First we introduce a new concept named strongly generalized \(( \phi,h,s )\)-preinvex functions in the second sense. It is defined as follows.
Definition 4
The function f on the invex set K is said to be strongly generalized \(( \phi,h,s )\)-preinvex in the second sense with modulus \(c>0\) if it is nonnegative, and for all \(u,v\in K\) and \(z\times s\in ( 0,1 ) \times ( 0,1 ] \), the following inequality holds:
$$ f \bigl( v+ze^{i\phi }\xi ( u,v ) \bigr) \leq h^{s} ( z ) f ( u ) +h^{s} ( 1-z ) f ( v ) -cz ( 1-z ) \bigl\Vert e^{i\phi }\xi ( u,v ) \bigr\Vert ^{2}. $$
Notation. Let \(f:I\subset R\rightarrow R\) be a differentiable mapping on \(I^{\circ }\) (the interior of I), from now on we will consider
$$\begin{aligned} &\varPsi _{f} \bigl( x,\delta,\sigma,\alpha,e^{i\phi }\xi ( u,v ) \bigr) \\ &\quad= \bigl( 1-\delta ^{\sigma } \bigr) \biggl[ \frac{ \xi ( v,x ) ^{\alpha }f ( v+e^{i\phi }\xi ( x,v ) ) +\xi ( x,u ) ^{\alpha }f ( u+e ^{i\phi }\xi ( x,u ) ) }{e^{i\phi }\xi ( v,u ) } \biggr] f ( x ) \\ &\qquad{}+\delta ^{\sigma } \biggl[ \frac{\xi ( v,x ) ^{\alpha }f ( v ) +\xi ( u,x ) ^{\alpha }f ( u ) }{e^{i\phi }\xi ( v,u ) } \biggr] \\ &\qquad{}-\frac{\varGamma ( \alpha +1 ) }{e^{i\alpha \phi }\xi ( v,u ) ^{\alpha }} \bigl[ J_{ ( u+e^{i\phi }\xi ( x,u ) ) +}^{\alpha }f ( u ) +J_{ ( v+e^{i\phi }\xi ( x,v ) ) +}^{\alpha }f ( v ) \bigr], \end{aligned}$$
where \(u< u+e^{i\phi }\xi ( v,u ) \), \(x\in [ u,u+e ^{i\phi }\xi ( v,u ) ] \), \(\delta \in[ 0,1]\), \(\alpha >0\) and Γ is Euler gamma function.
To get new integral inequalities, first we focus on proving the following lemma.
Lemma 1
Let
\(K_{\phi \xi }\subseteq R\)
be a
ϕ-invex subset with respect to
\(\phi (\cdot ) \)
and
ξ: \(K_{\phi \xi }\times K_{ \phi \xi }\subseteq R\)
with
\(u< u+e^{i\phi }\xi ( v,u ) \)
and
\(0\leq \phi \leq \frac{\pi }{2}\). Suppose that
\(f:K_{\phi \xi }\rightarrow R\)
is a differentiable mapping such that
\(f^{\prime }\in L ( [ u,u+e^{i\phi }\xi ( v,u ) ] ) \)
for all
\(x\in [ u,u+e^{i\phi }\xi ( v,u ) ] \), \(\delta \times \sigma \in [ 0,1 ] \), and
\(\alpha >0\), then we have
$$\begin{aligned} &\varPsi _{f} \bigl( x,\delta,\sigma,\alpha,e^{i\phi }\xi ( v,u ) \bigr) \\ &\quad=\frac{\xi ( x,u ) ^{\alpha +1}}{e ^{i\phi }\xi ( v,u ) } \int _{0}^{1} \bigl( z^{\alpha }- \delta ^{\sigma } \bigr) f^{\prime } \bigl( u+ze^{i\phi }\xi ( x,u ) \bigr) \,dz \\ &\qquad{}+\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \int _{0}^{1} \bigl( \delta ^{\sigma }-z^{\alpha } \bigr) f^{\prime } \bigl( v+ze^{i\phi }\xi ( x,v ) \bigr) \,dz. \end{aligned}$$
Proof
Using integration by parts, we get
$$\begin{aligned} & \int _{0}^{1} \bigl( z^{\alpha }-\delta ^{\sigma } \bigr) f^{{ \prime }} \bigl( u+ze^{i\phi }\xi ( x,u ) \bigr) \,dz \\ &\quad= \biggl[ \bigl( z^{\alpha }-\delta ^{\sigma } \bigr) \frac{f ( u+ze^{i\phi }\xi ( x,u ) ) }{e^{i\phi }\xi ( x,u ) }|_{0}^{1} - \alpha \int _{0}^{1}z^{ \alpha -1}\frac{f ( u+ze^{i\phi }\xi ( x,u ) ) }{e^{i\phi }\xi ( x,u ) }\,dz \biggr] \\ &\quad= \biggl[ \frac{ ( 1-\delta ^{\sigma } ) f ( u+e^{i \phi }\xi ( x,u ) ) +\delta ^{\sigma }f ( a ) }{e^{i\phi }\xi ( x,u ) }-\frac{\varGamma ( \alpha +1 ) }{e^{i\alpha \phi }\xi ( x,u ) ^{\alpha }}J_{ ( u+e^{i\phi }\xi ( x,u ) ) +}^{\alpha }f ( u ) \biggr]. \end{aligned}$$
Analogously, we have
$$\begin{aligned} & \int _{0}^{1} \bigl( \delta ^{\sigma }-z^{\alpha } \bigr) f^{{ \prime }} \bigl( v+ze^{i\phi }\xi ( x,v ) \bigr) \,dz \\ &\quad= \biggl[ \bigl( z^{\alpha }-\delta ^{\sigma } \bigr) \frac{f ( v+ze^{i\phi }\xi ( x,u ) ) }{e^{i\phi }\xi ( x,u ) }|_{0}^{1} + \alpha \int _{0}^{1}z^{ \alpha -1}\frac{f ( v+ze^{i\phi }\xi ( x,v ) ) }{e^{i\phi }\xi ( x,v ) }\,dz \biggr] \\ &\quad= \biggl[ \frac{ ( 1-\delta ^{\sigma } ) f ( v+e^{i \phi }\xi ( x,v ) ) +\delta ^{\sigma }f ( v ) }{e^{i\phi }\xi ( v,x ) }-\frac{\varGamma ( \alpha +1 ) }{e^{i\alpha \phi }\xi ( v,x ) ^{\alpha }}J_{ ( v+e^{i\phi }\xi ( x,v ) ) +}^{\alpha }f ( v ) \biggr]. \end{aligned}$$
Both sides of the above equalities are multiplied by \(\frac{\xi ( x,u ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) }\) and \(\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) }\) analogously, and then adding them, we obtain the required result. This completes the proof. □
Theorem 6
Let
\(K_{\phi \xi }\subseteq R\)
be a
ϕ-invex subset with respect to
\(\phi (\cdot ) \)
and
ξ: \(K_{\phi \xi }\times K_{ \phi \xi }\subseteq R\)
with
\(u< u+e^{i\phi }\xi ( v,u ) \)
and
\(0\leq \phi \leq \frac{\pi }{2}\). Suppose that
\(f:K_{\phi \xi }\rightarrow R\)
is a differentiable mapping such that
\(f^{\prime }\in L ( [ u,u+e^{i\phi }\xi ( v,u ) ] ) \). If
\(\vert f^{\prime } \vert \)
is strongly generalized
\(( \phi,h,s )\)-preinvex in the second sense and
\(\vert f^{\prime } ( x ) \vert \leq M\), then for all
\(x\in [ u,u+e^{i\phi }\xi ( v,u ) ] \), \(\delta \times \sigma \in [ 0,1 ] \), and
\(\alpha >0\), we have
$$\begin{aligned} & \bigl\vert \varPsi _{f} \bigl( x,\delta,\sigma,\alpha,e^{i\phi } \xi ( u,v ) \bigr) \bigr\vert \\ &\quad\leq \frac{\xi ( x,u ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \bigl[ M \bigl( \varPsi _{1} ( \delta, \sigma,\alpha,s ) +\varPsi _{2} ( \delta,\sigma, \alpha,s ) \bigr) -c \bigl\Vert e ^{i\phi }\xi ( u,x ) \bigr\Vert ^{2}\varPsi _{3} ( \delta,\sigma,\alpha ) \bigr] \\ &\qquad{}+\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \bigl[ M \bigl( \varPsi _{1} ( \delta, \sigma,\alpha ,s ) +\varPsi _{2} ( \delta,\sigma,\alpha,s ) \bigr) -c \bigl\Vert e^{i\phi }\xi ( v,x ) \bigr\Vert ^{2}\varPsi _{3} ( \delta,\sigma,\alpha ) \bigr]. \end{aligned}$$
(7)
Proof
Using Lemma 1, the property of modulus, and strongly generalized \(( \phi,h,s )\)-preinvexity in the second sense, we obtain
$$\begin{aligned} & \bigl\vert \varPsi _{f} \bigl( x,\delta,\sigma, \alpha,e^{i\phi } \xi ( u,v ) \bigr) \bigr\vert \\ &\quad\leq \frac{\xi ( x,u ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \int _{0}^{1} \bigl\vert z^{\alpha }-\delta ^{ \sigma } \bigr\vert \bigl\vert f^{\prime } \bigl( u+e^{i\phi }z \xi ( x,u ) \bigr) \bigr\vert \,dz \\ &\qquad{}+\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \int _{0}^{1} \bigl\vert z^{\alpha }-\delta ^{\sigma } \bigr\vert \bigl\vert f^{\prime } \bigl( v+e^{i\phi }z\xi ( x,v ) \bigr) \bigr\vert \,dz \\ &\quad\leq \frac{\xi ( x,u ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \int _{0}^{1} \bigl\vert z^{\alpha }-\delta ^{ \sigma } \bigr\vert \bigl( h^{s} ( z ) \bigl\vert f^{ \prime } ( x ) \bigr\vert +h^{s} ( 1-z ) \bigl\vert f^{\prime } ( u ) \bigr\vert -cz ( 1-z ) \bigl\Vert e^{i\phi }\xi ( u,x ) \bigr\Vert ^{2} \bigr) \,dz \\ &\qquad{}+\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \int _{0}^{1} \bigl( \delta ^{\sigma }-z^{\alpha } \bigr) \bigl( h^{s} ( z ) \bigl\vert f^{\prime } ( x ) \bigr\vert +h^{s} ( 1-z ) \bigl\vert f^{\prime } ( u ) \bigr\vert \\ &\qquad{}-cz ( 1-z ) \bigl\Vert e^{i\phi } \xi ( v,x ) \bigr\Vert ^{2} \bigr) \,dz \\ &\quad =\frac{\xi ( x,u ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \bigl[ \varPsi _{1} ( \delta,\sigma,\alpha,s ) \bigl\vert f^{\prime } ( x ) \bigr\vert +\varPsi _{2} ( \delta, \sigma,\alpha,s ) \bigl\vert f^{\prime } ( u ) \bigr\vert -c \bigl\Vert e^{i\phi }\xi ( u,x ) \bigr\Vert ^{2}\varPsi _{3} ( \delta,\sigma,\alpha ) \bigr] \\ &\qquad{}+\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \bigl[ \varPsi _{1} ( \delta,\sigma,\alpha,s ) \bigl\vert f^{\prime } ( x ) \bigr\vert +\varPsi _{2} ( \delta, \sigma,\alpha,s ) \bigl\vert f^{\prime } ( v ) \bigr\vert -c \bigl\Vert e^{i\phi }\xi ( v,x ) \bigr\Vert ^{2}\varPsi _{3} ( \delta,\sigma,\alpha ) \bigr], \end{aligned}$$
where we used the fact
$$\begin{aligned} &\varPsi _{1} ( \delta,\sigma,\alpha,s ) = \int _{0}^{1} \bigl\vert z^{\alpha }-\delta ^{\sigma } \bigr\vert h^{s} ( z ) \,dz, \\ &\varPsi _{2} ( \delta,\sigma,\alpha,s ) = \int _{0}^{1} \bigl\vert z^{\alpha }-\delta ^{\sigma } \bigr\vert h^{s} ( 1-z ) \,dz, \\ &\varPsi _{3} ( \delta,\sigma,\alpha ) = \int _{0}^{1} \bigl\vert z ^{\alpha }-\delta ^{\sigma } \bigr\vert z ( 1-z ) \,dz \\ &\phantom{\varPsi _{3} ( \delta,\sigma,\alpha )}=2\delta ^{\sigma }\beta \bigl( \delta ^{\frac{\sigma }{\alpha }},2,2 \bigr) -2\beta \bigl( \delta ^{\frac{\sigma }{\alpha }},\alpha +2,2 \bigr) +\beta ( \alpha +2,2 ) -\delta ^{\sigma }\beta ( 2,2 ). \end{aligned}$$
Hence the proof. □
Remark 1
On letting \(s=1\), \(\xi ( u,v ) =u-v\), \(\phi =c=\sigma =0\), \(x=\frac{u+v}{2}\), and \(h ( z ) =\frac{\sqrt{z}}{2 \sqrt{1-z}}\) in Theorem 6, then inequality (7) reduces to inequality (2).
Theorem 7
Let
\(K_{\phi \xi }\subseteq R\)
be a
ϕ-invex subset with respect to
\(\phi (\cdot ) \)
and
ξ: \(K_{\phi \xi }\times K_{ \phi \xi }\subseteq R\)
with
\(u< u+e^{i\phi }\xi ( v,u ) \)
for
\(0\leq \phi \leq \frac{\pi }{2}\). Suppose that
\(f:K_{\phi \xi }\rightarrow R\)
is a differentiable mapping such that
\(f^{\prime }\in L ( [ u,u+e^{i\phi }\xi ( v,u ) ] ) \). If
\(\vert f^{\prime } \vert \)
is strongly generalized
\(( \phi,h,s )\)-preinvex in the second sense with
\(q>1\), \(\frac{1}{p}+\frac{1}{q}=1\), and
\(\vert f^{\prime } ( x ) \vert \leq M\), then for all
\(x\in [ u,u+e^{i \phi }\xi ( v,u ) ] \), \(\delta \times \sigma \in ( 0,1 ] \times ( 0,1 ] \), and
\(\alpha >0\), we have
$$\begin{aligned} & \bigl\vert \varPsi _{f} \bigl( x,\delta,\sigma, \alpha,e^{i\phi } \xi ( u,v ) \bigr) \bigr\vert \\ &\quad\leq \frac{\xi ( x,u ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) }\varPsi _{6} ( \delta,\sigma,\alpha ) ^{\frac{1}{p}} \bigl[ M^{q} ( \varPsi _{4}+\varPsi _{5} ) -c \bigl\Vert e ^{i\phi }\xi ( u,x ) \bigr\Vert ^{2}\beta ( 2,2 ) \bigr] ^{\frac{1}{q}} \\ &\qquad{}+\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) }\varPsi _{6} ( \delta,\sigma,\alpha ) ^{ \frac{1}{p}} \bigl[ M^{q} ( \varPsi _{4}+\varPsi _{5} ) -c \bigl\Vert e ^{i\phi }\xi ( v,x ) \bigr\Vert ^{2}\beta ( 2,2 ) \bigr] ^{\frac{1}{q}}. \end{aligned}$$
(8)
Proof
Using Lemma 1 and the Holder integral inequality, we obtain
where we used the fact
$$\begin{aligned} &\varPsi _{4} = \int _{0}^{1}h^{s} ( z ) \,dz, \\ &\varPsi _{5} = \int _{0}^{1}h^{s} ( 1-z ) \,dz, \\ &\varPsi _{6} ( \delta,\sigma,\alpha ) = \int _{0}^{1} \bigl\vert z ^{\alpha }-\delta ^{\sigma } \bigr\vert \,dz=\frac{1+2\alpha \delta ^{\sigma ( \frac{1+\alpha }{\alpha } ) }}{1+\alpha }-\delta ^{\sigma }. \end{aligned}$$
This completes the proof. □
Remark 2
On letting \(\delta =s=1\), \(\xi ( u,v ) =u-v\), \(\phi =c=0\), and \(h ( z ) = \frac{\sqrt{z}}{2\sqrt{1-z}}\) in Theorem 7, inequality (8) reduces to inequality (3).
Remark 3
On letting \(s=1\), \(\xi ( u,v ) =u-v\), \(\phi =c=\delta =0\), and \(h ( z ) =\frac{\sqrt{z}}{2\sqrt{1-z}}\) in Theorem 7, inequality (8) reduces to inequality (4).
Remark 4
On letting \(\sigma =s=1\), \(\xi ( u,v ) =u-v\), \(\phi =c=0\), \(x=\frac{u+v}{2}\), and \(h ( z ) =\frac{ \sqrt{z}}{2\sqrt{1-z}}\) in Theorem 7, inequality (8) reduces to inequality (5).
Theorem 8
Let
\(f:I= [ u,u+e^{i\phi }\xi ( v,u ) ] \)
\(\subset [ 0,\infty ) \rightarrow R\)
be a differentiable mapping on
\(I^{\circ}\)
such that
\(f^{\prime } \in L_{1} ( [ u,u+e^{i\phi }\xi ( v,u ) ] ) \). If
\(\vert f^{\prime } \vert ^{q}\)
is strongly generalized
\(( \phi,h,s )\)-preinvex and
\(\vert f ^{\prime } ( x ) \vert \leq M\)
for all
\(x\in [ u,u+e^{i\phi }\xi ( v,u ) ] \), \(\delta \times \sigma \in [ 0,1 ] \times ( 0,1 ] \), and
\(\alpha >0\), we have
$$\begin{aligned} & \bigl\vert \varPsi _{f} \bigl( x,\delta,\sigma, \alpha,e^{i\phi } \xi ( u,v ) \bigr) \bigr\vert \\ &\quad\leq \frac{\xi ( x,u ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \bigl( \varPsi _{6} ( \delta,\sigma,\alpha ) \bigr) ^{\frac{1}{p}} \bigl[ M^{q} \bigl( \varPsi _{1} ( \delta,\sigma,\alpha,s ) +\varPsi _{2} ( \delta,\sigma,\alpha,s ) \bigr) \\ &\qquad{}-c \bigl\Vert e ^{i\phi }\xi ( u,x ) \bigr\Vert ^{2}\beta ( 2,2 ) \bigr] ^{\frac{1}{q}} \\ &\qquad{}+\frac{\xi ( v,x ) ^{\alpha +1}}{e^{i\phi }\xi ( v,u ) } \bigl( \varPsi _{6} ( \delta,\sigma,\alpha ) \bigr) ^{\frac{1}{p}} \bigl[ M^{q} \bigl( \varPsi _{1} ( \delta, \sigma,\alpha,s ) +\varPsi _{2} ( \delta,\sigma,\alpha,s ) \bigr) \\ &\qquad{}-c \bigl\Vert e^{i\phi } \xi ( v,x ) \bigr\Vert ^{2}\beta ( 2,2 ) \bigr] ^{\frac{1}{q}}. \end{aligned}$$
(9)
Proof
Using Lemma 1, the property of modulus and power mean inequality, we have
Hence the proof. □
Remark 5
On letting \(\sigma =s=1\), \(\xi ( u,v ) =u-v\), \(\phi =c=0\), \(x=\frac{u+v}{2}\), and \(h ( z ) =\frac{ \sqrt{z}}{2\sqrt{1-z}}\) in Theorem 8, inequality (9) reduces to inequality (6).