Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 1, Article 6, 2002
NORMS OF CERTAIN OPERATORS ON WEIGHTED `p SPACES AND LORENTZ
SEQUENCE SPACES
G.J.O. JAMESON AND R. LASHKARIPOUR
D EPARTMENT OF M ATHEMATICS AND S TATISTICS ,
L ANCASTER U NIVERSITY
L ANCASTER LA1 4YF,
G REAT B RITAIN .
g.jameson@lancaster.ac.uk
URL: http://www.maths.lancs.ac.uk/dept/people/gjameson.html
FACULTY OF S CIENCE ,
U NIVERSITY OF S ISTAN AND BALUCHISTAN
Z AHEDAN , I RAN .
lashkari@hamoon.usb.ac.ir
Received 07 May, 2001; accepted 21 September, 2001
Communicated by B. Opic
A BSTRACT. The problem addressed is the exact determination of the norms of the classical
Hilbert, Copson and averaging operators on weighted `p spaces and the corresponding Lorentz
sequence spaces d(w, p), with the power weighting sequence wn = n−α or the variant defined
by w1 + · · · + wn = n1−α . Exact values are found in each case except for the averaging operator
with wn = n−α , for which estimates deriving from various different methods are obtained and
compared.
Key words and phrases: Hardy’s inequality, Hilbert’s inequality, Norms of operators.
2000 Mathematics Subject Classification. 26D15, 15A60, 47B37.
1. I NTRODUCTION
In [13], the first author determined the norms and so-called “lower bounds" of the Hilbert,
Copson and averaging operators on `1 (w) and on the Lorentz sequence space d(w, 1), with the
power weighting sequence wn = 1/nα or the closely related sequence (equally natural in the
context of Lorentz spaces) given by Wn = n1−α , where Wn = w1 + · · · + wn . In the present
paper, we address the problem of finding the norms of these operators in the case p > 1. The
problem of lower bounds was considered in a companion paper [14].
The classical inequalities of Hilbert, Copson and Hardy describe the norms of these operators
on `p (where p > 1). Solutions to our problem need to reproduce these inequalities when we
take wn = 1, and the results of [13] when we take p = 1. The methods used for the case p = 1
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
039-01
2
G.J.O. JAMESON AND R. L ASHKARIPOUR
no longer apply. The norms of these operators on d(w, p) are determined by their action on
decreasing, non-negative sequences in `p (w): we denote this quantity by ∆p,w . In most cases,
it turns out to coincide with the norm on `p (w) itself. In the context of `p (w), we also consider
the increasing weight wn = nα , although such weights do not generate a Lorentz sequence
space. This case cannot always be treated together with 1/nα , because of the reversal of some
inequalities at α = 0.
Our two special choices of w are alternative analogues of the weighting function 1/xα in the
continuous case. The solutions of the continuous analogues of our problems are well known and
quite simple to establish. Best-constant estimations are notoriously harder for the discrete case,
essentially because discrete sums may be greater or less than their approximating integrals.
There is an extensive literature on boundedness of various classes of operators on `p spaces,
with or without weights. Less attention has been given to the exact evaluation of norms. Our
study aims to do this for the most “natural" operators and weights: as we shall see, the problem
is already quite hard enough for these specific cases without attempting anything more general.
Indeed, we fail to reach an exact solution in one important case. Problems involving two indices
p, q, or two weights, lead rapidly to intractable supremum evaluations. Though we do formulate
some estimates applying to general weights, our main objective is not to present new results of a
general nature. Rather, given the wealth of known results and methods, the task is to identify the
ones that lead to a solution, or at least a sharp estimate, for the problems under consideration.
Any particular theorem can be effective in one context and ineffective in another.
For the Hilbert operator H, the “Schur" method can be adapted to show that the value from
the continuous case is reproduced: for either choice of w, we have
kHkp,w = ∆p,w (H) =
π
.
sin[(1 − α)π/p]
The Copson operator C and the averaging operator A are triangular instead of symmetric, and
other methods are needed to deliver the right constant even when wn = 1. A better starting point
is Bennett’s systematic set of theorems on “factorable" triangular matrices [4, 5, 6]. For C, one
such theorem can be applied to show that (for general w), kCkp,w ≤ p supn≥1 (Wn /nwn ), and
hence that kCkp,w = ∆p,w (C) = p/(1 − α) for both our decreasing weights (reproducing the
value in the continuous case).
For the averaging operator A, a similar method gives the value p/(p − 1 − α) (reproducing
the continuous case) for the increasing weight nα (where α < p − 1). For Wn = n1−α , classical
methods can be adapted to show that ∆p,w (A) = p/(p−1+α), suggesting that this weight is the
“right" analogue of 1/xα in this context (though we do not know whether kAkp,w has the same
value). However, for wn = 1/nα , the problem is much more difficult. A simple example shows
that the above value is not correct. We can only identify and compare the estimates deriving
from the various theorems and methods available; different estimates are sharper in different
cases. The best estimate provided by the factorable-matrix theorems is pζ(p + α), and we show
that this can be replaced by the scale of estimates [rζ(r + α)]r/p for 1 ≤ r ≤ p. The case r = 1
occurs as a point on another scale of estimates derived by the Schur method. A precise solution
would have to reproduce the known values p∗ when α = 0 and ζ(1 + α) when p = 1: it seems
unlikely that it can be expressed by a single reasonably simple formula in terms of p and α.
2. P RELIMINARIES
Let w = (wn ) be a sequence of positive numbers. We write Wn = w1 + · · · + wn (and
similarly for sequences denoted by (xn ), (yn ), etc.). Let p ≥ 1. By `p (w) we mean the space of
J. Inequal. Pure and Appl. Math., 3(1) Art. 6, 2002
http://jipam.vu.edu.au/
O PERATORS ON WEIGHTED `p
sequences x = (xn ) with
Sp =
∞
X
3
SPACES
wn |xn |p
n=1
1/p
Sp .
convergent, with norm kxkp,w =
When wn = 1 for all n, we P
denote the norm by k.kp .
Now suppose that (wn ) is decreasing, limn→∞ wn = 0 and ∞
n=1 wn is divergent. The
Lorentz sequence space d(w, p) is then defined as follows. Given a null sequence x = (xn ), let
(x∗n ) be the decreasing rearrangement of |xn |. Then d(w, p) is the space of null sequences x for
which x∗ is in `p (w), with norm kxkd(w,p) = kx∗ kp,w .
We denote by en the sequence having 1 in place n and 0 elsewhere.
P∞
Let A be the operator defined by Ax = y, where yi =
j=1 ai,j xj . We write kAkp for
the norm of A as an operator on `p , and kAkp,w,v for its norm as an operator from `p (w) to
`p (v) (or just kAkp,w when v = w). This norm equates to the norm of another operator on
`p itself: by substitution, one has kAkp,w,v = kBkp , where B is the operator with matrix
1/p
−1/p
bi,j = vi ai,j wj .
We assume throughout that ai,j ≥ 0 for all i, j, which implies in each case that the norm is
determined by the action of A on non-negative sequences. Next, we establish conditions, adequate for the operators considered below, ensuring that kAkd(w,p) is determined by decreasing,
non-negative sequences (more general conditions are given in [13, Theorem 2]). Denote by
δp (w) the set of decreasing, non-negative sequences in `p (w), and define
∆p,w (A) = sup{kAxkp,w : x ∈ δp (w) : kxkp,w = 1}.
Lemma 2.1. Suppose
Pthat (wn ) is decreasing, that ai,j ≥ 0 for all i,j, and A maps δp (w) into
`p (w). Write cm,j = m
i=1 ai,j . Suppose further that:
(i) limj→∞ ai,j = 0 for each i;
and either
(ii) ai,j decreases with j for each i,
or (iii) ai,j decreases with i for each j and cm,j decreases with j for each m.
Then kA(x∗ )kd(w,p) ≥ kA(x)kd(w,p) for non-negative elements x of d(w, p). Hence kAkd(w,p) =
∆p,w (A).
Proof. Let y = Ax and z = Ax∗ . As before, write Xj = x1 + · · · + xj , etc. First, assume
condition (ii). By Abel summation and (ii), we have
∞
∞
X
X
yi =
ai,j xj =
(ai,j − ai,j+1 )Xj ,
j=1
j=1
and similarly for zi with Xj∗ replacing Xj . Since Xj ≤ Xj∗ for all j, we have yi ≤ zi for all i,
which implies that kykd(w,p) ≤ kzkd(w,p) .
Now assume (iii). Then yi and zi decrease with i, and
m X
∞
∞
∞
X
X
X
Ym =
ai,j xj =
cm,j xj =
(cm,j − cm,j+1 )Xj
i=1 j=1
j=1
j=1
and similarly for Zm . Hence
Zm for all m. By the majorization principle (e.g. [3,
P Ypm ≤ P
m
p
1.30]), this implies that m
y
≤
i=1 i
i=1 zi for all m, and hence by Abel summation that
kykd(w,p) ≤ kzkd(w,p) .
The evaluations in [13] are based on the property, special for p = 1, that kAk1,w is determined
by the elements en , and ∆1,w (A) by the elements e1 +· · ·+en . These statements fail when p > 1
(with or without weights). For kAkp this is very well known. For ∆p , let A be the averaging
J. Inequal. Pure and Appl. Math., 3(1) Art. 6, 2002
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4
G.J.O. JAMESON AND R. L ASHKARIPOUR
operator on `p . The lower-bound estimation in Hardy’s inequality shows that ∆p (A) = p∗ ,
while integral estimation shows that if xn = e1 + · · · + en , then
kAxn kp
sup
= (p∗ )1/p .
n≥1 kxn kp
The “Schur" method. By this (taking a slight historical liberty) we mean the following technique. It can be used to give a straightforward solution of the continuous analogues of all the
problems considered here (cf. [11, Sections 9.2 and 9.3]). We state a slightly generalized form
of the method for the discrete case.
Lemma 2.2. Let p > 1 and p∗ = p/(p − 1). Let B be the operator with matrix (bi,j ), where
bi,j ≥ 0 for all i, j. Suppose that (si ), (tj ) are two sequences of strictly positive numbers such
that for some C1 , C2 :
∞
∞
X
X
1/p
−1/p
1/p∗
−1/p∗
si
bi,j tj
≤ C1 for all i,
tj
bi,j si
≤ C2 for all j.
j=1
i=1
1/p∗ 1/p
Then kBkp ≤ C1 C2 .
P
Proof. Let yi = ∞
j=1 bi,j xj .
By Hölder’s inequality,
yi =
∞
X
1/p∗ −1/pp∗
(bi,j tj
1/p 1/pp∗
)(bi,j tj
xj )
j=1
∞
X
≤
!1/p∗
∞
X
−1/p
bi,j tj
≤
!1/p
j=1
j=1
1/p∗
bi,j tj xpj
−1/p
C1 si
∞
X
1/p∗
1/p∗
bi,j tj
!1/p
xpj
,
j=1
so
∞
X
yip
≤
p/p∗
C1
i=1
∞
X
1/p∗
xpj tj
j=1
p/p∗
≤ C1
C2
∞
X
−1/p∗
bi,j si
i=1
∞
X
xpj .
j=1
This result has usually been applied with sj = tj = j. As we will see, useful consequences
can be derived from other choices of sj , tj .
Theorems on “factorable" triangular matrices. These theorems appeared in [4, 5, 6], formulated for the more general case of operators from `p to `q . They can be summarized as follows.
In each case, operators S, T are defined by
(Sx)i = ai
∞
X
bj x j ,
(T x)i = ai
j=i
i
X
bj x j ,
j=1
where ai , bj ≥ 0 for all i, j. Note that S is upper-triangular, T lower-triangular. For p > 1, we
define
m
∞
m
∞
X
X
X
X
∗
p
p
p∗
αm =
ai , α̃m =
ai , βm =
bj , β̃m =
bpj .
i=1
J. Inequal. Pure and Appl. Math., 3(1) Art. 6, 2002
i=m
j=1
j=m
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O PERATORS ON WEIGHTED `p
5
SPACES
Proposition 2.3. (“head” version).
P
∗
p∗
(i) If m
≤ K1p αm for all m (in particular, if bj αj ≤ K1 ap−1
for all j),then
j
j=1 (bj αj )
kSkp ≤ pK1 .
P
p
p∗ −1
p
(ii) If m
for all j), then
i=1 (ai βi ) ≤ K1 βm for all m (in particular, if aj βj ≤ K1 bj
∗
kT kp ≤ p K1 .
Proposition 2.4. (“tail” version).
P
∗
p
(i) If ∞
≤ K2p β̃m for all m, (in particular, if aj β̃j ≤ K2 bpj −1 for all j),
i=m (ai β̃i )
then kSkp ≤ p∗ K2 .
P
∗
p∗
for all j)
(ii) If ∞
≤ K2p α̃m for all m, (in particular, if bj α̃j ≤ K2 ap−1
j
j=m (bj α̃j )
then kT kp ≤ pK2 .
Proposition 2.5. (“mixed” version).
1/p 1/p∗
∗
(i) If αm β̃m ≤ K3 for all m, then kSkp ≤ p1/p (p∗ )1/p K3 .
∗
1/p 1/p∗
(ii) If α̃m βm ≤ K3 for all m, then kT kp ≤ p1/p (p∗ )1/p K3 .
In each case, (ii) is equivalent to (i), by duality. Propositions 2.3 and 2.4 are [4, Theorem
2] and (with suitable substitutions) [5, Theorem 20 ] (note: the right-hand exponent in Bennett’s
formula (18) should be (r − 1)/(r − s)). For the reader’s convenience, we include here a direct
proof of Proposition 2.3(i), simplified from the more general theorem in [5]; the proof of 2.4(ii)
is almost the same.
Proof of Proposition 2.3. (i) Note first the following
proved by Abel summation: if
Pfact, easilyP
m
s
≤
sj , : tj (1 ≤ j ≤ N ) are real numbers such that m
j=1 tj for 1 ≤ m ≤ N , and
j=1 j
PN
PN
u1 ≥ . . . ≥ uN ≥ 0, then j=1 sj uj ≤ j=1 tj uj .
It isP
enough to consider finite sums with i, j ≤ N , and to take xi ≥ 0. Let y = Sx. Write
Ri = N
j=i bj xj , so that yi = ai Ri . By Abel summation,
N
X
yip
=
i=1
N
X
api Rip
=
i=1
N
−1
X
p
p
αi (Rip − Ri+1
) + αN RN
.
i=1
By the mean-value theorem, y p − xp ≤ p(y − x)y p−1 for any x, y ≥ 0. Since Ri − Ri+1 = bi xi
for 1 ≤ i ≤ N − 1, we have
p
Rip − Ri+1
≤ pbi xi Rip−1
p
p−1
for such i. Also, RN
≤ pbN xN RN
, since RN = bN xN . Hence
!1/p∗
!1/p N
N
N
N
X
X
X
X
∗
(bi αi )p Rip
yip ≤ p
αi bi xi Rip−1 ≤ p
xpi
i=1
i=1
i=1
i=1
(note that p∗ (p − 1) = p). By the “fact” noted above and our hypothesis (or directly from the
alternative hypothesis),
N
X
p∗
(bi αi )
i=1
Rip
≤
∗
K1p
N
X
api Rip
i=1
Together, these inequalities give kykp ≤ pK1 kxkp .
=
∗
K1p
N
X
yip .
i=1
Proposition 2.5 is [6, Theorem 9]; with standard substitutions, it is also a special case of [1,
Theorems 4.1 and 4.2]. The corresponding result for the continuous case was given in [16].
J. Inequal. Pure and Appl. Math., 3(1) Art. 6, 2002
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6
G.J.O. JAMESON AND R. L ASHKARIPOUR
We shall be particularly concerned with two choices of w, defined respectively by wn = 1/nα
(for α ≥ 0) and by Wn = n1−α (for 0 < α < 1). Note that in the second case,
Z n
1−α
1−α
1−α
wn = n
− (n − 1)
=
dt,
tα
n−1
and hence
1−α
1−α
≤
w
≤
.
n
nα
(n − 1)α
Several of our estimations will be expressed in terms of the zeta function. It will be helpful
to recall that ζ(1 + α) = 1/α + r(α) for α > 0, where 12 ≤ r(α) ≤ 1 and r(α) → γ (Euler’s
constant) as α → 0. Also, for the evaluation of various suprema that arise, we will need the
following lemmas.
P
Lemma 2.6. [9, Proposition 3]. Let An = nj=1 j α . Then An /n1+α decreases with n if α ≥ 0,
and increases if α < 0. If α > −1, it tends to 1/(1 + α) as n → ∞.
Lemma
2.7. [8, Remark 4.10] (without proof), [13, Proposition 6]. Let α > 0 and let Cn =
P∞
1+α
1/k
. Then nα Cn decreases with n, and (n − 1)α Cn increases.
k=n
Lemma 2.8. [7, Lemma 8], more simply in [9, Proposition 1]. The expression
n
X
1
kα
n(n + 1)α k=1
increases with n when α ≥ 1 or α ≤ 0, and decreases with n if 0 ≤ α ≤ 1.
3. T HE H ILBERT O PERATOR
We consider the Hilbert operator H, with matrix ai,j = 1/(i + j). This satisfies conditions
(i) and (ii) of Lemma 2.1. Hilbert’s classical inequality states that kHkp = π/ sin(π/p) for
p > 1. For the case p = 1, with either of our choices of w, it was shown in [13] that kHk1,w =
∆1,w (H) = π/ sin απ.
For the analogous operator in the continuous case, with w(x) = 1/xα , it is quite easily shown
by the Schur method that kHkp,w = π/ sin[(1 − α)π/p]. We show that the method adapts to the
discrete case, giving this value again for either of our choices of w. This is straightforward for
wn = 1/nα , but rather more delicate for Wn = n1−α .
Let 0 < a < 1. As with most studies of the Hilbert operator, we use the well-known integral
Z ∞
1
π
dt = a
.
a
t (t + c)
c sin aπ
0
Write
Z
n
gn (a) =
n−1
1
dt.
ta
a
Note that gn (a) ≥ 1/n .
Lemma 3.1. With this notation, we have for each j ≥ 1,
∞
∞
X
X
1
gi (a)
π
≤
≤ a
.
a
i (i + j)
i+j
j sin aπ
i=1
i=1
Proof. Clearly,
Z i
Z i
gi (a)
1
1
1
=
dt ≤
dt.
a
a
i+j
i + j i−1 t
i−1 t (t + j)
The statement follows, by the integral quoted above.
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O PERATORS ON WEIGHTED `p
7
SPACES
Now let 0 < α < 1. Write vn = gn (α) and vn0 = (1 − α)vn . In our previous terminology,
v 0 is our “second choice of w". Clearly, an operator will have the same norm on d(v 0 , p) and on
d(v, p). Note that by Hölder’s inequality for integrals, vnr ≤ gn (αr) for r > 1.
Theorem 3.2. Let H be the operator with matrix ai,j = 1/(i + j), and let p > 1. Let wn = n−α ,
where 1 − p < α < 1. Then
π
.
kHkp,w = ∆p,w (H) =
sin[(1 − α)π/p]
Rn
If 0 < α < 1 and vn = n−1 t−α dt, then kHkp,v and ∆p,v (H) also have the value stated.
Proof. Write M = π/ sin[(1 − α)π/p]. Let wn = n−α , where 1 − p < α < 1. Now kHkp,w =
kBkp , where B has matrix bi,j = (j/i)α/p /(i + j). In Lemma 2.2, take si = ti = i, and let
C1 , C2 be defined as before. Then
(1−α)/p
1
i
1/p −1/p
=
bi,j si tj
.
i+j j
By Lemma 3.1, it follows that C1 ≤ M . Similarly, C2 ≤ M .
Now let 0 < α < 1, and let v, w be as stated. We show in fact that kHkp,w,v ≤ M . It then
follows that kHkp,v ≤ M , since kxkp,w ≤ kxkp,v for all x. Note that kHkp,w,v = kBkp , where
now
1
bi,j =
(j α vi )1/p .
i+j
−1/α
Take si = vi
and tj = j. Then
bi,j (si /tj )1/p =
1
(j α vi )(α−1)/αp .
i+j
By Lemma 3.1,
∞
X
j (α−1)/p
i+j
j=1
1/α
(1−α)/αp
Since i−1 ≤ vi , we have i(α−1)/p ≤ vi
Also,
≤ i(α−1)/p M.
, and hence C1 ≤ M .
∗
bi,j (tj /si )1/p =
1
(j α vi )t ,
i+j
where
1
1
+ ∗.
p αp
Note that t > 1, so as remarked above, we have vit ≤ gi (αt). By Lemma 3.1 again,
t=
∞
X
gi (αt)
i=1
i+j
≤
M0
,
j αt
where M 0 = π/ sin αtπ. Now
αt =
α
1
1−α
+ ∗ =1−
,
p p
p
so that M 0 = M , and hence C2 ≤ M . The statement follows.
To show that ∆p,w (H) ≥ M , take r = (1 − α)/p, so that α + rp = 1. Fix N , and let
1/j r for j ≤ N,
xj =
0
for j > n.
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8
G.J.O. JAMESON AND R. L ASHKARIPOUR
Then (xj ) is decreasing and
the details), one finds that
P∞
j=1
wj xpj =
N
X
i=1
wi yip
PN
1
j=1 j .
≥M
p
Let y = Hx. By routine methods (we omit
N
X
1
i=1
i
− g(r),
where g(r) is independent of N . Clearly, the required statement follows. Minor modifications
give the same conclusion for v.
Note. A variant H0 of the discrete Hilbert operator has matrix 1/(i + j − 1). This is decidedly
more difficult! Already in the case p = 1, the norms do not coincide for our two weights, and it
seems unlikely that there is a simple formula for kH0 k1,w : see [13].
4. T HE C OPSON O PERATOR
The “Copson” operator C is defined by y = Cx, where yi =
transpose of the Cesaro matrix:
1/j for i ≤ j
ai,j =
.
0
for i > j
P∞
j=i (xj /j).
It is given by the
This matrix satisfies conditions (i) and (iii) of Lemma 2.1. Copson’s inequality [11, Theorem
331] states that kCkp ≤ p (Copson’s original result [10] was in fact the reverse inequality for
the case 0 < p < 1).
R∞
The corresponding operator in the continuous case is defined by (Cf )(x) = x [f (y)/y] dy.
When w(x) = 1/xα , it is quite easily shown, for example by the Schur method, that ∆p,w (C) =
kCkp,w = p/(1 − α).
For the discrete case, we will show that this value is correct for either decreasing choice
of w. However, the Schur method does not lead to the right constant in the discrete version of Copson’s inequality, and instead we use Proposition 2.3. First we formulate a result
for general weights. The 1-regularity constant of a sequence (wn ) is defined to be r1 (w) =
supn≥1 Wn /(nwn ).
Theorem 4.1. Suppose that (wn ) is 1-regular and p ≥ 1. Then the Copson operator C maps
d(w, p) into itself , and kCkp,w ≤ pr1 (w).
1/p
Proof. We have kCkp,w = kSkp , where S is as in Proposition 2.3, with ai = wi
1/p
1/(jwj ). In the notation of Proposition 2.3, αm = Wm , so
bj αj =
Wj
1/p
jwj
=
and bj =
Wj 1/p∗
1/p∗
p/p∗
wj ≤ r1 (w)wj = r1 (w)aj .
jwj
Since p/p∗ = p − 1, the simpler hypothesis of Proposition 2.3(i) holds with K1 = r1 (w).
Note. The same reasoning shows that kCkp,w,v ≤ pK1 , where K is such that Vn ≤
1/p 1/p∗
K1 nwn vn for all n.
An example in [15, Section 2.3] shows that kCkp,w is not necessarily equal to pr1 (w). However, equality does hold for our special choices of decreasing w, as we now show.
Theorem 4.2. Let C be the Copson operator. Suppose that p ≥ 1 and that w is defined either
by wn = 1/nα or by Wn = n1−α , where 0 ≤ α < 1. Then
p
∆p,w (C) = kCkp,w =
.
1−α
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O PERATORS ON WEIGHTED `p
9
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Proof. We show first that in either case, r1 (w) = 1/(1 − α), so that kCkp,w ≤ p/(1 − α). First,
consider wn = 1/nα . By comparison with the integrals of 1/tα on [1, n] and [0, n], we have
1
n1−α
(n1−α − 1) ≤ Wn ≤
,
1−α
1−α
from which the required statement follows easily. Now consider the case Wn = n1−α . Then
nwn /Wn = nα wn . By the inequalities for wn mentioned in Section 2,
1 − α ≤ nα wn ≤ (1 − α)
nα
,
(n − 1)α
from which it is clear that again r1 (w) = 1/(1 − α).
We now show that ∆p,w (C) ≥ 1 − α. Choose ε > 0 and define r by α + rp = 1 + ε. Let
xn = 1/nr for all n; note that (xn ) is decreasing. Then n−α xpn = 1/nα+rp = 1/n1+ε , so x is in
`p (w) for either choice of (wn ) (note that for the second choice, wn ≤ c/nα for a constant c).
Also, again by integral estimation,
∞
X
1
xn
1
yn =
≥ r =
1+r
k
rn
r
k=n
for all n, so
1
p
kykp,w ≥ kxkp,w =
kxkp,w .
r
1−α+ε
The statement follows.
Theorem 4.1 has a very simple consequence for increasing weights:
Proposition 4.3. Let (wn ) be any increasing weight sequence. Then kCkp,w ≤ p.
Proof. If (wn ) is increasing, then Wn ≤ nwn , with equality when n = 1. Hence r1 (w) = 1.
Clearly, ∆p,w (C) ≥ 1, since C(e1 ) = e1 . For the case wn = nα , the method of Theorem 4.2
shows that also ∆p,w (C) ≥ p/(1 + α). A simple example shows that ∆p,w (C) can be greater
than both 1 and p/(1 + α).
Example 4.1. Let p = 2 and α = 1, so that p/(1 + α) = 1. Take x = (4, 2, 0, 0, . . .). Then
y = (5, 1, 0, 0, . . .), and kxk22,w = 24, while kyk22,w = 27.
We leave further investigation of this case to another study; it is analogous to the problem of
the averaging operator with wn = 1/nα , considered in more detail below.
5. T HE AVERAGING O PERATOR : R ESULTS FOR G ENERAL W EIGHTS
Henceforth, A will mean the averaging operator, defined by y = Ax, where yn =
· · · + xn ). It is given by the Cesaro matrix
1/i for j ≤ i
ai,j =
,
0
for j > i
1
(x1
n
+
which satisfies conditions (i) and (ii) of Lemma 2.1. In this section, we give some results
for general weighting sequences. We consider upper estimates first. Hardy’s inequality states
that kAkp = p∗ for p > 1. It is easy to deduce that the same upper estimate applies for any
decreasing w:
Proposition 5.1. If (wn ) is any decreasing, non-negative sequence, then kAkp,w ≤ p∗ .
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G.J.O. JAMESON AND R. L ASHKARIPOUR
1/p
Proof. Let x be a non-negative element of `p (w), and let uj = wj xj . Then kukp = kxkp,w ,
and since (wj ) is decreasing,
wn1/p Xn
≤
n
X
1/p
wj xj = Un
j=1
(where, as usual, Xn means x1 + · · · + xn ). Hence
∞
X
wn p
kAxkpp,w =
Xn
p
n
n=1
∞
X
1 p
≤
U
p n
n
n=1
∗ p
≤ (p )
∞
X
upn
by Hardy’s inequality
n=1
= (p∗ )p kxkpp,w .
The next result records the estimates for the averaging operator derived from Propositions
2.3, 2.4 and 2.5. Instead of the fully general statements, we give simpler forms pertinent to our
objectives. For 1 ≤ r ≤ p, define
∞
X
wj
mr−1 Um (r)
,
V
(r)
=
sup
.
Um (r) =
r
j
w
m≥1
m
j=m
Proposition 5.2. Let A be the averaging operator and let p > 1. Then:
P
1−p∗
1−p∗
(i) kAkp,w ≤ p∗ K1 , where m
≤ K1 mwm
for all m;
j=1 wj
(ii) kAkp,w ≤ pV (p);
∗
(iii) if (wn ) is decreasing, then kAkp,w ≤ p1/p (p∗ )1/p V (p)1/p .
1/p
Proof. Recall that kAkp,w = kT kp , where T is as in Proposition 2.3, with ai = wi /i and bj =
P
P
−1/p
−p∗ /p
wj . With notation as before, we have α̃m = j≥m wj /j p = Um (p) and βm = m
.
j=1 wj
Noting that p∗ /p = p∗ − 1, one checks easily that Propositions 2.3(ii) and 2.4(ii) (with the
simpler, alternative hypotheses) translate into (i) and (ii).
−p∗ /p
p−1
For (iii), note that if (wn ) is decreasing, then βm ≤ mwm . Hence α̃m βm
≤
p−1
1/p
m Um (p)/wm , so the condition of Proposition 2.5(iii) is satisfied with K3 = V (p) .
For decreasing (wn ), (i) will give an estimate no less than p∗ , hence no advance on Proposition
5.1 (one need only take m = 1 to see that K1 is at least 1). Also, (iii) adds nothing to (ii), since
∗
[pV (p)]1/p (p∗ )1/p ≥ min[pV (p), p∗ ].
In the specific case we consider below, the supremum defining V (p) is attained at m = 1. In
such a case, nothing is lost by the inequality used in (iii) for βm . Furthermore, nothing would
be gained by using the more general [1, Theorem 4.1] instead of Proposition 2.5.
Another result relating to our V (p) is [12, Corollary of Theorem 3.4] (repeated, with an extra
condition removed, in [2, Corollary 4.3]). The quantity considered is C = supn≥1 np Un (p)/Wn .
Again, in our case, C coincides with V (p). It is shown in [12] that if C is finite, then so is
∆p,w (A), without giving an explicit relationship.
The next theorem improves on the estimate in (ii) by exhibiting it as one point in a scale of
estimates. We are not aware that it is a case of any known result.
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O PERATORS ON WEIGHTED `p
Theorem 5.3. Suppose that 1 ≤ r ≤ p and
Then kAkp,w ≤ [rV (r)]r/p .
P∞
n=1
11
SPACES
wn /nr is convergent. Define V (r) as above.
p/r
Proof. Write Un (r) = Un and V (r) = V . Let zn = xn and (as usual) Zn = z1 + · · · + zn . By
Hölder’s inequality,
Xn ≤ n1−r/p Znr/p ,
hence Xnp ≤ np−r Znr . If y = Ax, then yn = Xn /n, so ynp ≤ n−r Znr . For any N , we have
N
X
wn
n=1
Zr
nr n
=
N
X
(Un − Un+1 )Znr
n=1
=
N
X
r
Un (Znr − Zn−1
) − UN +1 ZNr
n=1
≤ r
N
X
Un zn Znr−1
by the mean-value theorem
n=1
N
X
wn
≤ rV
zn Znr−1 by the definition of V
r−1
n
n=1
!1/r N
!1/r∗
N
X
X wn
≤ rV
wn znr
Znr
by Hölder’s inequality.
r
n
n=1
n=1
Since znr = xpn , it follows that (for all N )
N
X
wn ynp
≤
n=1
N
X
wn
n=1
Zr
nr n
r
≤ (rV )
N
X
wn xpn .
n=1
The case r = 1 (for which the proof, of course, becomes simpler), gives kAkp,w ≤ V (1)1/p ,
in which
∞
1 X wj
V (1) = sup
.
j
m≥1 wm
j=m
Among the above results (including the Schur method) only Proposition 5.2(i) delivers the
right constant p∗ in Hardy’s original inequality. Another method that does so is the classical
one of [11, Theorem 326]. The next result shows what is obtained by adapting this method to
the weighted case. Note that it applies to ∆p,w (A), not kAkp,w .
Theorem 5.4. Let p ≥ 1, and let
nwn+1
c = sup
.
n≥1 Wn
Assume that c < p. Then
p
∆p,w (A) ≤
.
p−c
Proof. Let x = (xn ) be a decreasing, non-negative element of `p (w), and let y = Ax. Fix a
positive integer N . By Abel summation,
N
X
wn ynp
n=1
J. Inequal. Pure and Appl. Math., 3(1) Art. 6, 2002
=
N
X
p
p
Wn−1 (yn−1
− ynp ) + WN yN
.
n=2
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12
G.J.O. JAMESON AND R. L ASHKARIPOUR
p
By the mean-value theorem, yn−1
− ynp ≥ pynp−1 (yn−1 − yn ). Also, for n ≥ 2,
xn = nyn − (n − 1)yn−1 = yn − (n − 1)(yn−1 − yn ),
so
yn−1 − yn =
1
(yn − xn ).
n−1
Hence
N
X
wn ynp
≥ p
n=1
N
X
Wn−1 ynp−1 (yn−1 − yn )
n=2
= p
N
X
Wn−1
n=2
n−1
ynp−1 (yn − xn ).
Now since (xn ) is decreasing, yn ≥ xn for all n. Also, y1 = x1 . Since wn ≤ cWn−1 /(n − 1),
we deduce that
N
N
N
X
pX
pX
p
p−1
wn yn ≥
wn yn (yn − xn ) =
wn ynp−1 (yn − xn ),
c
c
n=1
n=2
n=1
so that
(p − c)
N
X
n=1
wn ynp
≤p
N
X
wn ynp−1 xn .
n=1
A standard application of Hölder’s inequality to the right-hand side finishes the proof.
We now consider lower estimates. First, an obvious one:
Proposition 5.5. We have
!1/p
∞
X
1
wn
.
∆p,w (A) ≥
w1 n=1 np
Proof. Take x = e1 . Then yn = 1/n for all n, so the statement follows. (In particular, the stated
series must converge in order for A(e1 ) to be in `p (w).)
Secondly, we formulate the lower estimate derived by the classical method of considering
xn = n−s for a suitable s.
P
Lemma 5.6. Suppose that an , bn are non-negative numbers such that ∞
n=1 an is divergent and
limn→∞ bn = 0. Then
PN
an b n
→ 0 as N → ∞.
Pn=1
N
n=1 an
Proof. Elementary.
Theorem 5.7. Let A be the P
averaging operator and let p > 1. Let (wn ) be any positive sequence,
q
and let q < p be such that ∞
n=1 wn /n is divergent. Then
p
∆w,p (A) ≥
.
p−q
Proof. Write p/(p − q) = r, so that 1/r = 1 − q/p. Fix N , and let
−q/p
n
for 1 ≤ n ≤ N
xn =
.
0
for n > N
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O PERATORS ON WEIGHTED `p
SPACES
13
Then
∞
X
(5.1)
wn xpn
=
n=1
N
X
wn
n=1
nq
.
Also, for n ≤ N ,
Z
n
Xn ≥
t−q/p dt = r(n1/r − 1),
1
so that
r
Xn
≥ q/p (1 − n−1/r ).
n
n
p
Since (1 − t) ≥ 1 − pt for 0 < t < 1, we have
yn =
ynp ≥
rp
(1 − pn−1/r ),
nq
and hence
wn ynp ≥ rp
(5.2)
wn
wn
− prp q+1/r (2).
q
n
n
Take ε > 0. By (5.1) and (5.2), together with Lemma 5.6 (with an = wn /nq and bn = n−1/r ),
it is clear that for all large enough N ,
N
X
wn ynp
≥ (1 − ε)r
n=1
p
∞
X
wn xpn .
n=1
Corollary 5.8. If
p∗ .
P∞
n=1
wn /n is divergent (in particular, if (wn ) is increasing), then ∆p,w (A) ≥
Remarks on the relation between kAkp,w and ∆p,w (A). If (wn ) is increasing, then kAkp,w =
∆p,w (A), for if x∗ = (x∗n ) is the decreasing rearrangement of x, then it is easily seen that
kx∗ kp,w ≤ kxkp,w , while kAx∗ kp,w ≥ kAxkp,w . The following example shows that this is not
true for decreasing weights in general (though it may possibly be true for 1/nα ), and indeed
that the estimate in Theorem 5.4 does not apply to kAkp,w .
Example
P 5.1. Let Ek = {i : k! ≤ i < (k +P1)!}, and let wi = 1/(k!k) for i ∈ Ek .
Then i∈Ek wi = 1 and, by integral estimation, i∈Ek (1/i) > log(k + 1). Now fix k, and
P
choose p close enough to 1 to have i∈Ek i−p > log k. Write k k for k kp,w . Take n = k!.
Then ken kp = wn = 1/(k!k), while
kAen kp >
X wi
log k
>
.
p
i
k!k
i∈E
k
Hence kAkpp,w > log k. We show that nwn+1 /Wn ≤ 32 for all n, so that ∆p,w (A) ≤ 3, by
Theorem 5.4. If k ≥ 3 and n ∈ Ek , then Wn ≥ k − 1, so
(k + 1)! 1
nwn+1
k+1
2
≤
=
≤ .
Wn
k!k k − 1
k(k − 1)
3
The required inequality is easily checked for n in E1 and E2 .
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14
G.J.O. JAMESON AND R. L ASHKARIPOUR
6. T HE AVERAGING O PERATOR :R ESULTS FOR S PECIFIC W EIGHTS
We now explore the extent to which the results of Section 5 solve the problem for our chosen
weighting
sequences. The analogous operator in the continuous case is given by (Af )(x) =
R
1 x
f
.
When
w(x) = x−α , one can show by the Schur method (or as in [17, Theorem 1.9.16])
x 0
that
kAkp,w = ∆p,w (A) = p/(p − 1 + α).
In the discrete case, when p = 1, it was shown in [13] that:
(i) if wn = 1/nα , then kAk1,w = ∆1,w (A) = ζ(1 + α),
(ii) if Wn = n1−α , then ∆1,w (A) = 1/α.
As suggested by (ii) and Hardy’s inequality, the continuous case is reproduced when Wn =
n1−α :
Theorem 6.1. Let A be the averaging operator and let p ≥ 1. Let (wn ) be defined by Wn =
n1−α , where 0 ≤ α < 1. Then
p
∆p,w (A) =
.
p−1+α
Proof. Recall that
wn+1 ≤
1−α
≤ wn .
nα
Hence
nwn+1
= nα wn+1 ≤ 1 − α
Wn
for n ≥ 1, so Theorem 5.4 applies with c = 1 − α. Also, Theorem 5.7 applies with q = 1 − α.
The stated equality follows.
We do not know whether kAkp,w has the same value; however, ∆p,w (A) is of more interest
for this choice of w, since it is motivated by Lorentz spaces.
For the increasing weight wn = nα , our problem is solved by the method of Proposition 2.3
(which, it will be recalled, was effective for the Copson operator with decreasing weights).
Theorem 6.2. Let wn = nα , where 0 ≤ α < p − 1. Then
p
kAkp,w = ∆p,w (A) =
.
p−1−α
Proof. By Theorem 5.7, we have ∆p,w (A) ≥ p/(p − 1 − α). We use Proposition 5.2(i) to prove
the reverse inequality. The condition is
m
X
j=1
1
j α(p∗ −1)
≤ K1
m
mα(p∗ −1)
.
Note that α(p∗ − 1) = α/(p − 1) < 1. By Lemma 2.6, the condition is satisfied with
K1 =
1
p−1
=
.
1 − α(p∗ − 1)
p−1−α
So kAkp,w ≤ p∗ K1 = p/(p − 1 − α).
Note. For α < 1, this result also follows from Theorem 5.4 and Lemma 2.8.
We now come to the hard case, wn = 1/nα . The trivial lower estimate 5.5 is enough to show
that ∆p,w (A) can be greater than p/(p − 1 + α). Indeed, we have at once from Proposition 5.5
and Theorem 5.7:
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O PERATORS ON WEIGHTED `p
15
SPACES
Proposition 6.3. Let p > 1 and wn = 1/nα , where α ≥ 0. Then ∆p,w (A) ≥ max(m1 , m2 ),
where
p
m1 =
, m2 = ζ(p + α)1/p .
p−1+α
Either lower estimate can be larger (even when α < 1), as the following table shows:
p
α
m1
m2
1.1 0.9 1.1 1.572
2 0.1 1.818 1.249
No improvement on m2 is obtained by considering elements of the form en or e1 + · · · + en .
However, the next example shows that ∆p,w (A) can be greater than both m1 and m2 .
Example 6.1. Let p = 2, α = 1. Then m1 = 1 and m2 = ζ(3)1/2 ≈ 1.096. Let x =
(2, 1, 0, 0, . . .). Then y1 = 2 and yn = 3/n for n ≥ 2. Hence kxk22,w = 4 12 , while kyk22,w =
4 + 9[ζ(3) − 1] ≈ 5.818, so that kyk2,w /kxk2,w ≈ 1.137.
(One could formulate another general lower estimate using this choice of x, but it is too
unpleasant to be worth stating explicitly.)
We now record the upper estimates derived from the various results above. Of course, by
Proposition 5.1, we have kAkp,w ≤ p∗ .
Proposition 6.4. Let p > 1, and let wn = 1/nα , where α ≥ 0. Then
p
∆p,w (A) ≤ M1 =:
.
p − 2−α
Proof. We have
n
Wn
(n + 1)α X 1
=
.
α
nwn+1
n
k
k=1
By Lemma 2.8, this expression increases with n, so has its least value 2α when n = 1. Hence
Theorem 5.4 applies with c = 2−α .
Proposition 6.5. Let p > 1 and 1 ≤ r ≤ p, and let wn = 1/nα , where α ≥ 0. Then
kAkp,w ≤ M2 (r) =: [rζ(r + α)]r/p .
Proof. Apply Theorem 5.3. In the notation used there, we have
∞
X
mr−1 Um (r)
1
r+α−1
=m
.
r+α
wm
j
j=m
By Lemma 2.7, this expression decreases with m, so is greatest when m = 1, with the value
ζ(r + α).
Note that in particular, M2 (1) = ζ(1 + α)1/p and M2 (p) = pζ(p + α). By the remark after
∗
Proposition 5.2, min[M2 (p), p∗ ] ≤ p1/p (p∗ )1/p m2 ≤ 2m2 .
To optimize the estimate in Proposition 6.5, we have to choose r (in the interval [1, p]) to
minimize [rζ(r + α)]r . Computations show that when α ≥ 0.4, the least value occurs when
r = 1, while for α = 0.1, it occurs when r ≈ 1.35.
The Schur method provides another scale of estimates, as follows.
Proposition 6.6. Let wn = 1/nα , where α > 0. Let α ≤ r < p + α. Then kAkp,w ≤ M3 (r),
where
1/p∗ 1/p
p
r
α
M3 (r) =
ζ 1+ ∗ +
.
p−r+α
p
p
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16
G.J.O. JAMESON AND R. L ASHKARIPOUR
Proof. In Lemma 2.2, take sj = tj = j r . We have bi,j = ai bj for j ≤ i, where ai = i−α/p−1 and
bj = j α/p . So C1 is the supremum (as i varies) of
i
(r−α)/p−1
i
X
j=1
1
j (r−α)/p
.
By Lemma 2.6 (or trivially when r = α),
p
1
=
.
1 − (r − α)/p
p−r+α
Also, C2 is the supremum (as j varies) of
∞
X
1
α/p+r/p∗
j
∗.
1+α/p+r/p
i
i=j
C1 =
By Lemma 2.7, this is greatest when j = 1, with value ζ(1 + r/p∗ + α/p).
Note that M3 (α) = M2 (1) = ζ(1 + α)1/p . Hence M3 will always be at least as good as M2
when α > 0.4.
The following table compares the estimates in some particular cases. We denote by m the
greater of the lower estimates m1 , m2 . Recall that M1 estimates ∆p,w (A). The values of r used
for M2 (r) and M3 (r) are indicated.
p
1.1
1.1
1.2
2
α
m
p∗
0.1 5.5 11
0.9 1.572 11
0.3 2.4
6
0.5 1.333 2
M1
6.587
1.950
3.095
1.547
M2 (r)
6.151 (1.1)
1.663 (1)
3.084 (1.1)
1.616 (1)
M3 (r)
6.031 (0.98)
1.663 (0.9)
2.886 (0.9)
1.593 (0.75)
Table 6.1: Numerical values of upper and lower estimates
In most, if not all cases, M3 is a better estimate than M2 , at the cost of being more complicated. Both M2 and M3 reproduce the correct value ζ(1 + α) when p = 1. On the other hand,
both can be larger than p∗ in other cases. Some product of the type M3 (1) would reproduce the
values ζ(1 + α) when p = 1 and p∗ when α = 0; the exponent would need to be of the form
f (α, p), where f (0, p) = 1 for p > 1 and f (α, 1) = 0 for α > 0.
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