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# A kind of extension of the famous Young inequality

- Luo Xianqiang
^{1, 2}Email author and - Xiong Zhiping
^{2}

**2013**:437

https://doi.org/10.1186/1029-242X-2013-437

© Xianqiang and Zhiping; licensee Springer. 2013

**Received:**5 February 2013**Accepted:**26 August 2013**Published:**15 September 2013

## Abstract

Young inequality, extended in (Geometry of Orlicz Spaces, 1986; Geometry of Orlicz Spaces, 1996), has extensive use and great effort in mathematical analysis. By the kind of extended Young inequality, we can get the famous Holder inequality and the Minkowski inequality. But until now, we have not found its strict proof of analysis. In (Geometry of Orlicz Spaces, 1986; Geometry of Orlicz Spaces, 1996), only the probable pattern description was found. In this paper, we will get the strict proof of analysis of a kind of extension of Young inequality with the approximation method.

**MSC:**46B20, 46B02, 46A22.

## Keywords

- Young inequality
*N*-function- strictly convex function

## 1 Introduction

where the equality holds if and only if $b=f(a)$.

Young inequality has an extensive use and a great effort in mathematical analysis. Now Young inequality was extended as follows.

Let $M(u)$ and $N(v)$ be complementary *N*-function with each other (see Definition 2.1 and Definition 2.2), then the kind of Young inequality $uv\le M(u)+N(v)$ holds, and the equality holds if and only if $u=q(|v|)signv$ or $v=p(|u|)signu$ for all $u,v\in (-\mathrm{\infty},+\mathrm{\infty})$.

By the kind of Young inequality, we can get the famous Holder inequality and the Minkowski inequality (see references [2] and [3]). But until now, we have not found its strict proof of analysis. In references [2] and [3], only the probable pattern description was found. Some other decisions can be found in [4–8].

In this paper, we will get its strict proof of analysis with the approximation method.

## 2 Preliminaries

**Definition 2.1** [2]

*N*-function if it has the following properties:

- (i)
$M(u)$ is even, continuous, convex and $M(0)=0$.

- (ii)
$M(u)>0$ for all $u\ne 0$.

- (iii)
${lim}_{u\to 0}\frac{M(u)}{u}=0$ and ${lim}_{u\to \mathrm{\infty}}\frac{M(u)}{u}=\mathrm{\infty}$.

**Lemma 2.1** [2]

*is an*

*N*-

*function if and only if there exists*$p(u):[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$

*with the following properties*:

- (i)
$p(u)$

*is right*-*continuous and nondecreasing*; - (ii)
$p(u)>0$

*whenever*$u\ne 0$; - (iii)
$p(0)=0$

*and*$p(\mathrm{\infty})=\mathrm{\infty}$, $M(u)={\int}_{0}^{|u|}p(t)\phantom{\rule{0.2em}{0ex}}dt$.

**Record 2.1** [2]

$p(u)$ is the right-derivative of *N*-function $M(u)$.

**Lemma 2.2** *Let* ${p}_{-}(u)$ *be the left*-*derivative of* *N*-*function* $M(u)$, *then* ${p}_{-}(u)={lim}_{h\to {0}^{+}}p(u-h)$, *and* ${\int}_{0}^{|u|}{p}_{-}(t)\phantom{\rule{0.2em}{0ex}}dt=M(u)$.

*Proof* From the proof process of Theorem 1.4 in reference [2], we know ${p}_{-}(u)$ is left continuous, and for all $0<u<v$, $p(u)\le {p}_{-}(v)\le p(v)$.

Hence, for $h>0$, we have $p(v-h)\le {p}_{-}(v)$ .

□

**Definition 2.2** [2]

Suppose that $M(u)$ is an *N*-function. Let $p(t)$ be the right derivative of $M(u)$. Let $q(s)={sup}_{p(t)\le s}t={inf}_{p(t)>s}t$, called the right-inverse function of $p(t)$. By Theorem 1.5 in reference [2], we know that $q(s)$ also satisfies the three properties of Lemma 2.1, and $N(v)={\int}_{0}^{|v|}q(s)\phantom{\rule{0.2em}{0ex}}ds$ is called the complementary *N*-function of $M(u)$. It is obvious, the left derivative ${q}_{-}(s)$ of $N(v)$ satisfies ${q}_{-}(s)={sup}_{p(t)<s}t={inf}_{p(t)\ge s}t$.

**Lemma 2.3** [2]

$q(p(t))\ge t$, $p(q(s))\ge s$; $q(p(t)-\epsilon )\le t$, $p(q(s)-\epsilon )\le s$.

**Lemma 2.4** [2]

$M(u)$ *is strictly convex if and only if* $p(t)$ *is strictly increasing*, *that is*, $q(s)$ *is continuous*.

**Lemma 2.5** [2]

*For any*

*N*-

*function*$M(u)$

*and*$\epsilon >0$,

*there exists a strictly convex*

*N*-

*function*${M}_{1}(u)$,

*such that*

*where* $p(t)$ *and* ${p}_{1}(t)$ *are the right derivatives of* $M(u)$ *and* ${M}_{1}(u)$, *respectively*.

**Record 2.2** Lemma 2.5 is Theorem 1.10 in reference [2], but it reverses the old conclusion ‘$M(u)\le {M}_{1}(u)\le (1+\epsilon )M(u)$,’ for the new conclusion ‘$(1-\epsilon )M(u)\le {M}_{1}(u)\le (1+\epsilon )M(u)$.’ From the construction process of ${p}_{1}(t)$, in the proof in reference [3], we know if $p(t)$ is continuous, then ${p}_{1}(t)$ is also continuous.

**Lemma 2.6** *Suppose that* $u\ge 0$ *and* $v\ge 0$, *then* $u=q(v)$ *or* $v=p(u)$ *if and only if* $u\in [{q}_{-}(v),q(v)]$. *By the symmetry*, *we get another necessary and sufficient condition*, *that is*, $v\in [{p}_{-}(u),p(u)]$.

*Proof* Sufficiency.

- (i)
If ${q}_{-}(v)=q(v)$, it is clear that $u={q}_{-}(v)=q(v)$.

- (ii)
If ${q}_{-}(v)\ne q(v)$, then ${q}_{-}(v)<q(v)$. If $u=q(v)$, then the conclusion holds.

If ${q}_{-}(v)\le u<q(v)$, we need only to prove that $p(u)=v$.

Necessity.

If $u=q(v)$, it is clearly established.

□

The next two lemmas are about the change of variable of integral and distribute integral.

**Lemma 2.7** [9]

*Suppose that*$f(x)$

*and*$g(x)$

*are defined on the interval*$[a,b]$,

*and the Stieltjes integral of*$f(x)$

*about*$g(x)$

*exists*.

*Suppose that*$x(t)$

*is a strictly increasing and continuous function on the interval*$[\alpha ,\beta ]$,

*and*$x(\alpha )=a$

*and*$x(\beta )=b$,

*then*

**Lemma 2.8** [9]

*Suppose that*$f(x)$

*and*$g(x)$

*are defined on the interval*$[a,b]$,

*and the Stieltjes integral of*$f(x)$

*about*$g(x)$

*exists*,

*then*

## 3 Main result

**Theorem 3.1** *Suppose that* $M(u)$ *is an* *N*-*function*, *and* $N(v)$ *is the complementary* *N*-*function of* $M(u)$, *then Young inequality* $uv\le M(u)+N(v)$ *holds*, *and* $uv=M(u)+N(v)$ *holds if and only if* $u=q(|v|)signv$ *or* $v=p(|u|)signu$.

*Proof* Suppose that $u\ge 0$ and $v\ge 0$.

Firstly, we will prove the necessity of the equality.

From Young inequality, we have learned that for all *u* and *v*, $F(u,v)\ge 0$.

From $M({u}_{0})+N({v}_{0})={u}_{0}{v}_{0}$, we have $F(u,{v}_{0})=M(u)+N({v}_{0})-u{v}_{0}$ and we can get the minimum 0 in ${u}_{0}$.

If ${u}_{0}=0$, from $M({u}_{0})+N({v}_{0})={u}_{0}{v}_{0}$, we get that ${v}_{0}=0$, then ${u}_{0}=q({v}_{0})=0$ or ${v}_{0}=p({u}_{0})=0$, that is, the necessity of the equality holds.

If ${u}_{0}\ne 0$, then $F({u}_{0},{v}_{0})$ is the minimum of the $F(u,{v}_{0})$ on the interval $(0,+\mathrm{\infty})$.

Therefore, the left derivative of $F(u,{v}_{0})$ is less than or equal to zero on the point ${u}_{0}$, and the right derivative of $F(u,{v}_{0})$ is more than or equal to zero on the point ${u}_{0}$.

From Lemma 2.6, we get ${u}_{0}=q({v}_{0})$ or ${v}_{0}=p({u}_{0})$.

That is, the necessity of the equality holds.

Secondly, we will get the proof of the Young inequality and the sufficiency of the equality in three steps.

Step I. Suppose that $M(u)$ and $N(v)$ are all strictly convex. From Lemma 2.4, the right derivative $p(t)$ and $q(s)$ are all strictly increasing, continuous, and are the right inverse-function of each other. From the reference [9], we have that the Stieltjes integral ${\int}_{0}^{q(v)}t\phantom{\rule{0.2em}{0ex}}dp(t)$ exists.

- (i)
If $u>q(v)$, then $p(u)>v$.

- (ii)
If $u<q(v)$, then $p(u)\le v$.

- (iii)
If $u=q(v)$, then $v=p(u)$.

From expression (1), we have $uv=M(u)+N(v)$.

That is, the sufficiency of the equality holds.

Step II. Suppose that $M(u)$ is strictly convex, then from Lemma 2.4, the right derivative $p(t)$ is strictly increasing, and the right-inverse function $q(s)$ is continuous and nondecreasing.

Let ${p}_{1}(t)$ be the right-inverse function of ${q}_{1}(s)$, then ${p}_{1}(t)$ is strictly increasing and continuous.

In the following, we will get the relation of ${p}_{1}(t)$ and $p(t)$.

In the following, we will prove the sufficiency of the equality.

In fact, if $s=p(u)$, from Definition 2.2, since $p(u)$ is strictly increasing, then we have $q(s)={sup}_{p(t)\le s}t={sup}_{p(t)\le p(u)}t=u$. If $s\in [{p}_{-}(u),p(u))$, from Lemma 2.6, we get $q(s)=u$. Therefore, we have ${\int}_{{p}_{1}(u)}^{p(u)}q(s)\phantom{\rule{0.2em}{0ex}}ds={\int}_{{p}_{1}(u)}^{p(u)}u\phantom{\rule{0.2em}{0ex}}ds=u(p(u)-{p}_{1}(u))$.

On the other hand, we have got the inequality $uv\le M(u)+N(v)$ .

That is, the sufficiency of the equality holds.

*N*-function $M(u)$, suppose that its complementary

*N*-function is $N(v)$, $p(t)$ is the right-inverse function of $M(u)$, and $q(s)$ is the right-inverse function of $N(v)$. From Lemma 2.5, for $0<\epsilon <\frac{1}{2}$ above, we can find a strictly convex

*N*-function ${M}_{1}(u)$ and its right-derivative ${p}_{1}(t)$ such that

Suppose that ${N}_{1}(v)$ is the complementary *N*-function of ${M}_{1}(u)$, ${q}_{1}(s)$ is the right derivative of ${N}_{1}(v)$.

In the following, we will prove sufficiency of the equality.

Let $\epsilon \to 0$, together with expression (13), we get ${p}_{1}(u)\to p(u)$, ${M}_{1}(u)\to M(u)$, and $N(\frac{{p}_{1}(u)}{1+2\epsilon})\to N(p(u))$ since $N(v)$ is continuous.

On the other hand, we have got the inequality $uv\le M(u)+N(v)$.

That is, the sufficiency of the equality holds. □

## Declarations

### Acknowledgements

This research was partially supported by the National Natural Science Foundation of China (Grant No: 11271245, and Grant No: 11301397), and the Natural Science Foundation Guangdong Province of China (2012KJCX0101).

## Authors’ Affiliations

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