In this section we obtain some new results about Hermite–Hadamard inequality for s-convex functions by using Hölder–İşcan integral inequality and improved power-mean integral inequality which provide better approach than the classical Hölder and power-mean integral inequalities, respectively.
In [5], Dragomir and Pearce obtained the following equality for differentiable functions.
Lemma 2.1
Let
\(f:I\subseteq \mathbb{R} \longrightarrow \mathbb{R} \)
be a differentiable function on
\(I^{\circ }\)
where
\(a,b\in I\)
with
\(a< b\). If
\(f^{\prime }\in L [ a,b ] \), then the following equality holds:
$$\begin{aligned} \frac{f ( a ) +f ( b ) }{2}-\frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx=\frac{b-a}{2} \int _{0}^{1} ( 1-2t ) f^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \,dt. \end{aligned}$$
In [17], Muddassar et al. obtained the following inequality for s-convex functions by using Lemma 2.1 and Hölder integral inequality.
Theorem 2.1
Let
\(f:I^{\circ }\subseteq \mathbb{R} \rightarrow \mathbb{R} \)
be a differentiable function on
\(I^{\circ }\), \(a,b\in I^{\circ }\)
with
\(a< b \)
and
\(f^{\prime }\in L [ a,b ] \). If
\(\vert f ^{\prime } \vert ^{q}\)
is s-convex on
\([ a,b ] \), \(p>1\)
such that
\(q=\frac{p}{p-1}\), then
$$\begin{aligned} \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \leq & \frac{b-a}{2 ( p+1 ) ^{\frac{1}{p}}} \biggl( \frac{ \vert f^{ \prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{s+1} \biggr) ^{\frac{1}{q}}. \end{aligned}$$
(3)
In the following theorem, we will obtain a new upper bound for the right-hand side of Hermite–Hadamard inequality for s-convex functions, which is better than inequality (3).
Theorem 2.2
Let
\(f:I^{\circ }\subseteq \mathbb{R} \rightarrow \mathbb{R} \)
be a differentiable function on
\(I^{\circ }\), \(a,b\in I^{\circ }\)
with
\(a< b \)
and
\(f^{\prime }\in L [ a,b ] \). If
\(\vert f ^{\prime } \vert ^{q}\)
is
s-convex on
\([ a,b ] \), where
\(p>1\)
such that
\(q=\frac{p}{p-1}\), then
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{2^{\frac{p+1}{p}} ( p+1 ) ^{\frac{1}{p}}}\cdot\frac{1}{ ( s+1 ) ^{\frac{1}{q}}} \\& \qquad {} \times \biggl\{ \biggl[ \frac{ ( s+1 ) \vert f^{ \prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{s+2} \biggr] ^{\frac{1}{q}}+ \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ ( s+1 ) \vert f^{\prime } ( b ) \vert ^{q}}{s+2} \biggr] ^{\frac{1}{q}} \biggr\} . \end{aligned}$$
(4)
Proof
From Lemma 2.1 and using Hölder–İşcan integral inequality (1), we have
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{2} \int _{0}^{1} \vert 1-2t \vert \bigl\vert f ^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert \,dt \\& \quad \leq \frac{b-a}{2} \biggl\{ \biggl( \int _{0}^{1} ( 1-t ) \vert 1-2t \vert ^{p}\,dt \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{1} ( 1-t ) \bigl\vert f^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{1} t \vert 1-2t \vert ^{p}\,dt \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{1} t \bigl\vert f^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt \biggr) ^{ \frac{1}{q}} \biggr\} . \end{aligned}$$
Since \(\vert f^{\prime } \vert ^{q}\) is s-convex on \([ a,b ] \), then
$$\begin{aligned} \int _{0}^{1} t \bigl\vert f^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt \leq & \int _{0}^{1} t \bigl[ t^{s} \bigl\vert f ^{\prime } ( a ) \bigr\vert ^{q}+ ( 1-t ) ^{s} \bigl\vert f^{\prime } ( b ) \bigr\vert ^{q} \bigr] \,dt \\ =&\frac{ ( s+1 ) \vert f^{\prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{ (s+1 ) ( s+2 )} \end{aligned}$$
(5)
and
$$\begin{aligned} \int _{0}^{1} ( 1-t ) \bigl\vert f^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt =& \int _{0}^{1} t \bigl\vert f^{\prime } \bigl( tb+ ( 1-t ) a \bigr) \bigr\vert ^{q}\,dt \\ =&\frac{ \vert f^{\prime } ( a ) \vert ^{q}+ ( s+1 ) \vert f^{\prime } ( b ) \vert ^{q}}{ ( s+1 ) ( s+2 ) }, \end{aligned}$$
(6)
and also
$$\begin{aligned} \int _{0}^{1} t \vert 1-2t \vert ^{p}\,dt =& \int _{0}^{1} ( 1-t ) \vert 1-2t \vert ^{p}\,dt \\ =&\frac{1}{2 ( p+1 )}. \end{aligned}$$
(7)
By inequalities (5), (6), and (7), we get inequality (4). □
Corollary 2.1
Let
\(f:I^{\circ }\subseteq \mathbb{R} \rightarrow \mathbb{R} \)
be a differentiable function on
\(I^{\circ }\), \(a,b\in I^{\circ }\)
with
\(a< b \)
and
\(f^{\prime }\in L [ a,b ] \). If
\(\vert f ^{\prime } \vert ^{q}\)
is s-convex on
\([ a,b ] \), where
\(p>1\)
such that
\(q=\frac{p}{p-1}\), then
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{2^{\frac{p+1}{p}} ( p+1 ) ^{\frac{1}{p}}} \frac{1}{(s+2)^{ \frac{1}{q}}} \biggl(1+\frac{1}{(s+1)^{\frac{1}{q}}} \biggr) \bigl( \bigl\vert f^{\prime } ( a ) \bigr\vert + \bigl\vert f ^{\prime } ( b ) \bigr\vert \bigr). \end{aligned}$$
(8)
Proof
Using the fact \(\sum_{i=1}^{n} ( a_{i}+b_{i} ) ^{k}\leq \sum ( a_{i} ) ^{k}+\sum ( b_{i} ) ^{k}\) for \(k\in ( 0,1 ) \) with \(p>1\) such that \(q=\frac{p}{p-1}\) completes the proof. □
Remark 2.1
Inequality (4) is better than inequality (3). Indeed, since the function \(g: [ 0,\infty ) \rightarrow \mathbb{R} \), \(g ( x ) =x^{r}\), \(r\in ( 0,1 ] \) is concave, we can write
$$\begin{aligned} \frac{\alpha ^{r}+\beta ^{r}}{2}=\frac{g ( \alpha ) +g ( \beta ) }{2}\leq g \biggl( \frac{\alpha +\beta }{2} \biggr)= \biggl( \frac{\alpha +\beta }{2} \biggr) ^{r} \end{aligned}$$
(9)
for all \(\alpha ,\beta \geq 0\). In inequality (9), if we choose
$$\begin{aligned} \alpha =\frac{ ( s+1 ) \vert f^{\prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{s+2} , \qquad \beta =\frac{ \vert f^{\prime } ( a ) \vert ^{q}+ ( s+1 ) \vert f^{\prime } ( b ) \vert ^{q}}{s+2} \end{aligned}$$
and \(r=\frac{1}{q}\), then we have
$$\begin{aligned}& \frac{1}{2} \biggl[ \frac{ ( s+1 ) \vert f^{\prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{s+2} \biggr] ^{\frac{1}{q}}+ \frac{1}{2} \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ ( s+1 ) \vert f^{\prime } ( b ) \vert ^{q}}{s+2} \biggr] ^{\frac{1}{q}} \\& \quad \leq \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{2} \biggr] ^{\frac{1}{q}}. \end{aligned}$$
Thus, we obtain the following inequality:
$$\begin{aligned}& \frac{b-a}{2^{\frac{p+1}{p}} ( p+1 ) ^{\frac{1}{p}}}\cdot\frac{1}{ ( s+1 ) ^{\frac{1}{q}}} \\& \qquad {}\times \biggl\{ \biggl[ \frac{ ( s+1 ) \vert f^{ \prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{s+2} \biggr] ^{\frac{1}{q}}+ \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ ( s+1 ) \vert f^{\prime } ( b ) \vert ^{q}}{s+2} \biggr] ^{\frac{1}{q}} \biggr\} \\& \quad \leq \frac{b-a}{2^{\frac{1}{p}} ( p+1 ) ^{\frac{1}{p}}}\cdot\frac{1}{ ( s+1 ) ^{\frac{1}{q}}} \biggl[ \frac{ \vert f^{ \prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{2} \biggr] ^{\frac{1}{q}} \\& \quad =\frac{b-a}{2 ( p+1 ) ^{\frac{1}{p}}}\cdot\frac{1}{ ( s+1 ) ^{\frac{1}{q}}} \bigl( \bigl\vert f^{\prime } ( a ) \bigr\vert ^{q}+ \bigl\vert f^{\prime } ( b ) \bigr\vert ^{q} \bigr) ^{\frac{1}{q}}. \end{aligned}$$
If we take \(s=1\) in Remark 2.1, we have the following result which was obtained by İşcan in [10].
Remark 2.2
Let \(f:I^{\circ }\subseteq \mathbb{R} \rightarrow \mathbb{R} \) be a differentiable function on \(I^{\circ }\), \(a,b\in I^{\circ }\) with \(a< b \) and \(f^{\prime }\in L [ a,b ] \). If \(\vert f ^{\prime } \vert ^{q}\) is convex on \([ a,b ] \), then
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{4 ( p+1 ) ^{\frac{1}{p}}} \biggl\{ \biggl[ \frac{2 \vert f^{\prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{3} \biggr] ^{\frac{1}{q}}+ \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+2 \vert f^{\prime } ( b ) \vert ^{q}}{3} \biggr] ^{\frac{1}{q}} \biggr\} \\& \quad \leq \frac{b-a}{2 ( p+1 ) ^{\frac{1}{p}}} \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ \vert f ^{\prime } ( b ) \vert ^{q}}{2} \biggr] ^{ \frac{1}{q}}, \end{aligned}$$
where \(\frac{1}{p}+\frac{1}{q}=1\).
In [17], Muddassar et al. obtained the following inequality by using Lemma 2.1 and power-mean integral inequality.
Theorem 2.3
Let
\(f:I^{\circ }\subseteq \mathbb{R} \longrightarrow \mathbb{R} \)
be a differentiable function on
\(I^{\circ }\)
where
\(a,b\in I\)
with
\(a< b\)
and
\(f^{\prime }\in L [ a,b ] \). If
\(\vert f^{\prime } \vert ^{q}\)
is
s-convex on
\([ a,b ] \)
for some
\(p>1\)
such that
\(q=\frac{p}{p-1}\), then the following inequality holds:
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{2^{\frac{p+1}{p}}} \biggl[ \biggl( \frac{s+2^{-s}}{ ( s+1 ) ( s+2 ) } \biggr) \bigl( \bigl\vert f ^{\prime } ( a ) \bigr\vert ^{q}+ \bigl\vert f^{\prime } ( b ) \bigr\vert ^{q} \bigr) \biggr] ^{\frac{1}{q}}. \end{aligned}$$
(10)
If Theorem 2.3 is proved again by using improved power-mean integral inequality, then we get the following result.
Theorem 2.4
Let
\(f:I^{\circ }\subseteq \mathbb{R} \longrightarrow \mathbb{R} \)
be a differentiable function on
\(I^{\circ }\)
where
\(a,b\in I\)
with
\(a< b\)
and
\(f^{\prime }\in L [ a,b ] \). If
\(\vert f^{\prime } \vert ^{q}\)
is
s-convex on
\([ a,b ] \)
for some
\(p>1\)
such that
\(q=\frac{p}{p-1}\), then the following inequality holds:
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{2.4^{\frac{1}{p}}}\cdot\frac{1}{ ( s+1 ) ^{\frac{1}{q}}} \\& \qquad {}\times \biggl\{ \biggl[ \frac{ ( s+1 ) ( s+1+2^{-s-1} ) \vert f^{\prime } ( a ) \vert ^{q}+ ( s-1+2^{-s-1} ( s+5 ) ) \vert f ^{\prime } ( b ) \vert ^{q}}{ ( s+2 ) ( s+3 ) } \biggr] ^{\frac{1}{q}} \\& \qquad {}+ \biggl[ \frac{ ( s-1+2^{-s-1} ( s+5 ) ) \vert f^{\prime } ( a ) \vert ^{q}+ ( s+1 ) ( s+1+2^{-s-1} ) \vert f^{\prime } ( b ) \vert ^{q}}{ ( s+2 ) ( s+3 ) } \biggr] ^{\frac{1}{q}} \biggr\} . \end{aligned}$$
(11)
Proof
From Lemma 2.1 and applying improved power-mean integral inequality (2) for \(q>1\), we have
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{2} \biggl\{ \biggl( \int _{0}^{1} ( 1-t ) \vert 1-2t \vert \,dt \biggr) ^{\frac{1}{p}} \biggl( \int _{0}^{1} ( 1-t ) \vert 1-2t \vert \bigl\vert f ^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt \biggr) ^{\frac{1}{q}} \\& \qquad {}+ \biggl( \int _{0}^{1} t \vert 1-2t \vert \,dt \biggr) ^{ \frac{1}{p}} \biggl( \int _{0}^{1} t \vert 1-2t \vert \bigl\vert f ^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt \biggr) ^{\frac{1}{q}} \biggr\} . \end{aligned}$$
By s-convexity of \(\vert f^{\prime } \vert ^{q}\) on \([ a,b ] \), we have
$$\begin{aligned}& \int _{0}^{1} t \vert 1-2t \vert \bigl\vert f^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt \\& \quad \leq \int _{0}^{1} t \vert 1-2t \vert \bigl[ t^{s} \bigl\vert f ^{\prime } ( a ) \bigr\vert ^{q}+ ( 1-t ) ^{s} \bigl\vert f^{\prime } ( b ) \bigr\vert ^{q} \bigr] \,dt \\& \quad =\frac{(s+1)(s+1+2^{-s-1}) \vert f^{\prime } ( a ) \vert ^{q}+ ( s-1+2^{-s-1}(s+5) ) \vert f^{\prime } ( b ) \vert ^{q}}{ ( s+1 ) ( s+2 ) ( s+3 ) } \end{aligned}$$
(12)
and
$$\begin{aligned}& \int _{0}^{1} ( 1-t ) \vert 1-2t \vert \bigl\vert f ^{\prime } \bigl( ta+ ( 1-t ) b \bigr) \bigr\vert ^{q}\,dt \\& \quad = \int _{0}^{1} t \vert 1-2t \vert \bigl\vert f^{\prime } \bigl( tb+ ( 1-t ) a \bigr) \bigr\vert ^{q}\,dt \\& \quad \leq \frac{ ( s-1+2^{-s-1}(s+5) ) \vert f^{\prime } ( a ) \vert ^{q}+(s+1)(s+1+2^{-s-1}) \vert f ^{\prime } ( b ) \vert ^{q}}{ ( s+1 ) ( s+2 ) ( s+3 ) }. \end{aligned}$$
Furthermore,
$$\begin{aligned} \int _{0}^{1} t \vert 1-2t \vert \,dt =& \int _{0}^{1} ( 1-t ) \vert 1-2t \vert \,dt \\ =&\frac{1}{4}. \end{aligned}$$
(13)
By using inequalities (12) and (13), we get inequality (11). □
Remark 2.3
Inequality (11) is better than inequality (10). Indeed, using inequality (9) in Remark 2.1, we have
$$\begin{aligned}& \frac{1}{2} \biggl[ \frac{ ( s+1 ) ( s+1+2^{-s-1} ) \vert f^{\prime } ( a ) \vert ^{q}+ ( s-1+2^{-s-1} ( s+5 ) ) \vert f ^{\prime } ( b ) \vert ^{q}}{ ( s+2 ) ( s+3 ) } \biggr] ^{\frac{1}{q}} \\& \qquad {}+\frac{1}{2} \biggl[ \frac{ ( s-1+2^{-s-1} ( s+5 ) ) \vert f^{\prime } ( a ) \vert ^{q}+ ( s+1 ) ( s+1+2^{-s-1} ) \vert f^{ \prime } ( b ) \vert ^{q}}{ ( s+2 ) ( s+3 ) } \biggr] ^{\frac{1}{q}} \\& \quad \leq \biggl[ \biggl( \frac{s+2^{-s}}{s+2} \biggr) \biggl( \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ \vert f ^{\prime } ( b ) \vert ^{q}}{2} \biggr) \biggr] ^{\frac{1}{q}}. \end{aligned}$$
Therefore, we obtain
$$\begin{aligned}& \frac{b-a}{2.4^{\frac{1}{p}}}\cdot\frac{1}{ ( s+1 ) ^{ \frac{1}{q}}} \\& \qquad {} \times \biggl\{ \biggl[ \frac{ ( s+1 ) ( s+1+2^{-s-1} ) \vert f^{\prime } ( a ) \vert ^{q}+ ( s-1+2^{-s-1} ( s+5 ) ) \vert f ^{\prime } ( b ) \vert ^{q}}{ ( s+2 ) ( s+3 ) } \biggr] ^{\frac{1}{q}} \\& \qquad {} + \biggl[ \frac{ ( s-1+2^{-s-1} ( s+5 ) ) \vert f ^{\prime } ( a ) \vert ^{q}+ ( s+1 ) ( s+1+2^{-s-1} ) \vert f^{\prime } ( b ) \vert ^{q}}{ ( s+2 ) ( s+3 ) } \biggr] ^{\frac{1}{q}} \biggr\} \\& \quad \leq \frac{b-a}{4^{\frac{1}{p}}}\cdot\frac{1}{ ( s+1 ) ^{ \frac{1}{q}}} \biggl[ \biggl( \frac{s+2^{-s}}{s+2} \biggr) \biggl( \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ \vert f ^{\prime } ( b ) \vert ^{q}}{2} \biggr) \biggr] ^{\frac{1}{q}} \\& \quad =\frac{b-a}{2^{\frac{p+1}{p}}} \biggl[ \biggl( \frac{s+2^{-s}}{ ( s+1 ) ( s+2 ) } \biggr) \bigl( \bigl\vert f^{ \prime } ( a ) \bigr\vert ^{q}+ \bigl\vert f^{\prime } ( b ) \bigr\vert ^{q} \bigr) \biggr] ^{\frac{1}{q}}. \end{aligned}$$
If we take \(s=1\) in Remark 2.3, we have the following result for convex functions which is better than the inequality given in [18, Theorem 1].
Remark 2.4
Let \(f:I^{\circ }\subseteq \mathbb{R} \rightarrow \mathbb{R} \) be a differentiable function on \(I^{\circ }\), \(a,b\in I^{\circ }\) with \(a< b \) and \(f^{\prime }\in L [ a,b ] \). If \(\vert f ^{\prime } \vert ^{q}\) is convex on \([ a,b ] \), then
$$\begin{aligned}& \biggl\vert \frac{f ( a ) +f ( b ) }{2}- \frac{1}{b-a} \int _{a}^{b} f ( x ) \,dx \biggr\vert \\& \quad \leq \frac{b-a}{4.2^{\frac{1}{p}}} \biggl\{ \biggl[ \frac{3 \vert f ^{\prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{4} \biggr] ^{\frac{1}{q}}+ \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+3 \vert f^{\prime } ( b ) \vert ^{q}}{4} \biggr] ^{\frac{1}{q}} \biggr\} \\& \quad \leq \frac{b-a}{4} \biggl[ \frac{ \vert f^{\prime } ( a ) \vert ^{q}+ \vert f^{\prime } ( b ) \vert ^{q}}{2} \biggr] ^{\frac{1}{q}}, \end{aligned}$$
where \(\frac{1}{p}+\frac{1}{q}=1\).