-
7/27/2019 Holder Ineq
1/3
International Mathematical Forum, 4, 2009, no. 40, 1993 -
1995
A Note on Holders Inequality
Marwan S. Abualrub
Department of Mathematics, University of JordanP.O. Box 11942,
Amman, Jordan
[email protected]
W. T. Sulaiman
Department of Mathematics, Mosul University
Mosul, Iraq
Abstract
In this note an easy method has been introduced to prove
Holders
Inequality.
Mathematics Subject Classification code: 26D15
Keywords: Holder Inequality, Rogers Inequality, Cauchy
Inequality, Minkowski
Inequality
1 Introduction
In 1888 Rogers proved the Inequality we will be considering in
this Note whichhas been proved later in 1889 by Holder and named
after him even thoughsome Mathematicians prefer to call it Rogers
Inequality.
In 1994 Hovenier [2] proved sharpening Cauchys Inequality; and
in 1995
Abramovich, Mond, and Pecaric [1] generalized the result of
Hovenier toHolders Inequality. Finally, it is vital to mention that
Holders Inequalityis used to prove Minkowskis Inequality.
In this Note we will give an easier proof of Holders
Inequality.
-
7/27/2019 Holder Ineq
2/3
1994 M. S. Abualrub and W. T. Sulaiman
2 Preliminary Notes
We will first consider the following Cauchys Inequality.
2.1 Cauchys InequalityLet a = (a1, a2,...,an) and b = (b1,
b2,...,bn) are complex vectors. In addi-
tion, let 1 p ; where1
p+ 1
q= 1, then
n
j=1
ajbj
(n
j=1
|aj|p)
1
p (n
j=1
|bj|q)
1
q (1)
In (1) equality holds if and only if the sequences {|aj|p} and
{|bj |
q} forj = 1, 2,...,n are proportional and arg(ajbj) is
independent of j.
3 Main Results
These are the main results of the paper.
Lemma 3.1 If a 0, b 1, s + t = 1, s 0, andt 0. Then
(1 a)s(1 b)t 1 asbt (2)
Proof: We will prove the above Lemma by two Methods.Method
1.Consider F(a) = (1 a)s(1 b)t + asbt ; where b is fixed. ThenF
(a) = 0 implies that ( 1a 1)1s = (1
b 1)t.
Which gives that a = b.Now[F
(a)]a=b = s(s 1)[(1 a)1 + a1] 0.
This means that F attains its maximum at a = b, and in turns we
have[F(a)]a=b = 1 a + 1 = 1.Hence F(a) 1 which gives inequality
(2).
Method 2.
Consider f(x) = xs1x1
; where x 0, s 1, s + t = 1, s 0, and t 0.Observe that f is
decreasing with limx1 f(x) = s.Therefore f(x) s.
-
7/27/2019 Holder Ineq
3/3
A note on Holders inequality 1995
Let x = ab
; and a b. (if a b, take x = ba
). Thereforeasbt as + bt.Now, replace a and b by 1 a and 1 b
respectively we obtain(1 a)s(1 b)t s(1 a) + t(1 b)
= 1 sa tb
1 asbt.
We will now proof Holders Inequality.
3.2 Holders Inequality
n1j=1
ajbj
(n1j=1
|aj|p)
1
p (n1j=1
|bj |q)
1
q (3)
Proof:
We want to prove that the above inequality (3) is true for j =
1, 2, ...,n.Using the above Lemma we havenj=1 ajbj
n1j=1 ajbj
+ |anbn| (
n1j=1 |aj|
p)1
p (n1
j=1 |bj |q)
1
q + |an| |bn|
= (n
j=1 |aj |p |an|
p)1
p (n
j=1 |bj |q |bn|
q)1
q + |an| |bn|
= (n
j=1 |aj |p)
1
p (n
j=1 |bj |q)
1
q
1 |an|
pnj=1
|aj |p
1p
1 |bn|qn
j=1|bj|
q
1q
+ |an| |bn|
(n
j=1 |aj|p)
1p (n
j=1 |bj |q)
1q
1 |an||bn|nj=1
|aj |p
1pn
j=1|bj |
q
1q
+ |an| |bn|
= (n
j=1 |aj |p)
1
p (n
j=1 |bj |q)
1
q
References
[1] Abramovich, S., Mond, B. and Pecaric, J.E., Sharpening
Holders Inequal-ity, J. Math. Anal. Appl. 196 (1995), 1131 -
1134.
[2] Hovenier, J.W., Sharpening Cauchys Inequality, J. Math.
Anal. Appl. 186(1994), 156 - 160.
Received: March, 2009
