Volume 2010, Article ID 461215, 9 pagesdoi:10.1155/2010/461215 Research Article A Converse of Minkowski’s Type Inequalities Romeo Me ˇstrovi ´c1 and David Kalaj2 1 Maritime Faculty, Univ
Trang 1Volume 2010, Article ID 461215, 9 pages
doi:10.1155/2010/461215
Research Article
A Converse of Minkowski’s Type Inequalities
Romeo Me ˇstrovi ´c1 and David Kalaj2
1 Maritime Faculty, University of Montenegro, Dobrota 36, 85330 Kotor, Montenegro
2 Faculty of Natural Sciences and Mathematics, University of Montenegro, Dˇzordˇza Vaˇsingtona BB,
81000 Podgorica, Montenegro
Correspondence should be addressed to Romeo Meˇstrovi´c,romeo@ac.me
Received 6 August 2010; Accepted 20 October 2010
Academic Editor: Jong Kim
Copyrightq 2010 R Meˇstrovi´c and D Kalaj This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
We formulate and prove a converse for a generalization of the classical Minkowski’s inequality
The case when 0 < p < 1 is also considered Applying the same technique, we obtain an analog
converse theorem for integral Minkowski’s type inequality
1 Introduction
If p > 1, a i ≥ 0, and b i ≥ 0 i 1, , n are real numbers, then by the classical Minkowski’s
inequality
n
i1
a i b ip
1/p
≤
n
i1
a p i
1/p
n
i1
b p i
1/p
This inequality was published by Minkowski1, pages 115–117 hundred years ago in his famous book “Geometrie der Zahlen.”
It is also knownsee 2 that for 0 < p < 1 the above inequality is satisfied with “≥”
instead of “≤”
Many extensions and generalizations of Minkowski’s inequality can be found in2,3
We want to point out the following inequality:
⎛
⎝n
j1
m
i1
a ij
p⎞
⎠
1/p
≤m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
Trang 2where p > 1 and a ij ≥ 0 i 1, , m; j 1, , n are real numbers Furthermore, if 0 <
p < 1, then the inequality 1.2 is satisfied with “≥” instead of “≤” 2, Theorem 24, page 30
In both cases, equality holds if and only if all columnsa1j , a 2j , , a mj , j 1, 2, , n, are
proportional
An extension of inequality1.2 was formulated by Ingham and Jessen see 2, pages 31-32 In 1948, Tˆoyama 4 published a converse of the inequality of Ingham and Jessen
see also a recent paper 5 for a weighted version of Tˆoyama’s inequality Namely, Tˆoyama
showed that if 0 < q < p and a ij ≥ 0 i 1, , m; j 1, , n are real numbers, then
⎛
⎜m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
q/p⎞
⎟
1/q
≤ minm, n 1/q−1/p
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
1/p
. 1.3
The main result of this paper gives a converse of inequality1.2 On the other hand, our result may be regarded as a nonsymmetric analogue of the above inequality, and it is given as follows
Theorem 1.1 Let p > 0, q > 0, and a ij ≥ 0 i 1, , m; j 1, , n be real numbers Then for
p ≥ 1 we have
m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
≤ C
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
1/p
where C is a positive constant given by
C
⎧
⎪
⎨
⎪
⎩
m1−1/q if 1 ≤ p ≤ q,
minm, n 1/q−1/p
m1−1/q if 1 ≤ q < p,
m1−1/p if 0 < q ≤ 1 ≤ p.
1.5
If 0 < p < 1, then
m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
≥ K
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
1/p
where K is a positive constant given by
K
⎧
⎪
⎨
⎪
⎩
m1−1/q if 0 < q ≤ p < 1,
minm, n 1/q−1/p
m1−1/q if 0 < p < q < 1,
m1−1/p if 0 < p < 1 ≤ q.
1.7
Trang 3Inequality1.4 with 1 ≤ p ≤ q and inequality 1.6 with 0 < q ≤ p < 1 are sharp for all m and n,
and they are attained for a ij a, i 1, , m, j 1, , n If m ≤ n, then inequality 1.4 is sharp in
the cases when 1 ≤ q < p and 0 < q ≤ 1 ≤ p In both cases the equalities are attained for
a ij
⎧
⎨
⎩
a, if i j,
0, if i / j. 1.8
When m ≤ n, the equalities in 1.6 concerned with 0 < p < q < 1 and 0 < p < 1 ≤ q are also attained
for previously defined values a ij
Remark 1.2 Note that, proceeding as in the proof of Theorem 1.1, we can prove similar inequalities to1.4 and 1.6 withn
j1m i1 instead of m
i1n j1 on the left-hand side
of these inequalities For example, such an inequality concerning the case when 1 ≤ q < p
i.e., 1.4 is
n
j1
m
i1
a p ij
1/p
≤ n1−1/p
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
1/p
The above inequality is sharp if n ≤ m, but it is not in spirit of a converse of Minkowski’s type
inequality
The following consequence ofTheorem 1.1for m 2 and q 2 can be viewed as a
converse of Minkowski’s inequality1.1
Corollary 1.3 Let n ≥ 1, p > 0, and let a j ≥ 0, b j ≥ 0 j 1, , n be real numbers Then for p ≥ 1
⎛
⎝n
j1
a p j
⎞
⎠
1/p
⎛
⎝n
j1
b p j
⎞
⎠
1/p
≤ 21−min{1/2,1/p}
⎛
⎝n
j1
a2j b2
j
p/2⎞
⎠
1/p
. 1.10
If 0 < p < 1, then
⎛
⎝n
j1
a p j
⎞
⎠
1/p
⎛
⎝n
j1
v j p
⎞
⎠
1/p
≥ 21−1/p
⎛
⎝n
j1
a2j b2
j
p/2⎞
⎠
1/p
. 1.11
Remark 1.4 It is well known that Minkowski’s inequality is also true for complex sequences
as well More precisely, if p ≥ 1 and u i , v i i 1, , n are arbitrary complex numbers, then
⎛
⎝n
j1
u j v jp
⎞
⎠
1/p
≤
⎛
⎝n
j1
u jp
⎞
⎠
1/p
⎛
⎝n
j1
v jp
⎞
⎠
1/p
. 1.12
Trang 4Note that the above inequality with u j a j ∈ R and v j ib j , b j ∈ R, for each j 1, 2, , n,
becomes
⎛
⎝n
j1
a2j b2
j
p/2⎞
⎠
1/p
≤
⎛
⎝n
j1
a p j
⎞
⎠
1/p
⎛
⎝n
j1
v p j
⎞
⎠
1/p
. 1.13
We see that the first inequality ofCorollary 1.3may be actually regarded as a converse of the previous inequality
Lemma 2.1 see 2, page 26 If u1, u2, u k , s, r are nonnegative real numbers and 0 < s < r, then
u s1 u s
2 · · · u s
k
1/s
≥u r1 u r
2 · · · u r
k
1/r
Proof of Theorem 1.1 In our proof we often use the well-known fact that the scale of power
means is nondecreasing see 2 More precisely, if a1 , a2, , a k are nonnegative integers
and 0 < α ≤ β < ∞, then
k
i1 a α i k
1/α
≤
⎛
⎝k i1 a β i k
⎞
⎠
1/β
In all the cases, for each i 1, 2, , m, we denote that
a i:
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
We will consider all the six cases related to the inequalities1.4 and 1.6
Case 1 1 ≤ p ≤ q The inequality between power means of orders q/p ≥ 1 and 1 for m positive numbers b i , i 1, 2, , m, states that
⎛
⎝m i1 b q/p i m
⎞
⎠
p/q
≥
m
i1 b i
whence for any fixed j 1, 2, n, after substitution of b i a p
ij , i 1, 2, m, we obtain
a q 1j a q
2j · · · a q
mj
p/q
≥ m p/q−1
a p 1j a p
2j · · · a p
mj
, 2.5
Trang 5whence after summation over j we find that
n
j1
a q 1j a q
2j · · · a q
mj
p/q
≥ m p/q−1n
j1
m
i1
a p ij
m p/q−1m
i1
a p i
2.6
Because p ≥ 1, the inequality between power means of orders p and 1 implies that
m
i1
a p i ≥ m1−p
m
i1
a i
p
The above inequality and2.6 immediately yield
m1−1/q
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
1/p
≥m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
Case 2 1 ≤ q < p If m ≤ n, then C m1−1/pin1.4, and a related proof is the same as that
for the following case when 0 < q ≤ 1 ≤ p.
Now suppose that m > n By the inequality for power means of orders p/q ≥ 1 and 1,
we obtain
⎛
⎜n j1
a q 1j a q
2j · · · a q
mj
p/q
n
⎞
⎟
q/p
≥
n
j1
a q 1j a q
2j · · · a q
mj
n ·
m
i1
a q i1 a q i2 · · · a q
in
2.9
Next, by the inequality for power meansof orders q ≥ 1 and 1, we obtain
m
i1
a q i1 a q i2 · · · a q
in
⎛
⎜m i1
a q i1 a q i2 · · · a q
in
1/q
m
⎞
⎟
q 2.10
For any fixed i ∈ {1, 2, , m} the inequality 2.1 ofLemma 2.1with s p > q r implies
that
a q i1 a q i2 · · · a q
in
1/q
≥a p i1 a p
i2 · · · a p
in
1/p
. 2.11
Trang 6Obviously, inequalities2.9, 2.10, and 2.11 immediately yield
n1−q/p· m q−1
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
q/p
≥
⎛
⎜m i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p⎞
⎟
q , 2.12
which is actually inequality1.4 with the constant C n 1/q−1/p · m1−1/q
Case 3 0 < q ≤ 1 ≤ p By inequality 2.1 with r q and s p, for each j 1, 2, , n, we
obtain
a q 1j a q
2j · · · a q
mj
p/q
≥ a p
1j a p
2j · · · a p
mj , 2.13
whence after summation over j, we have
n
j1
a q 1j a q
2j · · · a q
mj
p/q
≥n
j1
m
i1
a p ij m
i1
a p i1 a p i2 · · · a p
in
m
i1
a p i
2.14
By the inequality for power meansof orders p ≥ 1 and 1, we get
m
i1 a p i m
1/p
≥
m
i1 a i
or equivalently
m
i1
a p i
1/p
≥ m 1/p−1m
i1
a i m 1/p−1m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
. 2.16
The above inequality and2.14 immediately yield
m1−1/p
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
1/p
≥m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
, 2.17
as desired
Case 4 0 < q ≤ p < 1 The proof can be obtained from those of Case1, by replacing “≥” with
“≤” in each related inequality
Trang 7Case 5 0 < p < q < 1 If m ≤ n, then the proof is the same as that for Case6 If m > n, then
the proof can be obtained from those of Case 2, by replacing “≥” with “≤” in each related inequality
Case 6 0 < p < 1 ≤ q For any fixed j 1, 2, , n, inequality 2.1 ofLemma 2.1with r q and s p gives
a q 1j a q
2j · · · a q
mj
p/q
≤ a p
1j a p
2j · · · a p
mj , 2.18
whence after summation over j, we get
n
j1
a q 1j a q
2j · · · a q
mj
p/q
≤n
j1
m
i1
a p ijm
i1
a p i 2.19
As 1/p > 1, for positive integers b1 , b2, , b m, there holds
m
i1 b i
⎛
⎝m i1 b 1/p i m
⎞
⎠
p
whence for any fixed j 1, 2, n, after substitution of b i a p i , i 1, 2, m, we obtain
m
i1
a p i
1/p
≤ m 1/p−1m
i1
a i m 1/p−1m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
. 2.21
The above inequality and2.19 immediately yield
m1−1/p
⎛
⎝n
j1
m
i1
a q ij
p/q⎞
⎠
1/p
≤m
i1
⎛
⎝n
j1
a p ij
⎞
⎠
1/p
, 2.22
and the proof is completed
LetX, Σ, μ be a measure space with a positive Borel measure μ For any 0 < p < ∞ let
L p L p μ denote the usual Lebesgue space consisting of all μ-measurable complex-valued functions f : X → C such that
X
fp
Trang 8Recall that the usual norm · p of f ∈ L pis defined asf p X |f| p
dμ 1/p if p ≥ 1; f p
X |f| p
dμ if 0 < p < 1.
The following result is the integral analogue ofTheorem 1.1
Theorem 3.1 For given 0 < p < ∞ let u1, u2, , u m be arbitrary functions in L p Then, if 1 ≤ p <
∞, we have
u1 p · · · u mp ≤ m1−min{1/2,1/p}
|u1|2 · · · |u m|2
p 3.2
If 0 < p < 1, then
u1 p · · · u mp ≥ m1−1/p
|u1|2 · · · |u m|2
p
Both inequalities are sharp
For 1 < p ≤ 2 the equality in 3.2 and 3.3 is attained if u1 u2 · · · u m a.e on X If p > 2
or 0 < p < 1, then the equality is attained for u i χ E i , where E i are μ-measurable sets with
i 1, 2, , m, such that μE1 μE2 · · · μEn and E i ∩ E j ∅ whenever i / j.
Proof The proof of each inequality is completely similar to the corresponding one given in
Theorem 1.1with a fixed q 2 For clarity, we give here only a proof related to the case when
1 ≤ p ≤ 2 Applying the inequality between power means of orders 2/p ≥ 1 and 1 to the
functions|u i|p i 1, , m, we have
m
i1
|u i|2
p/2
≥ m p/2−1
m
i1
|u i|p
Integrating the above relation, we obtain
X
m
i1
|u i|2
p/2
dμ ≥ m p/2−1
m
i1
X
|u i|p dμ
which can be written in the form
|u1|2 · · · |u m|2
p
≥ m 1/2−1/p
m
i1
X
|u i|p dμ
1/p
√m
m
i1 u ip p m
1/p
≥√m ·
m
i1 u ip
3.6
Obviously, the above inequality yields3.2 for 1 < p ≤ 2.
Trang 9Corollary 3.2 Let p ≥ 1, and let w u iv be a complex function in L p Then there holds the sharp inequality
u p v p≤ 21−min1/2,1/pu iv p 3.7
References
1 H Minkowski, Geometrie der Zahlen, Teubner, Leipzig, Germany, 1910.
2 G H Hardy, J E Littlewood, and G P´olya, Inequalities, Cambridge Univerity Press, Cambridge, UK,
1952
3 E F Beckenbach and R Bellman, Inequalities, vol 30 of Ergebnisse der Mathematik und ihrer Grenzgebiete,
Springer, Berlin, Germany, 1961
4 H Tˆoyama, “On the inequality of Ingham and Jessen,” Proceedings of the Japan Academy, vol 24, no 9,
pp 10–12, 1948
5 H Alzer and S Ruscheweyh, “A converse of Minkowski’s inequality,” Discrete Mathematics, vol 216,
no 1–3, pp 253–256, 2000