalities for convex functions.
Corollary 4.2
Let
\(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be a convex function. Suppose that
\(f\in L^{1}(a,b)\). Then
$$f \biggl(\frac{a+b}{2} \biggr)\leq\frac{1}{b-a} \int_{a}^{b} f(x)\,dx\leq\frac {f(a)+f(b)}{2}. $$
Corollary 4.3
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\). Suppose that the function
\(|f'|\)
is
ε-convex on
\([a,b]\), \(\varepsilon\geq0\). Then
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(x)\,dx \biggr\vert \leq (b-a) \biggl[\frac{|f'(a)|+|f'(b)|}{8}+\frac{\varepsilon}{4} \biggr]. $$
Proof
Using (2.4) with \(w(t)=|1-2t|\), we obtain Using (2.3) with \(w(t)=|1-2t|\), we obtain for all \((u_{1},u_{2},u_{3})\in\mathbb{R}\times\mathbb{R}\times\mathbb{R}\). On the other hand, simple computations yield Therefore, we have Then Now, by Theorem 3.4, we have that is, which proves the desired inequality. □
$$\int_{0}^{1} L_{w(t)}\,dt=\varepsilon \int_{0}^{1} \bigl(1-w(t) \bigr)\,dt=\varepsilon \biggl(1- \int_{0}^{1} |1-2t|\,dt \biggr)=\frac{\varepsilon}{2}. $$
$$T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} t|1-2t|\,dt \biggr)u_{2}- \biggl( \int _{0}^{1}(1-t)|1-2t|\,dt \biggr)u_{3}- \varepsilon, $$
$$\int_{0}^{1} t|1-2t|\,dt= \int_{0}^{1} (1-t)|1-2t|\,dt=\frac{1}{4}. $$
$$T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \frac{u_{2}+u_{3}}{4}-\varepsilon,\quad (u_{1},u_{2},u_{3}) \in\mathbb{R}\times\mathbb{R}\times\mathbb{R}. $$
$$\begin{aligned} &T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}- \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr)+ \int_{0}^{1} L_{w(t)}\,dt \\ &\quad=\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\, dx \biggr\vert - \frac{|f'(a)|+|f'(b)|}{4}-\frac{\varepsilon}{2}. \end{aligned}$$
$$T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}- \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr)+ \int_{0}^{1} L_{w(t)}\,dt\leq0, $$
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(x)\,dx \biggr\vert \leq (b-a) \biggl[\frac{|f'(a)|+|f'(b)|}{8}+\frac{\varepsilon}{4} \biggr], $$
Corollary 4.4
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\). Suppose that the function
\(|f'|\)
is convex on
\([a,b]\). Then
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b} f(x)\,dx \biggr\vert \leq \frac{(b-a) (|f'(a)|+|f'(b)| )}{8}. $$
Corollary 4.5
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\), and let
\(p>1\). Suppose that
\(|f'|^{p/(p-1)}\)
is
ε-convex on
\([a,b]\), \(\varepsilon\geq0\), and
\(|f'|\in L^{p/(p-1)}(a,b)\). Then
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \leq \frac{(b-a)}{2(p+1)^{\frac{1}{p}}} \biggl(\frac{|f'(a)|^{\frac {p}{p-1}}+|f'(b)|^{\frac{p}{p-1}}}{2}+\varepsilon \biggr)^{\frac{p-1}{p}}. $$
Proof
Using (2.3) with \(w\equiv1\), we obtain Then where By Theorem 3.5, we have that is, which proves the desired inequality. □
$$T_{F,1}(u_{1},u_{2},u_{3})=u_{1}- \frac{u_{2}+u_{3}}{2}-\varepsilon,\quad (u_{1},u_{2},u_{3}) \in\mathbb{R}\times\mathbb{R}\times\mathbb{R}. $$
$$\begin{aligned} &T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr) \\ &\quad=A(p,f)-\frac{|f'(a)|^{\frac{p}{p-1}}+|f'(b)|^{\frac {p}{p-1}}}{2}-\varepsilon, \end{aligned}$$
$$A(p,f)= \biggl(\frac{2}{b-a} \biggr)^{\frac{p}{p-1}}(p+1)^{\frac {1}{p-1}} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert ^{\frac{p}{p-1}}. $$
$$T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr)\leq0, $$
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \leq \frac{(b-a)}{2(p+1)^{\frac{1}{p}}} \biggl(\frac{|f'(a)|^{\frac {p}{p-1}}+|f'(b)|^{\frac{p}{p-1}}}{2}+\varepsilon \biggr)^{\frac{p-1}{p}}, $$
Corollary 4.6
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\), and let
\(p>1\). Suppose that
\(|f'|^{p/(p-1)}\)
is convex on
\([a,b]\), and
\(|f'|\in L^{p/(p-1)}(a,b)\). Then
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \leq \frac{(b-a)}{2(p+1)^{\frac{1}{p}}} \biggl(\frac{|f'(a)|^{\frac {p}{p-1}}+|f'(b)|^{\frac{p}{p-1}}}{2} \biggr)^{\frac{p-1}{p}}. $$
4.2 The case of α-convexity
We have the following Hermite-Hadamard inequalities for α-convex functions.
Corollary 4.7
Let
\(f: [a,b]\to\mathbb{R}\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be an
α-convex function, \(\alpha\in(0,1]\). Suppose that
\(f\in L^{1}(a,b)\). Then
$$f \biggl(\frac{a+b}{2} \biggr)\leq\frac{1}{b-a} \int_{a}^{b} f(x)\,dx\leq\frac {f(a)+\alpha f(b)}{\alpha+1}. $$
Proof
From Example 2.5, we know that an α-convex function is F-convex. Using (2.5) and (2.6) with \(w\equiv1\), we have and So, applying Theorem 3.1, we obtain the desired result. □
$$F(u_{1},u_{2},u_{3},u_{4})=u_{1}-u_{4}^{\alpha}u_{2}- \bigl(1-u_{4}^{\alpha}\bigr)u_{3}, \quad (u_{1},u_{2},u_{3},u_{4})\in \mathbb{R}\times\mathbb{R}\times\mathbb{R}\times[0,1], $$
$$T_{F,1}(u_{1},u_{2},u_{3})=u_{1}- \frac{u_{2}}{\alpha+1}-\frac{\alpha}{\alpha+1} u_{3},\quad (u_{1},u_{2},u_{3}) \in\mathbb{R}\times\mathbb{R}\times\mathbb{R}. $$
Remark 4.8
Corollary 4.9
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\). Suppose that the function
\(|f'|\)
is
α-convex on
\([a,b]\), \(\alpha \in (0,1]\). Then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \\ &\quad\leq\frac{(b-a)}{2(\alpha+1)(\alpha+2)} \biggl( \bigl(2^{-\alpha }+\alpha \bigr)\bigl|f'(a)\bigr|+\frac{ [\alpha(\alpha+1)+2(1-2^{-\alpha }) ]}{2}\bigl|f'(b)\bigr| \biggr). \end{aligned}$$
Proof
Using (2.7), we have Using (2.6) with \(w(t)=|1-2t|\), we obtain for all \((u_{1},u_{2},u_{3})\in\mathbb{R}\times\mathbb{R}\times\mathbb{R}\). Simple computations yield and Therefore, for all \((u_{1},u_{2},u_{3})\in\mathbb{R}\times\mathbb{R}\times\mathbb{R}\). Now, we have So, applying Theorem 3.4, we obtain the desired result. □
$$\int_{0}^{1} L_{w(t)}\,dt=0. $$
$$T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} t^{\alpha}|1-2t|\,dt \biggr)u_{2}- \biggl( \int_{0}^{1} \bigl(1-t^{\alpha}\bigr)|1-2t| \,dt \biggr)u_{3}, $$
$$\int_{0}^{1} t^{\alpha}|1-2t|\,dt= \frac{1}{(\alpha+1)(\alpha+2)} \bigl(2^{-\alpha}+\alpha \bigr) $$
$$\int_{0}^{1} \bigl(1-t^{\alpha}\bigr)|1-2t| \,dt=\frac{\alpha(\alpha+1)+2(1-2^{-\alpha })}{2(\alpha+1)(\alpha+2)}. $$
$$T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \frac{ (2^{-\alpha}+\alpha )u_{2}}{(\alpha+1)(\alpha+2)}-\frac { [\alpha(\alpha+1)+2(1-2^{-\alpha}) ]u_{3}}{2(\alpha+1)(\alpha+2)}, $$
$$\begin{aligned} &T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}- \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr)+ \int_{0}^{1} L_{w(t)}\,dt \\ &\quad=T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}- \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr) \\ &\quad=\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\, dx \biggr\vert \\ &\qquad{}- \frac{ (2^{-\alpha}+\alpha )|f'(a)|}{(\alpha+1)(\alpha +2)}-\frac{ [\alpha(\alpha+1)+2(1-2^{-\alpha}) ]|f'(b)|}{2(\alpha+1)(\alpha+2)}. \end{aligned}$$
Corollary 4.11
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\), and let
\(p>1\). Suppose that
\(|f'|^{p/(p-1)}\)
is
α-convex on
\([a,b]\), \(\alpha\in(0,1]\), and
\(|f'|\in L^{p/(p-1)}(a,b)\). Then
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \leq \frac{(b-a)}{2(p+1)^{\frac{1}{p}}} \biggl(\frac{|f'(a)|^{\frac {p}{p-1}}+\alpha|f'(b)|^{\frac{p}{p-1}}}{\alpha+1} \biggr)^{\frac{p-1}{p}}. $$
Proof
Using (2.6) with \(w\equiv1\), we obtain Then where By Theorem 3.5, we have that is, which is the desired inequality. □
$$T_{F,1}(u_{1},u_{2},u_{3})=u_{1}- \frac{u_{2}}{\alpha+1}-\frac{\alpha}{\alpha+1} u_{3},\quad (u_{1},u_{2},u_{3}) \in\mathbb{R}\times\mathbb{R}\times\mathbb{R}. $$
$$\begin{aligned} &T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr) \\ &\quad=A(p,f)-\frac{|f'(a)|^{\frac{p}{p-1}}+\alpha|f'(b)|^{\frac {p}{p-1}}}{\alpha+1}, \end{aligned}$$
$$A(p,f)= \biggl(\frac{2}{b-a} \biggr)^{\frac{p}{p-1}}(p+1)^{\frac {1}{p-1}} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert ^{\frac{p}{p-1}}. $$
$$T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr)\leq0, $$
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \leq \frac{(b-a)}{2(p+1)^{\frac{1}{p}}} \biggl(\frac{|f'(a)|^{\frac {p}{p-1}}+\alpha|f'(b)|^{\frac{p}{p-1}}}{\alpha+1} \biggr)^{\frac{p-1}{p}}, $$
4.3 The case of h-convex functions
Let \(h: J\to[0,\infty)\) be a given function which is not identical to 0, where J is an interval in \(\mathbb{R}\) such that \((0,1)\subseteq J\). We suppose that \(h\in L^{1}(0,1)\) and \(h (\frac{1}{2} )\neq0\).
We have the following Hermite-Hadamard inequalities for h-convex functions (obtained by Sarikaya et al. [14]).
Corollary 4.13
Let
\(f: [a,b]\to[0,\infty)\), \((a,b)\in\mathbb{R}^{2}\), \(a< b\), be a
h-convex function. Suppose that
\(f\in L^{1}(a,b)\). Then
$$\frac{1}{2h (\frac{1}{2} )}f \biggl(\frac{a+b}{2} \biggr)\leq \frac{1}{b-a} \int_{a}^{b} f(x)\,dx \leq \biggl( \int_{0}^{1} h(t)\,dt \biggr) \bigl(f(a)+f(b) \bigr). $$
Proof
From Example 2.6, we know that a h-convex function is F-convex. Using (2.8) and (2.9) with \(w\equiv1\), we have and So, applying Theorem 3.1, we obtain the desired result. □
$$F(u_{1},u_{2},u_{3},u_{4})=u_{1}-h(u_{4}) u_{2}-h(1-u_{4})u_{3},\quad (u_{1},u_{2},u_{3},u_{4}) \in\mathbb{R}\times\mathbb{R}\times\mathbb{R}\times(0,1). $$
$$T_{F,1}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} h(t)\,dt \biggr) (u_{2}+u_{3}), \quad (u_{1},u_{2},u_{3})\in\mathbb{R}\times \mathbb{R}\times\mathbb{R}. $$
Corollary 4.14
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\). Suppose that the function
\(|f'|\)
is
h-convex on
\([a,b]\). Then
$$\biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \leq (b-a) \biggl( \int_{0}^{1} h(t) |1-2t|\,dt \biggr) \biggl( \frac {|f'(a)|+|f'(b)|}{2} \biggr). $$
Proof
Using (2.10), we have Using (2.9) with \(w(t)=|1-2t|\), we obtain Now, we have So, applying Theorem 3.4, we obtain the desired result. □
$$\int_{0}^{1} L_{w(t)}\,dt=0. $$
$$T_{F,w}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} h(t) |1-2t|\,dt \biggr) (u_{2}+u_{3}),\quad (u_{1},u_{2},u_{3}) \in\mathbb{R}\times\mathbb{R}\times \mathbb{R}. $$
$$\begin{aligned} &T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}- \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr)+ \int_{0}^{1} L_{w(t)}\,dt \\ &\quad=T_{F,w} \biggl(\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}- \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \biggr\vert ,\bigl|f'(a)\bigr|,\bigl|f'(b)\bigr| \biggr) \\ &\quad=\frac{2}{b-a} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\, dx \biggr\vert \\ &\qquad{}- \biggl( \int_{0}^{1} h(t) |1-2t|\,dt \biggr) \bigl(\bigl|f'(a)\bigr|+\bigl|f'(b)\bigr| \bigr). \end{aligned}$$
Corollary 4.15
Let
\(f: I^{\circ}\subseteq\mathbb{R}\to\mathbb{R}\)
be a differentiable mapping on
\(I^{\circ}\), \((a,b)\in I^{\circ}\times I^{\circ}\), \(a< b\), and let
\(p>1\). Suppose that
\(|f'|^{p/(p-1)}\)
is
h-convex on
\([a,b]\), and
\(|f'|\in L^{p/(p-1)}(a,b)\). Then
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \\ &\quad\leq \frac{(b-a)}{2(p+1)^{\frac{1}{p}}} \biggl( \int_{0}^{1} h(t)\,dt \biggr)^{\frac {p-1}{p}} \bigl(\bigl|f'(a)\bigr|^{\frac{p}{p-1}}+\bigl|f'(b)\bigr|^{\frac{p}{p-1}} \bigr)^{\frac{p-1}{p}}. \end{aligned}$$
Proof
Using (2.9) with \(w\equiv1\), we obtain Then where By Theorem 3.5, we have that is, which is the desired inequality. □
$$T_{F,1}(u_{1},u_{2},u_{3})=u_{1}- \biggl( \int_{0}^{1} h(t)\,dt \biggr) (u_{2}+u_{3}), \quad (u_{1},u_{2},u_{3})\in\mathbb{R}\times \mathbb{R}\times\mathbb{R}. $$
$$\begin{aligned} &T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr) =A(p,f)- \biggl( \int_{0}^{1} h(t)\,dt \biggr) \bigl(\bigl|f'(a)\bigr|^{\frac {p}{p-1}}+\bigl|f'(b)\bigr|^{\frac{p}{p-1}} \bigr), \end{aligned}$$
$$A(p,f)= \biggl(\frac{2}{b-a} \biggr)^{\frac{p}{p-1}}(p+1)^{\frac {1}{p-1}} \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert ^{\frac{p}{p-1}}. $$
$$T_{F,1} \bigl(A(p,f),\bigl|f'(a)\bigr|^{\frac{p}{p-1}},\bigl|f'(b)\bigr|^{\frac {p}{p-1}} \bigr)\leq0, $$
$$\begin{aligned} & \biggl\vert \frac{f(a)+f(b)}{2}-\frac{1}{b-a} \int_{a}^{b}f(x)\,dx \biggr\vert \\ &\quad\leq \frac{(b-a)}{2(p+1)^{\frac{1}{p}}} \biggl( \int_{0}^{1} h(t)\,dt \biggr)^{\frac {p-1}{p}} \bigl(\bigl|f'(a)\bigr|^{\frac{p}{p-1}}+\bigl|f'(b)\bigr|^{\frac{p}{p-1}} \bigr)^{\frac{p-1}{p}}, \end{aligned}$$
Acknowledgements
The author extends his appreciation to the Distinguished Scientist Fellowship Program (DSFP) at King Saud University (Saudi Arabia).
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The author declares to have no competing interests.