1 Introduction
Let \(a_{k} \ge0\), \(b_{k} \ge0\) (\(k=1,2,\ldots,n\)), \(p>1\), \(1/p+1/q=1\). The classical Hölder inequality [1] is stated as follows:
Similarly, the integral version of the Hölder inequality [1] is
where \(f(x)>0\), \(g(x)>0\), \(p>1\), \(1/p+1/q=1\), and \(f(x)\) and \(g(x)\) are continuous real-valued functions on \([a,b]\).
$$ \sum_{k=1}^{n} {a_{k} b_{k} } \le \Biggl( {\sum_{k=1}^{n} {a_{k}^{p} } } \Biggr)^{1/p} \Biggl( {\sum _{k=1}^{n} {b_{k}^{q} } } \Biggr)^{1/q}. $$
(1.1)
$$ \int_{a}^{b} {f(x)g(x)\, dx} \le \biggl( {\int _{a}^{b} {f^{p}(x)\, dx} } \biggr)^{1/p} \biggl( {\int_{a}^{b} {g^{q}(x)\, dx} } \biggr)^{1/q}, $$
(1.2)
If \(p=q=2\), then inequalities (1.1) and (1.2) reduce to the well-known Cauchy inequalities [2] of the discrete form and the continuous form, respectively.
Anzeige
The Hölder inequality and Cauchy inequality play an important role in many areas of pure and applied mathematics. A large number of generalizations, refinements, variations, and applications of these inequalities have been investigated in [3‐18] and references therein. Recently, Brito da Cruz et al. in [19] gave a \(\alpha,\beta \)-symmetric integral Hölder inequality as follows.
Let \(f, g: {\mathrm{R}}\to {\mathrm{R}}\) and \(a, b\in{\mathrm{R}}\) with \(a < b\). If \(\vert f\vert\) and \(\vert g\vert\) are \(\alpha,\beta\)-symmetric integrable on \([a,b]\), \(p>1\) with \(q = p/(p-1)\). Then
with equality if and only if the functions \(|f|\) and \(|g|\) are proportional.
$$ \int_{a}^{b} {\bigl\vert f(t)g(t)\bigr\vert \, d_{\alpha,\beta} t} \leq \biggl( {\int_{a}^{b} { \bigl\vert f(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p} \biggl( {\int_{a}^{b} {\bigl\vert g(t)\bigr\vert ^{q}\, d_{\alpha,\beta} t} } \biggr)^{1/q}, $$
(1.3)
The aim of this work is to establish a reversed version of the \(\alpha , \beta\)-symmetric integral Hölder inequality and some generalizations of the \(\alpha, \beta\)-symmetric integral Hölder inequality. Moreover, the obtained results will be applied to establish the \(\alpha, \beta\)-symmetric integral reverse Minkowski inequality, Dresher inequality, and their corresponding reverse versions. This paper is organized as follows. In Section 2, we recall some basic definitions and properties of \(\alpha,\beta\)-symmetric integral, which can also be found in [19, 20]; in Section 3, we establish a \(\alpha, \beta\)-symmetric integral reverse Hölder inequality and give some generalizations of the \(\alpha, \beta\)-symmetric integral Hölder inequality, we apply the obtained results to establish the reverse Minkowski inequality, Dresher inequality, and its reverse form involving \(\alpha, \beta\)-symmetric integral, some extensions of the Minkowski and Dresher inequalities are also given; in Section 4, we establish some further generalizations and refinements of the \(\alpha, \beta\)-symmetric integral Hölder inequality; in Section 5, we present a subdividing of the \(\alpha, \beta\)-symmetric integral Hölder inequality.
2 Preliminaries
In the section, we recall some basic definitions and properties of \(\alpha,\beta\)-symmetric integral.
Anzeige
The α-forward and β-backward differences are defined as follows (see [19]):
where \(\alpha>0\), \(\beta>0\).
$$\Delta_{\alpha}[f](t):=\frac{f(t+\alpha)-f(t)}{\alpha},\qquad \nabla_{\beta}[f](t):= \frac{f(t)-f(t-\beta)}{\beta}, $$
Definition 2.1
(see [19])
Assume that \(I \subseteq {\mathrm{R}}\) with \(\sup I=+\infty\) and let \(a, b\in I\) with \(a< b\). If \(f:I\to{\mathrm{R}}\) and \(\alpha>0\), then the α-forward integral of f is defined by
where \(\int_{x}^{+\infty} {f(t)\Delta_{\alpha}t} =\alpha \sum_{k=0}^{+\infty} {f(x+k\alpha)} \), provided the series converges at \(x = a\) and \(x = b\).
$$\int_{a}^{b} {f(t)\Delta_{\alpha}t} =\int _{a}^{\infty}{f(t)\Delta_{\alpha}t} -\int _{b}^{\infty}{f(t)\Delta_{\alpha}t} , $$
The α-forward integral has the following properties.
Proposition 2.1
(see [19])
Assume that
\(f, g:I\to {\mathrm{R}}\)
are
α-forward integrable on
\([a,b]\), \(c \in[a,b]\), \(k\in{\mathrm{R}}\), then:
1.
\(\int_{a}^{a} {f(t)\Delta_{\alpha}t} =0\);
2.
\(\int_{a}^{b} {f(t)\Delta_{\alpha}t} =\int_{a}^{c} {f(t)\Delta_{\alpha}t} +\int_{c}^{b} {f(t)\Delta_{\alpha}t} \), when the integrals exist;
3.
\(\int_{a}^{b} {f(t)\Delta_{\alpha}t} =-\int_{b}^{a} {f(t)\Delta _{\alpha}t} \);
4.
kf
is
α-forward integrable on
\([a,b]\)
and
$$\int_{a}^{b} {kf(t)\Delta_{\alpha}t} =k\int _{a}^{b} {f(t)\Delta_{\alpha}t} ; $$
5.
\(f +g \)
is
α-forward integrable on
\([a,b]\)
and
$$\int_{a}^{b} {(f+g) (t)\Delta_{\alpha}t} = \int_{a}^{b} {f(t)\Delta_{\alpha}t} +\int _{a}^{b} {g(t)\Delta_{\alpha}t} ; $$
6.
if
\(f \equiv0\), then
\(\int_{a}^{b} {f(t)\Delta_{\alpha}t} =0 \).
Proposition 2.2
(see [19])
Assume that
\(f:I\to{\mathrm{R}}\)
is
α-forward integrable on
\([a,b]\). Let
\(g:I \to{\mathrm{R}}\)
be a nonnegative
α-forward integrable function on
\([a,b]\). Then
fg
is
α-forward integrable on
\([a,b]\).
Proposition 2.3
(see [19])
Assume that
\(f:I \to{\mathrm{R}}\)
and let
\(|f| \)
be
α-forward integrable on
\([a,b]\). If
\(p>1\), then
\(\vert f\vert^{p}\)
is also
α-forward integrable on
\([a,b]\).
Proposition 2.4
(see [19])
Assume that
\(f, g:I \to {\mathrm{R}}\)
are
α-forward integrable on
\([a,b]\)
with
\(b=a+k\alpha\)
for some
\(k\in{\mathrm{N}}_{0}\). We have:
1.
If
\(f(t)\ge0\)
for all
\(t\in\{a+k\alpha:k\in{\mathrm{N}}_{0} \}\), then
\(\int_{a}^{b} {f(t)\Delta_{\alpha}t} \ge0 \).
2.
If
\(g(t)\ge f(t)\)
for all
\(t\in\{a+k\alpha:k\in{\mathrm{N}}_{0} \} \), then
\(\int_{a}^{b} {g(t)\Delta_{\alpha}t} \ge\int_{a}^{b} {f(t)\Delta_{\alpha}t}\).
Proposition 2.5
(Fundamental theorem of the α-forward integral, see [19])
Assume that
\(f:I \to{\mathrm{R}}\)
is
α-forward integrable over
I. Let
\(x\in I\)
and define
\(F(x)=\int_{a}^{x} {f(t)\Delta _{\alpha}t} \). Then
\(\Delta_{\alpha}[F](x)=f(x)\). Conversely, \(\int_{a}^{b} {\Delta_{\alpha}[f](t)\Delta_{\alpha}t} =f(b)-f(a)\).
Proposition 2.6
(α-Forward integration by parts, see [19])
Assume that
\(f, g:I \to{\mathrm{R}}\). Let
fg
and
\(f\Delta_{\alpha}[g]\)
be
α-forward integrable on
\([a,b]\). Then
$$\int_{a}^{b} {f(t)\Delta_{\alpha}[g](t) \Delta_{\alpha}t} =f(t)g(t)\vert_{a}^{b} -\int _{a}^{b} {\Delta_{\alpha}[f](t)g(t+\alpha) \Delta_{\alpha}[g](t)\Delta _{\alpha}t}. $$
Similarly, the β-backward integral is defined by Definition 2.2.
Definition 2.2
(see [19])
Assume that \(I\subseteq {\mathrm{R}}\) with \(\inf I=-\infty\) and let \(a, b \in I\) with \(a< b\). For \(f:I\to \mathrm{R}\) and \(\beta>0\), the β-backward integral of f is defined by
where \(\int_{-\infty}^{x} {f(t)\nabla_{\beta}t} =\beta \sum_{k=0}^{+\infty} {f(x-k\beta)} \), provided the series converges at \(x=a\) and \(x=b\).
$$\int_{a}^{b} {f(t)\nabla_{\beta}t} =\int _{-\infty}^{b} {f(t)\nabla_{\beta}t} -\int _{-\infty}^{a} {f(t)\nabla_{\beta}t} , $$
The β-backward integral has similar results and properties to the α-forward integral. In particular, the β-backward integral is the inverse operator of \(\nabla_{\beta}\).
We recall the \(\alpha,\beta\)-symmetric integral which is defined as a linear combination of the α-forward and the β-backward integrals.
Definition 2.3
(see [19])
Assume that \(f:{\mathrm{R}}\to {\mathrm{R}}\) and \(a, b \in \mathrm{R} \) with \(a< b\). If f is α-forward and β-backward integrable on \([a,b]\), \(\alpha,\beta\ge0\) with \(\alpha+\beta>0\), then we define the \(\alpha, \beta\)-symmetric integral of f from a to b by
The function f is \(\alpha,\beta\)-symmetric integrable if it is \(\alpha ,\beta\)-symmetric integrable for all \(a,b\in{\mathrm{R}}\).
$$\int_{a}^{b} {f(t)\, d_{\alpha,\beta} t} = \frac{\alpha}{\alpha+\beta}\int_{a}^{b} {f(t) \Delta_{\alpha}t} +\frac{\beta}{\alpha+\beta}\int_{a}^{b} {f(t)\nabla _{\beta}t}. $$
The \(\alpha, \beta\)-symmetric integral has the following properties.
Proposition 2.7
(see [19])
Suppose that
\(f, g:{\mathrm{R}} \to{\mathrm{R}}\)
are
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\). Let
\(c \in[a,b]\)
and
\(k \in \mathrm{R}\). Then:
1.
\(\int_{a}^{a} {f(t)\, d_{\alpha,\beta} t} =0\);
2.
\(\int_{a}^{b} {f(t)\, d_{\alpha,\beta} t} =\int_{a}^{c} {f(t)\, d_{\alpha ,\beta} t} +\int_{c}^{b} {f(t)\, d_{\alpha,\beta} t}\), when the integrals exist;
3.
\(\int_{a}^{b} {f(t)\, d_{\alpha,\beta} t} =-\int_{b}^{a} {f(t)\, d_{\alpha,\beta} t}\);
4.
kf
is
\(\alpha, \beta\)-symmetric integrable on
\([a,b]\)
and
$$\int_{a}^{b} {kf(t)\, d_{\alpha,\beta} t} =k\int _{a}^{b} {f(t)\, d_{\alpha,\beta} t} ; $$
5.
\(f +g\)
is
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\)
and
$$\int_{a}^{b} {(f+g) (t)\, d_{\alpha,\beta} t} = \int_{a}^{b} {f(t)\, d_{\alpha,\beta } t} +\int _{a}^{b} {g(t)\, d_{\alpha,\beta} t} ; $$
6.
fg
is
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\)
provided
g
is a nonnegative function.
Proposition 2.8
(see [19])
Suppose that
\(f:{\mathrm{R}}\to{\mathrm{R}}\)
and
\(p> 1\). Let
\(\vert f \vert\)
be symmetric
\(\alpha ,\beta\)-integrable on
\([a,b]\), then
\(\vert f\vert^{p}\)
is also
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\).
Proposition 2.9
(see [19])
Assume that
\(f, g:{\mathrm{R}}\to{\mathrm{R}}\)
are
\(\alpha,\beta\)-symmetric integrable functions on
\([a,b]\), \(A=\{ a+k\alpha :k\in{\mathrm{N}}_{0} \}\)
and
\(B=\{b-k\beta:k\in{\mathrm{N}}_{0} \}\). For
\(b \in A\)
and
\(a\in B\), we have:
1.
if
\(\vert f(t)\vert\le g(t)\)
for all
\(t \in A \cup B\), then
$$\biggl\vert {\int_{a}^{b} {f(t)\, d_{\alpha,\beta} t} } \biggr\vert \le\int_{a}^{b} {g(t)\, d_{\alpha,\beta} t} ; $$
2.
if
\(f(t)\ge0\)
for all
\(t\in A\cup B\), then
$$\int_{a}^{b} {f(t)\, d_{\alpha,\beta} t} \ge0; $$
3.
if
\(g(t)\ge f(t)\)
for all
\(t\in A \cup B\), then
$$\int_{a}^{b} {g(t)}\, d_{\alpha,\beta} t\ge\int _{a}^{b} {f(t)\, d_{\alpha,\beta } t} . $$
In Proposition 2.9 we assume that \(a, b\in{\mathrm{R}}\) with \(b\in A=\{ a+k\alpha :k\in{\mathrm{N}}_{0} \}\) and \(a\in B=\{b-k\beta:k\in{\mathrm{N}}_{0} \}\), where \(\alpha,\beta\in{\mathrm{R}}_{0}^{+}\), \(\alpha+\beta\ne0\).
Proposition 2.10
(Mean value theorem, see [19])
If
\(f, g:{\mathrm{R}}\to {\mathrm{R}}\)
are bounded and
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\)
with
g
nonnegative. Let
m
and
M
be the infimum and the supremum, respectively, of the function
f. Then there exists a real number
K
satisfying the inequalities
\(m\le K\le M\)
such that
$$\int_{a}^{b} {f(t)g(t)}\, d_{\alpha,\beta} t=K\int _{a}^{b} {g(t)\, d_{\alpha ,\beta} t} . $$
3 Main results
As before, let \(a, b\in{\mathrm{R}}\) with \(b\in A=\{a+k\alpha:k\in {\mathrm{N}}_{0} \}\) and \(a\in B=\{b-k\beta:k\in{\mathrm{N}}_{0} \}\), where \(\alpha ,\beta\in{\mathrm{R}}_{0}^{+} \), \(\alpha+\beta\ne0\).
Theorem 3.1
(Reverse Hölder inequality)
Let
\(f, g:{\mathrm{R}}\to {\mathrm{R}}\)
and
\(a, b\in{\mathrm{R}}\)
with
\(a < b\). If
\(\vert f\vert\)
and
\(\vert g\vert\)
are
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\), \(0< p<1\) (or
\(p<0\)) with
\(q = p/(p-1)\), then
with equality if and only if the functions
\(|f|\)
and
\(|g|\)
are proportional.
$$ \int_{a}^{b} {\bigl\vert f(t)g(t)\bigr\vert \, d_{\alpha,\beta} t} \ge \biggl( {\int_{a}^{b} { \bigl\vert f(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p} \biggl( {\int_{a}^{b} {\bigl\vert g(t)\bigr\vert ^{q}\, d_{\alpha,\beta} t} } \biggr)^{1/q}, $$
(3.1)
Proof
We assume that
and let
and
Since the two functions \(\xi(t)\) and \(\gamma(t)\) are symmetric \(\alpha,\beta\)-integrable on \([a,b]\), applying the following reverse Young inequality (see [21]):
with equality holding iff \(x=y\), we have
Therefore, we obtain the desired inequality. □
$$\biggl( {\int_{a}^{b} {\bigl\vert f(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p} \biggl( { \int_{a}^{b} {\bigl\vert g(t)\bigr\vert ^{q}\, d_{\alpha,\beta} t} } \biggr)^{1/q}\ne0, $$
$$\xi(t)=\bigl\vert f(t)\bigr\vert ^{p}\Big/\int_{a}^{b} {\bigl\vert f(\tau)\bigr\vert ^{p}\, d_{\alpha,\beta} \tau} $$
$$\gamma(t)=\bigl\vert g(t)\bigr\vert ^{q}\Big/\int_{a}^{b} {\bigl\vert f(\tau)\bigr\vert ^{q}\, d_{\alpha ,\beta} \tau} . $$
$$xy\ge \frac{1}{p}x^{p}+ \frac{1}{q}y^{q},\quad x,y>0, 0< p<1, 1/p+1/q=1, $$
$$\begin{aligned}& \int_{a}^{b} {\frac{\vert f(t)\vert}{ ( {\int_{a}^{b} {\vert f(\tau )\vert ^{p}\, d_{\alpha,\beta} \tau} } )^{1/p}}} \frac{\vert g(t)\vert }{ ( {\int_{a}^{b} {\vert g(\tau)\vert^{q}\, d_{\alpha,\beta} \tau} } )^{1/q}} \,d_{\alpha,\beta} t \\& \quad =\int_{a}^{b} {\xi^{1/p}(t) \gamma^{1/q}(t)\,d_{\alpha,\beta} t} \ge\int_{a}^{b} {\biggl(\frac{\xi(t)}{p}+\frac{\gamma(t)}{p}\biggr)\,d_{\alpha,\beta} t} \\& \quad =\frac{1}{p}\int_{a}^{b} {\biggl( \frac{\vert f(t)\vert^{p}}{\int_{a}^{b} {\vert f(\tau )\vert^{p}\, d_{\alpha,\beta} \tau} }\biggr)\,d_{\alpha,\beta} t} +\frac{1}{q}\int _{a}^{b} {\biggl(\frac{\vert g(t)\vert^{q}}{\int_{a}^{b} {\vert g(\tau )\vert^{q}\, d_{\alpha,\beta} \tau} }\biggr) \,d_{\alpha,\beta} t} =1. \end{aligned}$$
Corollary 3.1
Let
\(f_{j}:{\mathrm{R}}\to{\mathrm{R}}\), \(p_{j} \in {\mathrm{R}}\), \(j=1,2,\ldots, m\), \(\sum_{j=1}^{m} {1/p_{j} } =1\). If
\(\vert f_{j} \vert\)
is
\(\alpha, \beta\)-symmetric integrable on
\([a,b]\), then:
(1)
For
\(p_{j} >1\), we have
$$ \int_{a}^{b} {\prod_{j=1}^{m} {\bigl\vert {f_{j} (t)} \bigr\vert \,d_{\alpha ,\beta} t} } \le \prod_{j=1}^{m} { \biggl(\int _{a}^{b} {\bigl\vert {f_{j} (t)} \bigr\vert ^{p_{j} }\,d_{\alpha,\beta} t} \biggr)^{1/p_{j} }} . $$
(3.2)
(2)
For
\(0< p_{1} <1\), \(p_{j} <0\), \(j=2,\ldots, m\), we have
$$ \int_{a}^{b} {\prod_{j=1}^{m} {\bigl\vert {f_{j} (t)} \bigr\vert \,d_{\alpha ,\beta} t} } \ge \prod_{j=1}^{m} { \biggl(\int _{a}^{b} {\bigl\vert {f_{j} (t)} \bigr\vert ^{p_{j} }\,d_{\alpha,\beta} t} \biggr)^{1/p_{j} }} . $$
(3.3)
Theorem 3.2
(Reverse Minkowski inequality)
Let
\(f, g:{\mathrm{R}}\to {\mathrm{R}}\)
and
\(a, b, p\in{\mathrm{R}}\)
with
\(a< b\)
and
\(0< p< 1\) (or
\(p<0\)). If
\(|f|\)
and
\(|g|\)
are
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\), then
with equality if and only if the functions
\(|f|\)
and
\(|g|\)
are proportional.
$$\begin{aligned}& \biggl( {\int_{a}^{b} {\bigl\vert f(t)+g(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p} \\& \quad \ge \biggl( { \int_{a}^{b} {\bigl\vert f(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p}+ \biggl( {\int _{a}^{b} {\bigl\vert g(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p}, \end{aligned}$$
(3.4)
Proof
Let
By the \(\alpha,\beta\)-symmetric integral Hölder inequality [12], in view of \(0< p<1\), we have
By using (3.5), we immediately arrive at the Minkowski inequality and the theorem is completely proved. □
$$\begin{aligned}& M=\int_{a}^{b} {\bigl\vert f(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t},\qquad N=\int_{a}^{b} {\bigl\vert g(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t}, \\& W= \biggl( {\int_{a}^{b} {\bigl\vert f(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p}+ \biggl( { \int_{a}^{b} {\bigl\vert g(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p}. \end{aligned}$$
$$\begin{aligned} W =&\int_{a}^{b} {\bigl(\bigl\vert f(t)\bigr\vert ^{p}M^{1/p-1}+\bigl\vert g(t)\bigr\vert ^{p}N^{1/p-1}\bigr)\, d_{\alpha,\beta} t} \\ \le&\int_{a}^{b} {\bigl\vert f(t)+g(t)\bigr\vert ^{p}\bigl(M^{1/p}+N^{1/p}\bigr)^{1-p} \, d_{\alpha ,\beta} t} \\ =&W^{1-p}\int_{a}^{b} {\bigl\vert f(t)+g(t)\bigr\vert ^{p}\, d_{\alpha,\beta} t} . \end{aligned}$$
(3.5)
Theorem 3.3
Assume that
\(f, g:{\mathrm{R}}\to{\mathrm{R}}\)
and
\(a, b\in{\mathrm{R}}\)
with
\(a < b\). Assume that
\(\vert f \vert\)
and
\(\vert g \vert\)
are
\(\alpha, \beta\)-symmetric integrable on
\([a,b]\), \(p>0\), \(s,t\in {\mathrm{R}}\backslash\{0\}\), and
\(s\ne t\).
(1)
Suppose that
\(p, s, t\in{\mathrm{R}}\)
are different, such that
\(s, t>1\)
and
\((s-t)/(p-t)>1\), then
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{p}\,d_{\alpha,\beta} x} \\ & \quad \le \biggl[ { \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{s}\,d_{\alpha,\beta} x} } \biggr)^{\frac{1}{s}}+ \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{s}\,d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{s}}} \biggr]^{s(p-t)/(s-t)} \\ & \qquad {} \times \biggl[ { \biggl( {\int_{a}^{b} { \bigl\vert f(x)\bigr\vert ^{t}\,d_{\alpha,\beta} x} } \biggr)^{\frac{1}{t}}+ \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{t}\,d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{t}}} \biggr]^{t(s-p)/(s-t)}. \end{aligned}$$
(3.6)
(2)
Suppose that
\(p, s, t\in{\mathrm{R}}\)
are different, such that
\(0< s, t<1\)
and
\((s-t)/(p-t)<1\), then
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{p}\,d_{\alpha,\beta} x} \\ & \quad \ge \biggl[ { \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{s}\,d_{\alpha,\beta} x} } \biggr)^{\frac{1}{s}}+ \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{s}\,d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{s}}} \biggr]^{s(p-t)/(s-t)} \\ & \qquad {} \times \biggl[ { \biggl( {\int_{a}^{b} { \bigl\vert f(x)\bigr\vert ^{t}\,d_{\alpha,\beta} x} } \biggr)^{\frac{1}{t}}+ \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{t}\,d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{t}}} \biggr]^{t(s-p)/(s-t)}. \end{aligned}$$
(3.7)
Proof
(1) From the assumption, we have \((s-t)/(p-t)>1\), and it is obvious that
From the Hölder inequality (see [19]) with indices \((s-t)/(p-t)\) and \((s-t)/(s-p)\), it follows that
On the other hand, from the Minkowski inequality (see [19]) for \(s>1\) and \(t>1\), respectively, we obtain
and
From (3.8), (3.9), and (3.10), the desired result is obtained.
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{p}\, d_{\alpha,\beta} x} \\ & \quad =\int_{a}^{b} {\bigl(\bigl\vert f(x)+g(x) \bigr\vert ^{s}\bigr)^{(p-t)/(s-t)}\bigl(\bigl\vert f(x)+g(x)\bigr\vert ^{t}\bigr)^{(s-p)/(s-t)}\, d_{\alpha,\beta} x} . \end{aligned}$$
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{p}\, d_{\alpha,\beta} x} \\ & \quad \le \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} } \biggr)^{(p-t)/(s-t)} \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{t}\, d_{\alpha,\beta} x} } \biggr)^{(s-p)/(s-t)}. \end{aligned}$$
(3.8)
$$\begin{aligned}& \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} } \biggr)^{\frac{1}{s}} \\ & \quad \le \biggl( { \int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{s}\, d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{s}}+ \biggl( {\int _{a}^{b} {\bigl\vert g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} } \biggr)^{\frac{1}{s}} \end{aligned}$$
(3.9)
$$\begin{aligned}& \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{t}\, d_{\alpha,\beta} x} } \biggr)^{\frac{1}{t}} \\ & \quad \le \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{t}\, d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{t}}+ \biggl( {\int _{a}^{b} {\bigl\vert g(x)\bigr\vert ^{t}\, d_{\alpha,\beta} x} } \biggr)^{\frac{1}{t}}. \end{aligned}$$
(3.10)
(2) Based on the assumption, we have \((s-t)/(p-t)<1\) and in view of
by using inequality (3.1) with indices \((s-t)/(p-t)\) and \((s-t)/(s-p)\), we have
On the other hand, thanks to the Minkowski inequality (3.4) for the cases of \(0< s<1\) and \(0< t<1\),
and
It follows from (3.11), (3.12), and (3.13) that the desired result is obtained. □
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{p}\, d_{\alpha,\beta} x} \\& \quad =\int_{a}^{b} {\bigl(\bigl\vert f(x)+g(x) \bigr\vert ^{s}\bigr)^{(p-t)/(s-t)}\bigl(\bigl\vert f(x)+g(x)\bigr\vert ^{t}\bigr)^{(s-p)/(s-t)}\, d_{\alpha,\beta} x}, \end{aligned}$$
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{p}\, d_{\alpha,\beta}x} \\& \quad \ge \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} } \biggr)^{(p-t)/(s-t)} \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{t}\, d_{\alpha,\beta} x} } \biggr)^{(s-p)/(s-t)}. \end{aligned}$$
(3.11)
$$\begin{aligned}& \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} } \biggr)^{\frac{1}{s}} \\& \quad \ge \biggl( { \int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{s}\, d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{s}}+ \biggl( {\int _{a}^{b} {\bigl\vert g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} } \biggr)^{\frac{1}{s}} \end{aligned}$$
(3.12)
$$\begin{aligned}& \biggl( {\int_{a}^{b} {\bigl\vert f(x)+g(x)\bigr\vert ^{t}\, d_{\alpha,\beta} x} } \biggr)^{\frac{1}{t}} \\& \quad \ge \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{t}(dx)^{\alpha}} } \biggr)^{\frac{1}{t}}+ \biggl( {\int _{a}^{b} {\bigl\vert g(x)\bigr\vert ^{t}\, d_{\alpha ,\beta} x} } \biggr)^{\frac{1}{t}}. \end{aligned}$$
(3.13)
Remark 3.1
(1)
From the Minkowski inequality [19] and the reverse Minkowski inequality involving \(\alpha,\beta\)-symmetric integral, we can deduce the following generalization.
Corollary 3.2
Let
\(f_{j}:{\mathrm{R}}\to{\mathrm{R}}\), \(j=1,2,\ldots, m\). If
\(\vert f_{j} \vert\)
is
\(\alpha, \beta\)-symmetric integrable on
\([a,b]\), then:
(1)
For
\(p>1\), we have
$$ \Biggl( {\int_{a}^{b} {\Biggl\vert {\sum _{j=1}^{m} {f_{j} (x)} } \Biggr\vert ^{p}\, d_{\alpha ,\beta} x} } \Biggr)^{1/p}\le\sum _{j=1}^{m} { \biggl( {\int _{a}^{b} {\bigl\vert {f_{j}(x)} \bigr\vert ^{p}\, d_{\alpha,\beta} x} } \biggr)^{1/p}} . $$
(3.14)
(2)
For
\(0< p<1\), we have
$$ \Biggl( {\int_{a}^{b} {\Biggl\vert {\sum _{j=1}^{m} {f_{j} (x)} } \Biggr\vert ^{p}\, d_{\alpha ,\beta} x} } \Biggr)^{1/p}\ge\sum _{j=1}^{m} { \biggl( {\int _{a}^{b} {\bigl\vert {f_{j}(x)} \bigr\vert ^{p}\, d_{\alpha,\beta} x} } \biggr)^{1/p}} . $$
(3.15)
Corollary 3.3
Let
\(f_{j}:{\mathrm{R}}\to{\mathrm{R}}\), \(j=1,2,\ldots, m\). If
\(\vert f_{j} \vert\)
is
\(\alpha, \beta\)-symmetric integrable on
\([a,b]\), then:
(1)
For
\(p>1\), we have
$$ \int_{a}^{b} { \Biggl( {\sum _{j=1}^{m} {\bigl\vert f_{j} (x)\bigr\vert } } \Biggr)^{p}\, d_{\alpha ,\beta } x} \geq\sum _{j=1}^{m} {\int_{a}^{b} {\bigl\vert {f_{j} (x)} \bigr\vert ^{p}\, d_{\alpha ,\beta} x} } . $$
(3.16)
(2)
For
\(0< p<1\), we have
$$ \int_{a}^{b} { \Biggl( {\sum _{j=1}^{m} {\bigl\vert f_{j} (x)\bigr\vert } } \Biggr)^{p}\, d_{\alpha ,\beta } x} \leq\sum _{j=1}^{m} {\int_{a}^{b} {\bigl\vert {f_{j} (x)} \bigr\vert ^{p}\, d_{\alpha ,\beta} x} } . $$
(3.17)
Proof
(1) For \(p>1\), let \(s=p\), \(r=1\), by the Jensen inequality [3], it follows that
from the above inequality, we obtain
by integrating the above inequality with respect to x, we obtain the desired result.
$$\bigl\vert f_{1}(x)\bigr\vert +\bigl\vert f_{2}(x) \bigr\vert +\cdots+\bigl\vert f_{m}(x)\bigr\vert \geq \bigl(\bigl\vert f_{1}(x)\bigr\vert ^{p}+\bigl\vert f_{2}(x)\bigr\vert ^{p}+\cdots +\bigl\vert f_{m}(x)\bigr\vert ^{p} \bigr)^{1/p}, $$
$$\bigl(\bigl\vert f_{1}(x)\bigr\vert +\bigl\vert f_{2}(x)\bigr\vert +\cdots+\bigl\vert f_{m}(x)\bigr\vert \bigr)^{p}\geq \bigl\vert f_{1}(x)\bigr\vert ^{p}+\bigl\vert f_{2}(x)\bigr\vert ^{p}+ \cdots+\bigl\vert f_{m}(x)\bigr\vert ^{p}, $$
(2) For \(0< p<1\), let \(s=1\), \(r=p\), by the Jensen inequality [3], we have
it follows from the above inequality that
by integrating the above inequality with respect to x, the desired result is obtained. □
$$\bigl\vert f_{1}(x)\bigr\vert +\bigl\vert f_{2}(x) \bigr\vert +\cdots+\bigl\vert f_{m}(x)\bigr\vert \leq \bigl(\bigl\vert f_{1}(x)\bigr\vert ^{p}+\bigl\vert f_{2}(x)\bigr\vert ^{p}+\cdots +\bigl\vert f_{m}(x)\bigr\vert ^{p} \bigr)^{1/p}, $$
$$\bigl(\bigl\vert f_{1}(x)\bigr\vert +\bigl\vert f_{2}(x)\bigr\vert +\cdots+\bigl\vert f_{m}(x)\bigr\vert \bigr)^{p}\leq \bigl\vert f_{1}(x)\bigr\vert ^{p}+\bigl\vert f_{2}(x)\bigr\vert ^{p}+ \cdots+\bigl\vert f_{m}(x)\bigr\vert ^{p}, $$
Theorem 3.4
(Dresher inequality)
Let
\(f, g:{\mathrm{R}}\to {\mathrm{R}}\)
and
\(a, b\in {\mathrm{R}}\)
with
\(a< b\). If
\(\vert f \vert\)
and
\(\vert g\vert\)
are
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\), \(0< r<1<p\), then
with equality if and only if the functions
\(|f|\)
and
\(|g|\)
are proportional.
$$\begin{aligned}& \biggl( {\frac{\int_{a}^{b} {\vert {f(t)+g(t)} \vert ^{p}\,d_{\alpha,\beta } t} }{\int_{a}^{b} {\vert {f(t)+g(t)} \vert ^{r}\,d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)} \\& \quad \le \biggl( {\frac{\int_{a}^{b} {\vert {f(t)} \vert ^{p}\,d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert {f(t)} \vert ^{r}\,d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}+ \biggl( { \frac{\int_{a}^{b} {\vert {g(t)} \vert ^{p}\,d_{\alpha ,\beta} t} }{\int_{a}^{b} {\vert {g(t)} \vert ^{r}\,d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}, \end{aligned}$$
(3.18)
Proof
Based on the \(\alpha,\beta\)-symmetric integral Hölder and Minkowski inequalities [19], we have
From Theorem 3.2, we get
From (3.5) and (3.20), we get (3.18). Hence, the theorem is completely proved. □
$$\begin{aligned}& \biggl( \int_{a}^{b} \bigl\vert {f(t) + g(t)} \bigr\vert ^{p}\, d_{\alpha ,\beta}t \biggr)^{1/(p - r)} \\& \quad \le \biggl( \biggl( \int_{a}^{b} \bigl\vert {f(t)} \bigr\vert ^{p}\, d_{\alpha,\beta}t \biggr)^{1/p} + \biggl( \int_{a}^{b} \bigl\vert {g(t)} \bigr\vert ^{p}\, d_{\alpha,\beta}t \biggr)^{1/p} \biggr)^{p/(p - r)} \\& \quad = \biggl( \biggl( \frac{\int_{a}^{b} \vert {f(t)} \vert ^{p}\, d_{\alpha,\beta}t }{\int_{a}^{b} \vert {f(t)} \vert ^{r}\, d_{\alpha,\beta}t } \biggr)^{1/p} \biggl( \int _{a}^{b} \bigl\vert {f(t)} \bigr\vert ^{r}\, d_{\alpha,\beta}t \biggr)^{1/p} \\& \qquad \biggl.\biggl.{} + \biggl( \frac{\int_{a}^{b} \vert {g(t)} \vert ^{p}\, d_{\alpha,\beta}x }{\int_{a}^{b} \vert {g(t)} \vert ^{r}\, d_{\alpha,\beta}t } \biggr)^{1/p} \biggl( \int _{a}^{b} \bigl\vert {g(t)} \bigr\vert ^{r}\, d_{\alpha,\beta}t \biggr)^{1/p} \biggr)\biggr.^{p/(p - r)} \\& \quad \le \biggl( \biggl( \frac{\int_{a}^{b} \vert {f(t)} \vert ^{p}\, d_{\alpha,\beta}t }{\int_{a}^{b} \vert {f(t)} \vert ^{r}\, d_{\alpha,\beta}t } \biggr)^{1/(p - r)} + \biggl( \frac {\int_{a}^{b} \vert {g(t)} \vert ^{p}\, d_{\alpha,\beta}t }{\int_{a}^{b} \vert {g(t)} \vert ^{r}\, d_{\alpha,\beta}t } \biggr)^{1/(p - r)} \biggr) \\& \qquad {} \times \biggl( \biggl( \int_{a}^{b} \bigl\vert {f(t)} \bigr\vert ^{r}\, d_{\alpha,\beta}t \biggr)^{1/r} + \biggl( \int_{a}^{b} \bigl\vert {g(t)} \bigr\vert ^{r}\, d_{\alpha,\beta}t \biggr)^{1/r} \biggr)^{r/(p - r)}. \end{aligned}$$
(3.19)
$$\begin{aligned}& \biggl( { \biggl( {\int_{a}^{b} {\bigl\vert {f(t)} \bigr\vert ^{r}\, d_{\alpha,\beta} t} } \biggr)^{1/r}+ \biggl( {\int_{a}^{b} {\bigl\vert {g(t)} \bigr\vert ^{r}\, d_{\alpha,\beta } t} } \biggr)^{1/r}} \biggr)^{r} \\& \quad \le\int_{a}^{b} {\bigl\vert {f(t)+g(t)} \bigr\vert ^{r}\, d_{\alpha,\beta} t} . \end{aligned}$$
(3.20)
Corollary 3.4
Let
\(f_{j}:{\mathrm{R}}\to{\mathrm{R}}\), \(0< r<1<p\), \(j=1,2,\ldots, m\). If
\(\vert f_{j} \vert\)
is
\(\alpha, \beta \)-symmetric integrable on
\([a,b]\), then
$$ \biggl( {\frac{\int_{a}^{b} {\vert {\sum_{j=1}^{m} {f_{j} (t)} } \vert ^{p}\, d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert {\sum_{j=1}^{m} {f_{j} (t)} } \vert ^{r}\, d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}\le \sum _{j=1}^{m} { \biggl( {\frac{\int_{a}^{b} {\vert {f_{j} (t)} \vert ^{p}\, d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert {f_{j} (t)} \vert ^{r}\, d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}} . $$
(3.21)
Theorem 3.5
(Reverse Dresher inequality)
Let
\(f, g:{\mathrm{R}}\to {\mathrm{R}}\)
and
\(a,b\in{\mathrm{R}}\)
with
\(a< b\). If
\(\vert f \vert\)
and
\(\vert g\vert\)
are
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\), \(p\le 0\le r\le1\), then
with equality if and only if the functions
\(|f|\)
and
\(|g|\)
are proportional.
$$\begin{aligned}& \biggl( {\frac{\int_{a}^{b} {\vert {f(t)+g(t)} \vert ^{p}\, d_{\alpha,\beta } t} }{\int_{a}^{b} {\vert {f(t)+g(t)} \vert ^{r}\, d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)} \\& \quad \ge \biggl( {\frac{\int_{a}^{b} {\vert {f(t)} \vert ^{p}\, d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert {f(t)} \vert ^{r}\, d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}+ \biggl( { \frac{\int_{a}^{b} {\vert {g(t)} \vert ^{p}\, d_{\alpha ,\beta} t} }{\int_{a}^{b} {\vert {g(t)} \vert ^{r}\, d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}, \end{aligned}$$
(3.22)
Proof
Let \(\alpha_{1} \ge0\), \(\alpha_{2} \ge0\), \(\beta_{1} >0\), and \(\beta_{2} >0\), and \(-1<\lambda<0\), using the following Radon inequality (see [3]):
we have
with equality if and only if \((\alpha)\) and \((\beta)\) are proportional. Let
and set \(\lambda=\frac{r}{p-r}\). From (3.23)-(3.25), we have
Since \(-1<\lambda=\frac{r}{p-r}<0\), we may assume \(p<0<r\), and by Theorem 3.2 and \(0< r\le1\), we obtain, respectively,
with equality if and only if f and g are proportional, and
with equality if and only if \(|f|\) and \(|g|\) are proportional.
$$ \sum_{k=1}^{n} {\frac{a_{k}^{p} }{b_{k}^{p-1} }} < \frac{(\sum_{k=1}^{n} {a_{k} } )^{p}}{(\sum_{k=1}^{n} {b_{k} } )^{p-1}}, \quad x_{k} \ge0, a_{k} >0, 0<p<1, $$
$$ \frac{\alpha_{1}^{\lambda+1} }{\beta_{1}^{\lambda}}+\frac{\alpha _{2}^{\lambda +1} }{\beta_{2}^{\lambda}}\le\frac{(\alpha_{1} +\alpha_{2} )^{\lambda +1}}{(\beta_{1} +\beta_{2} )^{\lambda}}, $$
(3.23)
$$\begin{aligned}& \alpha_{1} = \biggl( {\int_{a}^{b} { \vert f\vert^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p}, \qquad \beta_{1} = \biggl( {\int_{a}^{b} {\vert f\vert ^{r}\, d_{\alpha,\beta } t} } \biggr)^{1/r}, \end{aligned}$$
(3.24)
$$\begin{aligned}& \alpha_{2} = \biggl( {\int_{a}^{b} { \vert g\vert^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p},\qquad \beta_{2} = \biggl( {\int_{a}^{b} {\vert g\vert ^{r}\, d_{\alpha,\beta } t} } \biggr)^{1/r}, \end{aligned}$$
(3.25)
$$\begin{aligned} \frac{\alpha_{1}^{\lambda+1} }{\beta_{1}^{\lambda}}+\frac{\alpha _{2}^{\lambda +1} }{\beta_{2}^{\lambda}} =&\frac{ ( {\int_{a}^{b} {\vert f\vert ^{p}\,d_{\alpha ,\beta} t} } )^{(\lambda+1)/p}}{ ( {\int_{a}^{b} {\vert f\vert ^{r}\,d_{\alpha,\beta} t} } )^{\lambda/r}}+\frac{ ( {\int_{a}^{b} {\vert g\vert^{p}\,d_{\alpha,\beta} t} } )^{(\lambda+1)/p}}{ ( {\int_{a}^{b} {\vert g\vert^{r}\,d_{\alpha,\beta} t} } )^{\lambda/r}} \\ =& \biggl( {\frac{\int_{a}^{b} {\vert f\vert^{p}\,d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert f\vert^{r}\,d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}+ \biggl( { \frac{\int_{a}^{b} {\vert g\vert^{p}\,d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert g\vert^{r}\,d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}\le\frac{(\alpha_{1} +\alpha_{2} )^{\lambda+1}}{(\beta_{1} +\beta_{2} )^{\lambda}} \\ =&\frac{[ ( {\int_{a}^{b} {\vert f\vert^{p}\,d_{\alpha,\beta} t} } )^{1/p}+ ( {\int_{a}^{b} {\vert g\vert^{p}\,d_{\alpha,\beta} t} } )^{1/p}]^{p/(p-r)}}{[ ( {\int_{a}^{b} {\vert f\vert^{r}\,d_{\alpha ,\beta } t} } )^{1/r}+ ( {\int_{a}^{b} {\vert g\vert^{r}\,d_{\alpha,\beta } t} } )^{1/r}]^{r/(p-r)}}. \end{aligned}$$
(3.26)
$$ \biggl[ { \biggl( {\int_{a}^{b} {\vert f \vert^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p}+ \biggl( {\int _{a}^{b} {\vert g\vert^{p}\, d_{\alpha,\beta} t} } \biggr)^{1/p}} \biggr]^{p}\ge\int _{a}^{b} {\vert f+g\vert^{p}\, d_{\alpha,\beta } t}, $$
(3.27)
$$ \biggl[ { \biggl( {\int_{a}^{b} {\vert f \vert^{r}\, d_{\alpha,\beta} t} } \biggr)^{1/r}+ \biggl( {\int _{a}^{b} {\vert g\vert^{r}\, d_{\alpha,\beta} t} } \biggr)^{1/r}} \biggr]^{r}\le\int _{a}^{b} {\vert f+g\vert^{r}\, d_{\alpha,\beta } t}, $$
(3.28)
Corollary 3.5
Let
\(f_{j}:{\mathrm{R}}\to{\mathrm{R}}\), \(p\le 0\le r<1\), \(j=1,2,\ldots, m\). If
\(\vert f_{j} \vert\)
is
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\), then
$$ \biggl( {\frac{\int_{a}^{b} {\vert {\sum_{j=1}^{m} {f_{j} (t)} } \vert ^{p}\, d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert {\sum_{j=1}^{m} {f_{j} (t)} } \vert ^{r}\, d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}\ge \sum _{j=1}^{m} { \biggl( {\frac{\int_{a}^{b} {\vert {f_{j} (t)} \vert ^{p}\, d_{\alpha,\beta} t} }{\int_{a}^{b} {\vert {f_{j} (t)} \vert ^{r}\, d_{\alpha,\beta} t} }} \biggr)^{1/(p-r)}} . $$
(3.29)
4 Some further generalizations of the Hölder inequality
Theorem 4.1
Suppose that
\(p_{k} >0\), \(\alpha_{kj} \in{\mathrm{R}}\) (\(j=1,2,\ldots ,m\), \(k=1,2,\ldots,s\)), \(\sum_{k}^{s} {\frac{1}{p_{k} }} =1\), \(\sum_{k=1}^{s} {\alpha_{kj} } =0\), \(f_{j} :{\mathrm{R}}\to{\mathrm{R}}\). If
\(\vert f_{j}\vert\)
is
\(\alpha,\beta\)-symmetric integrable on
\([a,b]\), then:
(1)
For
\(p_{k} >1\), we have
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \le\prod_{k=1}^{s} { \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+p_{k} \alpha_{kj} }} \, d_{\alpha,\beta} t} } \Biggr)^{1/p_{k} }}. $$
(4.1)
(2)
For
\(0< p_{s} <1\), \(p_{k} <0\) (\(k=1,2,\ldots,s-1\)), we have
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \ge\prod_{k=1}^{s} { \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+p_{k} \alpha_{kj} }}\, d_{\alpha,\beta} t} } \Biggr)^{1/p_{k} }}. $$
(4.2)
Proof
(1) Let
Applying the assumptions \(\sum_{k}^{s} {\frac{1}{p_{k} }} =1\) and \(\sum_{k=1}^{s} {\alpha_{kj} } =0\), it follows from a direct computation that
That is,
It is obvious that
From the Hölder inequality (3.2), it follows that
Substituting \(g_{k} (t)\) in (4.5), we have inequality (4.1) immediately.
$$ g_{k} (t)= \Biggl( {\prod_{j=1}^{m} {f_{j}^{1+p_{k} \alpha_{kj} } (t)} } \Biggr)^{1/p_{k} }. $$
(4.3)
$$\begin{aligned} \prod_{k=1}^{s} {g_{k} (t)} =&g_{1}(t) g_{2}(t) \cdots g_{s}(t) \\ =& \Biggl( {\prod_{j=1}^{m} {f_{j}^{1+p_{1} \alpha_{1j} } (t)} } \Biggr)^{1/p_{1} } \Biggl( {\prod _{j=1}^{m} {f_{j}^{1+p_{2} \alpha_{2j} } (t)} } \Biggr)^{1/p_{2} }\cdots \Biggl( {\prod_{j=1}^{m} {f_{j}^{1+p_{s} \alpha _{sj} } (t)} } \Biggr)^{1/p_{s} } \\ =&\prod_{j=1}^{m} {f_{j}^{1/p_{1} +\alpha_{1j} } (t)} \prod_{j=1}^{m} {f_{j}^{1/p_{2} +\alpha_{2j} } (t)} \cdots\prod_{j=1}^{m} {f_{j}^{1/p_{s} +\alpha_{sj} } (t)} \\ =&\prod_{j=1}^{m} {f_{j}^{1/p_{1} +1/p_{2} +\cdots1/p_{s} +\alpha_{1j} +\alpha_{2j} +\cdots+\alpha_{sj} } (t)} =\prod_{j=1}^{m} {f_{j} (t)} . \end{aligned}$$
$$\prod_{k=1}^{s} {g_{k} (t)} = \prod_{j=1}^{m} {f_{j} (t)} . $$
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} =\int_{a}^{b} {\Biggl\vert \prod_{k=1}^{s} {g_{k} (t)} \Biggr\vert \, d_{\alpha,\beta } t} . $$
(4.4)
$$ \int_{a}^{b} {\Biggl\vert \prod _{k=1}^{s} {g_{k} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \le\prod_{k=1}^{s} { \biggl( {\int_{a}^{b} {\bigl\vert g_{k} (t)\bigr\vert ^{p_{k} }\, d_{\alpha,\beta} t} } \biggr)^{1/p_{k} }} . $$
(4.5)
(2) The proof of inequality (4.2) is similar to the proof of inequality (4.1); by (4.3), (4.4), and (3.3), we obtain
Substitution of \(g_{k} (t)\) in Eq. (4.6) gives inequality (4.2) immediately. □
$$ \int_{a}^{b} {\Biggl\vert \prod _{k=1}^{s} {g_{k} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \ge\prod_{k=1}^{s} { \biggl( {\int_{a}^{b} {\bigl\vert g_{k} (t)\bigr\vert ^{p_{k} }\, d_{\alpha,\beta} t} } \biggr)^{1/p_{k} }} . $$
(4.6)
Remark 4.1
Many existing inequalities concerned with the Hölder inequality are special cases of the inequalities (4.1) and (4.2). For example, we have the following.
Corollary 4.1
Under the assumptions of Theorem
4.1, assume that
\(s=m\), \(\alpha_{kj} =-t/p_{k} \)
for
\(j\ne k\), and
\(\alpha_{kk} =t(1-1/p_{k} )\)
with
\(t\in \mathrm{R}\), then:
(1)
For
\(p_{k} >1\), one obtains
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \le\prod_{k=1}^{m} { \Biggl( {\int_{a}^{b} { \Biggl( {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert } } \Biggr)^{1-t}\bigl(\bigl\vert f_{k} (t)\bigr\vert ^{p_{k} }\bigr)^{t}\, d_{\alpha ,\beta} t} } \Biggr)^{1/p_{k} }} . $$
(4.7)
(2)
For
\(0< p_{m} <1\), \(p_{k} <0\) (\(k=1,2,\ldots,m-1\)), one obtains
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \ge\prod_{k=1}^{m} { \Biggl( {\int_{a}^{b} { \Biggl( {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert } } \Biggr)^{1-t}\bigl(\bigl\vert f_{k} (t)\bigr\vert ^{p_{k} }\bigr)^{t}\, d_{\alpha ,\beta} t} } \Biggr)^{1/p_{k} }} . $$
(4.8)
Theorem 4.2
Suppose that
\(p_{k} >0\), \(r\in{\mathrm{R}}\), \(\alpha_{kj} \in{\mathrm{R}}\) (\(j=1,2,\ldots,m\), \(k=1,2,\ldots,s\)), \(\sum_{k}^{s} {\frac{1}{p_{k} }} =r\), \(\sum_{k=1}^{s} {\alpha_{kj} } =0\), \(f_{j} :{\mathrm{R}}\to{\mathrm{R}}\). If
\(\vert f_{j} \vert\)
is
\(\alpha, \beta\)-symmetric integrable on
\([a,b]\), then:
(1)
For
\(rp_{k} >1\), we have
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \le\prod_{k=1}^{s} { \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp_{k} \alpha_{kj} }}\, d_{\alpha,\beta} t} } \Biggr)^{1/rp_{k} }}. $$
(4.9)
(2)
For
\(0< rp_{s} <1\), \(rp_{k} <0\) (\(k=1,2,\ldots,s-1\)), we have
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \ge\prod_{k=1}^{s} { \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp_{k} \alpha_{kj} }}\, d_{\alpha,\beta} t} } \Biggr)^{1/rp_{k} }}. $$
(4.10)
Proof
Corollary 4.2
Under the assumptions of Theorem
4.2, assume that
\(s=2\), \(p_{1} =p\), \(p_{2} =q\), \(\alpha_{1j} =-\alpha_{2j} =\alpha_{j} \), then:
(1)
For
\(rp>1\), we get
$$\begin{aligned}& \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta } t} \\& \quad \le \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp\alpha _{j} }}\, d_{\alpha,\beta} t} } \Biggr)^{1/rp} \Biggl( { \int_{a}^{b} {\prod_{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1-rq\alpha_{j} }}\, d_{\alpha ,\beta} t} } \Biggr)^{1/rq}. \end{aligned}$$
(4.11)
(2)
For
\(0< rp<1\), we get
$$\begin{aligned}& \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta } t} \\& \quad \ge \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp\alpha _{j} }} \, d_{\alpha,\beta} t} } \Biggr)^{1/rp} \Biggl( { \int_{a}^{b} {\prod_{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1-rq\alpha_{j} }} \, d_{\alpha ,\beta} t} } \Biggr)^{1/rq}. \end{aligned}$$
(4.12)
Theorem 4.3
Under the assumptions of Theorem
4.2:
(1)
For
\(rp_{k} >1\), one has
where
is a nonincreasing function with
\(a\le c\le b\).
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \le\varphi(c)\le\prod_{k=1}^{s} { \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp_{k} \alpha_{kj} }} \, d_{\alpha ,\beta} t} } \Biggr)^{1/rp_{k} }} , $$
(4.13)
$$\varphi(c)\equiv\int_{a}^{c} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert } \, d_{\alpha,\beta} t} +\prod_{k=1}^{s} { \Biggl( {\int_{c}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp_{k} \alpha_{kj} }} \, d_{\alpha ,\beta} t} } \Biggr)^{1/rp_{k} }} $$
(2)
For
\(0< rp_{s} <1\), \(rp_{k} <0\) (\(k=1,2,\ldots,s-1\)), one has
where
is a nondecreasing function with
\(a\le c\le b\).
$$ \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta} t} \ge\varphi(c)\ge\prod_{k=1}^{s} { \Biggl( {\int_{a}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp_{k} \alpha_{kj} }} \, d_{\alpha ,\beta} t} } \Biggr)^{1/rp_{k} }} , $$
(4.14)
$$\varphi(c)\equiv\int_{a}^{c} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert } \, d_{\alpha,\beta} t} +\prod_{k=1}^{s} { \Biggl( {\int_{c}^{b} {\prod _{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp_{k} \alpha_{kj} }} \, d_{\alpha ,\beta} t} } \Biggr)^{1/rp_{k} }} $$
Proof
(1) Let
By rearrangement and the assumptions of Theorem 4.2, it follows that
Then thanks to the Hölder inequality (3.2), we have
Therefore, we obtain the desired result.
$$g_{k} (t)= \Biggl( {\prod_{j=1}^{m} {f_{j}^{1+rp_{k} \alpha_{kj} } (t)} } \Biggr)^{1/rp_{k} }. $$
$$\prod_{j=1}^{m} {f_{j} (t)} = \prod_{k=1}^{s} {g_{k} (t)} . $$
$$\begin{aligned} \int_{a}^{b} {\Biggl\vert \prod _{j=1}^{m} {f_{j} (t)} \Biggr\vert \, d_{\alpha,\beta } t} =&\int_{a}^{b} {\Biggl\vert \prod_{k=1}^{s} {g_{k} (t)} \Biggr\vert \, d_{\alpha,\beta } t} \\ =&\int_{a}^{c} {\Biggl\vert \prod _{k=1}^{s} {g_{k} (t)} \Biggr\vert \, d_{\alpha ,\beta} t} +\int_{c}^{b} {\Biggl\vert \prod_{k=1}^{s} {g_{k} (t)} \Biggr\vert \, d_{\alpha,\beta } t} \\ \le&\int_{a}^{c} {\Biggl\vert \prod _{k=1}^{s} {g_{k} (t)} \Biggr\vert \, d_{\alpha ,\beta} t} +\prod_{k=1}^{s} { \biggl( {\int_{c}^{b} {\bigl\vert g_{k} (t)\bigr\vert ^{rp_{k} }\, d_{\alpha,\beta} t} } \biggr)^{1/rp_{k} }} \\ \le&\prod_{k=1}^{s} { \biggl( {\int _{a}^{c} {\bigl\vert g_{k} (t)\bigr\vert ^{rp_{k} }\, d_{\alpha,\beta} t} +\int_{c}^{b} {\bigl\vert g_{k} (t)\bigr\vert ^{rp_{k} }\, d_{\alpha ,\beta} t} } \biggr)^{1/rp_{k} }} \\ =&\prod_{k=1}^{s} { \biggl( {\int _{a}^{b} {\bigl\vert g_{k} (t)\bigr\vert ^{rp_{k} }\, d_{\alpha,\beta} t} } \biggr)^{1/rp_{k} }} \\ =&\prod_{k=1}^{s} { \Biggl( {\int _{a}^{b} {\prod_{j=1}^{m} {\bigl\vert f_{j} (t)\bigr\vert ^{1+rp_{k} \alpha_{kj} }} \, d_{\alpha,\beta} t} } \Biggr)^{1/rp_{k} }}. \end{aligned}$$
5 A subdividing of the Hölder inequality
Theorem 5.1
Let
\(f, g:{\mathrm{R}}\to{\mathrm{R}}\)
and
\(a,b\in{\mathrm{R}}\)
with
\(a < b\). Assume that
\(\vert f \vert\)
and
\(\vert g \vert\)
are
\(\alpha ,\beta\)-symmetric integrable on
\([a,b]\), and
\(s,t\in{\mathrm{R}}\), and let
\(p=(s-t)/1-t\), \(p=(s-t)/(s-1)\).
(1)
If
\(s<1<t\)
or
\(s>1>t\), then
with equality if and only if
\(f(x)\)
and
\(g(x)\)
are proportional.
$$\begin{aligned} \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert \, d_{\alpha,\beta} x} \le& \biggl( {\int_{a}^{b} { \bigl\vert f(x)\bigr\vert ^{sp}\, d_{\alpha,\beta} x} } \biggr)^{1/p^{2}} \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{tq}\, d_{\alpha,\beta } x} } \biggr)^{1/q^{2}} \\ &{}\times \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{tp}\, d_{\alpha,\beta} x} \int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{sq}\, d_{\alpha,\beta} x} } \biggr)^{1/pq} \end{aligned}$$
(5.1)
(2)
If
\(s>t>1\)
or
\(s< t<1\); \(t>s>1\)
or
\(t< s<1\), then
with equality if and only if
\(f(x)\)
and
\(g(x)\)
are proportional.
$$\begin{aligned} \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert \, d_{\alpha,\beta} x} \geq& \biggl( {\int_{a}^{b} { \bigl\vert f(x)\bigr\vert ^{sp}\, d_{\alpha,\beta} x} } \biggr)^{1/p^{2}} \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{tq}\, d_{\alpha,\beta } x} } \biggr)^{1/q^{2}} \\ &{}\times \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{tp}\, d_{\alpha,\beta} x} \int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{sq}\, d_{\alpha,\beta} x} } \biggr)^{1/pq} \end{aligned}$$
(5.2)
Proof
(1) Set \(p=\frac{s-t}{1-t}\), and it follows from \(s<1<t\) or \(s>1>t\) that
From inequality (1.3) with indices \(\frac{s-t}{1-t}\) and \(\frac{s-t}{s-1}\), it follows that
with equality if and only if \((fg)^{s}\) and \((fg)^{t}\) are proportional.
$$p=\frac{s-t}{1-t}>1. $$
$$\begin{aligned}& \int_{a}^{b} \bigl\vert f(x)g(x)\bigr\vert \, {d_{\alpha,\beta}}x \\& \quad = \int_{a}^{b} \bigl\vert f(x)g(x)\bigr\vert ^{s(1 - t)/(s - t)}\bigl\vert f(x)g(x)\bigr\vert ^{t(s - 1)/(s - t)}\, {d_{\alpha,\beta}}x \\& \quad \le \biggl( \int_{a}^{b} \bigl\vert f(x)g(x) \bigr\vert ^{s}\, {d_{\alpha,\beta}}x \biggr)^{(1 - t)/(s - t)} \biggl( \int_{a}^{b} \bigl\vert f(x)g(x)\bigr\vert ^{t}\, {d_{\alpha,\beta }}x \biggr)^{(s - 1)/(s - t)}, \end{aligned}$$
(5.3)
On the other hand, by the Hölder inequality again, for \(p=\frac{s-t}{1-t}>1\), the following two inequalities are obtained:
with equality if and only if \(f^{s(s-t)/(1-t)}\) and \(g^{s(s-t)/(s-1)}\) are proportional, and
with equality if and only if \(f^{t(s-t)/(1-t)}\) and \(g^{t(s-t)/(s-1)}\) are proportional.
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} \\& \quad \le \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{s(s-t)/(1-t)}\, d_{\alpha,\beta} x} } \biggr)^{(1-t)/(s-t)} \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{s(s-t)/(s-1)}\, d_{\alpha,\beta} x} } \biggr)^{(s-1)/(s-t)}, \end{aligned}$$
(5.4)
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert ^{t}\, d_{\alpha,\beta} x} \\& \quad \le \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{t(s-t)/(1-t)}\, d_{\alpha,\beta} x} } \biggr)^{(1-t)/(s-t)} \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{t(s-t)/(s-1)}\, d_{\alpha,\beta} x} } \biggr)^{(s-1)/(s-t)}, \end{aligned}$$
(5.5)
(2) Let \(p=\frac{s-t}{1-t}\), and in view of \(s>t>1\) or \(s< t<1\), we have
and \(t>s>1\) or \(t< s<1\), we have \(0<\frac{s-t}{1-t}<1\), by inequality (3.1) with indices \(\frac{s-t}{1-t}\) and \(\frac{s-t}{s-1}\), we have
with equality if and only if \((fg)^{s}\) and \((fg)^{t}\) are proportional.
$$p=\frac{s-t}{1-t}< 0, $$
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert \, {d_{\alpha,\beta}}x} \\& \quad = \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert ^{s(1 - t)/(s - t)}\bigl\vert f(x)g(x)\bigr\vert ^{t(s - 1)/(s - t)}\, {d_{\alpha,\beta}}x} \\& \quad \ge { \biggl( {\int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert ^{s}\, {d_{\alpha,\beta}}x} } \biggr)^{(1 - t)/(s - t)}} { \biggl( {\int_{a}^{b} { \bigl\vert f(x)g(x)\bigr\vert ^{t}\, {d_{\alpha,\beta }}x} } \biggr)^{(s - 1)/(s - t)}}, \end{aligned}$$
(5.6)
On the other hand, from the reverse Hölder inequality again for \(0< p=\frac{s-t}{1-t}<1\) or \(p=\frac{s-t}{1-t}<0\), we obtain the following two inequalities:
with equality if and only if \(f^{s(s-t)/(1-t)}\) and \(g^{s(s-t)/(s-1)}\) are proportional, and
with equality if and only if \(f^{t(s-t)/(1-t)}\) and \(g^{t(s-t)/(s-1)}\) are proportional.
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert ^{s}\, d_{\alpha,\beta} x} \\& \quad \ge \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{s(s-t)/(1-t)}\, d_{\alpha,\beta} x} } \biggr)^{(1-t)/(s-t)} \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{s(s-t)/(s-1)}\, d_{\alpha,\beta} x} } \biggr)^{(s-1)/(s-t)}, \end{aligned}$$
(5.7)
$$\begin{aligned}& \int_{a}^{b} {\bigl\vert f(x)g(x)\bigr\vert ^{t}\, d_{\alpha,\beta} x} \\& \quad \ge \biggl( {\int_{a}^{b} {\bigl\vert f(x)\bigr\vert ^{t(s-t)/(1-t)}\, d_{\alpha,\beta} x} } \biggr)^{(1-t)/(s-t)} \biggl( {\int_{a}^{b} {\bigl\vert g(x)\bigr\vert ^{t(s-t)/(s-1)}\, d_{\alpha,\beta} x} } \biggr)^{(s-1)/(s-t)}, \end{aligned}$$
(5.8)
Acknowledgements
The authors thank the editor and the referees for their valuable suggestions to improve the quality of this paper. This paper was partially supported by the Key Laboratory for Mixed and Missing Data Statistics of the Education Department of Guangxi Province (No. GXMMSL201404), the Scientific Research Project of Guangxi Education Department (Nos. YB2014560 and KY2015YB468), and the Natural Science Foundation of Guangxi Province (No. 2013JJAA10097).
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The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.