Proceedings of the 8th WSEAS International Conference on APPLIED MATHEMATICS, Tenerife, Spain, December 16-18, 2005 (pp309-314)
Transpose of the Weighted Mean Matrix on Weighted
Sequence Spaces
Rahmatollah Lashkaripour
Department of Mathematics, Faculty of Sciences,
Sistan and Baluchestan University,
Zahedan, Iran
Lashkari@hamoon.usb.ac.ir, Fax:+985412447166
Abstract: In this paper, we concern with transpose of the weighted mean matrix
(This is upper triangular matrix.) on weighted sequence spaces `p (w) and Lp (w) which
is considered by the author in [8] and [9] for special case of these operator, such as
Copson on `1 (w) and d(w, 1). Also, in a recent paper[7], the author has discovered the
upper bound for the Copson operator on the weighted sequence spaces d(w, p). Also, we
establish analogous upper bound for the continuous case. The weighted mean matrices
are considered by the author in [10].
Key Words:Transpose of Weighted Mean Matrix,Weighted Sequence Space.
2000 Mathematics subject classification. 26D15, 26A48, 47B37.
1. Introduction and Notations:
In this note, we consider the problem of finding the norm of transpose of the weighted mean
matrix Ad = (an,k ), denoted by Atd , where
(
an,k =
dk
Dn
for 1 ≤ k ≤ n
for k > n.
0
where the dn s are non-negative numbers with
partial sum Dn = d1 + . . . + dn (We insist that
d1 > 0, so that each Dn is positive.).
These results are extension of some results
which is considered by the author in [8] and [9]
P
and Bennett[2] and [4]. If rn = nk=1 wDk dnk , and
also Rn and Wn are defined as usual, then the
Rn
norm of Atd on `1 (w) is the supremum of W
.
n
Let w = (wn ) be a decreasing, non-negative
P
sequence with limn→∞ wn = 0 and ∞
n=1 wn divergent. Write Wn = w1 + . . . + wn . Then `p (w)
(and the Lorentz sequence space d(w, p) ), where
p ≥ 1, is the space of sequences x = (xn ) with
kxk`p (w) = (
∞
X
n=1
wn |xn |p )1/p ,
kxkw,p = (
∞
X
wn x∗n p )1/p
n=1
∗
(xn ) is
the decreasing rearconvergent, where
rangement of |xn |.
We now consider P
the operator A defined by
Ax = y, where yn = ∞
k=1 an,k xk . We shall write
kAk`p (w) for the norm of A when regarded as an
operator from `p (w) to `p (w), where
kAk`p (w) = sup{kAxk`p (w) : kxk`p (w) ≤ 1},
kAkw,p = sup{kAxkw,p : kxkw,p ≤ 1}.
Also, we define
Mw,p (A) = sup{kAxk`p (w) : kxk`p (w) = 1},
where x = (xn ) is regarded as a decreasing, nonnegative sequences in `p (w).
We assume that
1) an,k ≥ 0 for all n, k. This implies that
|Ax| ≤ A(|x|) for all x, and hence the nonnegative sequences x are sufficient to determine
kAk`p (w) .
We assume further that each A(ek ) is in `1 (w),
that is:
Proceedings of the 8th WSEAS International Conference on APPLIED MATHEMATICS, Tenerife, Spain, December 16-18, 2005 (pp309-314)
2) ∞
n=1 wn an,k is convergent for each k, that
garantte each A(ek ) is in `1 (w).
For two finite sequence x = (xn ) and y = (yn ),
write y x if
P
Yk ≤ Xk
(∀k),
where Xk = ki=1 xi and Yk = ki=1 yi .
Lemma 1: Suppose x, y ∈ IRn with x y
and (ai ) is decreasing. Suppose also
either an ≥ 0,
or Xn = Yn . Then
P
P
n
X
ak xk ≤
k=1
n
X
ak yk .
k=1
Proof: By Abel summation, it follows that
n
X
ak yk =
k=1
=
n
X
k=1
n−1
X
ak (Yk − Yk−1 )
(ii) ri,n ≥ rP
i+1,n (i, n = 1, 2, ...),
where ri,n = nj=1 ai,j .
Proof: (i) =⇒ (ii) follows by taking x to be
the sequence (1, ..., 1, 0, ...) of n ones followed by
zeros.
(ii) =⇒ (i): By Abel summation, it follows
that
(Y0 = 0)
yi =
∞
X
j=1
Yk (ak − ak+1 ) + an Yn .
k=1
Now, applying the hypothesis in both cases, we
deduce that
n−1
X
where x∗ is the decreasing rearrangement of |xn |.
Hence decreasing, non-negative sequences are
sufficient to determine kAk`1 (w) (kAkw,1 ).
Proposition 2([3], Lemma 9): Let A =
(ai,j )∞
i,j=1 be a matrix operator with non-negative
entrie s, and consider the associated
transformaP
tion, x → y, given by yi = ∞
a
x
i,j
j . Then the
j=1
following conditions are equivalent:
(i) y1 ≥ y2 ≥ . . . ≥ 0 whenever x1 ≥ x2 ≥
. . . ≥ 0.
ai,j xj =
∞
X
Since ri,n ≥ ri+1,n (∀i, n), and also (xn ) is decreasing, non-negative sequence, then
∞
X
ri,n (xn − xn+1 ) ≥
n=1
Yk (ak − ak+1 ) + an Yn ≥
=
k=1
n−1
X
n
X
k=1
k=1
Xk (ak − ak+1 ) + an Xn =
ak xk ≤
n
X
ak xk .
ak yk .
k=1
k=1
Proposition 1([8], Proposition 1.4.1):
Suppose that (1) holds, and that
(3) for all subsets M, N of IN having m, n elements respectively, we have
i∈M j∈N
ai,j ≤
m X
n
X
n=1
∞
X
ri+1,n (xn − xn+1 )
ai+1,j xj = yi+1 .
This completes the proof of the statement.
Lemma 2: Suppose u = (un ) and w = (wn )
are sequences of positive numbers.
Un
un
≤ M for all n, then m ≤ W
≤
(i) If m ≤ w
n
n
M for all n.
(ii) If
Corollary: Let x, y be decreasing, nonnegative elements of IRn (or `1 (w)) with x y.
Then
kxk`1 (w) ≤ kyk`1 (w) .
X X
∞
X
j=1
Therefore
n
X
ri,n (xn − xn+1 ).
n=1
ai,j .
i=1 j=1
Then
kAxk`1 (w) ≤ kAx∗ k`1 (w) (kAxkw,1 ≤ kAx∗ kw,1 )
for all non-negative elements x of `1 (w)(d(w, 1)),
so is
Un
Wn
un
wn
is increasing (or decreasing), then
.
un
(iii) If w
−→ U as n −→ ∞, then
n
as n −→ ∞ (also with U = ∞).
Proof. Elementary.
Un
Wn
−→ U
3. Transpose of the Weighted Mean
operator
Let now Ad be the weighted mean matrix with
properties (1), (2) and (3), and Atd be its transpose which is defined as follows:
(Atd x)(n) =
∞
X
dn xk
k=n
Dk
.
Proceedings of the 8th WSEAS International Conference on APPLIED MATHEMATICS, Tenerife, Spain, December 16-18, 2005 (pp309-314)
x1 ≥ x2 ≥ . . . ≥ 0. Then
This is an upper triangular matrix.
Recall that (wn ) is said to be 1-regular if
kAtd xkw,1
Wn
nw
n≥1
n
r1 (w) = sup
=
kAtd xk`1 (w)
rn =
=
kAtd k`p (w) ≤ pr1 (w) .
Proof: As mentioned before, it is sufficient
to consider decreasing, non-negative sequences.
Let x be in `p (w) or d(w, p) such that x1 ≥ x2 ≥
. . . ≥ 0, and so kxkw,p = kxk`p (w) and also the
same is true for the norm of Atd . Then applying
[11], Theorem 4.1.6, we deduce that:
kAtd kpw,p
=
≤
∞
X
n=1
∞
X
n=1
wn
wn
∞
X
dn xk
k=n
∞
X
∞
X
!p
= R
Dk
rn xn
Rn (xn − xn+1 )
n=1
∞
X
Wn (xn − xn+1 )
wn xn .
n=1
Hence kAtd kw,1 = Mw,1 (Atd ) ≤ R.
We have to show that this constant is the best
possible. We take x1 = x1 = . . . = xn = 1 and
xk = 0 for all k ≥ n + 1. Then
kxk`1 (w) = Wn
&
kAtd xk`1 (w) = Rn .
Therefore Mw,1 (Atd ) = R = kAtd kw,1 .
3. Copson Operator on Weighted
Sequence Spaces:
We now consider the Copson operator C on `1 (w)
and d(w, 1), which is defined by y = Cx, where
!p
wn xpn .
n=1
yi =
Hence kAtd kw,p ≤ pr1 (w). This completes the
proof.
Theorem 2: Suppose Atd is an operator on
`1 (w). If
Rn
< ∞,
R = sup
Wn
where rn = nk=1 wDk dnk , and Rn = r1 + . . . + rn
and Wn as usual. Then Atd is a bounded operator
from `1 (w) into itself, and we have Mw,1 (Atd ) =
R = kAtd kw,1 .
Proof: Since (Atd x)(n) ≤ (Atd x∗ )(n) for all
n, it is sufficient to consider decreasing, nonnegative sequences. Let x be in `1 (w) such that
∞
X
xj
j=i
j
.
It is given by the transpose of the matrix of the
Averaging operator A:
= (pr1 (w))p kxkpw,p .
P
n=1
∞
X
!
Dk
xk
k
k=n
≤ (pr1 (w))p
k=n
≤ R
Theorem 1: Suppose Atd is a weighted mean
operator defined as before and also d = (dn ) is
such that ndn ≤ Dk (∀k ≥ n). If w = (wn ) is
1-regular, then for p > 1, we have:
&
n=1
∞
X
n=1
∞
X
Wn
1
(w1 + . . . + wn ) =
.
n
n
kAtd kw,p ≤ pr1 (w)
∞
X
dn xk
=
=
is finite[11]. A pleasently simple statement can
also be made about the norm of the weighted
mean matrix operator for general w = (wn ).
With the previous notation,
∞
X
(
ai,j =
1
j
0
for i ≤ j
for i > j
This is an upper triangular matrix. The classical inequality of Copson [5] and [6] states that
kCkp = kAt kp = p(p > 1) as an operator on `p
spaces.
Proposition 3: If w = (wn ) is 1-regular, then
C maps `1 (w) into `1 (w). Also, we have
kCkw,1 = Mw,1 (C) ≤ kCk`1 (w) ≤ r1 (w).
Proof: Since
rn =
Wn
≤ r1 (w)wn
n
(∀n),
Proceedings of the 8th WSEAS International Conference on APPLIED MATHEMATICS, Tenerife, Spain, December 16-18, 2005 (pp309-314)
then by Lemma 2(i) and Theorem 2, it follows
that
kCkw,1 = Mw,1 (C) ≤ kCk`1 (w) ≤ r1 (w).
If
Corollary 1([8], Theorem 2.3.1): If
n
Wk
1 X
< ∞,
sup
Wn n=1 k
then the Copson operator is a bounded operator
from d(w, 1) into itself, and also we have
Mw,1 (C) = kCkw,1 = sup
Lemma 4: Suppose
un > 0
P∞ that vn >P0,
∞
for all n and that
v
and
n
n=1
n=1 un are
P∞
convergent. Let U(n) =
u
,
similarly
V(n) .
j=n j
n
Wk
1 X
.
Wn n=1 k
un
vn
U(n)
V(n)
is increasing (or decreasing), then so is
.
Proof: Elementary.
Proposition
P∞ 5:1 Let r >r 0 and let
U(n) =
Then n U(n) decreasj=n j 1+r .
r
ing, (n − 1) U(n) increasing. Both tend to 1r as
n −→ ∞.
Rn
1
1
and vn = n−1
dt.
Proof: Let un = n1+r
t1+r
1
Then V(n+1) = rnr . By the usual integral comparison,
1
1
≤ U(n) ≤
,
rnr
r(n − 1)r
Write
1
un = r ,
n
Z
(r > 0)
n
vn =
n−1
1
dt
tr
and(as usual) Un = u1 + . . . + un , etc. For r < 1,
the usual integral comparison gives
v2 + . . . + vn ≤ Un ≤ Vn ,
or
n1−r
1
(n1−r − 1) ≤ Un ≤
,
1−r
1−r
we need to know that UVnn is increasing. The following is the key lemma.
Lemma 3: With vn as above (for any r > 0),
nr vn decreases with n and nr vn+1 increases with
n.
Proof: Write tn = nr vn . Then
tn+1 = (n+1)r
Z
n+1
n
ds
= (n+1)r
sp
Z
n
n−1
ds
.
(s + 1)r
s+1
For n − 1 ≤ s ≤ n, we have n+1
n ≤
s , hence
r
(n+1)
nr
(∀n). Simi(s+1)r ≤ sr . Therefore tn+1 ≤ tn
larly for the second statement.
Proposition
4: Let 0 < r < 1 and let
Pn
Un
1
Un =
.
j=1 j p Then n1−r increases and tends
1
to 1−r
.
Proof: With vn as above, by Lemma 3, uvnn
increases with n, and so applying Lemma 1, we
deduce that UVnn is increasing. The limit follows
from the inequalities above.
We now consider the tail of the series for ζ(1 +
p). For the tail of a series, the analogous result
to Lemma 2(ii) is the following.
which implies the stated limits. By Lemma
un
3, ( vn+1
) is decreasing, so by Lemma 2(ii),
U
U(n)
V(n+1)
= rnr U(n) is decreasing. Similarly, V (n) is
(n)
increasing.
Remark: This is stated without proof in [1],
Remark 4.10.
Theorem 3: If w = ( n1p ), 0 < p ≤ 1, then
the Copson operator C is a bounded operator
from `1 (w)(d(w, 1)) into itself. Also, we have
Mw,1 (C) = kCkw,1 = kCk`1 (w) =
Proof: Since rn =
wn
n ,
1
.
1−p
then
Wn
Wn
rn
=
= 1−p .
wn
nwn
n
Also, since Wn is the Un of the Proposition 4,
1
n
then nW
1−p increases with n and tends to 1−p .
Hence applying Proposition 2, we deduce that
Mw,1 (C) = kCkw,1 = kCk`1 (w) =
1
.
1−p
Remark: When p = 1, so that wn =
have
rn
= Wn −→ ∞
wn
1
n,
we
(as n −→ ∞),
so the Copson operator C is not a bounded operator on d(w, 1), although of course is satisfies
condition (2).
Proceedings of the 8th WSEAS International Conference on APPLIED MATHEMATICS, Tenerife, Spain, December 16-18, 2005 (pp309-314)
Theorem 4: Let wn be defined by Wn =
n1−p , where 0 < p < 1. Then the Copson operator is a bounded operator from d(w, 1) into
itself. Also, we ahve
Hence, applying Holder’s inequality, we deduce
that:
x0
Z
0
1
= Mw,1 (C) =
.
1−p
kCkw,1
x0
Z
p
w(x)
0
x
x0
Z
p
Proof: We now have
0
Rn =
n
X
Wk
=
k
k=1
n
X
1
k
k=1
4. Continous Version of the Copson
Opreator:
In this section, we consider the analogous problem for the continuous case concern the space
Lp (w). In the continuous case, the Copson operator C is given by:
∞
(Cf )(x) =
x
f (t)
dt.
t
Let w(x) be a decreasing, non-negative
R function
on (0, ∞). We assume that W (x) = 0x w(t)dt is
finite for each x(Hence x1α is permited for 0 <
α < 1, but not for α = 1.). Then Lp (w) is the
space of functions f having
Z
∞
x0
Z
p
0
Z
x0
p
t
Z
w(x)dxdt =
0
Ax0 (t)p−1 a(t)W (t)dt =
0
is exactly the
of Theorem 3
so the new
and Proposition 4 again gives the statement.
Z
x0
p
,
p
Ax0 (t)p−1 a(t)dtdx =
Ax0 (t)p−1 a(t)
Z
rn
wn
Rn
Wn
x0
Z
w(x)Ax0 (x)p dx =
w(t)Ax0 (t)p−1 a(t)f (t)dt ≤
1/p Z
w(t)f (t)p dt
0
x0
0
1/p∗
w(t)Ax0 (t)p dt
.
Therefore
Z
0
x0
w(t)Ax0 (t)p dt
1/p
≤ pkf kLp (w) .
The above inequality is true for all x0 > 0, and so
true with x0 replacing by infinity. This completes
the proof.
(x)
Proposition 7: If W
(∀x > 0),
w(x) ≤ r1 (w)
then
kCkLp (w) ≤ pr1 (w).
Proof: We have
1
w(t)
≤ r1 (w)
,
t
W (t)
w(x)|f (x)|p dx
0
and so
convergent, with norm
∞
Z
kf kLP (w) =
(Cf )(x) ≤ r1 (w)
1/p
w(x)|f (x)|p dx
Proposition 6: Let f ≥ 0 be in Lp (w), a(x) =
R∞
w(x)
W (x) f (x), and also A∞ (x) = x a(t)dt. Then
A∞ (x) is finite and also we have:
kA∞ kLp (w) ≤ pkf kLp (w) .
Proof: Fix x0 . For any x < x0 , let xx0 a(t)dt =
d
Ax0 (x)p = −pAx0 (x)p−1 a(x),
Ax0 (x). Then dx
and so
R
Ax0 (x)p = Ax0 (x)p − Ax0 (x0 )p
Z
x0
= p
x
Ax0 (t)p−1 a(t)dt.
∞
x
.
0
Z
w(t)
f (t)dt = r1 (w)A∞ (x).
W (t)
This establishes the statement.
Theorem 5: If w(x) = x1α , where 0 ≤ α < 1,
then
p
kCkLp (w) =
.
1−α
Attained by action of C on decreasing positive
functions.
1−α
Proof: (i) We have W (x) = x1−α , and so
1
W (x)
=
xw(x)
1−α
Hence
(∀x > 0).
p
kCkLp (w) ≤
.
1−α
Proceedings of the 8th WSEAS International Conference on APPLIED MATHEMATICS, Tenerife, Spain, December 16-18, 2005 (pp309-314)
(ii) Now, by taking ε > 0, and define r by:
α + rp = 1 + ε, we deduce that:
f (x) =
1
xr
1
for x ≥ 1
for 0 ≤ x < 1.
Then f is decreasing and in Lp (w), since 01 x1α dx
R
1
dx are convergent. Also, we have:
and 0∞ xα+rp
R
Z
∞
(Cf )(x) =
x
1
1
dt = r
r+1
t
rx
for x ≥ 1,
and also we have:
(Cf )(x) ≥ (Cf )(1) =
1
r
for 0 < x < 1.
Hence (Cf )(x) ≥ 1r f (x) (∀x > 0), and so
1
kCf kLp (w) ≥ kf kLP (w) ,
r
p
where 1r = 1−α+ε
. Now, applying (i) and (ii) implies the statement.
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