On Pólya-Szegö’s inequality

Abstract

In the paper, we give some new improvements of Pólya-Szegö’s integral inequality which in a special case yield some of the recent results related with Pólya-Szegö’s inequality.

MSC:26D15.

1 Introduction

The well-known Pólya-Szegö’s inequality can be stated as follows ([1] or see [2], p.62).

If $0<m_1\le u_k\le M_1$ and $0<m_2\le v_k\le M_2$, where $k=1,2,\dots ,n$, then

$$\left(\sum _{k=1}^n{u}_k^2\right)\left(\sum _{k=1}^n{v}_k^2\right)\le \frac{1}{4}{(\sqrt{\frac{M_1{M}_2}{m_1{m}_2}}+\sqrt{\frac{m_1{m}_2}{M_1{M}_2}})}^2{\left(\sum _{k=1}^n{u}_k{v}_k\right)}^2.$$

An integral analogue of Pólya-Szegö’s inequality easy follows.

If $(E,\mathcal{A},x)$ is a measure space and $f(x)$, $g(x)$ are non-negative measurable functions and $f^2(x)$, $g^2(x)$ are integrable on E, if $0<m_1\le f(x)\le M_1$ and $0<m_2\le g(x)\le M_2$, then

$$({\int }_E{f}^2(x)dx)({\int }_E{g}^2(x)dx)\le \frac{1}{4}{(\sqrt{\frac{M_1{M}_2}{m_1{m}_2}}+\sqrt{\frac{m_1{m}_2}{M_1{M}_2}})}^2{({\int }_{E}f(x)g(x)dx)}^2.$$

(1.1)

Pólya-Szegö’s inequality was studied extensively and numerous variants, generalizations, and extensions appeared in the literatures (see [3–7] and the references cited therein). The aim of this paper is to give some new improvements of Pólya-Szegö’s integral inequality which are generalizations of Pólya-Szegö’s integral inequality and interrelated result.

Theorem 1.1 Let $(E,\mathcal{A},x)$ be a measure space and $f(x)$, $g(x)$, $u(x)$, $v(x)$ be non-negative measurable functions. Let $p,q>0$, $\frac{1}{p}+\frac{1}{q}=1$, and $f^{1/p}(x)g^{1/q}(x)$, $u^{1/p}(x)v^{1/q}(x)$ be integrable on E and $u(x)$ and $v(x)$ be proportional. If $0<m_1\le f(x),u(x)\le M_1$ and $0<m_2\le g(x),v(x)\le M_2$, and $f(x)>u(x)$, $g(x)>v(x)$, then

$$\begin{array}{c}{\left({\int }_E(f(x)-u(x))dx\right)}^{1/p}{\left({\int }_E(g(x)-v(x))dx\right)}^{1/q}\hfill \\ \le {\mathrm{\Gamma }}_{p,q}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\int }_E(f^{1/p}(x)g^{1/q}(x)-u^{1/p}(x)v^{1/q}(x))dx\hfill \end{array}$$

(1.2)

with equality if and only if $f(x)$ and $g(x)$ are proportional and

$$({\int }_{E}f(x)dx,{\int }_{E}u(x)dx)=\mu ({\int }_{E}g(x)dx,{\int }_{E}v(x)dx)$$

for some constant μ and where

$${\mathrm{\Gamma }}_{p,q}(\xi )={(\sqrt[p]{p}\cdot \sqrt[q]{q})}^{-1}\frac{1-\xi }{{(1-{\xi }^{1/p})}^{1/p}{(1-{\xi }^{1/q})}^{1/q}}\cdot {\xi }^{-1/pq}.$$

(1.3)

Remark 1.1 Taking for $p=q=2$ and $u(x)=v(x)\equiv 0$ in (1.2), (1.2) changes to the following result:

$${({\int }_{E}f(x)dx)}^{1/2}{({\int }_{E}g(x)dx)}^{1/2}\le \frac{1}{2}(\sqrt[4]{\frac{M_1{M}_2}{m_1{m}_2}}+\sqrt[4]{\frac{m_1{m}_2}{M_1{M}_2}}){\int }_E{f}^{1/2}(x)g^{1/2}(x)dx$$

(1.4)

with equality if and only if $f(x)$ and $g(x)$ are proportional.

Replace $f^{1/2}(x)$ and $g^{1/2}(x)$ by $f(x)$ and $g(x)$ in (1.4), respectively, and hence $m_i^{1/2}(x)$ and $M_i^{1/2}(x)$ are replaced by $m_i$ and $M_i$ ($i=1,2$), respectively. Therefore

$${({\int }_E{f}^{1/2}(x)dx)}^{1/2}{({\int }_E{g}^{1/2}(x)dx)}^{1/2}\le \frac{1}{2}(\sqrt{\frac{M_1{M}_2}{m_1{m}_2}}+\sqrt{\frac{m_1{m}_2}{M_1{M}_2}}){\int }_{E}f(x)g(x)dx.$$

This is just Pólya-Szegö integral inequality (1.1). In fact, Theorem 1.1 is just a special case of Theorem 2.1 stated in Section 2.

Theorem 1.2 Let $(E,\mathcal{A},x)$ be a measure space and $f(x)$, $g(x)$, $u(x)$, $v(x)$ be non-negative measurable functions, and let $f^{1/p}(x)$, $g^{1/p}(x)$, $u^{1/p}(x)$, $v^{1/p}(x)$ be integrable on E, and $u(x)$ and $v(x)$ be proportional. If $p>1$, $0<m_1\le \frac{f(x)}{{(f(x)+g(x))}^{p-1}},\frac{u(x)}{{(u(x)+v(x))}^{p-1}}\le M_1$ and $0<m_2\le \frac{g(x)}{{(f(x)+g(x))}^{p-1}},\frac{v(x)}{{(u(x)+v(x))}^{p-1}}\le M_2$, and $f(x)>u(x)$, $g(x)>v(x)$, then

$$\begin{array}{c}{\left({\int }_E[f^p(x)-u^p(x)]dx\right)}^{1/p}+{\left({\int }_E[g^p(x)-v^p(x)]dx\right)}^{1/p}\hfill \\ \le {\mathrm{\Gamma }}_{p,\frac{p}{p-1}}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\left({\int }_E({[f(x)+g(x)]}^p-{[u(x)+v(x)]}^p)dx\right)}^{1/p}\hfill \end{array}$$

(1.5)

with equality if and only if $f(x)$ and $g(x)$ are proportional and

$$({\int }_E{f}^p(x)dx,{\int }_E{u}^p(x)dx)=\mu ({\int }_E{g}^p(x)dx,{\int }_E{v}^p(x)dx)$$

for some constant μ and ${\mathrm{\Gamma }}_{p,\frac{p}{p-1}}(\xi )$ is as in (1.3).

Remark 1.2 Taking for $u(x)=v(x)\equiv 0$ in (1.5), (1.5) changes to the following inequality:

$${({\int }_E{f}^p(x)dx)}^{1/p}+{({\int }_E{g}^p(x)dx)}^{1/p}\le {\mathrm{\Gamma }}_{p,\frac{p}{p-1}}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\left({\int }_E{(f(x)+g(x))}^{p}dx\right)}^{1/p}$$

with equality if and only if $f(x)$ and $g(x)$ are proportional.

This is just the inequality in Lemma 2.2 (see Section 2). In fact, Theorem 1.2 is just a special case of Theorem 2.2 stated in Section 2.

2 Main results

We need the following lemmas to prove our main results.

Lemma 2.1 [8]

Let $(E,\mathcal{A},x)$ be a measure space and $f(x)$, $g(x)$ be non-negative measurable functions. Let $p,q>0$, $\frac{1}{p}+\frac{1}{q}=1$ and $f^{1/p}(x)g^{1/q}(x)$ be integrable on E. If $0<m_1\le f(x)\le M_1$ and $0<m_2\le g(x)\le M_2$, then

$${({\int }_{E}f(x)dx)}^{1/p}{({\int }_{E}g(x)dx)}^{1/q}\le {\mathrm{\Gamma }}_{p,q}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\int }_E{f}^{1/p}(x)g^{1/q}(x)dx$$

(2.1)

with equality if and only if $f(x)$ and $g(x)$ are proportional.

Lemma 2.2 [9]

Let $(E,\mathcal{A},x)$ be a measure space and $f(x)$, $g(x)$ be non-negative measurable functions, and $f^{1/p}(x)$, $g^{1/p}(x)$ be integrable on E. If $p>1$, $0<m_1\le \frac{f(x)}{{(f(x)+g(x))}^{p-1}}\le M_1$ and $0<m_2\le \frac{g(x)}{{(f(x)+g(x))}^{p-1}}\le M_2$, then

$${({\int }_E{f}^p(x)dx)}^{1/p}+{({\int }_E{g}^p(x)dx)}^{1/p}\le {\mathrm{\Gamma }}_{p,\frac{p}{p-1}}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\left({\int }_E{(f(x)+g(x))}^{p}dx\right)}^{1/p}$$

(2.2)

with equality if and only if $f(x)$ and $g(x)$ are proportional.

Lemma 2.3 (Bellman’s inequality [10])

If

$$\varphi (x)={(x_1^p-x_2^p-\cdots -x_n^p)}^{1/p},p>1$$

for $x_i$ in the region ℝ defined by

• (a) $x_i\ge 0$,

• (b) $x_1\ge {(x_2^p+x_3^p+\cdots +x_n^p)}^{1/p}$.

Then, for $x,y\in \mathbb{R}$, we have

$$\varphi (x+y)\ge \varphi (x)+\varphi (y),$$

(2.3)

with equality if and only if $x=\mu y$, where μ is a constant.

Lemma 2.4 [11]

Let $a,b,c,d>0$, $0<\alpha <1$, $0<\beta <1$ and $\alpha +\beta =1$. If $a>b$ and $c>d$, then

$$a^{\alpha }{c}^{\beta }-b^{\alpha }{d}^{\beta }\ge {(a-b)}^{\alpha }{(c-d)}^{\beta }$$

(2.4)

with equality if and only if $a/b=c/d$.

Our main results are given in the following theorems.

Theorem 2.1 Let $(E,\mathcal{A},x)$ be a measure space and $f(x)$, $g(x)$, $u(x)$, $v(x)$ be non-negative measurable functions. Let $p,q>0$, $\frac{1}{p}+\frac{1}{q}=1$, and $f^{1/p}(x)g^{1/q}(x)$, $u^{1/p}(x)v^{1/q}(x)$ be integrable on E, and $u(x)$ and $v(x)$ be proportional. If $0<m_1\le f(x)\le M_1$, $0<m_2\le g(x)\le M_2$, $0<n_1\le u(x)\le N_1$ and $0<n_2\le v(x)\le N_2$, and $f(x)>u(x)$, $g(x)>v(x)$, then

$$\begin{array}{c}{\left({\int }_E(f(x)-u(x))dx\right)}^{1/p}{\left({\int }_E(g(x)-v(x))dx\right)}^{1/q}\hfill \\ \le {\mathrm{\Gamma }}_{p,q}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\int }_E{f}^{1/p}(x)g^{1/q}(x)dx-{\mathrm{\Gamma }}_{p,q}\left(\frac{n_1{n}_2}{N_1{N}_2}\right){\int }_E{u}^{1/p}(x)v^{1/q}(x)dx\hfill \end{array}$$

(2.5)

with equality if and only if $f(x)$ and $g(x)$ are proportional and

$$({\int }_{E}f(x)dx,{\int }_{E}u(x)dx)=\mu ({\int }_{E}g(x)dx,{\int }_{E}v(x)dx)$$

for some constant μ.

Proof From the hypotheses and Lemma 2.1, we obtain

$${\mathrm{\Gamma }}_{p,q}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\int }_E{f}^{1/p}(x)g^{1/q}(x)dx\ge {({\int }_{E}f(x)dx)}^{1/p}{({\int }_{E}g(x)dx)}^{1/q}$$

(2.6)

with equality if and only if $f(x)$ and $g(x)$ are proportional, and

$${\mathrm{\Gamma }}_{p,q}\left(\frac{n_1{n}_2}{N_1{N}_2}\right){\int }_E{u}^{1/p}(x)v^{1/q}(x)dx={({\int }_{E}u(x)dx)}^{1/p}{({\int }_{E}v(x)dx)}^{1/q}.$$

(2.7)

From (2.6), (2.7) and in view of $1/p+1/q=1$, by using Lemma 2.4, we have

$$\begin{array}{c}{\mathrm{\Gamma }}_{p,q}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\int }_E{f}^{1/p}(x)g^{1/q}(x)dx-{\mathrm{\Gamma }}_{p,q}\left(\frac{n_1{n}_2}{N_1{N}_2}\right){\int }_E{u}^{1/p}(x)v^{1/q}(x)dx\hfill \\ \ge {({\int }_{E}f(x)dx)}^{1/p}{({\int }_{E}g(x)dx)}^{1/q}-{({\int }_{E}u(x)dx)}^{1/p}{({\int }_{E}v(x)dx)}^{1/q}\hfill \\ \ge {\left({\int }_E(f(x)-u(x))dx\right)}^{1/p}{\left({\int }_E(g(x)-v(x))dx\right)}^{1/q}.\hfill \end{array}$$

(2.8)

In view of the equality conditions of (2.4) and (2.6), it follows that the sign of equality in (2.5) holds if and only if $f(x)$ and $g(x)$ are proportional and

$$({\int }_{E}f(x)dx,{\int }_{E}u(x)dx)=\mu ({\int }_{E}g(x)dx,{\int }_{E}v(x)dx)$$

for some constant μ. □

Remark 2.1 If $0<n_2\le u(x)\le N_2$ and $0<n_2\le v(x)\le N_2$ change to $0<m_1\le u(x)\le M_1$ and $0<m_2\le v(x)\le M_2$, respectively, then (2.5) reduces to (1.2) stated in the Introduction.

Theorem 2.2 Let $(E,\mathcal{A},x)$ be a measure space and $f(x)$, $g(x)$, $u(x)$, $v(x)$ be non-negative measurable functions, and let $f^{1/p}(x)$, $g^{1/p}(x)$, $u^{1/p}(x)$, $v^{1/p}(x)$ be integrable on E and $u(x)$ and $v(x)$ be proportional. If $p>1$, $0<m_1\le \frac{f(x)}{{(f(x)+g(x))}^{p-1}}\le M_1$, $0<m_2\le \frac{g(x)}{{(f(x)+g(x))}^{p-1}}\le M_2$, $0<n_1\le \frac{u(x)}{{(u(x)+v(x))}^{p-1}}\le N_1$ and $0<n_2\le \frac{v(x)}{{(u(x)+v(x))}^{p-1}}\le N_2$, and $f(x)>u(x)$, $g(x)>v(x)$, then

$$\begin{array}{c}{\left({\int }_E[f^p(x)-u^p(x)]dx\right)}^{1/p}+{\left({\int }_E[g^p(x)-v^p(x)]dx\right)}^{1/p}\hfill \\ \le [{\mathrm{\Gamma }}_{p,\frac{p}{p-1}}^p\left(\frac{m_1{m}_2}{M_1{M}_2}\right)\left({\int }_E{(f(x)+g(x))}^{p}dx\right)\hfill \\ {-{\mathrm{\Gamma }}_{p,\frac{p}{p-1}}^p\left(\frac{n_1{n}_2}{N_1{N}_2}\right)\left({\int }_E{(u(x)+v(x))}^{p}dx\right)]}^{1/p}\hfill \end{array}$$

(2.9)

with equality if and only if $f(x)$ and $g(x)$ are proportional and

$$({\int }_E{f}^p(x)dx,{\int }_E{u}^p(x)dx)=\mu ({\int }_E{g}^p(x)dx,{\int }_E{v}^p(x)dx)$$

for some constant μ.

Proof From the hypotheses and Lemma 2.2, it is easy to obtain

$$\begin{array}{c}{\mathrm{\Gamma }}_{p,\frac{p}{p-1}}\left(\frac{m_1{m}_2}{M_1{M}_2}\right){\left({\int }_E{(f(x)+g(x))}^{p}dx\right)}^{1/p}\hfill \\ \ge {({\int }_E{f}^p(x)dx)}^{1/p}+{({\int }_E{g}^p(x)dx)}^{1/p}\hfill \end{array}$$

(2.10)

with equality if and only if f and g are proportional, and

$$\begin{array}{c}{\mathrm{\Gamma }}_{p,\frac{p}{p-1}}\left(\frac{n_1{n}_2}{N_1{N}_2}\right){\left({\int }_E{(u(x)+v(x))}^{p}dx\right)}^{1/p}\hfill \\ ={({\int }_E{u}^p(x)dx)}^{1/p}+{({\int }_E{v}^p(x)dx)}^{1/p}.\hfill \end{array}$$

(2.11)

From (2.10), (2.11) and by using Lemma 2.3, we have

$$\begin{array}{c}{[{\mathrm{\Gamma }}_{p,\frac{p}{p-1}}^p\left(\frac{m_1{m}_2}{M_1{M}_2}\right)\left({\int }_E{(f(x)+g(x))}^{p}dx\right)-{\mathrm{\Gamma }}_{p,\frac{p}{p-1}}^p\left(\frac{n_1{n}_2}{N_1{N}_2}\right)\left({\int }_E{(u(x)+v(x))}^{p}dx\right)]}^{1/p}\hfill \\ \ge \{{[{({\int }_E{f}^p(x)dx)}^{1/p}+{({\int }_E{g}^p(x)dx)}^{1/p}]}^p\hfill \\ {-{[{({\int }_E{u}^p(x)dx)}^{1/p}+{({\int }_E{v}^p(x)dx)}^{1/p}]}^p\}}^{1/p}\hfill \\ \ge {\left({\int }_E[f^p(x)-u^p(x)]dx\right)}^{1/p}+{\left({\int }_E[g^p(x)-v^p(x)]dx\right)}^{1/p}.\hfill \end{array}$$

(2.12)

In view of the equality conditions of (2.10) and (2.3), it follows that the sign of equality (2.9) holds if and only if f and g are proportional and

$$({\int }_E{f}^p(x)dx,{\int }_E{u}^p(x)dx)=\mu ({\int }_E{g}^p(x)dx,{\int }_E{v}^p(x)dx)$$

for some constant μ. □

Remark 2.2 If $0<n_1\le \frac{u(x)}{{(u(x)+v(x))}^{p-1}}\le N_1$, $0<n_2\le \frac{v(x)}{{(u(x)+v(x))}^{p-1}}\le N_2$ change to $0<m_1\le \frac{u(x)}{{(u(x)+v(x))}^{p-1}}\le M_1$, $0<m_2\le \frac{v(x)}{{(u(x)+v(x))}^{p-1}}\le M_2$, respectively, then (2.9) reduces to (1.5) stated in the Introduction.