A new improvement of Hölder inequality via isotonic linear functionals

1.   Introduction

The famous Young's inequality, as a classical result, state that: if $a, b > 0$ and $t\in \lbrack 0, 1]$, then

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

for any $a, b\geq 0$. 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 ^[11]). Let $p > 1$ and $1/p+1/q = 1$. If $f\ $and $g$ are real functions defined on $\left[a, b\right]$ and if $\left\vert f\right\vert ^{p}, \left\vert g\right\vert ^{q}$ are integrable functions on $\left[a, b\right]$ then

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

Theorem 2 (Hölder inequality for sums ^[11]). Let $a = \left(a_{1}, ..., a_{n}\right)$ and $b = \left(b_{1}, ..., b_{n}\right)$ be two positive n-tuples and $p, q > 1$ such that $1/p+1/q = 1.$ Then we have

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

In ^[10], İşcan gave new improvements for integral ans sum forms of the Hölder inequality as follow:

Theorem 3. Let $p > 1$ and $\frac{1}{p}+\frac{1}{q} = 1$. If $f$ and $g$ are real functions defined on interval $\left[a, b\right]$ and if $\left\vert f\right\vert ^{p}$, $\left\vert g\right\vert ^{q}$ are integrable functions on $\left[a, b\right]$ then

Theorem 4. Let $a = \left(a_{1}, ..., a_{n}\right)$ and $b = \left(b_{1}, ..., b_{n}\right)$ be two positive n-tuples and $p, q > 1$ such that $1/p+1/q = 1.$ Then

2.   Hölder's inequality for positive functionals

Let $E$ be a nonempty set and $L$ be a linear class of real valued functions on $E$ having the following properties

$L1:$ If $f, g\in L$ then $\left(\alpha f+\beta g\right) \in L$ for all $\alpha, \beta \in \mathbb{R}$;

$L2:$ $1\in L$, that is if $f(t) = 1, t\in E,$ then $f\in L;$

We also consider positive isotonic linear functionals $A:L\rightarrow \mathbb{R}$ is a functional satisfying the following properties:

$A1:$ $A\left(\alpha f+\beta g\right) = \alpha A\left(f\right) +\beta$ $A\left(g\right)$ for $f, g\in L$ and $\alpha, \beta \in \mathbb{R};$

$A2:$ If $f\in L,$ $f(t)\geq 0$ on $E$ then $A\left(f\right) \geq 0.$

Isotonic, that is, order-preserving, linear functionals are natural objects in analysis which enjoy a number of convenient properties. Functional versions of well-known inequalities and related results could be found in ^{[1,2,3,4,5,6,7,8,9,11,12]}.

Example 1. i.) If $E = \left[a, b\right] \subseteq \mathbb{R}$ and $L = L\left[a, b\right],$ then

is an isotonic linear functional.

ii.)If $E = \left[a, b\right] \times \left[c, d\right] \subseteq \mathbb{R} ^{2}$ and $L = L\left(\left[a, b\right] \times \left[c, d\right] \right),$ then

is an isotonic linear functional.

iii.)If $\left(E, \Sigma, \mu \right)$ is a measure space with $\mu$ positive measure on $E$ and $L = L(\mu)$ then

is an isotonic linear functional.

iv.)If $E$ is a subset of the natural numbers $\mathbb{N}$ with all $p_{k}\geq 0,$ then $A(f) = \sum_{k\in E}p_{k}f_{k}$ is an isotonic linear functional. For example; If $E = \left\{ 1, 2, ..., n\right\}$ and $f:E\rightarrow \mathbb{R}, f(k) = a_{k},$ then $A(f) = \sum_{k = 1}^{n}a_{k}$ is an isotonic linear functional. If $E = \left\{ 1, 2, ..., n\right\} \times \left\{ 1, 2, ..., m\right\}$ and $f:E\rightarrow \mathbb{R}, f(k, l) = a_{k, l},$ then $A(f) = \sum_{k = 1}^{n}\sum_{l = 1}^{m}a_{k, l}$ is an isotonic linear functional.

Theorem 5 (Hölder's inequality for isotonic functionals ^[13]). Let $L$ satisfy conditions $L1$, $L2$, and $A$ satisfy conditions $A1$, $A2$ on a base set $E$. Let $p > 1$ and $p^{-1}+q^{-1} = 1.$ If $w, f, g\geq 0$ on $E$ and $wf^{p}, wg^{q}, wfg\in L$ then we have

In the case $0 < p < 1$ and $A\left(wg^{q}\right) > 0$ (or $p < 0$ and $A\left(wf^{p}\right) > 0$), the inequality in (2.1) is reversed.

Remark 1. i.) If we choose $E = \left[a, b\right] \subseteq \mathbb{R}$, $L = L\left[a, b\right]$, $w = 1$ on $E$ and $A(f) = \int_{a}^{b}\left\vert f(t)\right\vert dt$ in the Theorem 5, then the inequality (2.1) reduce the inequality (1.3).

ii.) If we choose $E = \left\{ 1, 2, ..., n\right\},$ $w = 1$ on $E$, $f:E\rightarrow \left[0, \infty \right), f(k) = a_{k},$ and $A(f) = \sum_{k = 1}^{n}a_{k}$ in the Theorem 5, then the inequality (2.1) reduce the inequality (1.4).

iii.) If we choose $E = \left[a, b\right] \times \left[c, d\right], L = L(E)$, $w = 1$ on $E$ and $A(f) = \int_{a}^{b}\int_{c}^{d}\left\vert f(x, y)\right\vert dxdy$ in the Theorem 5, then the inequality (2.1) reduce the following inequality for double integrals:

The aim of this paper is to give a new general improvement of Hölder inequality for isotonic linear functional. As applications, this new inequality will be rewritten for several important particular cases of isotonic linear functionals. Also, we give an application to show that improvement is hold for double integrals.

3.   Main results

Theorem 6. Let $L$ satisfy conditions $L1$, $L2$, and $A$ satisfy conditions $A1$, $A2$ on a base set $E$. Let $p > 1$ and $p^{-1}+q^{-1} = 1.$ If $\alpha, \beta, w, f, g\geq 0$ on $E$, $\alpha wfg, \beta wfg, \alpha wf^{p}, \alpha wg^{q}, \beta wf^{p}, \beta wg^{q}, wfg\in L$ and $\alpha +\beta = 1$ on $E,$ then we have

i.)

ii.)

Proof. ⅰ.) By using of Hölder inequality for isotonic functionals in (2.1) and linearity of $A$, it is easily seen that

ⅱ.) Firstly, we assume that $A^{1/p}\left(wf^{p}\right) A^{1/q}\left(wg^{q}\right) \neq 0$. then

By the inequality (1.1) and linearity of $A$, we have

Finally, suppose that $A^{1/p}\left(wf^{p}\right) A^{1/q}\left(wg^{q}\right) = 0$. Then $A^{1/p}\left(wf^{p}\right) = 0$ or $A^{1/q}\left(wg^{q}\right) = 0$, i.e. $A\left(wf^{p}\right) = 0$ or $A\left(wg^{q}\right) = 0.$ We assume that $A\left(wf^{p}\right) = 0$. Then by using linearity of $A$ we have,

Since $A\left(\alpha wf\right), A\left(\beta wf\right) \geq 0$, we get $A\left(\alpha wf^{p}\right) = 0$ and $A\left(\beta wf^{p}\right) = 0.$ From here, it follows that

In case of $A\left(wg^{q}\right) = 0,$ the proof is done similarly. This completes the proof.

Remark 2. The inequality (3.2) shows that the inequality (3.1) is better than the inequality (2.1).

If we take $w = 1$ on $E$ in the Theorem 6, then we can give the following corollary:

Corollary 1. Let $L$ satisfy conditions $L1$, $L2$, and $A$ satisfy conditions $A1$, $A2$ on a base set $E$. Let $p > 1$ and $p^{-1}+q^{-1} = 1.$ If $\alpha, \beta, f, g\geq 0$ on $E$, $\alpha fg, \beta fg, \alpha f^{p}, \alpha g^{q}, \beta f^{p}, \beta g^{q}, fg\in L$ and $\alpha +\beta = 1$ on $E$, then we have

i.)

ii.)

Remark 3. i.) If we choose $E = \left[a, b\right] \subseteq \mathbb{R}$, $L = L\left[a, b\right]$, $\alpha (t) = \frac{b-t}{b-a}, \beta (t) = \frac{t-a}{ b-a}$ on $E$ and $A(f) = \int_{a}^{b}\left\vert f(t)\right\vert dt$ in the Corollary 1, then the inequality (3.3) reduce the inequality (1.5).

ii.) If we choose $E = \left\{ 1, 2, ..., n\right\},$ $\alpha (k) = \frac{k}{n}, \beta (k) = \frac{n-k}{n}$ on $E$, $f:E\rightarrow \left[0, \infty \right), f(k) = a_{k},$ and $A(f) = \sum_{k = 1}^{n}a_{k}$ in the Theorem1, then the inequality (3.3) reduce the inequality (1.6).

We can give more general form of the Theorem 6 as follows:

Theorem 7. Let $L$ satisfy conditions $L1$, $L2$, and $A$ satisfy conditions $A1$, $A2$ on a base set $E$. Let $p > 1$ and $p^{-1}+q^{-1} = 1.$ If $\alpha _{i}, w, f, g\geq 0$ on $E,$ $\alpha _{i}wfg, \alpha _{i}wf^{p}, \alpha _{i}wg^{q}, wfg\in L, i = 1, 2, ..., m,$ and $\sum_{i = 1}^{m}\alpha _{i} = 1$ on $E,$ then we have

i.)

ii.)

Proof. The proof can be easily done similarly to the proof of Theorem 6.

If we take $w = 1$ on $E$ in the Theorem 6, then we can give the following corollary:

Corollary 2. Let $L$ satisfy conditions $L1$, $L2$, and $A$ satisfy conditions $A1$, $A2$ on a base set $E$. Let $p > 1$ and $p^{-1}+q^{-1} = 1.$ If $\alpha _{i}, f, g\geq 0$ on $E,$ $\alpha _{i}fg, \alpha _{i}f^{p}, \alpha _{i}g^{q}, fg\in L, i = 1, 2, ..., m,$ and $\sum_{i = 1}^{m}\alpha _{i} = 1$ on $E,$ then we have

i.)

ii.)

Corollary 3 (Improvement of Hölder inequality for double integrals). Let $p, q > 1$ and $1/p+1/q = 1$. If $f\ $and $g$ are real functions defined on $E = \left[a, b\right] \times \left[c, d\right]$ and if $\left\vert f\right\vert ^{p}, \left\vert g\right\vert ^{q}\in L(E)$ then

where $\alpha _{1}(x, y) = \frac{\left(b-x\right) \left(d-y\right) }{\left(b-a\right) \left(d-c\right) }, \alpha _{2}(x, y) = \frac{\left(b-x\right) \left(y-c\right) }{\left(b-a\right) \left(d-c\right) }, \alpha _{3}(x, y) = \frac{\left(x-a\right) \left(y-c\right) }{\left(b-a\right) \left(d-c\right) }, , \alpha _{4}(x, y) = \frac{\left(x-a\right) \left(d-y\right) }{ \left(b-a\right) \left(d-c\right) }$ on $E$

Proof. If we choose $E = \left[a, b\right] \times \left[c, d\right] \subseteq \mathbb{R} ^{2}$, $L = L(E)$, $\alpha _{1}(x, y) = \frac{\left(b-x\right) \left(d-y\right) }{\left(b-a\right) \left(d-c\right) }, \alpha _{2}(x, y) = \frac{\left(b-x\right) \left(y-c\right) }{\left(b-a\right) \left(d-c\right) }, \alpha _{3}(x, y) = \frac{\left(x-a\right) \left(y-c\right) }{\left(b-a\right) \left(d-c\right) }, \alpha _{4}(x, y) = \frac{\left(x-a\right) \left(d-y\right) }{\left(b-a\right) \left(d-c\right) }$ on $E$ and $A(f) = \int_{a}^{b}\int_{c}^{d}\left\vert f(x, y)\right\vert dxdy$ in the Corollary 1, then we get the inequality (3.5).

Corollary 4. Let $\left(a_{k, l}\right)$ and $\left(b_{k, l}\right)$ be two tuples of positive numbers and $p, q > 1$ such that $1/p+1/q = 1.$ Then we have

where $\alpha _{1}(k, l) = \frac{kl}{nm}, \alpha _{2}(k, l) = \frac{\left(n-k\right) l}{nm}, \alpha _{3}(k, l) = \frac{\left(n-k\right) \left(m-l\right) }{nm}, \alpha _{4}(k, l) = \frac{k\left(m-l\right) }{nm}$ on $E.$

Proof. If we choose $E = \left\{ 1, 2, ..., n\right\} \times \left\{ 1, 2, ..., m\right\},$ $\alpha _{1}(k, l) = \frac{kl}{nm}, \alpha _{2}(k, l) = \frac{\left(n-k\right) l}{ nm}, \alpha _{3}(k, l) = \frac{\left(n-k\right) \left(m-l\right) }{nm}, \alpha _{4}(k, l) = \frac{k\left(m-l\right) }{nm}$ on $E$, $f:E\rightarrow \left[0, \infty \right), f(k, l) = a_{k, l},$ and $A(f) = \sum_{k = 1}^{n} \sum_{l = 1}^{m}a_{k, l}$ in the Theorem1, then we get the inequality (3.6).

4.   An application for double integrals

In ^[14], Sarıkaya et al. gave the following lemma for obtain main results.

Lemma 1. Let $f:\Delta \subseteq \mathbb{R} ^{2}\rightarrow \mathbb{R}$ be a partial differentiable mapping on $\Delta = \left[a, b\right] \times \left[c, d\right]$ in $\mathbb{R} ^{2}$with $a < b$ and $c < d.$ If $\frac{\partial ^{2}f}{\partial t\partial s} \in L(\Delta)$, then the following equality holds:

By using this equality and Hölder integral inequality for double integrals, Sar\i kaya et al. obtained the following inequality:

Theorem 8. Let $f:\Delta \subseteq \mathbb{R} ^{2}\rightarrow \mathbb{R}$ be a partial differentiable mapping on $\Delta = \left[a, b\right] \times \left[c, d\right]$ in $\mathbb{R} ^{2}$with $a < b$ and $c < d.$ If $\left\vert \frac{\partial ^{2}f}{\partial t\partial s}\right\vert ^{q}, q > 1,$ is convex function on the co-ordinates on $\Delta$, then one has the inequalities:

where

$1/p+1/q = 1$ and $f_{st} = \frac{\partial ^{2}f}{\partial t\partial s}.$

If Theorem 8 are resulted again by using the inequality (3.5), then we get the following result:

Theorem 9. Let $f:\Delta \subseteq \mathbb{R} ^{2}\rightarrow \mathbb{R}$ be a partial differentiable mapping on $\Delta = \left[a, b\right] \times \left[c, d\right]$ in $\mathbb{R} ^{2}$with $a < b$ and $c < d.$ If $\left\vert \frac{\partial ^{2}f}{\partial t\partial s}\right\vert ^{q}, q > 1,$ is convex function on the co-ordinates on $\Delta$, then one has the inequalities:

where

$1/p+1/q = 1$ and $f_{st} = \frac{\partial ^{2}f}{\partial t\partial s}.$

Proof. Using Lemma 1 and the inequality (3.5), we find

Since $\left\vert f_{st}\right\vert ^{q}$ is convex function on the co-ordinates on $\Delta$, we have for all $t, s\in \left[0, 1\right]$

for all $t, s\in \left[0, 1\right].$ Further since

a combination of (4.3) - (4.5) immediately gives the required inequality (4.2).

Remark 4. Since $\eta :\left[0, \infty \right) \rightarrow \mathbb{R}, \eta (x) = x^{s}, 0 < s\leq 1,$ is a concave function, for all $u, v\geq 0$ we have

From here, we get

Thus we obtain

This shows that the inequality (4.2) is better than the inequality (4.1).

5.   Conclusions

The aim of this paper is to give a new general improvement of Hölder inequality via isotonic linear functional. An important feature of the new inequality obtained here is that many existing inequalities related to the Hölder inequality can be improved. As applications, this new inequality will be rewritten for several important particular cases of isotonic linear functionals. Also, we give an application to show that improvement is hold for double integrals. Similar method can be applied to the different type of convex functions.

Acknowledgments

This research didn't receive any funding.

Conflict of interest

The author declares no conflicts of interest in this paper.