both divergent,

A TAUBERIAN THEOREM FOR STRETCHINGSDAVID F. DAWSONR. C. Buck fl] has shown that if a regular matrix sums every subsequence of a

sequence x, then x is convergent. I. J. Maddox [4] improved Buck's theorem byshowing that if a non-Schur matrix sums every subsequence of a sequence x, then xis convergent. Actually Maddox proved a stronger result: If x is divergent and Asums every subsequence of x, then A is a Schur matrix, i.e.,

(1) apq -> cq as p -*• oo, q ^ 1,00(2) £ \cq\ converges,

5 = 1

00(3) £ |aM-c,|-»Oasj>-> oo.

g = l

It should be remarked that a matrix is a Schur matrix if and only if it sums everybounded sequence.In [3] we worked with stretchings of sequences, and obtained a result similar toBuck's theorem. The purpose of the present paper is to give a result on stretchingswhich is analogous to Maddox's theorem. Thus we have the

THEOREM. // x is a divergent complex sequence and A is a complex matrix whichsums every stretching of x, then there exists N such that

(i) apq -> cq as p -» oo, q > N,00(ii) £ cq converges,

q=N+loo(iii) £ (apq-cq) -> 0 as p -» oo.q = N+lIf D denotes the set of all matrices A such that for some N(A), (i), (ii), and (iii)

hold for N = N(A), then our theorem can be formulated as a Tauberian theorem:If a non-D matrix sums every stretching of a sequence x, then x is convergent.We begin with a lemma which may be of independent interest.

LEMMA. // x is a sequence and A is a matrix with each

Z apq

q = lconvergent, then there exist a stretching x* of x and a row finite matrix A* such that if

y is any stretching of x*, then Ay exists and Ay and A*y are either both convergent or

Received 18 October, 1974.[J. LONDON MATH. SOC. (2), 13 (1976), 27-33]28both divergent,DAVID F. DAWSON

00 \ 00E aPq- E

l lp=l

is a null sequence, and, for each q, {apq—0*Jp°= i is a null sequence.Proof. Let«0 = 1, and let

Take nx < nQ such that

Mp=

L-i 19q=r

-l

for all r and 5 such that nt < r < s. In general, take nt> nt_i such that, for p =1,2, ...,*,

for all r and J such that n, < r < s. Let x* be the stretching of x induced (in the senseof [3; p. 455]) by {np}™=0. Define A* as follows: a*q = apq ifq < np, a*q = 0 otherwise.Let y be any stretching of x* and suppose {mp}p°=Q, with m0 = 1 and mp ^ np, p =1, 2, 3, ..., is an increasing sequence of integers which determines j; as a stretching of x.If p is a positive integer, then clearly

converges, andY a v - f a"00E «

q=np

mn-i — l

E aP

Zlv I

002J apq9 = 100S|Capq00< E

i = p

nj+i-1

2J flP99="i

Also

Thus ylj' and ^4*); are either both convergent or both divergent, and00 00 \ 00ZV"1 ni

pq i—i

1

!=1 9=1

is a null sequence. Clearly from the construction of A*, for eachg, {apq — fl*Jp°=i is anull sequence. This completes the proof.A TAUBERIAN THEOREM FOR STRETCHINGS 29Proof of theorem. Since x is divergent, there exists N such that xN # xN+1.Suppose n> N. Let x' and x" be stretchings of x defined as follows: xp' =xp" = xp

for p = 1,2, ...,JV-1, xp' = xp" = Xjy for p = AT, ...,n-l, xn' = xN+u xn" = xN,

XP = V = xN-n+P for P>n. Thus {apn}*=1 = (xJV+1-%)~1 A(x'-x") which

converges. Thus (i) holds.It was shown in [2] that each

004 = 1converges. Thus we could replace A and x with A* and (divergent) x*, respectively,of the lemma. Instead we will assume without loss of generality that A is row finite,and will continue to use x to denote a divergent sequence every stretching of whichis summed by A.Using the divergence of x and the Cauchy condition, we can find 8 > 0 and anincreasing sequence {«p}"=0 of integers with n0 = 1 such that if p is a positive integer,then \xn—xm\ > 2<5 for some pair n, m for which «p_t < n < m < np. Let z be thestretching of x determined by {«p}p°=o- Letj> = z—x. Then^ = 0 and y is divergent.Furthermore, A sums every stretching of y, since each stretching of y is the differenceof two stretchings of x. Let v be the stretching of y defined as follows: vp = yt forp = 1, 2, ..., N, and vN+p = yp for/? = 1, 2, 3, .... Then since A sums every stretchingof v, we see that the matrix B = (bpq), were bpq = ap>N+q, sums every stretching ofy.Now form the following two stretchings of y:

and let u be the difference of these stretchings, i.e.,« = {0,Aj>1>0,Ay2,0,A)>3,...},where Ayp = yp—yp+i. Let &€ = \impbpq, q = l,2,3 If we can show that00£*•

«=iconverges and «,limp £ 0M-*€) = 0,

« = i

then (ii) and (iii) will hold for the matrix A, and our proof will be complete, since wehave already shown that (i) holds. We will establish these properties of B by using thefact that B sums every stretching of the sequence u.Since B is row finite, there exists an increasing sequence {?p}p°= i of positive integerssuch that bpq =Oifq > tp.Suppose