In this paper, a new identity for functions defined on an open invex subset of set of real numbers is established, and by using the this identity and the Hölder and Power mean integral inequalities we present new type integral inequalities for functions whose powers of third derivatives in absolute value are preinvex and prequasiinvex functions.
Definition 1.1. A function \(f:I\subset \mathbb{R}\rightarrow\mathbb{R}\) is said to be convex if the inequality \(f\left( tx+(1-t)y\right) \leq tf\left( x\right) +(1-t)f\left( y\right)\) is valid for all \(x,y\in I\) and \( t\in \left[ 0,1\right]\). If this inequality reverses, then \(f\) is said to be concave on interval \(I\neq \varnothing\) .
This definition is well known in the literature. Convexity theory has appeared as a powerful technique to study a wide class of unrelated problems in pure and applied sciences.Definition 1.2. \(f:I\subset\mathbb{R}\rightarrow \mathbb{R} \) be a convex function on the interval \(I\) of real numbers and \(a,b\in I\) with \(a< b\). The following celebrated double inequality is well known in the literature as Hermite-Hadamard's inequality for convex functions [1]. Both inequalities hold in the reserved direction if \(f\) is concave.
The classical Hermite-Hadamard inequality provides estimates of the mean value of a continuous convex or concave function. Hadamard’s inequality for convex or concave functions has received renewed attention in recent years and a remarkable variety of refinements and generalizations have been found; for example see [1, 2, 3, 4, 5]. Hermite-Hadamard inequality [6] has been considered the most useful inequality in mathematical analysis. Some of the classical inequalities for means can be derived from Hermite-Hadamard inequality for particular choices of the function \(f\).Definition 1.3. A function \(f:I\subseteq \mathbb{R} \rightarrow \mathbb{R}\) is said to be quasi-convex if the inequality \(f\left( tx+(1-t)y\right) \leq \max \left\{ f\left( x\right) ,f\left( y\right) \right\}\) holds for all \(x,y\in I\) and \(t\in \left[ 0,1\right]\) . Clearly, any convex function is a quasi-convex function. Furthermore, there exist quasi-convex functions which are not convex [7].
Let us recall the notions of preinvexity and prequasiinvexity which are signicant generalizations of the notions of convexity and qusi-convexity respectively, and some related results.Definition 1.4.[8] Let \(K\) be a non-empty subset in \(\mathbb{R}^{n}\) and \(\eta :K\times K\rightarrow\mathbb{R}^{n}\). Let \(x\in K\), then the set \(K\) is said to be invex at \(x\) with respect to \(\eta \left( \cdot ,\cdot \right) \), if \(x+t\eta \left( y,x\right) \in K,~~~\forall x,y\in K~~~t\in \left[ 0,1\right] .\) \(K\) is said to be an invex set with respect to \(\eta \) if \(K\) is invex at each \(x\in K\). The invex set \(K\) is also called \(\eta \)-connected set.
Definition 1.4 essentially says that there is a path starting from a point \(x\) which is contained in \(K\). We do not require that the point \(y\) should be one of the end points of the path. This observation plays an important role in our analysis. Note that, if we demand that \(y\) should be an end point of the path for every pair of points \(x,y\in K\), then \(\eta \left( y,x\right)=y-x\), and consequently invexity reduces to convexity. Thus, it is true that every convex set is also an invex set with respect to \(\eta \left( y,x\right)=y-x\), but the converse is not necessarily true, see [9, 10] and the references therein. For the sake of simplicity, we always assume that \(K=\left[ x,x+t\eta \left( y,x\right) \right] \), unless otherwise specified [11].Definition 1.5.[8] A function \(f:K\rightarrow\mathbb{R}\) on an invex set \(K\subseteq\mathbb{R} \) is said to be preinvex with respect to \(\eta \), if \(f\left( u+t\eta (v,u)\right) \leq (1-t)f(u)+tf(v),~~~\forall u,v\in K,~~t\in \left[ 0,1% \right] .\) The function \(f\) is said to be preconcave if and only if \(-f\) is preinvex.
It is to be noted that every convex function is preinvex with respect to the map \(\eta \left( y,x\right) =x-y\) but the converse is not true see for instance.Definition 1.6.[12] A function \(f:K\rightarrow\mathbb{R}\) on an invex set \(K\subseteq\mathbb{R}\) is said to be prequasiinvex with respect to \(\eta \), if \(f\left( u+t\eta (v,u)\right) \leq \max \left\{ f(u),f(v)\right\} ,\forall u,v\in K,~~t\in % \left[ 0,1\right] .\)
Also every quasi-convex function is a prequasiinvex with respect to the map \( \eta (v,u)=v-u\) but the converse does not hold, see for example [12]. Mohan and Neogy [9] introduced Condition \(C\) defined as follows:Definition 1.7.[9] Let \(S\subseteq \mathbb{R}\) be an open invex subset with respect to the map \(\eta :S\times S\rightarrow\mathbb{R}\). We say that the function satisfies the Condition \(C\) if, for any \(x,y\in S \) and any \(t\in \left[ 0,1\right] \),
Theorem 1.8.[18] Let \(f:\left[ a,a+t\eta (b,a)\right] \rightarrow \left( 0,\infty \right) \) be a preinvex function on the interval of the real numbers \(K^{\circ }\) (the interior of \(K\)) and \(a,b\in K^{\circ }\) with \(\eta (b,a)>0\). Then the following inequalities holds
Lemma 2.1. Let \(K\subseteq\mathbb{R}\) be an open invex subset with respect to mapping \(\eta \left( \cdot ,\cdot \right) :K\times K\rightarrow\mathbb{R}^{n}\) and \(a,b\in K\) with \(\eta (b,a)>0\). Suppose that the function \(f:K\rightarrow\mathbb{R}\) is a three times differentiable function on \(K\) such that \(f^{\prime \prime \prime }\in L\left[ a,a+\eta (b,a)\right] .\) Then the following identity hold: \begin{eqnarray*} &&\frac{\eta ^{2}(b,a)}{2}\beta f^{\prime \prime }\left( a+\eta (b,a)\right) -\eta (b,a)\alpha f^{\prime }\left( a+\eta (b,a)\right) \\ &&+f\left( a+\eta (b,a)\right) \left( a+\eta (b,a)\right) -f(a)a-\int_{a}^{a+\eta (b,a)}f(x)dx \\ &=&\eta ^{3}(b,a)\int_{0}^{1}\frac{t^{2}}{2}\beta _{t}f^{\prime \prime \prime }(a+t\eta (b,a))dt. \end{eqnarray*}
Proof. Integrating three times by parts and then changing the variable, we obtain \begin{eqnarray*} &&\eta ^{3}(b,a)\int_{0}^{1}\frac{t^{2}}{2}\beta _{t}f^{\prime \prime \prime }(a+t\eta (b,a))dt \\ &=&\left. \eta ^{2}(b,a)\frac{t^{2}}{2}\beta _{t}f^{\prime \prime }\left( a+t\eta (b,a)\right) \right\vert _{0}^{1}-\left. \eta (b,a)t\alpha _{t}f^{\prime }\left( a+t\eta (b,a)\right) \right\vert _{0}^{1} \\ &&+\left. \left( a+t\eta (b,a)\right) f\left( a+t\eta (b,a)\right) \right\vert _{0}^{1}-\eta (b,a)\int_{0}^{1}f\left( a+t\eta (b,a)\right) dt \\ &=&\frac{\eta ^{2}(b,a)}{2}\beta f^{\prime \prime }\left( a+\eta (b,a)\right) -\eta (b,a)\alpha f^{\prime }\left( a+\eta (b,a)\right) \\ &&+\left( a+\eta (b,a)\right) f\left( a+\eta (b,a)\right) -af\left( a\right) -\eta (b,a)\int_{0}^{1}f\left( a+t\eta (b,a)\right) dt \\ &=&\frac{\eta ^{2}(b,a)}{2}\beta f^{\prime \prime }\left( a+\eta (b,a)\right) -\eta (b,a)\alpha f^{\prime }\left( a+\eta (b,a)\right) \\ &&+\left( a+\eta (b,a)\right) f\left( a+\eta (b,a)\right) -af\left( a\right) -\int_{a}^{a+\eta (b,a)}f(x)dx. \end{eqnarray*} This completes the proof of lemma.
Theorem 2.2. Let \(K\subseteq\mathbb{R}\) be an open invex subset with respect to mapping \(\eta \left( \cdot ,\cdot \right) :K\times K\rightarrow\mathbb{R}^{n}\) and \(a,b\in K\) with \(\eta (b,a)>0\). Suppose that the function \(f:K \rightarrow\mathbb{R}\) is a three times differentiable function on \(K\) such that \(f^{\prime \prime \prime }\in L\left[ a,a+\eta (b,a)\right] .\) If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is preinvex on \(K\) for \(q>1\), then the following inequality holds:
Proof. If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) for \(q>1\) is preinvex on \(\left[ a,a+\eta (b,a)\right] \), using Lemma 2.1, the Hölder integral inequality and \(\left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}\leq t\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}+(1-t)\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\)we get \begin{eqnarray*} &&\left\vert I_{f}(a,b,\eta )\right\vert \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert dt \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2p}dt\right) ^{\frac{1 }{p}}\left( \int_{0}^{1}\left\vert \beta _{t}\right\vert ^{q}\left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}dt\right) ^{\frac{1}{ q}} \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2p}dt\right) ^{\frac{1 }{p}}\left( \int_{0}^{1}\left\vert \beta _{t}\right\vert ^{q}\left[ t\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}+(1-t)\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\right] dt\right) ^{\frac{1}{q}} \\ &\leq &\frac{1}{2}\frac{\eta ^{3}(b,a)}{\left( 2p+1\right) ^{\frac{1}{p}}} \left( \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\int_{0}^{1}t\left\vert \beta _{t}\right\vert ^{q}dt+\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\int_{0}^{1}\left( 1-t\right) \left\vert \beta _{t}\right\vert ^{q}dt\right) ^{\frac{1}{q}} \\ &=&\frac{1}{2}\frac{\eta ^{3}(b,a)}{\left( 2p+1\right) ^{\frac{1}{p}}}\left( \frac{3\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}}{\eta ^{2}(b,a)}\int_{a}^{\beta }3\left( x-a\right) \left\vert x\right\vert ^{q}dx\right. \\ &&+\left. \frac{3\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}}{ \eta ^{2}(b,a)}\int_{a}^{\beta }\left( \eta (b,a)-3(x-a)\right) \left\vert x\right\vert ^{q}dx\right) ^{\frac{1}{q}} \\ &=&\frac{1}{2}3^{\frac{1}{q}}\frac{\eta ^{1+\frac{2}{p}}(b,a)}{\left( 2p+1\right) ^{\frac{1}{p}}}\left( \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\int_{a}^{\beta }3\left( x-a\right) \left\vert x\right\vert ^{q}dx\right. \\ &&+\left. \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\int_{a}^{\beta }\left( \eta (b,a)-3(x-a)\right) \left\vert x\right\vert ^{q}dx\right) ^{\frac{1}{q}} \\ &=&\frac{3^{\frac{1}{q}}}{2}\frac{\eta ^{1+\frac{2}{p}}(b,a)}{\left( 2p+1\right) ^{\frac{1}{p}}}\left[ \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}C_{1,\eta }\left( a,b\right) +\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}C_{2,\eta }\left( a,b\right) \right] ^{ \frac{1}{q}}. \end{eqnarray*} This completes the proof of theorem.
Corollary 2.3. Suppose that all the assumptions of Theorem 2.2 are satisfied. If we choose \(\eta (b,a)=b-a\) then when \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is convex on \(K\) for \(q>1\) we obtain following inequality: \begin{eqnarray*} &&\left\vert \frac{b-a}{2}\left( \frac{2a+b}{3}\right) f^{\prime \prime }(b)- \frac{a+b}{2}f^{\prime }(b)+\frac{f(b)b-f(a)a}{b-a}-\frac{1}{b-a} \int_{a}^{b}f(x)dx\right\vert \\ &\leq &\frac{1}{2}3^{\frac{1}{q}}\frac{(b-a)^{\frac{2}{p}}}{\left( 2p+1\right) ^{\frac{1}{p}}}\left[ \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}C_{1}\left( a,b\right) +\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}C_{2}\left( a,b\right) \right] ^{\frac{1}{q}}, \end{eqnarray*} where% \begin{equation*} C_{1}\left( a,b\right) =\left\{ \begin{array}{c} (b-a)\left[ L_{q+1}^{q+1}\left( \frac{2a+b}{3},a\right)-aL_{q}^{q}\left( \frac{2a+b}{3},a\right) \right] ,~\ a>0, \frac{2a+b}{3}>0, \\ \left( 5a+b\right) L_{q+1}^{q+1}\left( \frac{2a+b}{3},-a\right)\\ -\frac{6a}{ q+1}A\left( \left( \frac{2a+b}{3}\right) ^{q+1},(-a)^{q+1}\right) ,~\ \ \ \ \ \ \ \ \ \ \ \ a0, \\ -(b-a)\left[ L_{q+1}^{q+1}\left( -a,-\frac{2a+b}{3}\right) +aL_{q}^{q}\left( -a,-\frac{2a+b}{3}\right) \right],a< 0,\frac{2a+b}{3}0,\frac{2a+b}{3}>0, \\ -\left( 5a+b\right) L_{q+1}^{q+1}\left( \frac{2a+b}{3},-a\right)\\ +\frac{% 6\beta }{q+1}A\left( \left( \frac{2a+b}{3}\right) ^{q+1},(-a)^{q+1}\right) ,a0, \\ (b-a)\left[ L_{q+1}^{q+1}\left( -a,-\frac{2a+b}{3}\right) +\alpha L_{q}^{q}\left( -a,-\frac{2a+b}{3}\right) \right] ,~\ a< 0,\frac{2a+b}{3}< 0. \end{array} \right. \end{equation*}
Remark 2.5. If the mapping \(\eta \) satisfies condition C then by use of the preinvexity of \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) we obtain following inequality for every \(t\in \left[ 0,1\right] \):
Theorem 2.5. Let \(K\subseteq\mathbb{R}\) be an open invex subset with respect to mapping \(\eta \left( \cdot ,\cdot \right) :K\times K\rightarrow\mathbb{R}^{n}\) and \(a,b\in K\) with \(\eta (b,a)>0\). Suppose that the function \(f:K \rightarrow\mathbb{R}\) is a three times differentiable function on \(K\) such that \(f^{\prime \prime \prime }\in L\left[ a,a+\eta (b,a)\right] .\) If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is preinvex on \(K\) for \(q>1\), then the following inequality holds:
Proof. If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) for \(q>1\) is preinvex on \(\left[ a,a+\eta (b,a)\right] \), using Lemma 2.1, the Hölder integral inequality and \(\left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}\leq t\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}+(1-t)\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\) we obtain the following inequality: \begin{eqnarray*} &&\left\vert I_{f}(a,b,\eta )\right\vert \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert dt \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}\left\vert \beta _{t}\right\vert ^{p}dt\right) ^{\frac{1}{p}}\left( \int_{0}^{1}t^{2q}\left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}dt\right) ^{\frac{1}{q}} \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}\left\vert \beta _{t}\right\vert ^{p}dt\right) ^{\frac{1}{p}}\\ &&\times \left( \int_{0}^{1}t^{2q}\left[ t\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}+(1-t)\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\right] dt\right) ^{\frac{1}{q}} \\ &=&\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}\left\vert \beta _{t}\right\vert ^{p}dt\right) ^{\frac{1}{p}}\\ &&\times \left( \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\int_{0}^{1}t^{2q+1}dt+\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\int_{0}^{1}\left( t^{2q}-t^{2q+1}\right) dt\right) ^{\frac{1}{q}} \end{eqnarray*} \begin{eqnarray*} &=&\frac{1}{2}3^{\frac{1}{p}}\eta ^{2+\frac{1}{q}}(b,a)\left( \int_{a}^{\beta }\left\vert x\right\vert ^{p}dx\right) ^{\frac{1}{p}}\left[ \frac{\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}}{2q+2} +\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\left( \frac{1}{2q+1} -\frac{1}{2q+2}\right) \right] ^{\frac{1}{q}} \\ &=&\frac{1}{2}3^{\frac{1}{p}}\eta ^{2+\frac{1}{q}}(b,a)C_{3,\eta }^{\frac{1}{ p}}\left( a,b\right) \left[ \frac{\left( 2q+1\right) \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}+\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}}{\left( 2q+1\right) \left( 2q+2\right) }\right] ^{\frac{ 1}{q}}. \end{eqnarray*} This completes the proof of theorem.
Corollary 2.6. Suppose that all the assumptions of Theorem 2.5 are satisfied. If we choose \(\eta (b,a)=b-a\) then when \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is convex on \(K\) for \(q>1\) we have the following inequality: \begin{eqnarray*} &&\left\vert \frac{b-a}{2}\left( \frac{2a+b}{3}\right) f^{\prime \prime }(b)- \frac{a+b}{2}f^{\prime }(b)+\frac{f(b)b-f(a)a}{b-a}-\frac{1}{b-a}% \int_{a}^{b}f(x)dx\right\vert \\ &\leq &\frac{3^{\frac{1}{p}}}{2}(b-a)^{1+\frac{1}{q}}C_{3}^{\frac{1}{p}% }\left( a,b\right) \left[ \frac{\left( 2q+1\right) \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}+\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}}{\left( 2q+1\right) \left( 2q+2\right) }\right] ^{\frac{% 1}{q}}, \end{eqnarray*} where \begin{equation*} C_{3}\left( a,b\right) =\left\{ \begin{array}{c} \frac{b-a}{3}L_{p}^{p}\left( \frac{2a+b}{3},a\right) ,~\ \ \ \ \ \ \ \ \ \ \ \ \ \ a>0,~\frac{2a+b}{3}>0, \\ \frac{2}{p+1}A\left( \left( \frac{2a+b}{3}\right) ^{p+1},(-a)^{p+1}\right) ,a0, \\ \frac{b-a}{3}L_{p}^{p}\left( -a,-\frac{2a+b}{3}\right) ,~\ \ \ \ \ \ \ \ \ \ \ a< 0,~~\frac{2a+b}{3}< 0. \end{array} \right. \end{equation*}
Remark 2.7. If the mapping \(\eta \) satisfies condition C then using the inequality (5) in the proof of Theorem 2.5, then the inequality (7) becomes the following inequality:
Theorem 2.8. Let \(K\subseteq\mathbb{R}\) be an open invex subset with respect to mapping \(\eta \left( \cdot ,\cdot \right) :K\times K\rightarrow\mathbb{R}^{n}\) and \(a,b\in K\) with \(\eta (b,a)>0\). Suppose that the function \(f:K \rightarrow\mathbb{R}\) is a three times differentiable function on \(K\) such that \(f^{\prime \prime \prime }\in L\left[ a,a+\eta (b,a)\right] .\) If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is preinvex on \(K\) for \(q\geq 1\) , then the following inequality holds:
Proof. Using Lemma 2.1 and Power-mean integral inequality, we obtain \begin{eqnarray*} &&\left\vert I_{f}(a,b,\eta )\right\vert \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert dt \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert dt\right) ^{1-\frac{1}{q}}\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}dt\right) ^{\frac{1}{q}} \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert dt\right) ^{1-\frac{1}{q}}\\ &&\times\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left[ t\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}+(1-t)\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\right] dt\right) ^{\frac{1}{q}}\\ &=&\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert dt\right) ^{1-\frac{1}{q}}\\ &&\times\left( \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\int_{0}^{1}t^{3}\left\vert \beta _{t}\right\vert dt+\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\int_{0}^{1}t^{2}(1-t)\left\vert \beta _{t}\right\vert dt\right) ^{\frac{ 1}{q}}\\ &=&\frac{1}{2}\eta ^{3}(b,a)\left( \frac{27}{\eta ^{3}(b,a)}\right) ^{1- \frac{1}{q}}\left( \frac{27}{\eta ^{4}(b,a)}\right) ^{\frac{1}{q}}\left( \int_{a}^{\beta }(x-a)^{2}\left\vert x\right\vert dx\right) ^{1-\frac{1}{q}}\\ &&\times\left( \begin{array}{c} \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\int_{a}^{\beta }3(x-a)^{3}\left\vert x\right\vert dx\\+\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\int_{a}^{\beta }(x-a)^{2}\left[ \eta (b,a)-3(x-a)\right] \left\vert x\right\vert dx \end{array} \right)^{\frac{1}{q}} \\ &=&\frac{27}{2}\eta ^{-\frac{1}{q}}(b,a)\left( \int_{a}^{\beta }(x-a)^{2}\left\vert x\right\vert dx\right) ^{1-\frac{1}{q}} \\ &&\times \left( \begin{array}{c} \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\int_{a}^{\beta }3(x-a)^{3}\left\vert x\right\vert dx\\+\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}\int_{a}^{\beta }(x-a)^{2}\left[ \eta (b,a)-3(x-a)\right] \left\vert x\right\vert dx \end{array} \right) ^{\frac{1}{q}} \\ &=&\frac{27}{2}\eta ^{-\frac{1}{q}}(b,a)D_{1,\eta }^{1-\frac{1}{q}}\left( a,b\right) \left[ \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}D_{2,\eta }\left( a,b\right) +\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}D_{3,\eta }\left( a,b\right) \right] ^{\frac{1}{q}}. \end{eqnarray*} This completes the proof of theorem.
Corollary 2.9. Suppose that all the assumptions of Theorem 3 are satisfied. If we choose \( \eta (b,a)=b-a\) then when \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is convex on \(K\) for \(q\geq 1\) we get \begin{eqnarray*} &&\left\vert \frac{b-a}{2}\left( \frac{2a+b}{3}\right) f^{\prime \prime }(b)- \frac{a+b}{2}f^{\prime }(b)+\frac{f(b)b-f(a)a}{b-a}-\frac{1}{b-a}% \int_{a}^{b}f(x)dx\right\vert \\ &\leq &\frac{27}{2}(b-a)^{-1-\frac{1}{q}}D_{1}^{1-\frac{1}{q}}\left( a,b\right) \left[ \left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}D_{2}\left( a,b\right) +\left\vert f^{\prime \prime \prime }(a)\right\vert ^{q}D_{3}\left( a,b\right) \right] ^{\frac{1}{q}}, \end{eqnarray*} where \begin{eqnarray*} D_{1}(a,b) &=&\left\{ \begin{array}{c} \frac{(b-a)^{3}}{27}\frac{b+3a}{12},~\ \ \ \ \ \ \ a>0,\beta >0~ \\ \frac{(b-a)^{3}}{27}\frac{b+3a}{12}+\frac{a^{4}}{6},a< 0,~\beta >0 \\ -\frac{(b-a)^{3}}{27}\frac{b+3a}{12},~\ \ \ \ \ a< 0,\beta < 0 \end{array} \right. , \\ D_{2}(a,b) &=&\left\{ \begin{array}{c} \frac{(b-a)^{4}}{27}\frac{4b+11a}{60},~\ \ \ \ \ \ a>0,\beta >0 \\ \frac{(b-a)^{4}}{27}\frac{4b+11a}{60}-3\frac{a^{5}}{10},a< 0,\beta >0 \\ -\frac{(b-a)^{4}}{27}\frac{4b+11a}{60},~\ \ \ \ \ a< 0,\beta < 0 \end{array} \right. , \end{eqnarray*} and \begin{equation*} D_{3}(a,b)=\left\{ \begin{array}{c} \frac{(b-a)^{4}}{27}\frac{b+4a}{60},~\ \ \ \ \ \ \ \ ~\ \ \ \ \ \ \ \ \ \ \ \ \ a>0,\beta >0 \\ \frac{(b-a)^{4}}{27}\frac{b+4a}{60}+(b-a)\frac{a^{4}}{6}+3\frac{a^{5}}{10} ,a< 0,\beta >0 \\ -\frac{(b-a)^{4}}{27}\frac{b+4a}{60},~\ \ \ \ \ \ ~\ \ \ \ \ \ \ \ \ \ \ \ \ a< 0,\beta< 0. \end{array} \right. \end{equation*}
Remark 2.10. If the mapping \(\eta \) satisfies condition C then using the inequality (5) in the proof of Theorem 2.8, then the inequality (9) becomes the following inequality:
Corollary 2.11. If we take \(q=1\) in Theorem 2.8, then we have the following inequality: \begin{equation*} \left\vert I_{f}(a,b,\eta )\right\vert \leq \frac{27}{2\eta (b,a)}\left[ \left\vert f^{\prime \prime \prime }(b)\right\vert D_{2,\eta }\left( a,b\right) +\left\vert f^{\prime \prime \prime }(a)\right\vert D_{3,\eta }\left( a,b\right) \right] \end{equation*}
Now we will give our results for prequasiinvex functions by using Lemma 2.1.Theorem 2.12. Let \(K\subseteq\mathbb{R}\) be an open invex subset with respect to mapping \(\eta \left( \cdot ,\cdot \right) :K\times K\rightarrow\mathbb{R}^{n}\) and \(a,b\in K\) with \(\eta (b,a)>0\). Suppose that the function \(f:K \rightarrow\mathbb{R}\) is a three times differentiable function on \(K\) such that \(f^{\prime \prime \prime }\in L\left[ a,a+\eta (b,a)\right] .\) If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is prequasiinvex on \(K\) for \(q>1\) , then the following inequality holds:
Proof. If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) for \(q>1\) is prequasiinvex on \(\left[ a,a+\eta (b,a)\right] \), using Lemma 2.1, the Hölder integral inequality and \(\left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}\leq \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} \) we obtain \begin{eqnarray*} &&\left\vert I_{f}(a,b,\eta )\right\vert \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert dt \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2p}dt\right) ^{\frac{1% }{p}}\left( \int_{0}^{1}\left\vert \beta _{t}\right\vert ^{q}\left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}dt\right) ^{\frac{1}{% q}} \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2p}dt\right) ^{\frac{1% }{p}}\left( \int_{0}^{1}\left\vert \beta _{t}\right\vert ^{q}\max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} dt\right) ^{\frac{1}{q}} \\ &=&\frac{3^{\frac{1}{q}}}{2}\eta ^{2+\frac{1}{p}}(b,a)\left( \frac{1}{2p+1}% \right) ^{\frac{1}{p}}\left( \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} \right) ^{\frac{1}{q}}\left( \int_{a}^{\beta }\left\vert x\right\vert ^{q}dx\right) ^{\frac{1}{q}} \\ &=&\frac{3^{\frac{1}{q}}}{2}\eta ^{2+\frac{1}{p}}(b,a)\left( \frac{1}{2p+1}% \right) ^{\frac{1}{p}}\left( \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} \right) ^{\frac{1}{q}}C_{\eta }^{\frac{1}{q}}(q,a,b) \end{eqnarray*} This completes the proof of theorem.
Corollary 2.13. Suppose that all the assumptions of Theorem 2.12 are satisfied. If we choose \(\eta (b,a)=b-a\) then when \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is prequasiinvex on \(K\) for \(q>1\) we have \begin{eqnarray*} &&\left\vert \frac{b-a}{2}\left( \frac{2a+b}{3}\right) f^{\prime \prime }(b)-% \frac{a+b}{2}f^{\prime }(b)+\frac{f(b)b-f(a)a}{b-a}-\frac{1}{b-a}% \int_{a}^{b}f(x)dx\right\vert \\ &\leq &\frac{3^{\frac{1}{q}}}{2}(b-a)^{1+\frac{1}{p}}\left( \frac{1}{2p+1}% \right) ^{\frac{1}{p}}\left( \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} \right) ^{\frac{1}{q}}C^{\frac{1}{q}}(q,a,b) \end{eqnarray*} where \begin{equation*} C(q,a,b)=\left\{ \begin{array}{c} \frac{b-a}{3}L_{q}^{q}\left( \frac{b+2a}{3},a\right) ,~\ \ \ \ \ \ \ \ \ \ a>0,\frac{b+2a}{3}>0, \\ \frac{2}{q+1}A\left[ \left( \frac{b+2a}{3}\right) ^{q+1},\left( -a\right) ^{q+1}\right] ,a0, \\ \frac{b-a}{3}L_{q}^{q}\left( -a,-\frac{b+2a}{3}\right) ,~\ \ \ \ \ \ a< 0, \frac{b+2a}{3}< 0. \end{array} \right. \end{equation*}
Remark 2.14. If the mapping \(\eta \) satisfies condition C then by use of the prequasiinvexity of \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) we get
Theorem 2.15. Let \(K\subseteq\mathbb{R}\) be an open invex subset with respect to mapping \(\eta \left(\cdot ,\cdot \right) :K\times K\rightarrow\mathbb{R}^{n}\) and \(a,b\in K\) with \(\eta (b,a)>0\). Suppose that the function \(f:K\rightarrow\mathbb{R}\) is a three times differentiable function on \(K\) such that \(f^{\prime \prime \prime }\in L\left[ a,a+\eta (b,a)\right] .\) If \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is prequasiinvex on \(K\) for \( q\geq 1\), then the following inequality holds:
Proof. From Lemma 2.1 and Power-mean integral inequality, we obtain \begin{eqnarray*} &&\left\vert I_{f}(a,b,\eta )\right\vert \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert dt \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert dt\right) ^{1-\frac{1}{q}}\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \left\vert f^{\prime \prime \prime }(a+t\eta (b,a))\right\vert ^{q}dt\right) ^{\frac{1}{q}} \\ &\leq &\frac{1}{2}\eta ^{3}(b,a)\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert dt\right) ^{1-\frac{1}{q}}\left( \int_{0}^{1}t^{2}\left\vert \beta _{t}\right\vert \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} dt\right) ^{\frac{1}{q}} \\ &=&\frac{1}{2}\eta^{3}(b,a)\left( \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime}(b)\right\vert ^{q}\right\} \right) ^{\frac{1}{q}}\int_{0}^{1}t^{2}\left \vert \beta _{t}\right\vert dt \\ &=&\frac{27}{2}\left( \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} \right) ^{\frac{1}{q}}\int_{a}^{\beta }(x-a)^{2}\left\vert x\right\vert dx \\ &=&\frac{27}{2}\left( \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} \right) ^{\frac{1}{q}}D_{1,\eta }(a,b). \end{eqnarray*} This completes the proof of theorem.
Corollary 2.16. Suppose that all the assumptions of Theorem 2.15 are satisfied. If we choose \(\eta (b,a)=b-a\) then when \(\left\vert f^{\prime \prime \prime }\right\vert ^{q}\) is prequasiinvex on \(K\) for \(q\geq 1\) we have \begin{eqnarray*} &&\left\vert \frac{b-a}{2}\left( \frac{2a+b}{3}\right) f^{\prime \prime }(b)- \frac{a+b}{2}f^{\prime }(b)+\frac{f(b)b-f(a)a}{b-a}-\frac{1}{b-a} \int_{a}^{b}f(x)dx\right\vert \\ &\leq &\frac{27}{2}\frac{D_{1}(a,b)}{b-a}\left[ \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} \right] ^{\frac{1}{q}} \end{eqnarray*} where \begin{equation*} D_{1}(a,b)=\left\{ \begin{array}{c} \frac{(b-a)^{3}}{27}\frac{b+3a}{12},~\ \ \ \ \ \ \ a>0,\frac{b+2a}{3}>0~ \\ \frac{(b-a)^{3}}{27}\frac{b+3a}{12}+\frac{a^{4}}{6},a0 \\ -\frac{(b-a)^{3}}{27}\frac{b+3a}{12},~\ \ \ \ \ a< 0,\frac{b+2a}{3}< 0 \end{array} \right. , \end{equation*}
Remark 2.17. If we use the inequality (12) in the proof of Theorem 2.15, then the inequality (14) becomes the following inequality: \begin{equation*} \left\vert I_{f}(a,b,\eta )\right\vert \leq \frac{27}{2}\left( \max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(a+\eta (b,a))\right\vert ^{q}\right\} \right) ^{\frac{1}{q} }D_{1,\eta }(a,b) \end{equation*} This inequality is better than the inequality (14).
Corollary 2.18. If we take \(q=1\) in Theorem 2.15, then we have the following inequality: \begin{equation*} \left\vert I_{f}(a,b,\eta )\right\vert \leq \frac{27}{2}\max \left\{ \left\vert f^{\prime \prime \prime }(a)\right\vert ^{q},\left\vert f^{\prime \prime \prime }(b)\right\vert ^{q}\right\} D_{1,\eta }(a,b) \end{equation*}
