Citation: Yuanyuan Huang, Yiping Hao, Min Wang, Wen Zhou, Zhijun Wu. Optimality and stability of symmetric evolutionary games with applications in genetic selection[J]. Mathematical Biosciences and Engineering, 2015, 12(3): 503-523. doi: 10.3934/mbe.2015.12.503
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The famous Young's inequality, as a classical result, state that: if $ a, b > 0 $ and $ t\in \lbrack 0, 1] $, then
$ atb1−t≤ta+(1−t)b $
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(1.1) |
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
$ ab≤app+bqq $
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(1.2) |
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
$ ∫ba|f(x)g(x)|dx≤(∫ba|f(x)|pdx)1/p(∫ba|g(x)|qdx)1/q, $
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(1.3) |
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
$ n∑k=1akbk≤(n∑k=1apk)1/p(n∑k=1bqk)1/q. $
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(1.4) |
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
$ ∫ba|f(x)g(x)|dx≤1b−a{(∫ba(b−x)|f(x)|pdx)1p(∫ba(b−x)|g(x)|qdx)1q+(∫ba(x−a)|f(x)|pdx)1p(∫ba(x−a)|g(x)|qdx)1q} $
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(1.5) |
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
$ n∑k=1akbk≤1n{(n∑k=1kapk)1/p(n∑k=1kbqk)1/q+(n∑k=1(n−k)apk)1/p(n∑k=1(n−k)bqk)1/q}. $
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(1.6) |
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
$ A(f)=∫baf(t)dt $
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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
$ A(f)=∫ba∫dcf(x,y)dxdy $
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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
$ A(f)=∫Efdμ $
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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
$ A(wfg)≤A1/p(wfp)A1/q(wgq). $
|
(2.1) |
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:
$ ∫ba∫dc|f(x,y)||g(x,y)|dxdy≤(∫ba∫dc|f(x,y)|pdx)1/p(∫ba∫dc|g(x,y)|qdx)1/q. $
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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.
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.)
$ A(wfg)≤A1/p(αwfp)A1/q(αwgq)+A1/p(βwfq)A1/q(βwgq) $
|
(3.1) |
ii.)
$ A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)≤A1/p(wfp)A1/q(wgq). $
|
(3.2) |
Proof. ⅰ.) By using of Hölder inequality for isotonic functionals in (2.1) and linearity of $ A $, it is easily seen that
$ A(wfg)=A(αwfg+βwfg)=A(αwfg)+A(βwfg)≤A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq). $
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ⅱ.) Firstly, we assume that $ A^{1/p}\left(wf^{p}\right) A^{1/q}\left(wg^{q}\right) \neq 0 $. then
$ A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)A1/p(wfp)A1/q(wgq)=(A(αwfp)A(wfp))1/p(A(αwgq)A(wgq))1/q+(A(βwfp)A(wfp))1/p(A(βwgq)A(wgq))1/q, $
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By the inequality (1.1) and linearity of $ A $, we have
$ A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)A1/p(wfp)A1/q(wgq)≤1p[A(αwfp)A(wfp)+A(βwfp)A(wfp)]+1q[A(αwgq)A(wgq)+A(βwgq)A(wgq)]=1. $
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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,
$ 0=A(wfp)=A(αwfp+βwfp)=A(αwfp)+A(βwfp). $
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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
$ A1/p(αwfp)A1/q(αwgq)+A1/p(βwfp)A1/q(βwgq)=0≤0=A1/p(wfp)A1/q(wgq). $
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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.)
$ A(fg)≤A1/p(αfp)A1/q(αgq)+A1/p(βfq)A1/q(βgq) $
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(3.3) |
ii.)
$ A1/p(αfp)A1/q(αgq)+A1/p(βfp)A1/q(βgq)≤A1/p(fp)A1/q(gq). $
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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.)
$ A(wfg)≤m∑i=1A1/p(αiwfp)A1/q(αiwgq) $
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ii.)
$ m∑i=1A1/p(αiwfp)A1/q(αiwgq)≤A1/p(wfp)A1/q(wgq). $
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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.)
$ A(fg)≤m∑i=1A1/p(αifp)A1/q(αigq) $
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(3.4) |
ii.)
$ m∑i=1A1/p(αifp)A1/q(αigq)≤A1/p(fp)A1/q(gq). $
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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
$ ∫ba∫dc|f(x,y)||g(x,y)|dxdy≤4∑i=1(∫ba∫dcαi(x,y)|f(x,y)|pdx)1/p(∫ba∫dcαi(x,y)|g(x,y)|qdx)1/q, $
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(3.5) |
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
$ n∑k=1m∑l=1ak,lbk,l≤4∑i=1(n∑k=1m∑l=1αi(k,l)apk,l)1/p(n∑k=1m∑l=1αi(k,l)bqk,l)1/q, $
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(3.6) |
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).
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:
$ f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−12[1b−a∫ba[f(x,c)+f(x,d)]dx+1d−c∫dc[f(a,y)+f(b,y)]dy]=(b−a)(d−c)4∫10∫10(1−2t)(1−2s)∂2f∂t∂s(ta+(1−t)b,sc+(1−s)d)dtds. $
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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:
$ |f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−A|≤(b−a)(d−c)4(p+1)2/p[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q4]1/q, $
|
(4.1) |
where
$ A=12[1b−a∫ba[f(x,c)+f(x,d)]dx+1d−c∫dc[f(a,y)+f(b,y)]dy], $
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$ 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:
$ |f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−A|≤(b−a)(d−c)41+1/p(p+1)2/p{[4|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+|fst(b,d)|q36]1/q+[2|fst(a,c)|q+|fst(a,d)|q+4|fst(b,c)|q+2|fst(b,d)|q36]1/q+[2|fst(a,c)|q+4|fst(a,d)|q+|fst(b,c)|q+2|fst(b,d)|q36]1/q+[|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+4|fst(b,d)|q36]1/q}, $
|
(4.2) |
where
$ A=12[1b−a∫ba[f(x,c)+f(x,d)]dx+1d−c∫dc[f(a,y)+f(b,y)]dy], $
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$ 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
$ |f(a,c)+f(a,d)+f(b,c)+f(b,d)4−1(b−a)(d−c)∫ba∫dcf(x,y)dxdy−A|≤(b−a)(d−c)4∫10∫10|1−2t||1−2s||fst(ta+(1−t)b,sc+(1−s))|dtds≤(b−a)(d−c)4{(∫10∫10ts|1−2t|p|1−2s|pdtds)1/p×(∫10∫10ts|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q+(∫10∫10t(1−s)|1−2t|p|1−2s|pdtds)1/p×(∫10∫10t(1−s)|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q+(∫10∫10(1−t)s|1−2t|p|1−2s|pdtds)1/p×(∫10∫10(1−t)s|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q+(∫10∫10(1−t)(1−s)|1−2t|p|1−2s|pdtds)1/p×(∫10∫10(1−t)(1−s)|fst(ta+(1−t)b,sc+(1−s))|qdtds)1/q}. $
|
(4.3) |
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] $
$ |fst(ta+(1−t)b,sc+(1−s))|q≤ts|fst(a,c)|q+t(1−s)|fst(a,d)|q+(1−t)s|fst(a,c)|q+(1−t)(1−s)|fst(a,c)|q $
|
(4.4) |
for all $ t, s\in \left[0, 1\right]. $ Further since
$ ∫10∫10ts|1−2t|p|1−2s|pdtds=∫10∫10t(1−s)|1−2t|p|1−2s|pdtds=∫10∫10(1−t)s|1−2t|p|1−2s|pdtds $
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(4.5) |
$ =∫10∫10(1−t)(1−s)|1−2t|p|1−2s|pdtds=14(p+1)2, $
|
(4.6) |
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
$ η(u+v2)=(u+v2)s≥η(u)+η(v)2=us+vs2. $
|
From here, we get
$ I={[4|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+|fst(b,d)|q36]1/q+[2|fst(a,c)|q+|fst(a,d)|q+4|fst(b,c)|q+2|fst(b,d)|q36]1/q+[2|fst(a,c)|q+4|fst(a,d)|q+|fst(b,c)|q+2|fst(b,d)|q36]1/q+[|fst(a,c)|q+2|fst(a,d)|q+2|fst(b,c)|q+4|fst(b,d)|q36]1/q}≤2{[6|fst(a,c)|q+3|fst(a,d)|q+6|fst(b,c)|q+3|fst(b,d)|q72]1/q+[3|fst(a,c)|q+6|fst(a,d)|q+3|fst(b,c)|q+6|fst(b,d)|q72]1/q} $
|
$ ≤4{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q16]1/q $
|
Thus we obtain
$ (b−a)(d−c)41+1/p(p+1)2/pI≤(b−a)(d−c)41+1/p(p+1)2/p4{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q16]1/q}≤(b−a)(d−c)4(p+1)2/p{[|fst(a,c)|q+|fst(a,d)|q+|fst(b,c)|q+|fst(b,d)|q4]1/q}. $
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This shows that the inequality (4.2) is better than the inequality (4.1).
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.
This research didn't receive any funding.
The author declares no conflicts of interest in this paper.
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