Extensions of Copson's inequalities

0.13 MB·English

by  Peng Gao

EXTENSIONS OF COPSON’S INEQUALITIES
PENG GAO
Abstract. WeextendtheclassicalCopson’sinequalitiessothatthevaluesofparametersinvolved
go beyond what is currently known.
2
1
0
2

n 1. Introduction
a
J Let p > 0 and x= (x ) be a non-negative sequence. Let (λ ) be a non-negative sequence
n n≥1 n n≥1
4 with λ > 0 and let Λ = n λ . The classical Copson’s inequalities are referred as the following
1 n i=1 i
ones [3, Theorem 1.1, 2.1]:
]
A P
C ∞ n p p p ∞
(1.1) λ Λ−c λ x ≤ λ Λp−cxp, 1 < c≤ p;
h. n n k k! c−1 n n n
n=1 k=1 (cid:18) (cid:19) n=1
t X X X
a ∞ ∞ p p ∞
m (1.2) λ Λ−c λ x ≤ p λ Λp−cxp, 0 ≤ c< 1.
n n k k 1−c n n n
[ !
nX=1 kX=n (cid:18) (cid:19) nX=1

1
When λ = 1 for all k and c = p, inequality (1.1) becomes the celebrated Hardy’s inequality ([8,
v k
3 Theorem 326]). We note that the reversed inequality of (1.2) holds when c ≤ 0 < p < 1 and the
9 constants are best possible in all these cases.
8
It is easy to show that inequalities (1.1) and (1.2) are equivalent to each other by the duality
0
. principle [10, Lemma 2] for the norms of linear operators. It’s observed by Bennett [1, p. 411]
1
that inequality (1.1) continues to hold for c > p with constant (p/(p −1))p. A natural question
0
2 to ask now is whether inequality (1.1) itself continues to hold for c > p. Note that in this case
1 the constant (p/(c−1))p is best possible by considering the case λ = 1, x = n(c−p−1−ǫ)/p with
: n n
v ǫ → 0+.
i
X As analogues to Copson’s inequalities, the following inequalities are due to Leindler [9, (1)]:
r
a ∞ n p p ∞
p
(1.3) λ Λ∗−c λ x ≤ λ Λ∗p−cxp, 0 ≤ c< 1;
n n k k 1−c n n n
!
n=1 k=1 (cid:18) (cid:19) n=1
X X X
∞ ∞ p p ∞
p
(1.4) λ Λ∗−c λ x ≤ λ Λ∗p−cxp, 1 < c≤ p,
n n k k c−1 n n n
!
n=1 k=n (cid:18) (cid:19) n=1
X X X
where we assume ∞ λ < ∞ and we set Λ∗ = ∞ λ . We point out here that Leindler’s
n=1 n n k=n k
result corresponds to case c = 0 of inequality (1.3), after a change of variables. Inequalities (1.3)
P P
and (1.4) are given in [1, Corollary 5, 6, p. 412]. Again it is easy to see that inequalities (1.3)
and (1.4) are equivalent to each other by the duality principle. Moreover, the constants are best
possible.
2000 Mathematics Subject Classification. Primary 26D15.
Key words and phrases. Copson’s inequalities.
1
2 PENGGAO
As an application of Copson’s inequalities, we note the following result of Bennett and Grosse-
Erdmann [2, Theorem 8] that asserts for p ≥ 1,α ≥ 1,
∞ ∞ n ∞ ∞ p
(1.5) λ Λαx ≤ (αp+1)p λ Λαp x .
n k k n n k
! !
n=1 k=n n=1 k=n
X X X X
Here the constant is best possible. They also conjectured [2, p. 579] that inequality (1.5) (resp.
its reverse) remains valid with the same best possible constant when p ≥ 1,0 < a < 1 (resp.
−1/p < a < 0). Weaker constants are given for these cases in [2, Theorem 9, 10].
Itis ourgoal inthis paperto showinthenextsection thatthemethoddeveloped in [4]-[7]can be
applied to extend Copson’s inequality (1.1) to some c > p (or equivalently, by the duality principle,
to extend Copson’s inequality (1.2) to some c < 0). In Section 3, we extend inequality (1.5) to
some 0< α < 1.
2. Main Result
Before we prove our main result, we need a lemma first.
Lemma 2.1. Let p > 0 be fixed. In order for the following inequality (resp. its reverse)
1−p
1−c 1−c
(2.1) x ≤ 1+ x −(1−x)1−c
p p
(cid:18) (cid:19)
to be valid when c < 0,p > 1 (resp. when 0< c < 1,0 < p < 1) for all 0 ≤ x ≤ 1, it suffices that it
is valid when x = 1.
Proof. As the proofs for both cases are similar, we only consider the case c< 0,p > 1 here. Let
1−p
1−c 1−c
f (x)= 1+ x −(1−x)1−c − x.
p,c
p p
(cid:18) (cid:19)
Note that f (0) = 0 and we have
p,c
−p
(1−c)(1−p) 1−c 1−c
f′ (x) = 1+ x +(1−c)(1−x)−c− ,
p,c p p p
(cid:18) (cid:19)
(1−c)2(p−1) 1−c −p−1
f′′ (x) = 1+ x +(1−c)c(1−x)−c−1.
p,c p p
(cid:18) (cid:19)
It is easy to see that f′′ (x) = 0 is equivalent to the equation g (x) = 0 where
p,c p,c
−1/(p+1)
−pc 1−c
g (x)= (1−x)(1+c)/(p+1) −1− x.
p,c
(p−1)(1−c) p
(cid:18) (cid:19)
If 0 > c > −1, then it is easy to see that f′′′(x) < 0 so that f′′ (x) = 0 has at most one root
p,c p,c
in (0,1). As limx→1−fp′′,c(x) = −∞, it follows that if fp′′,c(x) ≤ 0 for all 0 ≤ x < 1, then fp,c(x)
is concave down and the assertion of the lemma follows. Otherwise we have f′′ (0) > 0 and this
p,c
combined with the observation that f′ (0) = 0,f′ (1) < 0 implies that there exists an x ∈ [0,1]
p,c p,c 0
such that f′ (x) ≥ 0 for 0 ≤ x ≤ x and f′ (x) ≤ 0 for x ≤ x ≤ 1 and the assertion of the lemma
p,c 0 p,c 0
follows. The case c = −1 can be similarly discussed.
If c < −1, then g′ (x) = 0 has at most one root in (0,1) so that f′′ (x) = 0 has at most two
p,c p,c
roots in (0,1). If f′′ (0) < 0, then as f′′ (1) > 0, it follows that f′′ (x) = 0 has exactly one root in
p,c p,c p,c
(0,1), and as f′ (0) = 0,f′ (1) < 0, it follows that f′ (x) < 0 for all x ∈ [0,1] and the assertion
p,c p,c p,c
of the lemma follows. If f′′ (0) > 0, then f′′ (x) = 0 has either no root or two roots in (0,1). If
p,c p,c
f′′ (x) = 0 has no root in (0,1), then f′′ (x) ≥ 0 for x ∈ [0,1]. As f′ (0) = 0,f′ (1) < 0, we
p,c p,c p,c p,c
see that this is not possible. If f′′ (x) = 0 has two roots in (0,1), it follows that f (x) is first
p,c p,c
increasing, then decreasing and then increasing again for x ∈ [0,1] and it follows from f′ (1) < 0
p,c
EXTENSIONS OF COPSON’S INEQUALITIES 3
that there exists an x′ ∈ [0,1] such that f′ (x) ≥ 0 for 0 ≤ x ≤ x′ and f′ (x) ≤ 0 for x′ ≤ x ≤ 1
0 p,c 0 p,c 0
and the assertion of the lemma again follows. The case f′′ (0) = 0 can be discussed similarly as
p,c
above and this completes the proof. (cid:3)
We now consider extending inequality (1.2) to c < 0. For two fixed two positive sequences
{a },{b }, we recall that it is shown in [5, Section 6] that we have the following inequality:
n n
wp−1 ∞ 1−p N ∞ −(p−1) wp−1 wp−1 N
(2.2) 1 w apAp+ w n − n−1 apAp ≤ xp,
bp k 1 1 k bp bp n n n
1 (cid:16)Xk=1 (cid:17) nX=2(cid:16)kX=n (cid:17) (cid:16) n n−1 (cid:17) nX=1
where {w } is a positive sequence, N is a large integer and for 1 ≤ n≤ N, we set S = N b x
n n k=n k k
and A = S /a .
n n n
P
We now recast inequality (1.2) as
∞ ∞ p p ∞
p
(2.3) λ1/pΛ−c/p λ1−1/pΛ−(1−c/p)x ≤ xp.
n n k k k 1−c n
!
n=1 k=n (cid:18) (cid:19) n=1
X X X
It remains to establish inequality (2.3). For this, it suffices to establish inequality (2.3) with the
infinite sums replaced by finite sums from 1 to N. We may also assume λ > 0 for all n. We then
n
set
a = λ−1/pΛc/p, b = λ1−1/pΛ−(1−c/p)
n n n n n n
in inequality (2.2) to see that in order to establish inequality (2.3), it suffices to find a positive
sequence {w } such that
n
∞ p−1 p p wp−1Λp−c wp−1Λp−c
w ≤ λ−1Λc n n − n−1 n−1 , n ≥ 2;
k 1−c n n λp−1 λp−1
(cid:16)kX=n (cid:17) (cid:18) (cid:19) (cid:16) n n−1 (cid:17)
∞ p−1 p p wp−1Λp−c p p wp−1Λp−1
w ≤ λ−1Λc 1 1 = 1 1 .
k 1−c 1 1 λp−1 1−c λp−1
(cid:16)Xk=1 (cid:17) (cid:18) (cid:19) 1 (cid:18) (cid:19) 1
Upon a change of variables: w → λ w , we can recast the above inequalities as
n n n
∞ p−1 p p−c
1 p Λ Λ
(2.4) λ w ≤ n wp−1−wp−1 n−1 , n ≥ 2;
Λ k k 1−c λ n n−1 Λ
n k=n ! (cid:18) (cid:19) n (cid:18) n (cid:19) !
X
∞ p−1 p
1 p
p−1
(2.5) λ w ≤ w .
Λ k k 1−c 1
1 k=1 ! (cid:18) (cid:19)
X
We now define the sequence {w } inductively by setting w = 1 and for n ≥ 2,
n 1
∞
p
λ w = Λ w .
k k n−1 n−1
1−c
k=n
X
This implies that for n ≥ 2,
−1
Λ 1−cλ
n−1 n
w = 1+ w .
n n−1
Λ p Λ
n (cid:18) n(cid:19)
Using the above relations, we can simplify inequalities (2.4), (2.5) to see that inequality (2.4) is
equivalent to inequality (2.1) with x = λ /Λ while inequality (2.5) is equivalent to
n n
1−p
1−c 1−c
(2.6) 1+ − ≥ 0.
p p
(cid:18) (cid:19)
4 PENGGAO
Itis easy to see thattheabove inequality is justthecase x = 1of inequality (2.1), wethen conclude
from Lemma 2.1 that inequality (1.2) is valid for c < 0 as long as the above inequality holds.
Next, we consider extending inequality (1.3) to c < 0. For two fixed two positive sequences
{a },{b }, we recall that it is shown in [6, (3.6)] (see also the discussion in Section 5 of [7]) that
n n
in order for the following inequality
∞ n p ∞
a b x ≤ U xp.
n k k p n
!
n=1 k=1 n=1
X X X
to be valid for a given constant U ,p > 1, it suffices to find a positive sequence {w } such that
p n
n p−1 wp−1 wp−1
(2.7) w ≤ U ap n − n+1 .
k p n bp bp
(cid:16)Xk=1 (cid:17) (cid:16) n n+1 (cid:17)
Without loss of generality, we may assume λ > 0 for all n. By a change of variables, we recast
n
inequality (1.3) as
∞ n p p ∞
p
λ1/pΛ∗−c/p λ1−1/pΛ∗−(1−c/p)x ≤ xp.
n n k k k 1−c n
!
n=1 k=1 (cid:18) (cid:19) n=1
X X X
It follows from (2.7) that in order to establish the above inequality, it suffices to find a positive
sequence {w } such that
n
n p−1 p p Λ∗c wp−1Λ∗p−c wp−1Λ∗p−c
w ≤ n n n − n+1 n+1 .
(cid:16)Xk=1 k(cid:17) (cid:18)1−c(cid:19) λn (cid:16) λpn−1 λpn−+11 (cid:17)
By a change of variables: w 7→ λ w , we can recast the above inequality as
n n n
1 n p−1 p p Λ∗ Λ∗ p−c
(2.8) λ w ≤ n wp−1−wp−1 n+1 .
Λ∗ k k 1−c λ n n+1 Λ∗
n k=1 ! (cid:18) (cid:19) n (cid:18) n (cid:19) !
X
We now define the sequence {w } inductively by setting w = 1 and for n≥ 1,
n 1
n
p
λ w = Λ∗ w .
k k 1−c n+1 n+1
k=1
X
This implies that for n ≥ 2,
Λ∗ 1−cλ −1
w = n+1 1+ n w .
n Λ∗ p Λ∗ n+1
n (cid:18) n(cid:19)
Usingtheabove relations, wecansimplify inequality (2.8)toseethatthen ≥2cases areequivalent
to inequality (2.1) with x = λ /Λ∗. It is also easy to see that the n = 1 case of (2.8) corresponds
n n
to the following inequality:
1−p
1−c 1−c
x ≤ x −(1−x)1−c.
p p
(cid:18) (cid:19)
It is easy to see that the above inequality is implied by inequality (2.1), we then conclude from
Lemma 2.1 that inequality (1.3) holds for c < 0 as long as inequality (2.6) holds.
Note that for fixed p > 0, the function (1+x)1−p −x is a decreasing function of x. Moreover,
it is easy to see that inequality (2.6) (resp. its reverse) always holds with c = 0 when p > 1 (resp.
when 0 <p < 1). We note that our discussions above for inequality (1.2) can be carried out for the
case 0 < p < 1,0 < c< 1 with the related inequalities reversed. We therefore obtain the following
EXTENSIONS OF COPSON’S INEQUALITIES 5
Theorem 2.1. Let p > 0 be fixed. Let c denote the unique number satisfying
0
1−p
1−c 1−c
0 0
1+ − = 0.
p p
(cid:18) (cid:19)
Then inequalities (1.2) and (1.3) hold for all c ≤ c < 1 when p > 1 and the reversed inequality
0
(1.2) holds for all c < c when 0 < p < 1.
0
We leave it to the reader for the corresponding extensions to c> p of inequalities (1.1) and (1.4)
by the duality principle.
3. Some related results
In this section we first consider the conjecture of Bennett and Grosse-Erdmann on inequality
(1.5) for the case 0< α < 1. We may assume λ >0 for all n. We note here that it is shown in [2,
n
(153), (156)] that it suffices to show that
∞ ∞ p ∞
λ Λα−Λα x ≤ (αp)p λ (Λαx )p,
n k k−1 k n n n
!
n=1 k=n n=1
X X(cid:0) (cid:1) X
where we set Λ = 0. By the duality principle, it is easy to see that the above inequality is
0
equivalent to
∞ Λα−Λα n p αp p ∞
n n−1 λ1−1/px ≤ xp.
λ1−1/pΛα k k! p−1 n
n=1 n n k=1 (cid:18) (cid:19) n=1
X X X
It follows from (2.7) that in order to establish the above inequality, it suffices to find a positive
sequence {w } such that
n
n p−1 αp p Λα−Λα −p wp−1 wp−1
w ≤ n n−1 n − n+1 .
(cid:16)Xk=1 k(cid:17) (cid:18)p−1(cid:19) λ1n−1/pΛαn ! (cid:16)λpn−1 λpn−+11(cid:17)
By a change of variables: w 7→ λ w , we can recast the above inequality as
n n n
1 n p−1 p p αλ Λα−1 p Λ
(3.1) λ w ≤ n n n wp−1−wp−1 .
Λ k k p−1 Λα−Λα λ n n+1
n Xk=1 ! (cid:18) (cid:19) (cid:18) n n−1(cid:19) n (cid:16) (cid:17)
We now define the sequence {w } inductively by setting w = 1 and for n ≥ 1,
n 1
n
p
λ w = Λ w .
k k n n+1
p−1
k=1
X
This implies that
1 λ
n
w = 1− w .
n+1 n
pΛ
(cid:18) n(cid:19)
Usingtheaboverelations,wecansimplifyinequality(3.1)toseethatitisequivalenttothefollowing:
p x 1−p 1−(1−x)α p
(3.2) 1− −1 ≥ x ,
p−1 p αx
!
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
where we set x = λ /Λ so that 0 ≤ x ≤ 1.
n n
By Hadamard’s inequality, which asserts for a continuous convex function h(u) on [a,b],
1 b a+b
h(u)du ≥ h( ),
b−a 2
Za
6 PENGGAO
we see that
p/x x 1−p 1 1 x −p
1− −1 = u−pdu ≥ 1− .
p−1 p 1−(1−x/p) 2p
(cid:18) (cid:19) (cid:18) (cid:19) ! Z1−x/p (cid:18) (cid:19)
Thus, it remains to show that
x −1 1−(1−x)α
1− ≥ ,
2p αx
(cid:18) (cid:19)
Equivalently, we need to show f (x) ≥ 0 where
α,p
x
f (x) = αx− 1− (1−(1−x)α).
α,p
2p
(cid:18) (cid:19)
It’s easy to see that f (0) = f′ (0) = 0 and f′′ (x) has a most one root in (0,1). It follows that
α,p α,p α,p
f′ (x) has a most one root in (0,1). Suppose α > 1−1/p so that f′′ (0) > 0. This together with
α,p α,p
the observation that limx→1−fα′,p(x) = −∞ implies that in order for fα,p(x) ≥ 0 for all x ∈ [0,1],
it suffices to have f (1) ≥ 0. We then deduce that we need to have
α,p
1
α≥ 1− .
2p
We then obtain the following
Theorem 3.1. Inequality (1.5) is valid for p > 1,α ≥ 1− 1 .
2p
We now consider the following analogue to inequality (1.5):
∞ n p ∞ n p
(3.3) λ Λ∗αx ≤ (αp+1)p λ Λ∗αp x .
n k k n n n
! !
n=1 k=1 n=1 k=1
X X X X
Again we may assume λ > 0 for all n. We set
n
n
y = x
n k
k=1
X
to recast inequality (3.3) as
∞ n−1 p ∞
λ Λ∗α−Λ∗α y +Λ∗αy ≤ (αp+1)p λ Λ∗αpyp.
n k k+1 k n n n n n
!
n=1 k=1 n=1
X X(cid:0) (cid:1) X
By Minkowskis inequality, we have
1
∞ n−1 p p
λ Λ∗α−Λ∗α y +Λ∗αy
n k k+1 k n n
! !
n=1 k=1
X X(cid:0) (cid:1)
∞ n−1 p p1 ∞ p1
≤ λ Λ∗α−Λ∗α y + λ (Λ∗αy )p .
n k k+1 k n n n
! ! !
n=1 k=1 n=1
X X(cid:0) (cid:1) X
Thus, it suffices to show that
∞ n p ∞
(3.4) λ Λ∗α−Λ∗α y ≤ (αp)p λ Λ∗αpyp.
n k k+1 k n n n
!
n=1 k=1 n=1
X X(cid:0) (cid:1) X
When 0< α ≤ 1, we note that we have
∞ n p ∞ n p
λ Λ∗α−Λ∗α y ≤ λ αλ Λ∗α−1 y .
n k k+1 k n k k k
! !
nX=1 Xk=1(cid:0) (cid:1) nX=1 Xk=1(cid:16) (cid:17)
EXTENSIONS OF COPSON’S INEQUALITIES 7
It then follows from inequality (1.3) with c = 0,x = Λ∗α−1y that
k k k
∞ n p ∞
λ λ Λ∗α−1y ≤ pp λ Λ∗αpyp.
n k k k n n n
!
n=1 k=1 n=1
X X X
Thus, inequality (3.4) is valid when 0 < α ≤ 1.
We now consider the case α ≥ 1. By the duality principle, it is easy to see that inequality (3.4)
is equivalent to
∞ Λ∗α−Λ∗α ∞ p αp p ∞
(3.5) n n+1 λ1−1/py ≤ yp.
λ1−1/pΛ∗α k k! p−1 n
n=1 n n k=n (cid:18) (cid:19) n=1
X X X
We then see that upon setting
λ1−1/pΛ∗α
a = n n , b = λ1−1/p
n Λ∗α−Λ∗α n n
n n+1
in inequality (2.2)that one can establish inequality (3.5)as long as one can finda positive sequence
{w } such that
n
∞ w p−1 ≤ αp p Λ∗α−Λ∗α −pλp−1Λ∗αp wnp−1 − wnp−−11 , n ≥ 2;
k p−1 n n+1 n n λp−1 λp−1
(cid:16)kX=n (cid:17) (cid:18) (cid:19) (cid:0) (cid:1) (cid:16) n n−1(cid:17)
∞ p−1 αp p wp−1
w ≤ (Λ∗α−Λ∗α)−pλp−1Λ∗αp 1 .
k p−1 1 2 1 1 λp−1
(cid:16)Xk=1 (cid:17) (cid:18) (cid:19) 1
Upon a change of variables: w → λ w , we can recast the above inequalities as
n n n
1 ∞ p−1 p p αλ Λ∗α−1 p Λ∗
(3.6) λ w ≤ n n n wp−1−wp−1 , n ≥ 2;
Λ∗ k k p−1 Λ∗α−Λ∗α λ n n−1
(cid:16) n kX=n (cid:17) (cid:18) (cid:19) (cid:18) n n+1(cid:19) n (cid:16) (cid:17)
1 ∞ p−1 p p αλ Λ∗α−1 p Λ∗
(3.7) λ w ≤ 1 1 1wp−1.
Λ∗ k k p−1 Λ∗α−Λ∗α λ 1
(cid:16) 1 Xk=1 (cid:17) (cid:18) (cid:19) (cid:18) 1 2(cid:19) 1
We now define the sequence {w } inductively by setting w = 1 and for n ≥ 2,
n 1
∞
p
λ w = Λ∗w .
k k p−1 n n−1
k=n
X
This implies that for n ≥ 2,
−1
1λ
n
w = 1− w .
n pΛ∗ n−1
(cid:18) n(cid:19)
Using the above relations, we can simplify inequality (3.6) to see that it is equivalent to inequality
(3.2) with x = λ /Λ∗ while inequality (3.7) is equivalent to the following inequality
n n
p x 1−p 1−(1−x)α p
1− ≥ x .
p−1 p αx
(cid:18) (cid:19)(cid:18) (cid:19) (cid:18) (cid:19)
As the above inequality is implied by inequality (3.2), it suffices to establish inequality (3.2) for
all α ≥ 1. For this, we note that it is easy to show that the right-hand side expression of (3.2)
is a decreasing function of α and inequality (3.2) is valid when α = 1. It therefore follows that
inequality (3.2) is valid for all α ≥ 1. As it is easy to check that the constant in (3.3) is best
possible by considering λ = n−a,a > 1,x = nb,b = ((a−1)(αp+1)−ǫ)/p−1 with ǫ → 0+, we
n k
conclude the paper with the following
Theorem 3.2. Inequality (3.3) is valid for p > 1,α > 0. The constant is best possible.
8 PENGGAO
References
[1] G. Bennett, Some elementary inequalities, Quart. J. Math. Oxford Ser. (2) 38 (1987), 401–425.
[2] G. Bennett and K.-G. Grosse-Erdmann, On series of positive terms, Houston J. Math., 31 (2005), 541-586.
[3] E. T. Copson, Note on series of positive terms, J. London Math. Soc., 3 (1928), 49-51.
[4] P. Gao, Hardy-typeinequalities via auxiliary sequences, J. Math. Anal. Appl.,343 (2008), 48-57.
[5] P. Gao, On weighted remainder form of Hardy-typeinequalities, arXiv:0907.5285.
[6] P. Gao, On lp norms of weighted mean matrices, Math. Z.,264 (2010), 829-848.
[7] P. Gao, On a result of Levin and Steˇckin,Int. J. Math. Math. Sci.,2011 (2011), Art.ID 534391, 15pp.
[8] G. H. Hardy,J. E. Littlewood and G. Po´lya, Inequalities, Cambridge Univ.Press, 1952.
[9] L.Leindler,GeneralizationofinequalitiesofHardyandLittlewood,ActaSci.Math.(Szeged),31(1970),279–285.
[10] H.L. Montgomery, Theanalytic principle of thelarge sieve, Bull. Amer. Math. Soc. 84 (1978), 547–567.
DivisionofMathematicalSciences,SchoolofPhysicalandMathematicalSciences,NanyangTech-
nological University, 637371 Singapore
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