Now we introduce exponentially p-convex functions.
Definition 2.1
Consider an interval \(\mathcal{K}\subset (0, \infty )=\mathbb{R}_{+}\) and \(p\in \mathbb{R\setminus }\{0\}\). A function \(\psi :\mathcal{K} \rightarrow \mathbb{R}\) is called exponentially p-convex, if
$$ \psi \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2}^{p} \bigr]^{ \frac{1}{p}} \bigr) \leq r\frac{\psi (u_{1})}{e^{\alpha u_{1}}}+(1-r)\frac{ \psi (u_{2})}{e^{\alpha u_{2}}}, $$
(2.1)
for all \(u_{1},u_{2}\in \mathcal{K}\), \(r\in [0,1]\) and \(\alpha \in \mathbb{R}\). If the inequality in (2.1) is reversed, then ψ is called exponentially p-concave.
It is easy to note that, by taking \(\alpha =0\), an exponentially p-convex function becomes p-convex.
Example 2.1
Consider a function \(\psi :(\sqrt{2},\infty )\rightarrow \mathbb{R}\), defined by \(\psi (u)=(\ln (u))^{p}\) for \(p\geq 2\). Then ψ is exponentially p-convex for all \(\alpha <0\), and not p-convex.
Note that ψ satisfies inequality (2.1) for all \(\alpha <0\). But for \(u_{1}=2\), \(u_{2}=3\) and \(p=5\), inequality (1.2) does not hold.
2.1 Integral inequalities
Throughout this section, we denote by \(\mathcal{K}\subset (0, \infty )=\mathbb{R}_{+}\) an interval with interior \(\mathcal{K}^{\circ }\) and \(p\in \mathbb{R\setminus }\{0\}\). We start with our results for exponentially p-convex functions.
Theorem 2.2
Let
\(\psi :\mathcal{K}\rightarrow \mathbb{R}\)
be an integrable exponentially
p-convex function. Let
\(u_{1},u_{2}\in \mathcal{K}\)
with
\(u_{1}< u_{2}\). Then for
\(\alpha \in \mathbb{R}\), we have
$$ \psi \biggl( \biggl[ \frac{u_{1}^{p}+u_{2}^{p}}{2} \biggr] ^{\frac{1}{p}} \biggr)\leq \frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}} ^{u_{2}}\frac{\psi (w)}{w^{1-p}e^{\alpha w}}\,dw\leq A_{1}(r)\frac{ \psi (u_{1})}{e^{\alpha u_{1}}}+A_{2}(r)\frac{\psi (u_{2})}{e^{\alpha u_{2}}}, $$
(2.2)
where
$$ A_{1}(r)= \int _{0}^{1}\frac{rdr}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}} \quad \textit{and} \quad A_{2}(r)= \int _{0}^{1}\frac{(1-r)\,dr}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2} ^{p})^{\frac{1}{p}}}}. $$
Proof
By using the exponential p-convexity of ψ, we have
$$ 2\psi \biggl( \biggl[ \frac{w^{p}+z^{p}}{2} \biggr]^{\frac{1}{p}} \biggr) \leq \frac{\psi (w)}{e^{\alpha w}}+\frac{\psi (z)}{e^{\alpha z}}. $$
(2.3)
Letting \(w^{p}=ru_{1}^{p}+(1-r)u_{2}^{p}\) and \(z^{p}=(1-r)u_{1}^{p}+ru _{2}^{p}\), we get
$$ 2\psi \biggl( \biggl[ \frac{u_{1}^{p}+u_{2}^{p}}{2} \biggr]^{ \frac{1}{p}} \biggr)\leq \frac{\psi ( [ru_{1}^{p}+(1-r)u _{2}^{p} ]^{\frac{1}{p}} ) }{e^{\alpha (ru_{1}^{p}+(1-r)u _{2}^{p})^{\frac{1}{p}}}}+\frac{\psi ( [(1-r)u_{1}^{p}+ru _{2}^{p} ]^{\frac{1}{p}} ) }{e^{\alpha ((1-r)u_{1}^{p}+ru _{2}^{p})^{\frac{1}{p}}}}. $$
(2.4)
Integrating with respect to \(r\in [0,1]\) and applying a change of variable, we find
$$ \psi \biggl( \biggl[ \frac{u_{1}^{p}+u_{2}^{p}}{2} \biggr] ^{\frac{1}{p}} \biggr)\leq \frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}} ^{u_{2}}\frac{\psi (w)}{w^{1-p}e^{\alpha w}}\,dw. $$
(2.5)
Hence the first inequality of (2.2) has been established. For the next inequality, again using the exponential p-convexity of ψ, we have
$$ \frac{\psi ( [ru_{1}^{p}+(1-r)u_{2}^{p} ]^{ \frac{1}{p}} ) }{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}} \leq \frac{r\frac{\psi (u_{1})}{e^{\alpha u_{1}}}+(1-r)\frac{ \psi (u_{2})}{e^{\alpha u_{2}}}}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}}. $$
(2.6)
Integrating with respect to \(r\in [0,1]\), we get
$$\begin{aligned}& \frac{p}{u_{2}^{p}-u_{1}^{p}} \int _{u_{1}}^{u_{2}}\frac{ \psi (w)}{w^{1-p}e^{\alpha w}}\,dw \\& \quad \leq \frac{\psi (u_{1})}{e^{\alpha u _{1}}} \int _{0}^{1}\frac{rdr}{e^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{ \frac{1}{p}}}}+\frac{\psi (u_{2})}{e^{\alpha u_{2}}} \int _{0}^{1}\frac{(1-r)\,dr}{e ^{\alpha (ru_{1}^{p}+(1-r)u_{2}^{p})^{\frac{1}{p}}}}. \end{aligned}$$
(2.7)
By combining (2.5) and (2.7), we get (2.2). □
Remark 2.1
In Theorem 2.2, by taking \(\alpha =0\), we attain inequality (1.3) in Theorem 1.2.
Theorem 2.3
Let
\(\psi :\mathcal{K}\rightarrow \mathbb{R}\)
be a differentiable function on
\(\mathcal{K}^{\circ }\)
and
\(u_{1},u_{2} \in \mathcal{K}\)
with
\(u_{1}< u_{2}\)
and
\(\psi '\in L_{1}[u_{1},u_{2}]\). If
\(|\psi '|^{q}\)
is exponentially
p-convex on
\([u_{1},u_{2}]\)
for
\(q\geq 1\)
and
\(\alpha \in \mathbb{R}\), then
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} B_{1}^{1-\frac{1}{q}} \biggl[B _{2} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert ^{q}+B_{3} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr] ^{\frac{1}{q}} , \end{aligned}$$
(2.8)
where
$$\begin{aligned}& \begin{aligned} B_{1}&=B_{1}(u_{1},u_{2};p)= \frac{1}{4} \biggl(\frac{u_{1}^{p}+u_{2} ^{p}}{2} \biggr)^{\frac{1}{p}-1} \\ &\quad {}\times \biggl[{}_{2}F_{1} \biggl(1- \frac{1}{p},2;3;\frac{u_{1}^{p}-u _{2}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr)+{}_{2}F_{1} \biggl(1- \frac{1}{p},2;3;\frac{u_{2}^{p}-u_{1}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr) \biggr] , \end{aligned} \\& \begin{aligned} B_{2}&=B_{2}(u_{1},u_{2};p)= \frac{1}{24} \biggl(\frac{u_{1}^{p}+u_{2} ^{p}}{2} \biggr)^{\frac{1}{p}-1} \biggl[{}_{2}F_{1} \biggl(1- \frac{1}{p},2;4; \frac{u_{1}^{p}-u_{2}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr) \\ &\quad {}+6\,{}_{2}F_{1} \biggl(1-\frac{1}{p},2;3; \frac{u_{2}^{p}-u_{1}^{p}}{u _{1}^{p}+u_{2}^{p}} \biggr)+{}_{2}F_{1} \biggl(1- \frac{1}{p},2;4;\frac{u _{2}^{p}-u_{1}^{p}}{u_{1}^{p}+u_{2}^{p}} \biggr) \biggr] , \end{aligned} \\& B_{3}=B_{3}(u_{1},u_{2};p)=B_{1}-B_{2}. \end{aligned}$$
Proof
Applying the power mean inequality to (1.4) of Lemma 1.1, we get
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \int _{0}^{1} \biggl\vert \frac{1-2r}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}} \biggr\vert \bigl\vert \psi ' \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2}^{p} \bigr]^{\frac{1}{p}} \bigr) \bigr\vert \,dr \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1} \frac{|1-2r|}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}}\,dr \biggr)^{1-\frac{1}{q}} \\ &\qquad {}\times \biggl( \int _{0}^{1} \frac{|1-2r|}{[ru_{1}^{p}+(1-r)u_{2} ^{p}]^{1-\frac{1}{p}}} \bigl\vert \psi ' \bigl( \bigl[ru_{1}^{p}+(1-r)u _{2}^{p} \bigr]^{\frac{1}{p}} \bigr) \bigr\vert ^{q} \,dr \biggr)^{ \frac{1}{q}} . \end{aligned}$$
(2.9)
Since \(|\psi '|^{q}\) is exponentially p-convex on \([u_{1},u_{2}]\), we have
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1} \frac{|1-2r|}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}}\,dr \biggr)^{1-\frac{1}{q}} \\ &\qquad {}\times \biggl( \int _{0}^{1} \frac{||1-2r|| [r \vert \frac{ \psi '(u_{1})}{e^{\alpha u_{1}}} \vert ^{q} +(1-r) \vert \frac{ \psi '(u_{2})}{e^{\alpha u_{2}}} \vert ^{q} ] }{[ru_{1}^{p}+(1-r)u _{2}^{p}]^{1-\frac{1}{p}}} \,dr \biggr)^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} B_{1}^{1-\frac{1}{q}} \biggl[B _{2} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u_{1}}} \biggr\vert ^{q}+B_{3} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr] ^{\frac{1}{q}}. \end{aligned}$$
(2.10)
It is easy to note that
$$\begin{aligned}& \int _{0}^{1} \frac{|1-2r|}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{1-\frac{1}{p}}} \,dr=B_{1}(u _{1},u_{2};p), \\& \int _{0}^{1} \frac{|1-2r|r}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{1- \frac{1}{p}}} \,dr=B_{2}(u_{1},u_{2};p), \\& \int _{0}^{1} \frac{|1-2r|(1-r)}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{1- \frac{1}{p}}} \,dr=B_{1}(u_{1},u_{2};p)-B_{2}(u_{1},u_{2};p). \end{aligned}$$
Hence the proof is completed. □
Remark 2.2
In Theorem 2.3,
-
(a)
by taking \(\alpha =0\), we attain Theorem 7 in [19];
-
(b)
by taking \(p=1\), we attain Theorem 5 in [4].
Corollary 2.4
Let
\(\psi :\mathcal{K}\rightarrow \mathbb{R}\)
be a differentiable function on
\(\mathcal{K}^{\circ }\)
and
\(u_{1},u_{2} \in \mathcal{K}\), \(u_{1}< u_{2}\), and
\(\psi '\in L_{1}[u_{1},u_{2}]\). If
\(|\psi '|\)
is exponentially
p-convex on
\([u_{1},u_{2}]\), then
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl[B_{2} \biggl\vert \frac{\psi '(u _{1})}{e^{\alpha u_{1}}} \biggr\vert +B_{3} \biggl\vert \frac{\psi '(u_{2})}{e ^{\alpha u_{2}}} \biggr\vert \biggr] , \end{aligned}$$
(2.11)
where
\(B_{2}\)
and
\(B_{3}\)
are given in Theorem 2.3.
Remark 2.3
In Corollary 2.4,
-
(a)
by taking \(\alpha =0\), we attain Corollary 1 in [19];
-
(b)
by taking \(p=1\), we attain Theorem 3 in [4].
Theorem 2.5
Let
\(\psi :\mathcal{K}\rightarrow \mathbb{R}\)
be a differentiable function on
\(\mathcal{K}^{\circ }\). Let
\(u_{1},u_{2} \in \mathcal{K}\), \(u_{1}< u_{2}\), and
\(\psi '\in L_{1}[u_{1},u_{2}]\). If
\(|\psi '|^{q}\)
is exponentially
p-convex on
\([u_{1},u_{2}]\), and
\(q,l>1\), \(1/q+1/l=1\), and
\(\alpha \in \mathbb{R}\), then
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl(\frac{1}{l+1} \biggr) ^{\frac{1}{l}} \biggl[B_{4} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \biggr\vert ^{q}+B_{5} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr]^{\frac{1}{q}} , \end{aligned}$$
(2.12)
where
$$\begin{aligned}& \begin{aligned} B_{4} &=B_{4}(u_{1},u_{2};p;q) \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0, } \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & { p>0 ,} \end{cases}\displaystyle \end{aligned} \\& \begin{aligned} B_{5} &=B_{5}(u_{1},u_{2};p;q) \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0 }, \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & { p>0 .} \end{cases}\displaystyle \end{aligned} \end{aligned}$$
Proof
Using Hölder’s inequality on (1.4) of Lemma 1.1 and then applying the exponential p-convexity of \(|\psi '|^{q}\) on \([u_{1},u_{2}]\), we get
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1} |1-2r|^{l}\,dr \biggr) ^{\frac{1}{l}} \\ &\qquad {}\times \biggl( \int _{0}^{1} \frac{1}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q(1- \frac{1}{p})}} \bigl\vert \psi ' \bigl( \bigl[ru_{1}^{p}+(1-r)u_{2}^{p} \bigr] ^{\frac{1}{p}} \bigr) \bigr\vert ^{q} \,dr \biggr)^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl(\frac{1}{l+1} \biggr) ^{\frac{1}{l}} \biggl( \int _{0}^{1} \frac{r \vert \frac{\psi '(u_{1})}{e ^{\alpha u_{1}}} \vert ^{q} +(1-r) \vert \frac{\psi '(u_{2})}{e^{ \alpha u_{2}}} \vert ^{q} }{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q- \frac{q}{p}}} \,dr \biggr)^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl(\frac{1}{l+1} \biggr) ^{\frac{1}{l}} \biggl[B_{4} \biggl\vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \biggr\vert ^{q}+B_{5} \biggl\vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr]^{\frac{1}{q}} , \end{aligned}$$
(2.13)
where after calculations, we have
$$\begin{aligned}& \begin{aligned} B_{4}&= \int _{0}^{1}\frac{r}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q- \frac{q}{p}}}\,dr \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}, \end{cases}\displaystyle \end{aligned} \\& \begin{aligned} B_{5}&= \int _{0}^{1}\frac{1-r}{[ru_{1}^{p}+(1-r)u_{2}^{p}]^{q- \frac{q}{p}}}\,dr \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},2;3;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{qp-q}}\,{}_{2}F_{1} (q-\frac{q}{p},1;3;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}. \end{cases}\displaystyle \end{aligned} \end{aligned}$$
□
Remark 2.4
In Theorem 2.5,
-
(a)
by letting \(\alpha =0\), we attain Theorem 8 in [19];
-
(b)
by letting \(p=1\), we attain Theorem 4 in [4].
Theorem 2.6
Let
\(\psi :\mathcal{K}\rightarrow \mathbb{R}\)
be a differentiable function on
\(\mathcal{K}^{\circ }\)
and
\(u_{1},u_{2} \in \mathcal{K}\), \(u_{1}< u_{2}\), and
\(\psi '\in L_{1}[u_{1},u_{2}]\). If
\(|\psi '|^{q}\)
is exponentially
p-convex on
\([u_{1},u_{2}]\), and
\(q,l>1\), \(1/q+1/l=1\), and
\(\alpha \in \mathbb{R}\), then
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr) ^{\frac{1}{q}} \biggl(\frac{ \vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \vert ^{q}+ \vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \vert ^{q} }{2} \biggr)^{\frac{1}{q}}, \end{aligned}$$
(2.14)
where
$$\begin{aligned} B_{6} &=B_{6}(u_{1},u_{2};p;l) \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}. \end{cases}\displaystyle \end{aligned}$$
Proof
Using Hölder’s inequality on (1.4) of Lemma 1.1 and then applying the exponential p-convexity of \(|\psi '|^{q}\) on \([u_{1},u_{2}]\), we get
$$\begin{aligned} & \biggl\vert \frac{\psi (u_{1})+\psi (u_{2})}{2}-\frac{p}{u_{2}^{p}-u_{1} ^{p}} \int _{u_{1}}^{u_{2}}\frac{\psi (w)}{w^{1-p}}\,dw \biggr\vert \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p} \biggl( \int _{0}^{1}\frac{1}{[ru _{1}^{p}+(1-r)u_{2}^{p}]^{l-\frac{l}{p}}} \,dr \biggr)^{\frac{1}{l}} \\ &\qquad {}\times \biggl( \int _{0}^{1} |1-2r|^{q} \bigl\vert \psi ' \bigl( \bigl[ru _{1}^{p}+(1-r)u_{2}^{p} \bigr]^{\frac{1}{p}} \bigr) \bigr\vert ^{q} \,dr \biggr) ^{\frac{1}{q}} \\ &\quad \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr) ^{\frac{1}{q}} \biggl(\frac{ \vert \frac{\psi '(u_{1})}{e^{\alpha u _{1}}} \vert ^{q}+ \vert \frac{\psi '(u_{2})}{e^{\alpha u_{2}}} \vert ^{q} }{2} \biggr)^{\frac{1}{q}} , \end{aligned}$$
(2.15)
where a simple calculation implies
$$\begin{aligned} B_{6}(u_{1},u_{2};p;l) &= \int _{0}^{1}\frac{1}{[ru_{1}^{p}+(1-r)u_{1} ^{p}]^{l-\frac{l}{p}}} \,dr \\ &= \textstyle\begin{cases} \frac{1}{2u_{1}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{2}}{u_{1}})^{p} ) , & {p< 0}, \\ \frac{1}{2u_{2}^{pl-l}}\,{}_{2}F_{1} (l-\frac{l}{p},1;2;1-(\frac{u _{1}}{u_{2}})^{p} ), & {p>0}, \end{cases}\displaystyle \end{aligned}$$
(2.16)
and
$$ \int _{0}^{1} r|1-2r|^{q} \,dr= \int _{0}^{1} (1-r)|1-2r|^{q} \,dr= \frac{1}{2(q+1)}. $$
(2.17)
By substituting (2.16) and (2.17) into (2.15), we get (2.14). □
Remark 2.5
In Theorem 2.6, by letting \(\alpha =0\), we attain Theorem 9 in [19].
2.2 Applications
Consider some special means of two positive numbers \(u_{1}\), \(u_{2}\), \(u_{1}< u_{2}\):
-
(1)
The arithmetic mean
$$ A=A(u_{1},u_{2})=\frac{u_{1}+u_{2}}{2}; $$
-
(2)
The harmonic mean
$$ H=H(u_{1},u_{2})=\frac{2u_{1}u_{2}}{u_{1}+u_{2}}; $$
-
(3)
The p-logarithmic mean
$$ L_{p}=L_{p}(u_{1},u_{2})= \biggl( \frac{u_{2}^{p+1}-u_{1}^{p+1}}{(p+1)(u _{2}-u_{1})} \biggr)^{\frac{1}{p}} ,\quad p\in \mathbb{R}\setminus \{-1,0 \}. $$
In the next three propositions we consider \(0< u_{1}< u_{2}\) and \(q>1\).
Proposition 2.1
Let
\(\alpha \in \mathbb{R}\)
and
\(p<1\). Then we have
$$ \bigl\vert L^{p-1}_{p-1}-HL^{p-2}_{p-2} \bigr\vert \leq \frac{u_{2}^{p}-u _{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr)^{ \frac{1}{q}} A^{\frac{1}{q}} \biggl( \biggl\vert \frac{1}{u_{1}^{2}e^{ \alpha u_{1}}} \biggr\vert ^{q}, \biggl\vert \frac{1}{u_{2}^{2}e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr)HL^{p-1}_{p-1}, $$
where
\(B_{6}\)
is defined as in Theorem 2.6.
Proof
The proof ensues from Theorem 2.6, for a function \(\psi :(0, \infty )\rightarrow \mathbb{R}\), \(\psi (w)=\frac{1}{w}\). Here note that \(|\psi '(w)|^{q}=|\frac{1}{w^{2}}|^{q}\) is exponentially p-convex for all \(p<1\) and \(\alpha \in \mathbb{R}\). □
Proposition 2.2
Let
\(\alpha \leq 0\)
and
\(p>1\). Then we have
$$ \bigl\vert L^{p-1}_{p-1}A\bigl(u_{1}^{p},u_{2}^{p} \bigr)-L_{2p-1}^{2p-1} \bigr\vert \leq \frac{u_{2}^{p}-u_{1}^{p}}{2}B_{6}^{\frac{1}{l}} \biggl(\frac{1}{q+1} \biggr)^{\frac{1}{q}} A^{\frac{1}{q}} \biggl( \biggl\vert \frac{1}{u_{1} ^{p-1}e^{\alpha u_{1}}} \biggr\vert ^{q}, \biggl\vert \frac{1}{u_{2}^{p-1}e^{ \alpha u_{2}}} \biggr\vert ^{q} \biggr)L_{p-1}^{p-1} , $$
where
\(B_{6}\)
is defined as in Theorem 2.6.
Proof
The proof ensues from Theorem 2.6, for \(\psi :(0,\infty )\rightarrow \mathbb{R}\), \(\psi (w)=w^{p}\). Here note that \(|\psi '(w)|^{q}=|pw ^{p-1}|^{q}\) is exponentially p-convex for all \(p>1\) and \(\alpha \leq 0\). □
Proposition 2.3
Let
\(\alpha \leq 0\)
and
\(p>1\). Then we have
$$ \bigl\vert L^{p-1}_{p-1}A-L^{p}_{p} \bigr\vert \leq \frac{u_{2}^{p}-u_{1}^{p}}{2p}B_{6}^{\frac{1}{l}} \biggl( \frac{1}{q+1} \biggr)^{\frac{1}{q}} A^{\frac{1}{q}} \biggl( \biggl\vert \frac{1}{e^{ \alpha u_{1}}} \biggr\vert ^{q}, \biggl\vert \frac{1}{e^{\alpha u_{2}}} \biggr\vert ^{q} \biggr) L^{p-1}_{p-1}, $$
where
\(B_{6}\)
is given as in Theorem 2.6.
Proof
The proof ensues from Theorem 2.6, for \(\psi :(0,\infty )\rightarrow \mathbb{R}\), \(\psi (w)=w\). Here note that \(|\psi '(w)|^{q}=1\) is exponentially p-convex for all \(p>1\) and \(\alpha \leq 0\). □