New Refinements for integral and sum forms of H\"older inequality

TL;DR

This paper introduces new refinements for the integral and sum forms of H"older inequality, enhancing existing inequalities and demonstrating their improvements through applications.

Contribution

The paper presents novel refinements of H"older inequality for both integral and sum forms, improving upon existing inequalities and showcasing their practical applications.

Findings

New refined inequalities for H"older inequality

Improved bounds for related inequalities

Practical applications demonstrating the refinements

Abstract

In this paper, new refinements for integral and sum forms of H\"older inequality are established. We note that many existing inequalities related to the H\"older inequality can be improved via obtained new inequalities in here, we show this in an application.

Equations

$$x^t{y}^{1-t}\le tx+(1-t)y$$

$$xy\le \frac{x^p}{p}+\frac{y^q}{q}$$

$${\int }_a^b\left|f(x)g(x)\right|dx\le {\left({\int }_a^b{\left|f(x)\right|}^{p}dx\right)}^{1/p}{\left({\int }_a^b{\left|g(x)\right|}^{q}dx\right)}^{1/q},$$

$$\sum _{k=1}^n{a}_k{b}_k\le {\left(\sum _{k=1}^n{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^n{b}_k^q\right)}^{1/q}.$$

$${\int }_a^b\left|f(x)g(x)\right|dx$$

$$\frac{1}{b-a}\{{\left({\int }_a^b(b-x){\left|f(x)\right|}^{p}dx\right)}^{1/p}{\left({\int }_a^b(b-x){\left|g(x)\right|}^{q}dx\right)}^{1/q}$$

$${\int }_a^b\left|f(x)g(x)\right|dx$$

$${\int }_a^{\lambda b+(1-\lambda )a}\left|f(x)g(x)\right|dx\le {\left({\int }_a^{\lambda b+(1-\lambda )a}{\left|f(x)\right|}^{p}dx\right)}^{1/p}{\left({\int }_a^{\lambda b+(1-\lambda )a}{\left|g(x)\right|}^{q}dx\right)}^{1/q}$$

$${\int }_{\lambda b+(1-\lambda )a}^b\left|f(x)g(x)\right|dx\le {\left({\int }_{\lambda b+(1-\lambda )a}^b{\left|f(x)\right|}^{p}dx\right)}^{1/p}{\left({\int }_{\lambda b+(1-\lambda )a}^b{\left|g(x)\right|}^{q}dx\right)}^{1/q}.$$

$${\int }_a^b\left|f(x)g(x)\right|dx$$

$$=$$

$${\left({\int }_a^b{\left|f(x)\right|}^{p}dx\right)}^{1/p}{\left({\int }_a^b{\left|g(x)\right|}^{q}dx\right)}^{1/q}=0.$$

$${\int }_a^b\left|f(x)g(x)\right|dx=0.$$

$$I={\left({\int }_a^b{\left|f(x)\right|}^{p}dx\right)}^{1/p}{\left({\int }_a^b{\left|g(x)\right|}^{q}dx\right)}^{1/q}\ne 0.$$

$$\frac{1}{\left(b-a\right)I}\{{\left({\int }_a^b(b-x){\left|f(x)\right|}^{p}dx\right)}^{1/p}{\left({\int }_a^b(b-x){\left|g(x)\right|}^{q}dx\right)}^{1/q}$$

$${\int }_a^b\left|f(x)g(x)\right|dx$$

$$\sum _{k=1}^n{a}_k{b}_k$$

$$\frac{1}{n}\left\{{\left(\sum _{k=1}^{n}k{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^{n}k{b}_k^q\right)}^{1/q}+{\left(\sum _{k=1}^n\left(n-k\right)a_k^p\right)}^{1/p}{\left(\sum _{k=1}^n\left(n-k\right)b_k^q\right)}^{1/q}\right\}$$

$$\sum _{k=1}^n{a}_k{b}_k$$

$${\left(\sum _{k=1}^n{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^n{b}_k^q\right)}^{1/q}=0.$$

$$\sum _{k=1}^n{a}_k{b}_k=0.$$

$$S={\left(\sum _{k=1}^n{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^n{b}_k^q\right)}^{1/q}\ne 0.$$

$$\frac{1}{nS}\left\{{\left(\sum _{k=1}^{n}k{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^{n}k{b}_k^q\right)}^{1/q}+{\left(\sum _{k=1}^n\left(n-k\right)a_k^p\right)}^{1/p}{\left(\sum _{k=1}^n\left(n-k\right)b_k^q\right)}^{1/q}\right\}$$

Full text

1. Introduction

The famous Young inequality for two scalars is the $t$-weighted arithmetic-geometric mean inequality. This inequality says that if $x,y>0$ and $t\in[0,1]$, then

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with equality if and only if $a=b.$ Let $p,q>1$ such that $1/p+1/q=1$. The inequality (1.1) can be written as

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for any $x,y\geq 0$. In this form, the inequality (1.2) was used to prove the celebrated Hölder inequality. One of the most important inequalities of analysis is Hölder’s inequality. It contributes wide area of pure and applied mathematics and plays a key role in resolving many problems in social science and cultural science as well as in natural science.

Theorem 1** (Hölder Inequality for Integrals [8]).**

Let $p>1$ and $1/p+1/q=1$. If f\and $g$ are real functions defined on $\left[a,b\right]$ and if $\left|f\right|^{p},\left|g\right|^{q}$ are integrable functions on $\left[a,b\right]$ then

[TABLE]

with equality holding if and only if $A\left|f(x)\right|^{p}=B\left|g(x)\right|^{q}$ almost everywhere, where $A$ and $B$ are constants.

Theorem 2** (Hölder Inequality for Sums [8]).**

Let $a=\left(a_{1},...,a_{n}\right)$ and $b=\left(b_{1},...,b_{n}\right)$ be two positive n-tuples and $p,q>0$ such that $1/p+1/q=1.$ Then we have

[TABLE]

Equality hold in (1.4) if and only if $a^{p}$ and $b^{q}$ are proportional.

Of course the Hölder’s inequality has been extensively explored and tested to a new situation by a number of scientists. Many generalizations and refinements for Hölder’s inequality have been obtained so far. See, for example, [1, 3, 4, 5, 6, 7, 8, 9, 10, 11] and the references therein. In this paper, by using a simple proof method some new refinements for integral and sum forms of Hölder’s inequality are obtained.

2. Main Results

Theorem 3**.**

Let $p>1$ and $1/p+1/q=1$. If f\and $g$ are real functions defined on $\left[a,b\right]$ and if $\left|f\right|^{p},\left|g\right|^{q}$ are integrable functions on $\left[a,b\right]$, then

• i.)

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• ii.)

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Proof.

i.)First method for Proof (Short method): By using of Hölder inequality in (1.3), it is easily seen that

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Second method for Proof (Long method): Applying (1.3) on the subinterval $[a,\lambda b+(1-\lambda)a]$ and on the subinterval $[\lambda b+(1-\lambda)a,b]$, respectively, we get

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and

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Adding the resulting inequalities, we get:

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By the change of variable $x=ub+(1-u)a$; on the right hand sides integrals in (2.3), we have

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Integrating both sides of this inequality over $\left[0,1\right]$ with respect to $\lambda$ we obtain that

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Then, By applying of Hölder inequality for the right hand sides integrals in the last inequality, we have

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By Fubini theorem and the change of variable $u=(x-a)/(b-a)$ we get

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[TABLE]

b) First let us consider the case

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Then, $f(x)=0$ for almost everywhere $x\in\left[a,b\right]$ or $g(x)=0$ for almost every where $x\in\left[a,b\right].$ Thus, we have

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Therefore the inequality (2.2) is trivial in this case.

Finally, we consider the case

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Then

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Applying (1.1) on the right hand sides integrals of the last inequality

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This completes the proof.

Remark 1**.**

The inequality (2.2) show that the inequality (i.)) is better than the inequality (1.3).

The more general versions of Theorem 3 can be given as follow:

Theorem 4**.**

Let $p>1$ and $1/p+1/q=1$. If f\and $g$ are real functions defined on $\left[a,b\right]$ and if $\left|f\right|^{p},\left|g\right|^{q}$ are integrable functions on $\left[a,b\right]$, then

i.)

[TABLE]

where $\alpha,\beta:\left[a,b\right]\rightarrow\left[0,\infty\right)$ are continuous functions such that $\alpha(x)+\beta(x)=1,\ x\in\left[a,b\right].$

ii.)

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where $\alpha_{i}:\left[a,b\right]\rightarrow\left[0,\infty\right),i=1,2,...n,$ are continuous functions such that $\sum_{i=1}^{n}\alpha_{i}(x)=1,\ x\in\left[a,b\right].$

Proof.

The proof of Theorem is easily seen by using similar method the proof of Theorem 3.

Remark 2**.**

It is easily seen that the inequalities obtained in Theorem 4 are the best than the inequality (1.3).

Remark 3**.**

i.) In the inequality (4) of Theorem 4, if we take $\alpha(x)=\sin^{2}x$ and $\beta(x)=\cos^{2}x$, then we have

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ii.) In the inequality (4) of Theorem 4, if we take $\alpha(x)=\frac{b-x}{b-a}$ and $\beta(x)=\frac{x-a}{b-a}$, then we have the inequality (i.)).

Theorem 5**.**

Let $a=\left(a_{1},...,a_{n}\right)$ and $b=\left(b_{1},...,b_{n}\right)$ be two positive n-tuples and $p,q>0$ such that $1/p+1/q=1.$ Then

• i.)

[TABLE]

• ii.)

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Proof.

i.) By using of Hölder inequality in (1.4), it is easily seen that

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ii.) First let us consider the case

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Then $a_{k}=0$ for $k=1,2,..,n$ or $b_{k}=0$ for $k=1,2,..,n.$ Thus, we have

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Therefore the inequality (ii.)) is trivial in this case.

Finally, we consider the case

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Then

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Applying (1.1) on the right hand sides sums of the last inequality

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This completes the proof.

Remark 4**.**

The inequality (ii.)) show that the inequality (i.)) is better than the inequality (1.4).

The more general versions of Theorem 3 can be given as follow:

Theorem 6**.**

Let $a=\left(a_{1},...,a_{n}\right)$ and $b=\left(b_{1},...,b_{n}\right)$ be two positive n-tuples and $p,q>0$ such that $1/p+1/q=1.$

i.) If $c=\left(c_{1},...,c_{n}\right)$ and $d=\left(d_{1},...,d_{n}\right)$ be two positive n-tuples such that $c_{k}+d_{k}=1,k=1,2,...,n.$ Then

[TABLE]

ii.) If $c^{(i)}=\left(c_{1}^{(i)},...,c_{n}^{(i)}\right),i=1,2,...,m$ be positive n-tuples such that $\sum_{i=1}^{m}c_{k}^{(i)}=1,k=1,2,...,n.$ Then

[TABLE]

Proof.

The proof of Theorem is easily seen by using similar method the proof of Theorem 3.

Remark 5**.**

It is easily seen that the inequalities obtained in Theorem 6 are the best than the inequality (1.4).

Remark 6**.**

i.) In the inequality (6) of Theorem 6, if we take $c=\left(\sin^{2}1,...,\sin^{2}n\right)$ and $d=\left(\cos^{2}1,...,\cos^{2}n\right)$, then we have

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ii.) In the inequality (6) of Theorem 6, if we take $c=\left(\frac{1}{n},\frac{2}{n}...,1\right)$ and $d=\left(\frac{n-1}{n},\frac{n-2}{n}...,0\right)$ , then we have the inequality (i.)).

3. An Application

In [2], Dragomir et al. gave the following lemma for obtain main results.

Lemma 1**.**

Let $f:I^{\circ}\subseteq\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$, $a,b\in I^{\circ}$ with $a<b$ and $q>1.$ If $f\in L\left[a,b\right]$, then the following equality holds:

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By using this equality and Hölder integral inequality Dragomir et al. obtained the following inequality:

Theorem 7**.**

Let $f:I^{\circ}\subseteq\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$, $a,b\in I^{\circ}$ with $a<b$. If the new mapping $\left|f^{\prime}\right|^{q}$ is convex on $\left[a,b\right]$, then the following inequality holds:

[TABLE]

where $1/p+1/q=1.$

If Theorem 7 are resulted again by using the inequality (i.)) in Theorem 3, then we get the following result:

Theorem 8**.**

Let $f:I^{\circ}\subseteq\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable mapping on $I^{\circ}$, $a,b\in I^{\circ}$ with $a<b$. If the new mapping $\left|f^{\prime}\right|^{q}$ convex on $\left[a,b\right]$, then the following inequality holds:

[TABLE]

where $1/p+1/q=1.$

Proof.

Using Lemma 1 and the inequality (i.)), we find

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Using the convexity of $\left|f^{\prime}\right|^{q}$, we have

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and

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Further, since

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a combination of (3)-(3.5) immediately gives the required inequality (8).

Remark 7**.**

Since $h:\left[0,\infty\right)\rightarrow\mathbb{R},h(x)=x^{s},0<s\leq 1,$ is a concave function, for  all $u,v\geq 0$ we have

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From here, we get

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Thus we obtain

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This show us that the inequality (8) is the best than the inequality (3.1).