Internat. J. Math. & Math. Sci.
Vol. 24, No. 1 (2000) 59–61
S0161171200003537
© Hindawi Publishing Corp.
ON AN ISOLATION AND A GENERALIZATION OF
HÖLDER’S INEQUALITY
XIAOJING YANG
(Received 23 December 1998 and in revised form 16 June 1999)
Abstract. We generalize the well-known Hölder inequality and give a condition at which
the equality holds.
Keywords and phrases. Hölder inequality, equality.
2000 Mathematics Subject Classification. Primary 34C10; Secondary 34L05.
Let us consider the famous Hölder inequality [1]
n
m Xij ≤
i=1 j=1
n
j=1
m
i=1
pj
Xij
1/pj
,
(1)
n
where Xij > 0 (1 ≤ i ≤ m, 1 ≤ j ≤ n), pj > 1, j=1 (1/pj ) = 1.
We give a generalization of (1), first, we need the following lemmas.
n
n
m
Lemma 1. Let yi = j=1 Xij , i = 1, 2, . . . , m, then k=1 1/pk ( i=1 yi ln zik ) = 0, where
pk
zik = Xik /yi , i = 1, 2, . . . , m.
n
n
Proof. Since yi = j=1 Xij , we have ln yi = j=i ln Xij and
pk n
n
n
Xik
1
1
1 ln zik =
ln
pk ln Xik − ln yi
=
p
p
y
p
k
k
i
k
k=1
k=1
k=1
n
n
n
1
ln Xik −
ln yi = ln
Xik − ln yi = 0,
=
p
k=1
k=1 k
k=1
here we have used the fact
m
i=1
yi
n
k=1 (1/pk )
n
1
ln zik
p
k=1 k
(2)
i = 1, 2, . . . , m
= 1, from the above calculations we obtain
n
1
=
p
k=1 k
m
yi ln zik
= 0.
(3)
i=1
Lemma 2. If a, b are positive numbers, then
(ln a − ln b) at − bt ≥ 0,
(ln a − ln b) at − bt ≤ 0,
and the equalities hold if and only if (a − b)t = 0.
if t ≥ 0,
if t ≤ 0,
(4)
60
XIAOJING YANG
Proof. If a = b, (4) holds, let a ≠ b, by the mean value theorem f (a) − f (b) =
(a − b)f (z) with z ∈ (a, b), we have
1
ln a − ln b = (a − b) , z1 ∈ (a, b),
z1
t
a − bt = (a − b)tz2t−1 , z2 ∈ (a, b)
hence
(ln a − ln b) at − b
t

(a − b)2 tz2t−1 ≥ 0,
=
≤ 0,
z1
(5)
if t ≥ 0
(6)
if t ≤ 0
and the equality holds if and only if t = 0.
The main result of our paper is the following theorem.
Theorem 3. If Xij > 0 (1 ≤ i ≤ m, 1 ≤ j ≤ n), pi > 1,
h(t) =

n

k=1
m
yi
i=1
p
Xikk
yi
n
i=1 (1/pi )
= 1, let

1/pk
t 1/pk
n
(1−t)
n
m
pk t


=
Xij
.
Xik 
k=1
i=1
(7)
j=1
Then
h (t) ≥ 0,
h (t) ≤ 0,
if t ≥ 0,
(8)
if t ≤ 0,
n
p
p n
the equality holds if and only if t = 0, or Xikk / τ=1 Xiτ = Xjkk / τ=1 Xjτ for 1 ≤ i, j ≤ m,
p
p
p k n
n
n
/ j=1 Xmj .
k = 1, 2, . . . , n, that is, X1kk / j=1 X1j = X2kk / j=1 X2j = · · · = Xmk
Proof. Let H(t) = ln h(t), t ∈ R, then by Lemmas 1 and 2 we obtain
m
n
yi zt ln zik
1
h (t)
i=1
m ik t =
H (t) =
h(t)
p
i=1 yi zik
k=1 k
m
m
n
n
t
1
1
i=1 yi zik ln zik
i=1 yi ln zik
m
=
−
m
t
p
p
i=1 yi
i=1 yi zik
k=1 k
k=1 k

t
t
n
≥ 0, if t ≥ 0,
− zjk
zik
1
1≤i<j≤m yi yj ln zik − lnjk
m
m
=
t
≤ 0, if t ≤ 0.
p
i=1 yi
i=1 yi zik
k=1 k
(9)
The equality holds if and only if t = 0 or
z1k = z2k = · · · = zmk ,
where yi =
the proof .
n
j=1 Xij ,
p
zik = Xikk /
n
j=1 Xij ,
k = 1, 2, . . . , n,
(10)
i = 1, 2, . . . , m, k = 1, 2, . . . , n. This completes
Corollary 4. Let t2 < t1 ≤ 0 or 0 ≤ t1 < t2 , we have
n
k=1


m
i=1
n
j=1
1−t1
Xij
1/pk
p t
Xikk 1 
≤
n
k=1


m
i=1
n
j=1
1−t2
Xij
1/pk
p t
Xikk 2 
,
(1 )
61
ON AN ISOLATION AND A GENERALIZATION OF . . .
equality holds if and only if for any k = 1, 2, . . . , n,
pk
p
Xjk
X k
n ik
= n
,
τ=1 Xiτ
τ=1 Xjτ
(2 )
for 1 ≤ i ≤ j ≤ m,
and for all 0 < t < 1,
n
m i=1 j=1
Xij ≤
n
k=1

m

i=1
i
j=1
1−t
Xij
1/pk
p t
Xikk 
≤
n
i=1
m
j=1
pj
Xij
1/pj
(3 )
equality holds only (2 ) holds.
References
[1]
E. F. Beckenbach and R. Bellman, Inequalities, Springer-Verlag, Berlin, 1961. MR 28#1266.
Zbl 097.26502.
Xiaojing Yang: Department of Mathematics, Tsinghua University, Beijing, China