PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 99(113) (2016), 211–216 DOI: 10.2298/PIM1613211Z

HÖLDER’S REVERSE INEQUALITY AND ITS APPLICATIONS

Chang-Jian Zhao and Wing Sum Cheung

Abstract. We establish a new reverse Hölder integral inequality and its discrete version. As applications, we prove Radon’s, Jensen’s reverse and weighted power mean inequalities and their discrete versions.

1. Introduction

The well-known classical Hölder inequality can be stated as follows.

Theorem 1.1. Let a_{i} and b_{i} (i = 1, …, n) be positive real sequences. If p > 1 and \frac{1}{p} + \frac{1}{q} = 1, then

(1.1)  ( ∑ a_{i}^{p})^{1/p}( ∑ b_{i}^{q})^{1/q} ⩾ ∑ a_{i}b_{i}.

Here and in what follows ∑ means ∑ _{i=1}^{n} .

The integral version of (1.1) is the following.

Theorem 1.2. Let f(x) and g(x) be positive continuous functions on [a, b]. If p > 1 and \frac{1}{p} + \frac{1}{q} = 1, then

(1.2)  ( ∫_{a}^{b}f^{p}(x) dx)^{1/p}( ∫_{a}^{b}g^{q}(x) dx)^{1/q} ⩾ ∫_{a}^{b}f(x) g(x) dx.

Hölder’s inequality plays an important role in different branches of modern mathematics such as classical real and complex analysis, numerical analysis, probability and statistics, differential equations and et al. In recent years some authors [1,2,6,7,9,10,16–19] have given considerable attention to Hölder’s inequality together by its integral version and various generalizations. Some reverse versions were recent established [4, 13, 15, 18].

2010 Mathematics Subject Classification: 26D15.

Key words and phrases: Hölder’s inequality, Radon’s inequality, Jensen’s inequality, power mean inequality.

The first author’s research is supported by National Natural Science Foundation of China (11371334). The second author’s research is supported by National Natural Science Foundation of China (11371334) and a HKU Seed Grant for Basic Research.

Communicated by Gradimir Milovanović.

We establish a new reverse Hölder integral inequality and with its discrete form. As applications, we prove Radon’s, Jensen’s reverse and weighted power mean inequalities.

2. Hölder’s reverse inequalities

Lemma 2.1. If a, b are positive real numbers and \frac{1}{p} + \frac{1}{q} = 1 and p > 1, then (see [14])

(2.1)  S( \frac{a}{b}) a^{1/p}b^{1/q} ⩾ \frac{a}{p} + \frac{b}{q},

where

S(h) = \frac{h^{1/(h−1)}}{e logh^{1/(h−1)}} , h ≠ 1.

Theorem 2.1 (Hölder’s reverse inequality). Let \frac{1}{p} + \frac{1}{q} = 1 and p > 1. If f(x)

and g(x) are non-negative continuous functions and f^{1/p}(x) g^{1/q}(x) is integrable on [a, b], then

(2.2)  ( ∫_{a}^{b}f(x)^{p}dx)^{1/p}( ∫_{a}^{b}g(x)^{q}dx)^{1/q} ⩽ ∫_{a}^{b}S( \frac{Y f^{p}(x)}{Xg^{q}(x)}) · f(x)g(x) dx,

where

X = ∫_{a}^{b}f^{p}(x) dx, Y = ∫_{a}^{b}q^{q}(x) dx,

and S(h) is as in Lemma 2.1.

Proof. Let

a = \frac{f^{p}(x)}{X} , b = \frac{g^{q}(x)}{Y}.

By using Lemma 2.1, we have

S( \frac{Y f^{p}(x)}{Xg^{q}(x)}) · \frac{f(x)g(x)}{X^{1/p}Y^{1/q}} ⩾ \frac{1}{p} \frac{f^{p}(x)}{X} + \frac{1}{q} \frac{g^{p}(x)}{Y}.

Therefore

\frac{∫_{a}^{b}S( \frac{Y f^{p} (x)}{Xg^{q}(x)} )f(x)g(x)dx}{X^{1/p}Y^{1/q}} ⩾ \frac{1}{p} \frac{∥f(x)∥_{p}^{p}}{X} + \frac{1}{q} \frac{∥g(x)∥_{p}^{p}}{Y} = 1.

This proof is completed. □

Remark 2.1. Obviously, inequality (2.2) is just an inverse of inequality (1.2). Moreover, let f(x) and g(x) reduce to positive real sequences a_{i} and b_{i} (i = 1, …, n), respectively and with appropriate changes in the proof of (2.1), we have

( ∑ a_{i}^{p})^{1/p}( ∑ b_{i}^{q})^{1/q} ⩽ ∑ S( \frac{Y′a_{i}^{p}}{X′b_{i}^{q}}) a_{i}b_{i},

where X′ = ∑ a_{i}^{p}, Y′ = ∑ b_{i}^{q}.

This is just an inverse of the well-known Hölder’s inequality (1.1).

HÖLDER’S REVERSE INEQUALITY AND ITS APPLICATIONS 213

3. Applications of Hölder’s reverse inequality

Theorem 3.1 (Radon’s reverse integral inequality). Let f(x) and g(x) be positive and continuous functions. If m > 0, then

(3.1)  ∫_{a}^{b} \frac{f^{m+1} (x)}{g^{m}(x)} dx ⩽ \frac{( ∫_{a}^{b}S( \frac{Ỹf^{m+1} (x)}{X̃ g^{m+1} (x)} ) f(x) dx)^{m+1}}{( ∫_{a}^{b}g(x) dx)^{m}} ,

where

X̃ = ∫_{a}^{b} \frac{f^{m+1} (x)}{g^{m}(x)} dx, Ỹ = ∫_{a}^{b}g(x) dx.

Proof. Let p = m + 1, q = (m + 1)/m and replacing f(x) and g(x) by u(x) and v(x) in (2.2), respectively, we have ( ∫_{a}^{b}u(x)^{m+1} dx) ^{1/(m+1)} ( ∫_{a}^{b}v(x)^{(m+1)/m} dx) ^{m/(m+1)} (3.2)

⩽ ∫_{a}^{b}S( \frac{Ŷu^{m+1} (x)}{X̂v^{(m+1)/m} (x)}) u(x)v(x) dx,

where X̂ = ∫_{a}^{b}u^{m+1} (x) dx, Ŷ = ∫_{a}^{b}v^{(m+1)/m} (x) dx. Taking u(x) = ( \frac{f(x)}{g(x)} )^{1/(m+1)} , v(x) = f^{m/(m+1)} (x) g^{1/(m+1)} (x) in (3.2), we obtain

∫_{a}^{b}S( \frac{Ȳ}{X̄g^{(m+1)/m} (x)}) f(x) dx

⩾ ( ∫_{a}^{b} \frac{f(x)}{g(x)} dx) ^{1/(m+1)} ( ∫_{a}^{b}f(x) g^{1/m} (x) dx) ^{m/(m+1)} ,

where

X̄ = ∫_{a}^{b} \frac{f(x)}{g(x)} dx, Ȳ = ∫_{a}^{b}f(x) g^{1/m} (x) dx.

Hence

∫_{a}^{b} \frac{f(x)}{g(x)} dx ⩽ \frac{( ∫_{a}^{b}S( \frac{Ȳ}{X̄ g^{(m+1)/m} (x)} ) f(x)dx)^{m+1}}{( ∫_{a}^{b}f(x)g^{1/m} (x) dx)^{m}} .

Replacing f(x) and g(x) by u(x) and v(x), respectively, and leting u(x) = f(x) and v(x) = ( \frac{g(x)}{f(x)} )^{m}, we get

∫_{a}^{b} \frac{f^{m+1} (x)}{g^{m}(x)} dx ⩽ \frac{( ∫_{a}^{b}S( \frac{Ỹf^{m+1} (x)}{X̃ g^{m+1} (x)} ) f(x) dx)^{m+1}}{( ∫_{a}^{b}g(x) dx)^{m}} ,

where

X̃ = ∫_{a}^{b} \frac{f^{m+1} (x)}{g^{m}(x)} dx, Ỹ = ∫_{a}^{b}g(x) dx.  □

Remark 3.1. Let f(x) and g(x) reduce to positive real sequences a_{i} and b_{i} (i = 1, …, n), respectively and with appropriate changes in the proof of (3.1), we have

∑ \frac{a_{i}^{m+1}}{b_{i}^{m}} ⩽ \frac{( ∑ S( \frac{Ỹ′a_{i}^{m+1}}{X̃′ b_{i}^{m+1}} ) a_{i})^{m+1}}{( ∑ b_{i})^{m}} ,

where X̃′ = ∑ a_{i}^{m+1} /b_{i}^{m}, and Ỹ′= ∑ b_{i}.

This is just an inverse of following well-known the Radon inequality [5]

∑ \frac{a_{i}^{m+1}}{b_{i}^{m}} ⩾ \frac{( ∑ a_{i})^{m+1}}{( ∑ b_{i})^{m}} .

Theorem 3.2 (Jensen’s reverse integral inequality). Let f(x) and p(x) be positive continuous functions and ∫_{a}^{b}p(x) dx = 1. If 0 < s < t, then

(3.3)  ( ∫_{a}^{b}S( \frac{f^{t}(x)}{X̌} ) f^{s}(x)p(x) dx)^{1/s} ⩾ ( ∫_{a}^{b}p(x)f^{t}(x) dx)^{1/t},

where

X̌ = ∫_{a}^{b}p(x)f^{t}(x)t dx.

Proof. From the hypotheses, we have

(3.4) ∫_{a}^{b}S( \frac{f^{t}(x)}{X̌} ) f^{s}(x)p(x) dx

= ∫_{a}^{b}S( \frac{Y̌[p^{s/t} (x)f^{s}(x)]^{t/s}}{X̌[p^{1−s/t} (x)]^{t/(t−s)}} ) p^{s/t} (x)f^{s}(x) · p^{1−s/t} (x) dx,

where X̌ = ∫_{a}^{b}p(x)f^{t}(x) dx and Y̌ = ∫_{a}^{b}p(x) dx. In view if p = \frac{t}{s} and then q = \frac{t}{t−s} , and by using (2.2) on the right-hand side of (3.4), we have

(3.5) ∫_{a}^{b}S( \frac{Y̌[p^{s/t} (x)f^{s}(x)]^{t/s}}{X̌[p^{1−s/t} (x)]^{t/(t−s)}} ) p^{s/t} (x)f^{s}(x) · p^{1−s/t} (x) dx

⩾ ( ∫_{a}^{b}(p^{s/t} (x)f^{s}(x))^{t/s} dx)^{s/t} ( ∫_{a}^{b}[(p(x))^{1−s/t} ]^{t/(t−s)} dx)^{(t−s)/t} .

From (3.4), (3.5) and in view of ∫_{a}^{b}p(x) dx = 1, we obtain

∫_{a}^{b}S( \frac{f^{t}(x)}{X̌} ) f^{s}(x) p(x) dx ⩾ ( ∫_{a}^{b}p(x)f^{t}(x) dx)^{s/t} .

Hence

( ∫_{a}^{b}S( \frac{f^{t}(x)}{X̌} ) f^{s}(x) p(x) dx)^{1/s} ⩾ ( ∫_{a}^{b}p(x)f^{t}(x) dx)^{1/t}.  □

Remark 3.2. If f(x) and p(x) reduce to positive real sequences a_{i} and λ_{i} (i = 1, …, n), respectively and with appropriate changes in (3.3), we have

( ∑ S( \frac{a_{i}^{t}}{X̌′}) a_{i}^{s}λ_{i})^{1/s} ⩾ ( ∑ λ_{i}a_{i}^{t})^{1/t},

where X̌′ = ∑ λ_{i}a_{i}^{t}.

This is just an inverse of the following well-known Jensen’s inequality [3]

( ∑ a_{i}^{s}λ_{i}) ^{1/s} ⩽ ( ∑ λ_{i}a_{i}^{t}) ^{1/t} ,

Theorem 3.3 (Reverse weighted power mean integral inequality). Let f(x) and p(x) be positive and continuous functions. If r > 0 and ∫_{a}^{b}p(x) dx = 1, then

(3.6) ( ∫_{a}^{b}S( \frac{∫_{a}^{b}p(x)f^{2r} (x) dx}{f^{2r} (x)} ) p(x)f^{r}(x) dx)^{1/r} ⩾ ( ∫_{a}^{b}p(x)f^{2r} (x) dx)^{1/2r} .

Proof. From the hypotheses, we have

(3.7) ∫_{a}^{b}S( \frac{∫_{a}^{b}p(x)f^{2r} (x) dx}{f^{2r} (x)} ) p(x)f^{r}(x) dx

= ∫_{a}^{b}S( \frac{p(x) ∫_{a}^{b}p(x)f^{2r} (x) dx}{p(x)f^{2r} (x) ∫_{a}^{b}p(x) dx} ) p^{1/2} (x) · p^{1/2} (x)f^{r}(x) dx.

From (2.2), we obtain

(3.8)  ( ∫_{a}^{b}f(x)^{2}dx)^{1/2} ( ∫_{a}^{b}g(x)^{2}dx)^{1/2} ⩽ ∫_{a}^{b}S( \frac{Y̆f^{2}(x)}{X̆g^{2}(x)}) f(x) g(x) dx,

where X̆ = ∫_{a}^{b}f^{2}(x) dx, Y̆ = ∫_{a}^{b}g^{2}(x) dx. By using (3.8) on the right-hand side of (3.7), we have

( ∫_{a}^{b}S( \frac{∫_{a}^{b}p(x)f^{2r} (x) dx}{f^{2r} (x)} ) p(x)f^{r}(x) dx)^{2}

⩾ ∫_{a}^{b}p(x) dx · ∫_{a}^{b}p(x)f^{2r} (x) dx = ∫_{a}^{b}p(x)f^{2r} (x) dx.

Hence

( ∫_{a}^{b}S( \frac{∫_{a}^{b}p(x)f^{2r} (x) dx}{f^{2r} (x)} ) p(x)f^{r}(x) dx)^{1/r} ⩾ ( ∫_{a}^{b}p(x)f^{2r} (x) dx)^{1/2r} .  □

Remark 3.3. If f(x) and p(x) reduce to positive real sequences a_{i} and λ_{i} (i = 1, …, n), respectively and with appropriate changes in (3.6), we have

( ∑ S( \frac{∑ λ_{i}a_{i}^{2r}}{a_{i}^{2r}} ) λ_{i}a_{i}^{r})^{1/r} ⩾ ( ∑ λ_{i}a_{i}^{2r} )^{1/2r} .

This is just an inverse of the following well-known weighted power mean inequality [5]

( ∑ λ_{i}a_{i}^{r})^{1/r} ⩽ ( ∑ λ_{i}a_{i}^{2r} )^{1/2r} .

It is worth noting that literature [11] is relevant in spirit and methodology to the present paper, but for the Hilbert-type integral inequality and its connection with analytic number theory.

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Department of Mathematics, China Jiliang University (Received 24 04 2013) Hangzhou, P. R. China (Revised 11 01 2016)

 chjzhao@163.com, chjzhao@cjlu.edu.cn

Department of Mathematics, The University of Hong Kong, Hong Kong

 wscheung@hku.hk