Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 110, 2006
A q-ANALOGUE OF AN INEQUALITY DUE TO KONRAD KNOPP
L. D. ABREU
D EPARTMENT OF M ATHEMATICS
U NIVERSIDADE DE C OIMBRA
A PARTADO 3008, 3001-454 C OIMBRA
P ORTUGAL
daniel@mat.uc.pt
URL: http://www.mat.uc.pt/faculty/daniel.html
Received 25 January, 2006; accepted 22 March, 2006
Communicated by H.M. Srivastava
A BSTRACT. We derive a simple q-analogue of Konrad Knopp’s inequality for Euler-Knopp
means, using the finite and infinite q-binomial theorems.
Key words and phrases: Knopp’s inequality, q-analogue.
2000 Mathematics Subject Classification. 05A30.
1. I NTRODUCTION
Given a sequence (an ) and a number t such that 0 < t < 1, its Euler-Knopp mean en (t) is
defined by
n X
n
en (t) =
(1 − t)n−m tm am .
m
m=0
Knopp’s inequality [2, Section 3] states that, if p > 1 and 0 < t < 1 then
∞
X
∞
1X
[am ]p
[en (t)] ≤
t m=0
n=0
(1.1)
p
if the series on the right hand side is convergent.
The aim of this short note is to prove the following q-extension of Knopp’s inequality (1.1),
valid if p > 1, 0 < t < 1 and 0 < q < 1:
(1.2)
∞
X
[en (t; q)]p ≤
n=0
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
023-06
∞
1 X m(m−1)
q 2 [am ]p ,
t m=0
2
L. D. A BREU
provided the series on the right hand side of (1.2) converges. The sequence (en (t; q)) is
n X
m(m−1)
n
en (t; q) =
q 2 (1 − t)n−m tm am
m q
m=0
n
with the q-analogue of the numbers m
defined by
q
n
(q; q)n
,
=
m q (q; q)m (q; q)n−m
where
(a; q)0 = 1,
(a; q)n =
n
Y
(1 − aq k−1 ),
k=1
(a; q)∞ = lim (a; q)n .
n→∞
n
n
When q → 1 then clearly m q → m and both members in (1.2) tend to (1.1).
2. P ROOF OF (1.2)
We will follow the notations in [1].
The proof of our result uses the q-binomial theorem and the finite q-binomial theorem. The
finite q-binomial theorem states that
n
X
m(m−1)
n
m n
(2.1)
[x − a]q =
(−1)
q 2 xn−m am ,
m q
m=0
where
[x − a]nq = (x − a)(x − aq)...(x − aq n−1 ).
The q-binomial theorem states that
∞
(az; q)∞ X (a; q)n n
(2.2)
=
z ,
|z| < 1.
(z; q)∞
(q;
q)
n
n=0
Using Hölder’s inequality in the form
hX
ip X
hX ip−1
ak b k ≤
bk apk
bk
,
p>1
we have, if p > 1,
[en (t; q)]p ≤
"
n X
n
m=0
m
q
m(m−1)
2
(1 − t)n−m tm apm
q
#p−1
n X
n
m=0
m
q
m(m−1)
2
(1 − t)n−m tm
.
q
The finite q-binomial theorem (2.1) gives
n
n X
X
m(m−1)
m(m−1)
n
n−m m
m n
2
q
(1 − t)
t =
(−1)
q 2 (1 − t)n−m (−t)m
m q
m q
m=0
m=0
= [(1 − t) + t]nq
= (1 − t + t)(1 − t + qt)...(1 − t + q n−1 t) < 1
Therefore,
(2.3)
n X
n
p
[en (t; q)] ≤
m=0
J. Inequal. Pure and Appl. Math., 7(3) Art. 110, 2006
m
q
m(m−1)
2
(1 − t)n−m tm apm .
q
http://jipam.vu.edu.au/
q-A NALOGUE OF K NOPP ’ S I NEQUALITY
3
Now, if 0 < t < 1, then for every positive k we have (1 − q k )t < 1 − q k . This implies
t < (1 − (1 − t)q k ) and consequently, tm < ((1 − t)q; q)m . Using this last estimate on (2.3) and
summing in n from 0 to ∞, the result is
∞
∞ X
n X
X
m(m−1)
n
p
[en (t; q)] ≤
q 2 (1 − t)n−m ((1 − t) q; q)m apm
m q
n=0
n=0 m=0
∞ ∞
X
X
m(m−1)
n
p
2
(2.4)
(1 − t)n−m .
=
((1 − t) q; q)m q
am
m
q
m=0
n≥m
To evaluate the sum in n, observe that
∞
∞ X
X
(q; q)n
n
(1 − t)n−m =
(1 − t)n−m
m
(q;
q)
(q;
q)
n−m
m
q
n>m
n≥m
=
∞
X
k=0
=
(q; q)m+k
(1 − t)k
(q; q)k (q; q)m
∞
X
(q m+1 ; q)k
k=0
m+1
(q; q)k
(1 − t)k
(q
(1 − t); q)∞
(q(1 − t); q)∞
1
=
,
((1 − t); q)m+1
where the q-binomial formula (2.2) was used in the fourth identity. Substituting in (2.4) we
finally have
∞
n
n
X
X
X
((1 − t) q; q)m m(m−1) p
1 m(m−1) p
p
[en (t; q)] ≤
q 2 am =
q 2 am
((1
−
t);
q)
t
m+1
n=0
m=0
m=0
=
for every n. Taking the limit as n → ∞ gives (1.2).
R EFERENCES
[1] V. KAC AND P. CHEUNG, Quantum Calculus, Springer, 2002.
[2] K. KNOPP, Über reihen mit positiven gliedern, J. Lond. Math. Soc., 18 (1930), 13–21.
J. Inequal. Pure and Appl. Math., 7(3) Art. 110, 2006
http://jipam.vu.edu.au/