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On a generalized Hölder inequality
Journal of Inequalities and Applications volume 2015, Article number: 88 (2015)
Abstract
Refining the Hölder inequality, a result of S Wu is extended to the case of multiple sequences.
1 Introduction
There are a lot of generalizations and extensions of the classical Hölder inequality. Among them is the existence of a function
for which
Here \(a= \{ a _{i} \} \), \(b= \{ b _{i} \} \) are positive sequences and \(1 \leq p, q < \infty\), \(\frac{1}{p} + \frac{1}{q} = 1\). Hu and Wu published several results [1–3] on this topic. Among them is the following theorem.
Theorem A
([3], Theorem 1)
Let \(a_{i} >0\), \(b_{i} >0\) (\(i=1,2, \ldots,n\)), \(1-e_{i} +e_{j} \geq0\) (\(i, j=1, 2, \ldots,n\)), and let \(p \geq q > 0\). Then
Our purpose of this paper is to extend Theorem A to the case of multiple sequences. The following are the main results of this paper.
Theorem 1.1
Let \(a_{j i} >0\), \(p_{j}>0\) (\(i=1,2, \ldots,n \); \(j=1,2, \ldots ,m\)), \(1-e_{i} +e_{j} \geq0\) (\(i, j=1,2, \ldots,n \)), \(\Lambda:=\sum_{j=1} ^{m} \frac{1}{p_{j}}\) and \(\{p_{j_{1}}, p_{j_{2}} \} \subset \{p_{1}, p_{2},\ldots, p_{m} \}\). Then
Theorem 1.2
Let \(a_{j i} >0\) (\(i=1,2, \ldots,n \); \(j=1,2, \ldots,m\)), \(1-e_{i} +e_{j} \geq0\) (\(i, j=1,2, \ldots,n\)), \(p_{j}<0\) (\(j=1,2, \ldots,m-1\)), \(0<\sum_{j=1}^{m} \frac{1}{p_{j}}\leq 1\) and \(\{p_{j_{1}}, p_{j_{2}} \} \subset \{p_{1}, p_{2},\ldots, p_{m} \}\). Then
If \(m=2\), then Theorem 1.1 reduces to Theorem A. The case \(\Lambda=1\) and \(\frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } = \frac{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} }\) of Theorem 1.1 is the well-known Hölder inequality.
Theorem 1.2 is a reversed version of Theorem 1.1. The case \(m=2\) with \(\Lambda=1\) and \(\frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } = \frac{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} }\) of Theorem 1.2 reduces to the well-known reversed Hölder inequality.
The idea of proofs of Theorem 1.1 and Theorem 1.2 are essentially the same as that of Theorem A. The point is that a variant of the methods of Wu in [3] works for the case of multiple sequences. The process will be done in Section 3 and Section 4 after the preparing lemmas in Section 2. In Section 5, as an application of Theorem 1.1, we refine a reversed Hölder inequality different from Theorem 1.2. We refer to [4, 5] for the general theory of inequalities.
2 Lemmas
In order to prove Theorem 1.1 and Theorem 1.2, we need the following lemmas.
Lemma 2.1
([2], Lemma 1(1))
Let \(a _{j i} >0\), \(\lambda _{j} > 0\) (\(i=1, 2, \ldots, n \); \(j=1, 2, \ldots, m\)) and \(\Lambda:= \sum_{j=1} ^{m} \lambda _{j} \). Then
with equality holding if and only if \(a _{j 1} =a _{j2} = \cdots=a _{j n}\) (\(j=1, 2, \ldots, m\)) for \(\Lambda <1\), or \(\frac{a _{1 i}}{\sum_{i=1} ^{n} a _{1 i}} = \frac{a _{2i}}{\sum_{i=1} ^{n} a _{2 i}} = \cdots= \frac{a _{m i}}{\sum_{i=1} ^{n} a _{m i}}\) (\(i=1, 2, \ldots, n\)) for \(\Lambda=1\).
Lemma 2.2
([2], Lemma 1(2))
Let \(a _{j i} >0\) (\(i=1, 2, \ldots, n \); \(j=1, 2, \ldots, m \)), \(\lambda _{j} <0\) (\(j=1, 2, \ldots, m-1\)), \(\lambda _{m}\) be a real number, and \(\Lambda:= \sum_{j=1} ^{m} \lambda _{j}\). Then
with equality holding if and only if \(a _{j 1} =a _{j2} = \cdots=a _{j n}\) (\(j=1, 2, \ldots, m\)) for \(\Lambda>1\), or \(\frac{a _{1 i}}{\sum_{i=1} ^{n} a _{1 i}} = \frac{a _{2i}}{\sum_{i=1} ^{n} a _{2 i}} = \cdots= \frac{a _{m i}}{\sum_{i=1} ^{n} a _{m i}}\) (\(i=1, 2, \ldots, n\)) for \(\Lambda=1\).
3 Proof of Theorem 1.1
We may suppose \(p _{j _{1}} \leq p _{j _{2}}\).
Case (1): \(\Lambda> 1\). We show
It is simple to see that
by applying Lemma 2.2 and Lemma 2.1. On the other hand, it is straightforward to see that
and
where \(A _{j l} =A _{j l} ( i, k ) =a _{j k} ^{p _{j}} a _{l i} ^{p _{l}} ( 1-e _{i} +e _{k} )\) (\(j,l=1,2,3,\ldots,m\)). Hence
by applying Lemma 2.1 again. On the other hand, it is also straightforward to see that
and, for \(j=1, 2, 3, \ldots, m\), that
Hence,
and
Therefore (3.1) follows.
Case (2): \(0<\Lambda \leq1\). We show
Since \(\sum_{j=1} ^{m} \frac{1}{p _{j} \Lambda} =1\), we have
by applying Lemma 2.1. On the other hand, it is straightforward to see that
and
where \(B _{jl} =B _{jl} ( i,k ) =a _{jk} ^{p _{j} \Lambda} a _{li} ^{p _{l} \Lambda} ( 1-e _{i} +e _{k} )\) (\(j,l=1,2,3,\ldots,m\)). Hence
by applying Lemma 2.1. Applying Lemma 2.1 again, we have
and
Hence
By the same process as in the proof of Case (1),
Therefore (3.2) follows. The proof is complete.
4 Proof of Theorem 1.2
We may suppose \(p _{j _{2}} \leq p _{j _{1}}\) and let \(\sum_{j=1} ^{m} \frac{1}{p _{j}}=\Lambda\). We show
It is simple to see that
by applying Lemma 2.1 and Lemma 2.2. On the other hand, it is straightforward to see that
and
where \(A _{j l} =A _{j l} ( i, k ) =a _{j k} ^{p _{j}} a _{l i} ^{p _{l}} ( 1-e _{i} +e _{k} )\) (\(j,l=1,2,3,\ldots,m\)). Since only one of the numbers \(\frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}}\), \(\frac{1}{p _{j _{2}}}\), \(\frac{1}{p _{j}}\) (\(j \ne j _{1} ,j _{2}\)) is positive, by applying Lemma 2.2 again,
On the other hand, it is also straightforward to see that
and, for \(j=1, 2, 3, \ldots, m\), that
Hence,
Therefore (4.1) follows. The proof is complete.
5 Further observations
As a generalization of the reversed Hölder inequality, JF Tian obtained the following.
Theorem B
([6], Theorem 2.2)
Let \(a_{i}>0\), \(b_{i}>0\) (\(i=1,2, \ldots, n\)), \(1-e_{i}+e_{j}\geq0\) (\(i, j=1,2, \ldots, n\)), \(q<0\), and \(\frac {1}{p}+\frac{1}{q}\geq0\). Then
We extended Theorem B to multiple sequences as follows. The idea of the proof is essentially the same as that of Theorem B and is different from that of Theorem 1.2. Note that there is no implication between Theorem 1.2 and Theorem 5.1.
Theorem 5.1
Let \(a_{j i}>0\) (\(i=1, 2, \ldots, n\); \(j=1, 2, \ldots, m\)), \(1-e_{i}+e_{j}\geq0\) (\(i, j=1, 2, \ldots, n\)), \(p_{j}<0\) (\(j=1, 2, \ldots, m-1\)), \(\Lambda:=\sum_{j=1}^{m}\frac{1}{p_{j}} >0\), and \(\{ p_{j_{1}}\} \subset\{p _{1}, p _{2} , \ldots, p _{m-1}\}\). Then
Proof
Case (1): \(-1 \leq p_{j_{1}} <0\). Since \(\sum_{j=1}^{m}\frac{1}{p_{j}} >0\), we have \(p_{m}>0\). Let \(q_{j}=-\frac{p_{j}}{p_{m}}>0\) (\(j=1, 2, \ldots, m-1\)), and \(q _{m} = \frac{{1}}{ {p _{m}}}>0\), then \(q _{m} \geq q _{j_{1}} >0\).
When \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we obtain
by using Theorem 1.1. That is,
Therefore
Hence
From \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we have \(\sum_{j=1} ^{m} \frac{1}{p _{j}} >1\). Consequently
When \(\sum_{j=1} ^{m} \frac{1}{q _{j}} \geq1\), then \(\sum_{j=1} ^{m} \frac{1}{p _{j}} \leq1\), so that similarly we obtain
Case (2): \(p _{j_{1}} < -1\). Since \(\sum_{j=1} ^{m} \frac{1}{p _{j}} >0\), we have \(p _{m}>0\). Let \(q _{j} =- \frac{p _{j}}{p _{m}} >0\) (\(j=1, 2, \ldots, m-1\)), and \(q _{m} = \frac{1}{p _{m}} >0\), then \(q _{j_{1}} > q _{m} >0\).
When \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we obtain
by using Theorem 1.1 again. That is,
Hence
From \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we have \(\sum_{j=1} ^{m} \frac{1}{p _{j}} >1\). Consequently
When \(\sum_{j=1} ^{m} \frac{1}{q _{j}} \geq1\), then \(\sum_{j=1} ^{m} \frac{1}{p _{j}} \leq1\), so that similarly we obtain
The proof is complete. □
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Acknowledgements
This work was supported by NRF-2010-0021986.
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EGK has been involved in drafting the manuscript and JEB helped to draft the manuscript. All authors read and approved the final manuscript.
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Kwon, E.G., Bae, J.E. On a generalized Hölder inequality. J Inequal Appl 2015, 88 (2015). https://doi.org/10.1186/s13660-015-0612-9
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DOI: https://doi.org/10.1186/s13660-015-0612-9