M. Si.
Internal. J. Math.
Vol. 4 No. 2 (1981) 353-364
353
AN ORDERED SET OF NORLUND MEANS
J. DEFRANZA
Department of Mathematics
St. Lawrence Unlverslty
13617 U.S.A.
Canton, New York
(Received March 7, 1980 and in revised form April 28, 1980)
ABSTRACT.
Norlund methods of s-m.bility are studied as mappings from
Those Norlund methods that map
41
into
i
are characterized.
i
into
i"
Inclusion results
are given and a class of Norlund methods is shown to form an ordered abelian semi-
group.
KEF WORDS AN PHRASES. Inlion Theorem, 1-1 mthod, ordered abian emigroup,
6rd ’md.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
i.
Primary 40D25, 40G05.
INTRODUCTION.
Let p be a complex sequence
the n+l-st partial sum of’ p.
P0 + 0,
n
and let P
Suppose the sequence P
K be the least positive integer so that P
method of summability N
P
by
n
Z p
k=0
k
nffi0 1 2
denote
is eventually non-zero.
0 for all n > K.
Let
Define the Norlund
354
J. DEFRANZA
Pn_k/Po,
if 0_<_ n < K,
Pn-k/Pn
if
k_<_ n,
P
0
n_>_ K, k_<_n
and
otherwise.
Let N denote the collection of all such Norlund methods N
n
>= 0,
that is K
p
If P
+
n
0 for all
0, then
Pn_k/Pn,
if k
<__
n and
Np[n,k]
0
Throughout we let
n
P0
otherwise.
if 0 < n < K and
a sequence x is then given by
NpX,
n
P
n
if n > K.
The N
p
transform of
where
n
(NpX) n (i/n)
kZffi0Pn_kXk
for all n > O.
A matrix summability method is called an -E method if and only if it maps
the space
-
I -=
method A is
sup
k
into itself.
In [4], Knopp and Lorentz proved that the matrix
if and only if there exists some M > 0 such that
Z
n=0
lank
< M.
-
In the special case of a Norlund method N we have:
P
THEOREH 1.
(i)
(ii)
PROOF.
lira
n
n
The Norlund method N
P
p e J, and
n
-+
0 as n
/
=.
is
if and only if
First suppose that (i) and (ii) hold.
exists and by (ii) is non-zero.
Since p
it implies
So there exist strictly positive numbers
NRLUND MEANS
ORDERED SET OF
H and
rIpnl<H
such that
Z
n=k
INP [n,k]l-<
and
lnl
Z
IPn_k[
>
Cl/)
Thus by the Knopp-Lorentz Theorem N
-.
Now suppose that N is
P
r.
IPn_k/’nl.
P
is
-.
Then
0(i).
In particular if n=k it implies
.
Now suppose that p
For sufficiently large
li,nl
Thus for each fixed k,
Z
n--k
n--k
n=k
for all n > 0.
355
0(i) and hence
We assert that the series
n
@
0 as n
/-.
l([pn
n
n
z pk
[Pnl
[1/n]
k=O
<
):
k--O
Ipk[
sn
Then for sufficiently large N
N+m
Z
n=N+l
(]Pn II) >= (Iss+m)
1-
But
as m
Z
(Ip.I/li’n I)
e=
n=N+l
-
/
COROLLARY 1 [i,
Then N
2.
=.
SN+m +
P
is
N+m
z
n=N+l
Ip.l
SN/SN+m.
So choose m large enough such that
>
.
"
SN+m > 2S.N.
The theorem now follows.
Theorem 4].
if and only if p e
.
Let N
We make the following definitions.
p
be a Norlund method with
pn >
0
P0 >
0.
J. DEFRANZA
356
DEFINITION.
Let N
DEFINITION.
Let
DEFINITION.
Given two Norlund methods
E(Np)C_ E(Nq).
(Np) E(Nq), and Np and Nq
The method
Given N
DEFINITION.
(i)
.
p(z)
p
n= 0
(ll)
(ili)
q(z)
a(z)
n
Z
n
r.
n
n= 0
.Z
r. qn Zn ,Q(z)
n=0
p(z)lq(z)
and N
N
q
q
is in
E.
is k-stronger than N
Np
(Np) (Nq).
is strictly k-stronger than
P
pro-
define formally:
q
P(z)
Nq
Np
are k-equlvalent provided
and N
P
NpX
consist of all sequences x such that
E(Np)
if and only if
vided
that are E- methods.
denote the collection of all N
p
E
n
z
n
and
n-0
Z a Z
n
n-0
n
b(z)
r. b z n
n-O n
q(z)Ip(z)
The next propositions follow by an argument similar to the one used in
Theorem 18 of [3]
PROPOSITION 1.
If N
PROPOSITION 2.
If
P
E’
then the series r.
n
pn z
n
and r.
n
n
z
n
converge
for
b(z)
N
Np, Nq
nr anZ
then the series a(z)
n
and
r. b zn have positive radii of convergence and, moreover
n
n
(i)
(i)
(iii)
(iv)
Pn anqo
+
+ aoqn’
anQo
+
+
aO’
qn bnP0
+
+
b0Pn’
bn’O +
+
bon
n
n
PROPOSITION 3.
the series S(z)
SuPpose
r. S z
n
n
n
and
N v
p
E and the sequence S
{S } is in
has positive radius of convergence.
n
(Np).
Then
357
ORDERED SET OF NORLUND MEANS
PROOF
I/p(z).
Let h(z)
z[<L
function for
Since
Po
a e (0,i) such that p(z)
By Proposition I, p(z) defines an analytic
+ 0,
by the continuity of p(z) there exists some
0 for all z e (-a,a).
Therefore h(z)
Z h zn has
n
n
positive radius of convergence.
n
E
k=0
[h-kk(NpS)k]n
(Np),
Since S
Therefore Z S z
n
n
E S zn
n
n= 0
3.
S
n
n
say N
g
N
{k.0k(NpS)kzk}.
+ Pnq0 for all
p*q of the sequences p and q is defined by
n > 0.
Given Norlund methods N
p
is the symmetric product of N
P*q
Pk(NpS)kzk converges for
has positive radius of convergence since
{h(z)}
P0qn +
I
it follows that the series
The symmetric product g
gn
O,
Now for n
p
and N
q
and N
in
q
N
we
provided
In order to prove an inclusion result for two Norlund methods in
N we need
the following lemma.
Let the complex sequences p and q be given and define r
Lemma i.
Suppose
Np, Nr
e h/.
Then in order that
(Np) C_ (Nr)
it is necessary and
sufficient that there is some M > 0, independent of k, such that
[kl Z ]qn_k/n[
Proof.
Let x
< M.
{x } be any sequence.
n
Then
p*q.
58
Let
5. DEFRANZA
enk- qn_kPk/Rn
’(Np) (Nr)
if k
<__ n,
and 0 if k > n.
Then by the Knopp-Lorentz Theorem
if and only if there exists some M > O such that
. leak
sup {
k
n-k
That is,
} < 14.
-
sp{[k[
[qn_k/n[
k
n-k
< 14.
The lemma can now be used to get the desired inclusion result.
Theorem 2.
Np, gq
Suppose
N.
The
(gp)_ (Nq)
if and only if
b- {b }z4.
Proof.
In the previous 1emma replace the sequence q with the sequence b
which implies r-= p*b
some M > 0,
ndependent
[k[.Z.i bn_k/
But since
q.
Np, Nq
positive constants.
Then
if and only if there exists
of k, such that
< 14"
, lkl
n
(Np)_ (Nr) (Nq)
and
lkl
respectively are bounded by two strictly
[k[ Y. Ibn_k/n < M
Thus
independent of k, is equivalent to
b4.
Corollary 2.
Suppose
Np, Nq 4"
The
(1)
4(Np) 4(Nq)
if and only if both a
(ll)
4(Np)4(Nq)
if and only if a
{an } 4
and b
{bn }
4
and
Corollary 3.
only if h e 4.
Suppose
Np 4
and h(z)
{an }
4 and b
1/p(z).
Then
{bn }
4(Np)
4.
4 if and
ORDERED SET OF NORLUND MEANS
Proof.
.
I(z)
n-^U
i z
n
Let I be the identity matrix so that t(I)
n
I; that is i0
p(z)II(z)
a(z)
1 and i
p(z) and b(z)
(bnk)
if k-n-i or k-n, n > 1, and 0 otherwise.
fined by
P0-1-Pl, Pn-0
for n > 2.
.
Then
0 for all n > I.
n
I(z)/p(z)
The binary matrix B
Exmaple.
359
h(z).
Therefore
The corollary now follows.
is given by
1, if n=k=O, 1/2,
bnk
Thus g is the t- Norlund method de-
It now follows by Corollary 3 that
(B).
The next result addresses the question of when two Norlund methods
Np, Nq
N are comparable.
.
generating sequence N
t(Np) 6 t(Nq)
P
can result in a method N
N t. Let q
Np
(P0 Pl
If the sequence q satisfies lim
.
t(Sp) f (Nq)
Proof.
Nt satisfying
q
That is, the methods are not comparable.
Suppose
Theorem 3.
We prove that changing only the first term in the
Since q
P0
+ P0
q
n-
, Nq
..) with
/, [nl
and
lnl
respectively lle between two
Now for any sequence x
strictly positive constants.
(Po P0)Xn/n +[’(SpX)n] [n/n
(SqX)
d hence
I(qX)n[
[Fn/nll(pX)nl
+
t(Np),
erefore, if x
(N),
Slarly if x
Corollary 4.
q
.
(Po’ Pl ’’’’)’
l(Np)/ (Nq)
x
x
t, then x
,
t(Np).
the. x
Suppose p is a positive number sequence in
where
PO
> 0 and
PO PO"
Then N
P
and N
q
.
Let
are -t with
J. DEFRANZA
360
Suppose p is a sequence in
Theorem 4.
q
(pl,P2,...).
Qn
If
,
with P
n
eventually non-zer Let
Np, Nq
is eventually non-zero, then
e
N with
The next theorem is a special case of Theorem 2.
If N
Theorem 5.
(i)
(ii)
(iii)
then
and N
are Norlund methods with
q
Pn+i/Pn >= pn/Pn_l n
qn > 0, q0 I, and
> 0,
(Np)
(Nq).
Proof.
By Theorem 22 of [3]
i- c.z
-+/-
Then for small
I.
Y0
So
PO
i,
Pn
>
]z[,
c2z2-
q(z)/p(z)-
where c
> 0 for all n > 0 and Ec
(qnY0 +...+q0Yn)Z
n=0
Z
Z b z
n
where
Yn
we have
n-0
?-[bn a
Z
(}q0l[Yn[
+
n=0
n-0
(Np) (Nq).
Therefore by Theorem 2,
Corollary 6.
+
If N
P
and
Nq
are 1-1 Norlund methods such that
(i)
Pn+l/Pn pn/Pn_i
n > 0,
p0=i, Pn
> 0,
(ii)
qn+I/qn >__ qn/qn_l
n > 0,
q0=l, qn
> 0,
and
(Np)= (Nq).
< i.
n
that if q(z)/p(z)
n= 0
then
O,
,
p, q e
(p(z)’-1}--
and
P
-c
n
for n > 0
361
ORDERED SET OF NORLOND MEANS
" and qn
Moreover by Corollary 3, (Np
(Np) (Ny).
For example if
4.
Pn
T
0
" where 0
(Nq)
< p, T <
If
I, then
Yn
1/nk, k > 1 and n > 0, then
N forms an ordered abelian semigroup.
We now show that
(Sq ).
(sp)
The order relation
is set inclusion between the absolute summabillty fields, and the binary operation
We need the following
is the symmetric product of the generating sequences.
Lemma 2.
Let r
Suppose N
p*q.
Then llm
n.-o
(R_/PQ
n),,
We assert that N
Proof.
Theorem, see for example
qn-9 /^Qn
(1)
N and p is a sequence for which
e
q
[6],
q
n
+
P
0.
1.
is a regular method.
it suffices
By the Silverman-Toeplltz
to show that
n-- for each fixed
+ 0 as
lim P
n-=
> 0, and
n
,
--0
But since N
q
e
(1) and (ii) follow.
n
n/n
(iIn)
which is the N
q
n
(iIn)
1,..z__O rk
kXoqn_k’k
transform of the convergent sequence {P }.
n
If N
Lemma 3.
Now for all n sufficiently large
p’
Nq
e
N and r
p’q, then Nr e
+
N.
Thus
,
O. In order to have N e
By Lemma i, we have lim n
r
n
method. By Theorem I, it suffices to show that r
need to show N is an
Proof.
r
But
n= 0
n
n=O
n=O
we
4.
362
Lemma 4.
__
J. DEFRANZA
_
Np, Nq N,
If
p’q, then
r
Nr
is
E(Np) k3 E(Nq) (Nr).
By Lemma 2, N
Proof.
E(Np) k3 (Nq)
(i)
If
(Np)
(ii)
If
(Np)
Proof of (i).
b
E.
E
p’s, and
Np, Nq, N s E E,
E(Nq), then E(N) E(Nv).
(Nq), then (N;) (N).
. qn
n
Z v z ={
n
n= 0
n= 0
Z
v
q*s.
q(z)/p(z) and c(z)fu(z)/(z).
Let b(z)
We need to show that c
v(z)
By Theorem 2 it follows that
(Nr).
Suppose
Lemma 5.
N.
r
-
with
n
E.
}
Now for
By Theorem 2,
zl<l,
E s Zn }
n
n= 0
q(z)s(z).
p(z)s(z).
Similarly V(z)
E c z
n= 0
n
v(z)/v(z)
c(z)
n
Therefore
r. b z n
n--O n
b(z)
(Nv) _C (Nv).
Thus
The proof of (ii) follows by Theorem 2 and Corollary 2.
A semigroup with order relation < is an ordered semigroup provided (i) a<b
and b<c implies a<c, and (ii) a<b implies ac<bc for all c.
We now have the
following theorem.
Theorem 6.
binary operation,
Proposition 4.
If
Qn
Izl<l.
as the order relation and
*
as the
is an ordered abelian semigroup.
Let.
Np
P-I 0 and qn Pn-i + Pn
(Np) (Nq).
Define
E
is eventually non-zero, then
Proof.
for
..
"strictly -weaker than
With
First note that N
q
Then by Corollary 2,
E
(Np)
for n
=> 0.
Moreover, it follows that q(z)-(l+z)p(z)
(Nq).
ORDERED SET OF NORLUND MEANS
Proposition 5.
P
> 0. Then N
P (i)
p
(n)
(l+z)
(z)
.
p(1)(z)
Let
Proof.
There exists infinite chains of Nrlund methods from
e
n-1
I
n= 0
p
(n)e
(n-1) k
7.
for n > i.
where
_(i)
{p(1)}
n
Pn
.
> 0 for n > 0, and
Define
k--O
Then N
pn(1)zn,
Pk
z
for n
>= 2.
Moreover by Proposition 4 we have
(Np(1)) (Np(2))
(Np(n))
The next theorem is a different version of Lemma 3.
Theorem 7.
If
Nq
e
’
n-lim Pn
P
+ 0,
and r
p’q, then
(Np)_. (Nr).
By Lemma i it suffices to show there exists some M > 0, independent
Proof.
of k, such that
lkl
Since N
q
e
,
[n_k/n
7.
n--k
qn_k/nl
< M.
there exists some H > 0 such that
< H
for all n > k and for all k > 0.
for all n > k.
Then by Lemma i, we have
Thus the result follows.
The author is indebted to the referee whose suggestions improved the
exposition of these results.
363
J. DEFRANZA
364
REFERENCES
i.
J. A. Frldy, Absolute Summabillty Matrices That Are Stronger Than the
Identity Mapping, P_roc. Amer. Math. Soc., 47, 1(1975), 112-118.
2.
C. Goffman and H. V. Huneke, The Ordered Set of Norlund Means, Math Z.,
iI1970),
i-8.
3.
G. H. Hardy, Divergent Series, Oxford University Press (1949).
4.
K. Knopp and G. G. Lorentz,
.2(1949), 10-16.
5.
B. Kwee, Some Theorems on Norlund Summability, Proc. London Math. Soc. (3),
14(1964), 353-368.
6.
R. E. Powell and S. M. Shah, Summabillty Theory and Applications, Van Nostrand
Reinhold Co., (1972).
7.
B. Thorpe, On Inclusion Theorems and Consistency of Real Regular Norlund
Methods of S,Imma_billty, J.:. London .Math. Soc.. (2), (1972), 519-525.
Beltrge
zur absoluten Limltlerung, Arch.
Math.,