C OMPOSITIO M ATHEMATICA
B. K UTTNER
A tauberian theorem for Cesàro summability
Compositio Mathematica, tome 35, n
o1 (1977), p. 49-56
<http://www.numdam.org/item?id=CM_1977__35_1_49_0>
© Foundation Compositio Mathematica, 1977, tous droits réservés.
L’accès aux archives de la revue « Compositio Mathematica » (http:
//http://www.compositio.nl/) implique l’accord avec les conditions géné- rales d’utilisation (http://www.numdam.org/conditions). Toute utilisation commerciale ou impression systématique est constitutive d’une infrac- tion pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
49
A TAUBERIAN THEOREM FOR
CESÀRO
SUMMABILITYB. Kuttner Noordhoff International Publishing
Printed in the Netherlands
1
The "standard" Tauberian theorems
applicable
to Cesàro sum-mability (C, a)
continue to hold when(C, a)
isreplaced by
Abelsummability, though
theproof
for Abelsummability
is often harder.It is therefore of some interest to obtain a Tauberian theorem for
summability (C, a)
which becomes false when(C, a)
isreplaced by
Abel
summability.
Such theorems have beengiven
in[1], [2];
in thepresent
paper wegive
such a result with a Tauberian condition of asomewhat different nature.
We consider a series
(in general,
ofcomplex numbers),
and will write
throughout
THEOREM 1: Let a > 0.
Suppose
thatand that
(1) is
bounded(C, a).
Then(1)
converges.We note that we need assume
only
the(C, a)
boundedness of(1),
and not its
(C, a) summability.
1 obtained this result in the course of certain
investigations
intoNôrlund
summability.
It is familiar that the Nôrlund transformation(N, an )
isregular
if andonly
if(2)
holds and50
Thus
(2)
is relevant.However,
the result appears to be of someinterest for its own
sake,
and thus we shall not be concerned with Nôrlundsummability
in the present paper.This theorem becomes false when
(C, a)
isreplaced by
Abelsummability,
even if we assume Abelsummability (instead
ofonly boundedness),
and even if weimpose
the additional restriction(3).
Inother
words,
we have thefollowing
result.THEOREM 2: There is a series
satisfying (2)
and(3)
which is Abel summable but not convergent.The results for Abel
summability
become somewhat different if werestrict ourselves to real series.
THEOREM 3:
Suppose
that an is real.Suppose
that(2)
and(3) hold,
and that
(1)
is Abel bounded. Then(1)
converges. This result becomesfalse
when(3)
isomitted,
evenif
wereplace
Abel boundednessby
Abelsummability.
We let
Sn
denote the(C, a)
sum of(1);
that is to say,where,
as isusual, Aa
denotes the binomial coefficientFor any
positive integer
m, letSm
denote theoperator
definedby
if a is a
positive integer,
let8:
denote the result ofapplying
theoperator 8m a
times. With thisnotation,
we have thefollowing
lemma.LEMMA: Let a be a
positive integer.
Thenand where
PROOF: The
proof
isby
induction on a. The result is evident whena = 1, since
Now assume the result true f or a
(where a ? 1),
and prove it true fora + 1. We have
Here we
adopt
the convention thatKv m
is to be taken to mean 0 whenv a or v > am. Thus
(4)
holds with areplaced by a
+1,
and withThis
gives
us(4)
and(5).
In order to prove(6),
we needonly
observethat, by (4)
the sum on the left of(6)
isequal
to the value ofS03B1mS03B1n
inthe
special
case inwhich sk
= 1 for all k.But,
in thisspecial
case,S"
=An,
and(6)
followseasily.
3
We can now prove Theorem 1. In view of the well known result
that,
if a’> a> - 1,
any series bounded(C, a)
is also bounded(C, a’),
there is no loss of
generality
insupposing
that a is aninteger.
We willmake this
assumption throughout
in what follows.We note that it follows at once from
(2)
that there is a constantc >
0,
which will bekept
fixedthroughout,
suchthat,
for all suffi-ciently large
n,52
It follows that for
sufficiently large n
and all m > n,It is
enough
to provethat ISnl
isbounded,
for the convergence(indeed, absolute)
of(1)
will then follow from(2).
So we suppose thatISnl
isunbounded,
and show that this leads to a contradiction. Since|Sn| |
isunbounded,
it follows from(8)
thatas n ---> oo.
Now,
for each n definek(n)
as the least value of k > n such thatWe will show
that,
for allsufficiently large
n,k(n )
exists(that
is tosay, there is some k > n
satisfying (10)) and, further,
thatas n - 00. This is
equivalent
toshowing that,
if 5 > 0 isgiven, then,
for allsufliciently large
n, there is some k with n k(1
+5)n satisfying (10).
To provethis,
suppose the assertionfalse;
thusfor all k with n k ~
(1
+ 03B4) n. Now letand define m =
m(n) by
m =[n8Ia]. Then, by
the lemmatogether
with
(12)
and(8),
as n --->~
(8 being fixed). But,
since(1)
is bounded(C, a),
there is aconstant A such that
for all n a
1,
m ? 1. In view of(9),
this contradicts(13)
whenever n issufficiently large.
Now choose any
sufficiently large
fixedinteger
no, and define{nr}
inductively by
We note
that, by (8)
wehave,
for r ~1,
[Here
andelsewhere,
in order to avoidcomplicated sufhxes,
we writes (n )
inplace of sn
whenever n isreplaced by
a morecomplicated expression].
In view of the definition of nr, it follows at once thatand hence that
Now let p be a fixed
positive integer
so chosen thatThen, by (7)
and(14)
Applying
a similar argument with noreplaced by
nP, we find thatand so on.
Thus,
for allpositive integers v,
Now it follows from
(11)
thatas v - 00, so that
54
Thus, given
any q > 0 we havefor all
sufficiently large
v. But the(C, a)
boundedness of(1) implies
that there is a constant B such
that,
for allsufficiently large n
Thus,
for allsufficiently large v,
If we choose Ty so that
this contradicts
(15)
whenever v issufficiently large.
Theproof
ofTheorem 1 is thus
completed.
In order to prove Theorem
2,
let(1)
be chosen sothat,
for|z| 1,
It is
immediately
evident that(1)
is Abel summable to 0.By (5.1.9)
of[3]
wehave,
with the notation for theLaguerre polynomials
used in[3],
It now follows from
(5.6.1)
of[3]
thatWe deduce from
(16)
with the aid of Theorem 8.22.7 of[3]
andStirling’s
formulathat,
forlarge
n,where À denotes a non-zero constant
(which
is different at eachoccurrence).
We deduce from(17)
thatand it follows
easily
from(17)
and(18)
that(2)
and(3)
hold. Thiscompletes
theproof
of Theorem 2.The first
part
of Theorem 3 istrivial,
but it is stated for the sake ofcompleteness. Suppose
thehypothesis
of the theorem satisfied.Then,
as in the case of Theorem
1,
it isenough
to provethat sn
is bounded.Supposing
thisfalse,
it follows that(9)
holds.But, since an
isreal, (3) implies that sn
isultimately
of constantsign.
Henceeither sn
z + 00 asn ---> oo or sn ~ 2014 ~ as n - or. This
implies
thattends to either +00 or - 00 as x -->
1-,
in contradiction to the assump- tionthat 03A6 (x)
is bounded.For the second
part
of Theorem3,
let 8 be a fixedpositive
numberwith
el)
2. DefineThus
It is clear that
(2)
holds. AlsoNow if
then ~~ n=0 bn
converges and is thuscertainly
Abel summable.Hence,
as x -->
1-,
where c is a constant, and where
With the notation for the theta functions
used,
forexample,
in[4],
wecan write
(19)
as56
By
Jacobi’sproduct
for the theta function(see,
e.g.,[4], §§21.3
and21.42),
we haveNow for
0 x 1,
Since e03B4 2,
the first factor lies between ±1;
the secondclearly
liesbetween 0 and 1. Thus
Since G - 0 as x -
1-,
it is clear thatli(x)
- 0 as x - 1-. Thus(1)
is Abel summable. Since(1)
isclearly
not convergent, the theorem isproved.
REFERENCES
[1] P. L. BUTZER and H. G. NEUHEUSER: Auf Cesàro-Limitierungsverfahren ein- geschränkte Tauberbedingungen. Monatsh. für Math. 69 (1965) 1-17.
[2] H. R. PITT: A note on Tauberian conditions for Abel and Cesàro summability.
Proc. American Math. Soc. 6 (1955) 616-619.
[3] G. SZEGÖ: Orthogonal Polynomials. American Math. Soc. Colloquium publi- cations, 23 (New York, 1939).
[4] E. J. WHITTAKER and G. N. WATSON: Modem Analysis (fourth edition) Cambridge, 1927).
(Oblatum 10-IX-1975) The University of Birmingham Birmingham, B.15 2TT England