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[HESIS

 

This is to certify that the

thesis entitled

On a Generalization of the

Lototsky Summability Method
presented by

Herbert B. Skerry

has been accepted towards fulfillment
of the requirements for

flLQL—degree inmflic S

fi7WI

T V Majorv! o ssor

Date Z/Z 0// 6 7

0-169

ABSTRACT
ON A GENERALIZATION .OF THE LOTOTSKY SUMMABILITY'METHOD

by Herbert B. Skerry

A. Jakimovski defined his (F,dn) summability method
as a generalization of the Lototsky method. Subsequently,
G. Smith generalized the (F,dn) method to the (f,dn) method,
and later generalized this to the (f,dn,zl) method. Let
f be analytic at the origin and suppose 21 is a point
in its disc of convergence. Let Coo = 1, c0k = 0 for
k > 0, and

n f(zzl) + dV
f(zfl + dv

 

CI)
2: 2k for n51.

H=l

Then the (f,dn,zl) method is defined by the matrix (an)'
The (f,dn), (F,dn), and Lototsky methods are, respectively,
the methods (f,dn,1), (z,dn,l), and (z,n-1,l).

Chapter 1 is concerned with the generalization (f,dn,zl)*

of (f,dn); it is defined by the matrix (c* where

nk)’

cfio = 1, czk = O for k > O, and

n f(z) + dv co * k

W = Z c z for n,: 1.
1 f(sz+ dV k=0 nk

 

Various properties of this method, which behaves quite
differently from Qf,dn,zll are extracted.

Sufficient conditions for the regularity of (f,dn) have
been given by Smith, but all require that f .have real,
non-negative Taylor coefficients. Chapter 2 presents some

sufficiency conditions for the regularity of (f,dn,zl)

Herbert B. Skerry

under different restrictions on f.

Chapter 3 concerns itself with the coincidence of the
(f,dn,zl) method with various other methods, including the
Sonnenschein, Hausdorff, quasi-Hausdorff, Norlund, and
Riesz methods, and the modified quasi—Hausdorff method of
M. S. Ramanujan.

The last chapter deals with questions of inclusion
between two (f,dn,zl) methods and between (f,dn,zl) and
several other methods, e.g., the Riesz, Abel, Y, and (E,p)

methods.

ON A GENERALIZATION OF THE-LOTOTSKY SUMMABILITY METHOD

By

<3

Herbert BE Skerry

A THESIS

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Department of Mathematics

1967

ACKNOWLEDGMENT

The author is indebted to Professor W. T.-Sledd
for his patience and encouragement during the prepara-

tion of the thesis.

ii

TABLE OF CONTENTS

INTRODUCTION . . . . . . . . . . . .
Chapter
I. THE METHOD (f,dn.zl)* . . . . . . .

II. REGULARITY CONDITIONS FOR (f,dn,zl)
III. COINCIDENCE OF METHODS . . . . . .
IV. INCLUSION RELATIONS . . . . . . . .

BIBLIOGRAPHY . . . . . .

iii

INTRODUCTION

A sequence summability method S is a way of associ—
ating a unique number with each of a class of sequences.
The largest Class of sequences for which S performs this
function is called the summability field of S. If S
associates L with x = {xn] we say S sums x to L.

If S sums every convergent sequence, then S is £227
servative, and if, moreover, S sums each convergent se-
quence to its limit, then .S is regular. We will be con—
cerned only with complex sequences.

Every complex matrix having infinitely many rows and
columns defines a sequence summability method. If A is
such a matrix, then for suitable sequences x the A-trans-
form, t = Ax; determined by matrix multiplication is a
sequence; if it converges we say A sums: x to lim' tn.
It is clear that the summability field of A is the class
of sequences whose A-transform exists and converges.

Necessary and sufficient conditions for such a matrix

A = (ank) to be conservative are well-known ([8], p. 43).
They are
(0.1) sup 2 lankl < co ,
n k
(0.2) 3 lim 2 a = C exists,
n k nk
(0.3) 13m ank = ak ex1sts for each R.

2
A is regular if and only if it is conservative with C = 1
and ak = 0 for each k ([8], p. 43).
Given the methods 5 and S‘, we will say S is a;
least as strong as 81 if the summability field of S’
is a subset of that of S. Under these conditions the

two methods are consistent if each element of the smaller

 

summability field is summed to the same number by both
methods.

A number of the classical sequence summability methods
are matrix methods. The best-known among them are probably
the Cesaro methods (see [8]). In [10] Lototsky defined a
method which Agnew considers to rival the Cesaro methods
in importance (see [4]). This method was subsequently
generalized by Jakimovski [9] to his (F,dn) method, and
this in turn was generalized by Smith [15] to the (f,dn)
method. Finally, in a paper to appear [16], Smith general-
ized (f,dn) to (f,dn,zl). We will concern ourselves with
the last three methods.

The following definitions and conventions will be used.

Definition 0.4: Let f be holomorphic at the origin
and let [dn)io be a sequence of complex numbers with
dn # -f(zl), where 21 is in the disc of convergence of f.
Let

(M) 71r(f(z)+dk)= z: pnkz , nel.

Then the method (f,dn,zl) is defined by the matrix C = (cnk),

where

 

k
1, k:0 pnkzl
(0.6) c k = and C k = , r1.i 1.
° 0"”0 n F<f<z>+>
1 1 dk

In terms of the above definition, the Lototsky method
iS'flmnmethod (z, n-1, 1) and (F’dn) is the.method (z,dn,1).
The (f,dn) method is defined to be (f,dn,1).

We will assume throughout that 21 # 0. For conveni-

ence we will use Jakimovski's notation:

we:

(f(z) + dk) = (f(z) + an): .

The definition of (f,dn,zl) insures that the condition
(0.2) holds with Q = 1, so we need only consider the re-
maining two conditions in any questions of regularity or
conservativity that arise in connection with (f,dn,zl)

and its various special cases.

CHAPTER I
THE METHOD (f.dn:zi)*

We have remarked that Smith first defined and in-
vestigated the method (f,dn) and only later generalized it
to (f,dn,z1). The question arises as to whether some
other generalization is not equally as natural and as use-

ful. In this light, consider the

a)
Definition 1.1: Let f,{dn}1 , and 21 be as in

 

definition 0.4. Then the method (f,dn,zl)* is given by

 

the matrix C* = (chk)’ where
1 k = 0 p
' k _
(1.2) C* = and c* = n . , n‘il
0k 0,k > 0 nk (f(zl)+dn).
k
It seems clear that Smith inserted the factor 21 in
the expression for an to preserve the property
E cnk = 1 which obtains for (f,dn). We have dropped this

factor in the above definition and have accordingly lost
this convenient property, but it is not obvious that any-
thing more than convenience has been lost.

The following necessity conditions for regularity are
restatements of theorems in [15] with f(zl) substituted
for f(l). -For the sake of completeness we include the
proofs, which are essentially Smith's. We will use the
notation

ien

(1.3) Re f(z) = u, Im f(z) = v, dn = xn + 1yn = pne .

f(zl) = a + ib.
4

5
Theorem 1.4: Let dn # -f(0) Then a necessary
condition that (f,dn,zl)* be regular is that there exists
a strictly increasing sequence [nk] of positive integers

such that
f(O) + dn 2
k

f(zl) + dnk

 

Z (1 -
k-1

) = 00.

 

Proof: By setting 2 = 0 in (0.5) and using (1.2),

we see that

* n f(0) + (1k
Cno = ¥ f(zl)‘+ dk '

Since the method is regular, lim CEO = 0. It follows
n

 

immediately that

 

 

. * 2 _oof(0)+dk 2 00 f(o)+d.k 2
13m lcnol - T; m = [Til-[1' f 21 +dk H

(1 - a = 0. where a < 1.

0° >
T,
1 k

Clearly, an infinite number of '5 must be positive in

ak
cm
order that w(1 - ak) = 0; let {an } be the sequence of
1 k
00 00
positive ak's. Then v (1 — an ) = 0, whence 2 an = a)
k=1 k k=1 k

as claimed.

Corollary 1.5: Let dn # —f(0) and let f(O) and
f(zl) be real. Then a necessary condition for the regu-
larity of (f,dn,zl)* is the existence of a strictly in-
creasing sequence {nk} of natural numbers satisfying

f(O) + f(zl) + 2xn
k

2 2 = i a).
kzl ‘f(zl) + an l

6
Proof: Since f(0) and f(zl) are real we have

f(O) + dn 2 f(zl) + f(O) + 2x
k

 

 

(1 6) 1- = [f(z ) - f(0)1
f(Z1) + dnk 1 [f(21)+ dnkIZ

The above theorem then says that

a) f(zl) + f(O) + 2x

[f(zi) - f(0)] Z 2 = a>
k-1 [f(zl) + d I

 

“k

so the result follows.

Corollary 1.7: Let dn # —f(0). Then a necessary

 

condition for the regularity of (f,dn,zl)* is that
[f(zl) + dnl > [f(O) + dn| for infinitely many n. Fur-
thermore, if f(O) and f(zl) are real, a necessary con—

dition is that

xn > --%[f(0) + f(zl)] > -f(zl) when f(zl) > f(O)

and

xn < —-%[f(0) + f(zl)] < -f(zl) when f(zl) < f(0)

for infinitely many n.

Proof: In the proof of theorem 1.4 it was observed

 

 

 

 

f(0) + dn 2
that 0 < 1 - f(zl) + dn < 1 for Infinitely many n,
f(O) + dn ‘2
i.e., 0 < f(ZI) + an < 1 for infinitely many n. The

 

 

first statement of the corollary follows. This last in-
equality together with (1.6) says that

f(zl) + f(0) + 2xn > 0 when f(zl) > f(O), i.e.,

7

xn > — %[f(0) + f(zl)] > -f(zl).

The remaining statement of the corollary is proved in a

similar manner.

Corollary 1.8: If dn # -f(0), a necessary condition

 

for the regularity of (f,dn,zl)* is that f(zl) # f(0).

Proof: This is an immediate consequence of Theorem 1.4.

 

Theorem 1.9: If (f,ddz1)* is regular, then

“3 1
§_ [f(0)+ dnl = 00'

 

Egggfi: If f(O) + dn = 0 for some n, the theorem
follows trivially. Hence, suppose f(O? + dn # 0. Let
{nk} be the sequence of positive integers whose existence
is assured by Corollary 1.7; then ‘f(zl)+dnk| > |f(0)+dnk|.

It follows that

 

 

 

 

 

 

f(0) + d 2 |f(zl) + dn [2 - |f(0) + d [2
1 k = ._o k “k
f(zi) + dnk |f(zi) + dnklz
2 (|f(zl) + dnk|-|f(0) + dnk|)(|f(zl) + dnk|+|f(0) + dnkl)
[f(zl) + dnk"2
.: |(f(zi) + dEk)-(f(0) + dnk)|(|f(zi)§+dnkl+lf(0) + dnkl)
|f(21)+ an |
k
[f(zl) + dn |+|f(0) + d | z _
= |f<zl> - f(o)| k 2 k . 2'f(1)f(°)'
[f(zl) + d l |f(zl)+d I

2|f(zl) - f(o)| . oo 1
[f(Ol + dnkT . Thus if kil Tf(07»+ dnkF converges,

<

 

 

2
OD f(O) + dn

so also does E1 (1- f(2[)+ dn ) in Violation of Theorem

 

 

 

1.4.
Corollary 1.10: If dn # -f(0), then (f,dn,zl)* is
co
regular only if 2 JL-= a) for arbitrary N.
N Pn
Corollary 1.11: If dn # -f(0), then (f,dn,zl)* is
00 1
regular only if 2 = 00, N arbitrary.
NTf<Zly+ dnl

For reference purposes we include the following three
lemmas, which occur in essence in [15] either as results

or as portions of proofs,

 

Lemma 1.12: Let dn 3.0 and let the power series
expansion of f about the origin have real, non-negative
coefficients. Then (f’dn) is regular if and only if
00 1

{3 f(1)+ dn

 

=03.

Proof: This is Lemma 2.2 in [15].

Lemma 1.13: In the notation of (1.3),

 

 

f(z) + dnl exp {’pn[(u-a)cos on + (v-b)sin 9n]

f(21)+ dnl :. [f(zl) + dntz

u2+v2-a2—b2}

+ 2|f(zl) + dnjz

9

 

 

 

Re([f(E)-f(zl)]dn) u2 + V2 _ a2 _ b2
= exp 2 + 2
[f(zl) + an] -2|f(zl) + an
f(z) + dnjz
— 1 and use the fact

 

that 1 + x i.ex to get the result by a short calculation.

Lemma 1.14: Let 2 be fixed. In the notation of

 

(1.3), let

2 ——-= d), e -——> e, pn -—> oo.

Then
OOf(z) + d
F n

1 f(zl)+“dn =

 

if (u-a)cos 9+(v-b)sin 6 =

Re([f(z§ — (215]eie) < 0

and

(D f(z) + dn

w = a) if u-a cos 9+ v-b sin 9 =

 

Re([f(z) - 12151eie) > 0.

Proof: Let

 

 

Q = pn[(u-a)cos 6n+(v-b)sin 9n] + u2 + v2 _ a2 _ b2
n 2 . 2
[f(z1)+dnl 2]f(z1)+dnl
and Q = (u-a)cos 6 + (v-b)sin 9. It is clear that

pn n ——> Q. If Q < 0, then there is a K > 0 such that

for all large n, so -§—-> Qn for large n.
n

-K > ann

f(z) + dn
By Lemma 1.13, f(21)+ anI: exp [Qn] < exp [_

 

K
9n}

for

10

large n. Then

 

co f(z) + (1n CD 1
v .2 exp {-K 2 -—-} = 0.
N f(z1)+ dn N pn

 

 

If Q > O we may use essentially the same argument, inter-

changing the roles of z and 21 in Lemma 1.13, to show

 

 

 

 

 

ODJf(Zl) + dnl oo f(z) + dn
that v f(z) + d‘ = 0, whence w f(z )+ d oo.
N nl 1 1 n

,ie
Theorem 1.15: Let Tn = Im[(f(215 — (1))e n] and
ien
let dn — pne g -f(0). Suppose
(i) a + ib = f(zl) f f(1) = o + 16,

(ii) an is bounded away from 9* + 23w for large n,
and either

(iii) lim inf Tn > 0
or

(iv) lim sup Tn < 0.

Then (f,dnzl)* is not regular.

Proof: Suppose the contrary. Then, in the notation

 

of (1.2),
n f(1) + d
2 c* = w m -—> 1 as n ——¢ oo.
k nk 1 f(zl)+ dm

As a consequence,

 

 

 

 

 

f(l) + dm f(l) — f(zl);
" 1 -—> 0 as m'-—> oo,
_f(zl)+ dm f(zl)+ dm I
50 pm -—> 00. From (ii) we may suppose that 9* + 5.: 9n":

6* + 2H - O for large n. Since pm'-> 00, we may use this

same branch of the argument for Wm = arg[f(1) + dm] and

11

 

' f(1)+ dm
Tm = arg[f(zl) + dm] for large m. Let 3m = arg f(21)+ dm

f(l) + am
The factors f(z1)+ dm -—> 1, so we may use the prinCIpal

 

branch of the argument for Tm = Tm - YA for large m.

00 f(l) + d

The conver ence of N
g 1 f(zl)+ dm

 

implies the convergence

00
of Z ¢h. For a given large m, if neither a + xm nor
1

a + xm is zero, we have, by choosing the appropriate branch

of the inverse tangent function for each of Wm and TA,

6 + Y B + Y
_ -1 m I: —1 m
that Tm — tan E-Iji— and Wm tan 3—:j;—. Then
m m
B+ym_b+ym
_ I_C1+X a+x =
tan ¢m tan (Tm Wm) — m m

 

B+ym b+ym

1+*._
a+X a+x
m

(a + xm)(5 + ym> - (a + xm><b + ym>
(a + mea + me+ (6 5,71% + ym)= Qm ' so

 

(3+X )(B‘ty ) '- (a+X’)(b+y )
-1 _1 h ’
("'16) ("m = Tan Qm = Ta“ Fa+x:)(a+x:) + “3+1!er (Ma/:3

 

where Tan”1 denotes the principal branch.
,It can be easily shown by routine calculation that
(1.16) is still valid if d + xm = 0 or a + xm = 0. Now

write

(a+pmcosem)(6+pmsin6m)-(a+pmcosem)(b+pmsin9m)

p Q I O
(a+pmcos9m)(a+pmcosem)+(3+pmSin9m)(b+pmSIn9m)

mm Pm

 

(aB—ab)+pm(B-b)cosem+pm(a—a)sinem

=prr . _2
"(aa+bB)+pm(a+a)cosem+pm(b+a)Sin9m+pm

 

12

o(1)+(B-b)cos 6m + (a-a)sin em 0(1) + Tm
= 0(I7 + 1 = o(1) + 1

 

 

If (iii) holds, then for all large m we have mem': O > 0.

so, in particular, Qm > 0. Then (1.16) implies Tm > 0

and thus ¢m"__> 0+. But now, since Qm -—¢ 0+ from (1.16),

¢m _1 Tan Qm >
:WmTan Qm='T—'PQO-s>°
Pm m

for large m, so pgl 2.5-1¢m. Then the convergence of

Z ¢m implies that of 2 p;1 in violation of Corollary 1.10.
If, on the other hand, (iv) holds, the argument proceeds

in a similar manner.

We remark here that the above proof depends only on
the conservativity condition (0.2) with C # O and on
Corollary 1.10. Inspection of the proofs of this corollary
and its antecedents shows that the only regularity condition

used is lim C20 = 0. We may thus state
n

Corollary 1.17: Under the hypotheses of Theorem 1.15,

the (f,dnzq)*-method cannot satisfy both lim Cho = 0 and
n

The following corollary appeared in the proof of

Theorem 1.15.

Corollary 1.18: Let lim 2 czk = g r 0. Then
' n

pn -—> a).
We will now prove a result which is not a direct cor-

ollary of Theorem 1.15, but which is closely allied to it.

 

u
'o
m
5
Wk
I
m
’2‘
m
+
p.
o‘
ll

Theorem 1.19: Let dn

f(z1) # f(1) = a + iB, and en -—9 6. Then (f,dn,zl) is

not regular.

Proof: By Corollaries 1.10 and 1.18 the result follows

00
immediately unless 2 .1; : a) for arbitrary N and
n

pn‘—-> a). Thus suppose these conditions met. Then
Lemma 1.14 gives

(1.20) (a - a) cos 9 + (B - b) sin 6 = 0 ,

assuming (f,dn,zl)* is regular. If also

(1.21) (B - b) cos 9 + (a — a) sin 9 = 0

and e # i %-+ 22w, then tan 6 = —-§—E€S . But from (1.20),
= _ a - a B - b : a - a _ 2 = _ _ 2
tan 9- 5—:TE , so a _ a B—:—b' and (B b) (a a) .

It follows that o a and B = b, violating the hypothesis.

On the other hand, if e = .t 12; + 2% and (1.21) holds,
then it follows from (1.20) and 1.21) together that a = a
and B = b, again violating the hypotheses. Hence, the

assumption of regularity implies that (1.21) cannot be true,

so lim T
n n

1%m[(B - b)cos 9n + (a — a) sin 9n] =
= (B - b) cos 6 + (a - a) sin 9. is either
positive or negative. Theorem 1.15 now gives a contradiction.
This last theorem is in marked contrast to Lemma 1.12,
so (f,dn,zl)*'behaves quite differently from (f,dn,zl) if
f(zi) # f(l)-

We have seen that (1.20) is a consequence of regularity

if dn # —f(0), and, in fact, it is a consequence of the

14

conservativity conditions lim C* = 0 and lim 2 c*

= C # 0. If (1.20) is interpreted as a dot product of

vectors, it says

Corollary 1.22: Let dn # —f(0) and suppose

lim c* = o and lim 2 c* = g e 0. Then f(zl) and f(l)
n no n k nk
lie on a normal to the ray arg z = 6.

In order to consider the relationship between (f,dn,zl)*

and (f,dn,zz)*, let yn = (f(zl) + dn)1 and on =

(f(zz) + dn)1, and suppose the elements of the matrices

corresponding to the two methods are, respectively, Chk

and b; From (1.2),

k.

y
*_n*
(1.21) bnk - 3; an°
From this it is clear that if a sequence is (f,dn,zl)*-

summable to s, it is (f,dn,zz)*-summable if and only if

L = l m

i exists; in that event, it is summable to Ls.
n .

50';

If L f 0, summability (f,dn,22)* of a sequence to 3 im-
plies summability (f,dn.zl)* to L713, so the summability
fields of the two methods are the same.

The following definition is Agnew's [3].

Definition 1.22: Given the sequence-to-sequence
transforms S and T and the sequence x, let Sx = {Sn}
and Tx = {Tn}. Then the transforms S and T are

equiconvergent for the class C of sequences if

15

lim(Sn - Tn) = 0 for every x e C.
* * _
Let x be a sequence. Then i (cnk bnk)xk
yn * .
(1 - 5—) Z ansk, so if L = 1 the two methods are equi-
n k

convergent on the class of sequences for which the
(f,dn,zl)* transform is bounded whether or not these
sequences are summable by either of the methods. In
particular, if (f,dn,zl)* satisfies (0.1), then the
methods are equicOnvergent on at least the space SB of

bounded sequences.

n f(zi) + dk
Theorem 1.23: Let g f(225 + dk -¢ L. Then a

 

sequence which is (f,dn,zl)*-summable to s is (f,dn,zz)*-
summable to Ls. If L # 0, the summability fields are

the same. If L = 1, the methods are consistent on their
common summability field and are equiconvergent for all
sequences for which either transform is bounded. In par-
ticular, if either transform satisfies the conservativity
condition (0.1), then the methods are equiconvergent at

least on SB.

Corollary 1.24: A necessary condition that (f,dn,zl)*
be conservative with C # 0 (in the notation of (0.2)) is
that (f,dn) be conservative and have the same summability

field.

*-

Proof: Let the (f,dn,zl). matrix be (Chk)' Then

16

 

 

* n f(l) + dm n f(zi)+ dm -—1
em§%k-%mgnnnam-Ctmsofimqnn+am=é

Then Theorem 1.23 says that the summability fields are the
same.

Agnew [1] formulated the following

 

Definition 1.25: A sequence summation method is
multiplicative with multiplier L if every sequence con-
vergent to s is summed to Ls by the method.

It is known [8] that necessary and sufficient condi-
tions for a matrix A = (ank) to be multiplicative are
the conservativity conditions (0.1), (0.2), and (0.3) with

ak = 0. The multiplier is then Q.

Corollary 1.26: Necessary conditions that(f,dn,zl)*

 

be multiplicative with non-zero multiplier C are that

(f,dn) be regular and have the same summability field.

Proof: In the notation of the above corollary and
n f(zl)+ dm

-1
its roof, the relation lim = 0 im lies,
9 T f(l) + dm C F p

 

by Theorem 1.23, that (f,dn) is multiplicative with multi-
plier 1, i.e., it is regular, and it has the same field

of summation.

Corollaries 1.24 and 1.26 show, in effect, that for
a given f and a given sequence [dn], the entire class
of conservative (f,dn,zl)* methods for which C # 0 is
no stronger than the single method (f,dn).

With the machinery now at hand we can deal with the

17
question of regularity for a large class of (f,dn,zl)*

methods. -We need the following lemma (Theorem 4.6 in [15]).

Lemma 1.27: Let f(z) = azm, where a > 0 and m

 

is a positive integer. Let a be given with 0 < a < g»
and suppose there exist a > 0 and N > 0 such that if
i6
n

dn = pn e , then r :.en > a and pn > e for all n > N;

Then the method (f,dn) is not regular.

Theorem 1.28: Let f(z) = azm, where a > 0 and m
is a positive integer. Let a be given with 0 < a < %V
and suppose there exist 8 > 0 and -N > 0 such that if
16
_. n .
dn - pn e , then T :.en > a and pn > g for n > N.
It follows that (f,dn,zl)* is not regular or even multi-

plicative with non-zero multiplier.

Proof: Were the contrary true, Corollary 1.26 would
imply the regularity of (azm,dn), thereby violating
Lemma 1.27.

We will now pass to considerations of a different
nature.

We need the following lemma, proved by Agnew in [2]

(Lemma 3.1).

00
Lemma 1.22: For every n.i 0 let 2 |a

 

woo.
_ nv
v-o
-___ co
and let lim 2 |a | = A < oo. Then if lim a = 0
_ nv nv
.n v-o n

. OD
for each v it follows that 1im|2 anvsv I f. A.1im|s'v|.
n v=o

18
Moreover, A is the smallest such constant in the sense
that there is a bounded sequence not converging to 0 for

which the equality holds.

Corollary 1.30: Let

 

 

n “a“ + dV 1 001* I (1)
Q = w - Z c = O; ,
n 1 f(zz) + dv v=o nv
. n f(z1) + dV *
let 11m (q f(2;7_+ dv — 1)cnv = 0 for each v, and let
(1) (2) *
[3v] be bounded. Let {tn } and [tn ] be the (f,dnz1)

and (f,dn,zz)* transforms, respectively, of [sv]. Then if
_ _ 2 __
Q = lim Q it follows that lim|t( ) - t(1)| 2.0 limls I.
n n n n n v v
Moreover, Q is the smallest such constant in the sense

that there is a bounded sequence not converging to 0 for

which the equality holds.

Proof. tn — tn = § bnvsv E nv v
v-o v-o
CD * *
= 2 (b - C )s
v=o nv nv v
(2) (1) = y1'1 _ * - -
By (1.21). tn — tn ((5n 1)? CnvSv. An application

of Lemma 1.29 now gives the result.

In view of Theorem 1.23, we give a set of conditions
under which the (f,dn,z1)*-transform is bounded on SB: in
fact, the conditions are sufficient for (0.1) to hold for

the (f,dn,z1)* matrix.

19

Lemma 1.31: In the notation (1.3), let pn ——> a)

 

and Z FL-< oo. Then if f is holomorphic on the closed
pnt° pn

unit disc, the (f,dn,z1)* transform is bounded on SB’

Proof: Let the (f,dn,z1)* matrix have entries Chk’

and let {3k} € SE. Then, in view of (1.2) and (0.5),

1 (f(t) + d )!
c = , - . f n dt ,
k (£321) + dn). ZWIC tk+1

 

 

 

where C is the curve )t1 = r > 1. If we let

n f( + dkl

M (r) = sup N
n . t€C 1|f(:1y+ dkl

Mn(r)

ICnkl "_—1?_- ’ so
r

then Cauchy's estimate gives

00
,1 _
(1.32) 12 C*ksk|:0(1i [onk|_<_0(1).Mn(r) >5}...

r0(1)rMn(r)

r _ 1 a 0(1)‘ Mn(r)°

 

 

 

 

_ n f(t) + dkl n f(t)- f(zi)
But Mn”) " :23 New «511‘ " :23? 1 "' femdk
fi.?(1 + |f(g:%)+ dkT)=1), since the con-

vergence of Z -l- implies the convergence of the infi-
pn#o in

nite product. The result follows.

One class of (f,dn,z1)* methods is easily discerned.

20
Theorem 1.33: .In the notation of (1.3), let

pn—>00.9n->9, and >3 -1--=oo. Let f beholo—

pn#o pn
morphic on the closed unit disc and let C be the circle
about the origin of radius r > 1. Then (f,dn,z1)* sums
every bounded sequence to zero if

sup Re ([f(t) - (21)] eie) < 0.

teC

Proof: By (1.32), [i Chk Ski : 0(1)-Mn(r). We may

show Mn(r) -—> 0 by proceeding in a manner analogous to
that used in the proof of Lemma 1.14.
V Let f(t) = u(t) + iv(t). Define

t) = pk[(u(t)-a)cos 9k+(v(t)—b)sin 9k] +

|f(zl) ‘3ku

 

u2(t)+ v2(t) - a2 - b2

2|f(z1) + aky"

 

Now, u(t) and v(t) are bounded on C and pk -> a),

so pk0k(t) -—> (u(t) - a)cos9 + (v(t) - b)sin 9 uni-

formly on C. Since C is compact, there is a t e C

 

l

vergence, kak(tn) ——> (u(tn) - a)cos 9 + (v(tn) - b)sin 9

= Re([f(tn) - f(zm e‘ie) i :“8 Re([f(t) - f(zrhe’ie) =
€

-26.<.0, so that for k > N we have kak(tn) < -O, or

Qk(tn) < -O/pk, and N is independent of n. Then, by

21

Lemma 1.13,

)+dk|

 

 

 

 

 

 

 

 

 

:(:§7»+ dkl —-exp(Qk(tn )] < exp{- %;J for k > N.
Nf(t.)+dk nf(tn) +dk
It follows that Mn(r) q f(zl) + dk .N+1 f(z1) + dk
Nf(t )+d.k| n1
w exp[-O Z -:J, whence lim M (r) = 0.
1 f(z1) + dkl N+1pk n-Too’

To show that the hypotheses of Theorem 1.33 can be

satisfied, consider the example f(z) 3 exp{ze-1 argpz1}.
Let a = arg z1 and let t = rel¢. Simple calculations

r cos(O-a):

show that f(t) = e [cos(r sin(¢-d)) +

x
i sin(r sin(¢-a))], so u Z'Re f(t) : er COS(¢-a)

cos(r sin(¢-a)).fi er for all t e C. On the other hand,

f(z1) = e|z1i = a. Then, if 6 = 0,
sup Re([f(t) - f(z1)]e-le)= sup(u - a) i.er — elz1I < 0
teC teC

if r < [21]. Hence, we have only to Choose 1 < r < [21]
and 6 = 0 to have the above choice of f satisfy the
conditions of Theorem 1.33.

We conclude this Chapter with two examples. In view
of Theorems 1.15, 1.19, and 1.28, the question as to
whether or not there is a regular (f,dn,z1)* method for
which f(z1) # f(1) arises naturally. The first example
exhibits such a method. The second example shows that
Theorems 1.15, 1.19, and 1.28 fail if the regularity in

the conclusion is weakened to conservativity.

22
Example 1.34: The following result is due to Miracle

([12], Theorem 2.1):
]oo

Let Mn 1

be a sequence of positive numbers such

-1“ =' =_'
that 2 kn - oo. Let d2n—1 rdln and d2n LJkn.

Then (z,dn) is regular.

Let d be defined as in Miracle's theorem and suppose
k -1+ dk

where Arg denotes the principal branch. Let

Kn -—> oo monotonically. Let P - Arg P

k 'k’

Yk = Arg (1 + dk) and Yfi = Arg (-1 + dk)' ‘We may suppose

11 is so large that ¢ = Wk - Yi for every k.. From

k
the monotonicity of {kn}, it is geometrically clear that

 

 

 

 

k odd k even

|¢kl -v 0 monotonically. Moreover, d2m = —d2m-1' so

-1)] = yam-1 - 7’ and

I6
ll

2m Arg(-1 + dzm) = Arg[-(1 + d2m

W

2m Arg(1 + d2m) = Arg[—(-1 - dam)] = Arg[-(-1 + dzm- )]

1

_ ” _ =: - ' = . - = -
— Y2m-1 w. Then ¢2m Yzm 92m Yam-1 Y2m+1 ¢2m-1°

¢

= 0, and 3

2n
s
§ k 2n-1 2n 2n

It follows that San =
a)
-¢2n--¢ 0, so § ¢k - 0. As a consequence&)if ll 13 as-

sumed to be sufficiently large, we have V Pk = 1. Now,
. 1

23
(z,dn) is regular by Miracle's theorem, and Theorem 1.23

gives the regularity of (z,dn,—1)*.
Example 1.35: Let f(z) = 2. Then from (0.5) follows

'-' Z d. d. d. if n > k, and = 1.
pnk _ . 31 32 In;k pnn
L131‘32<

... ' 2
<Jn-k‘—n

Similarly, if dk is replaced by ldkl, then

n n k
+ =

{(2 ldkl) k=o anz ,

where
q = 2 d. d. ... d. if n > k, and q = 1.
nk 1:j%<j2< l 31 32 Jn-k‘ nn
...<jn_k§n
n k
<

It follows that lpnki _.an , so kgo lpnklolzl .2
n

n
2 anlzlk = ¥(|z| + [dk]). Then, letting z = 1,
k-o

<

11:1:

n
2

(1 + ldkl). From this it is easily seen that
k 0 .

Ipnkl

1 + [dkl

n n
:Z IC ! 5 w T——————T- if c is the typical element in

the (z,dn) matrix.

Now consider the (z,in2) method. For this method,

n n 2 n . f 2 n 00
z |ckli1r1+kivr1+k =1T(1+l.)<1r(1+]{-2)<oo,
k=o “ 1 |1+ik2| 1 k2 1 k2 1

so (0.1) is true, and, of course, (0.2) follows automatically

24

for any (f,dn) method. Now,

1 1 n t+iv2

1
C 17' . = ———. f
nk 2w; C tk+1 1 1+1v2 2w1 C tk+1

‘1'"? )d
1+1v2

 

 

 

n
‘n'(1+
1
where C is a circle about the origin. Since the product

m . t-l o
w (1 + , ) converges absolutely and uniformly on the
1 1+1v2

 

compact set C. it is clear that lim cnk exists, i.e.,
n

(0.3) is valid. Thus (z,in2) is conservative, although

Corollary 1.10 shows it is not regular. The product

 

00 1+1v2 CD .
-—————— = n-(l + ) is absolute convergent, hence
1- -1+iv2 1 —1+iv2

convergent, so Theorem 1.23 says (z,in2,-1)* is also con-
servative. But this method satisfies the hypotheses of
Theorems 1.15, 1.19, and 1.28. In connection with Theorem
1.28 it should be observed that although the method is

conservative, it is not multiplicative. By (0.5)

 

 

 

-n f (12) t * -n* dk

pno - g dk’ so rom . we ge cno - g -1+dk . Then
_1+ n n

1 ’1 dk 1 < 1

———=.1r =Tr 1——_7r(1+T—T)=

CEO! 1 dk 1 dk 1 dk
n .1._ co 1.
T{(1+k2)<77'1(1+k2)<00

for each n 2.1, so lim CEO # O.
n

CHAPTER II

REGULARITY CONDITIONS FOR (f,dn,zl)

All of the sufficiency conditions for regularity given
in [15] by Smith require that the power series expansion
of f about the origin have real, non-negative coeffici—
ents. We now give some sufficiency conditions which do
not so require. Of course there are compensating additionald
hypotheses. We will need the following theorem of Bajsanski

[5] and Clunie and Vermes [7]:

Theorem 2.1: Let f be holomorphic on the disc
lzi < R, R > 1. Let [f(z)] < 1 for (2| = 1 except at

a finite number of points g at which |f(§)[ - 1. Then,

k

if fn(z) = 2 an 2 , it follows that Z (a
k

k nkI

k = 0(1) if

and only if ReAC # O for each such C, where AC[i(z—1)]p(C)

is the lead term of the Taylor expansion about 1 of

hc(z) - za(§), and where hc(z) = £é%%% and u(C) = h&(1).

Theorem 2.2: Let f satisfy the conditions of

Theorem 2.1, (21] = 1, f(z1) = 1, and
a) (Imfdh)3
2 -—-—-—;— <
1 11+dn|
19 19

If dn 3 pne n, let lim sup Re[(f(05 - 1)e
n->oo , dnyéo

' a) 1
(f,d .21) is regular if 2 = d).
n 1 I]. + dnl

25

n .
Proof: (f(z) + d )1 = 2 p zk ='Z o. fn-J(z),
-————- n k‘ nk j=o j
where 00 = 1 and 0. = Z d ... d for j > O.
..<vjjn
Now,
r1 oo 00 n
2 o fn 3(z) = 2 o 2 an_ kzk - 2 ( z 0 an_ k)zk,
3‘0 j=o 3 k-o 3’ k=o 3'0 3 3
so
( n
2.3) p - Z a .
nk j=o j n-j,k
~Since f(z1) = l,
n
(2.4) (f(z1) + dn). - (1-+dn). = g oj.
Substituting pn for dn in (2.4) gives (1 + pn)! =
n
2 0!, where o! = 1 and o! = ,2 p ... p for
0 J 3 1<V1< V1 vj
..<v.
n n
3 > 0 Then .E [o I :.Z 05 = (1 + pn)! It follows that
o
_ 1 k _ 1 _
i ian1 ‘ Itifi'l' i lpnkzl |1+d | Z lpnkl
1 n 1

H
:1
Q:
th
L4—
II M
Q
U
:3
I
LJ.
WI
H
E]
0.:
M
7?
LJ
M
75
:3
I
U
W

.1 KB— 2 lo I 4 B , +Pn 1 where B = sup 2 [a ]
T11&;1: o 3 - 2|1+dn 1' n k nk '
1+pv <v)(1+p < (1+PV )2
But T—d—[- “ma—r: -eXP {-1 + 121.1“:
-]1+dv]2 + (1+pv)2 4(Im'Jd v)2

 

ex“ 1111,11 ’ I ”x10 {W

27
(1 + pn)1 n (Imwfd—)2

so . i.exp {4 2 V ] = 0(1). The regularity
[I + dn—[. 1 I]. + dviz

 

 

condition (0.1) is thus verified, and (0.2) is always true.

We need only show that lim cnk = 0 for each k. Suppose
n
i6
lim 8UP Re[(f(0) - 1)e n] = -35, 6 > 0. By the maximum
n->dmdh#0
modulus principle, if(0)| < 1-»251 for some 51 > 0. Let

52 = min(51,6). Now let P be a circle about the origin
so small that (f(t) - f(0)| < 52 on P. Then [f(t)|.i
'6

1
if(0)] + 52 < 1 - 51 on P. |Re[(f(0) - 1)e n] -
ie

16
Re[(f(t) - 1)e “1| = |Re[(£Zo) - (t))e “1|.i |£(o)-f(t)|
< 82 uniformly in t and n if dn # 0. Consequently,
ien ien
] —36 => Re[(f(0) - 1)e ] < -26
iGn <
for large n, so Re[(f(t) - 1)e ] < -25 + 52 --—6 for

lim sup Re[(f(0) - 1)e
n—>oo,dn#0

large n. If we let f(t) = u(t) + iv(t). then
19
(2.5) Re[(f(E7 - 1)e n1 = (u - 1)cos en + vsin en < -a

on P for n > N, dn # 0.

Let 0 < w < 1 and choose 8 so that 0 < 5 <

81 ‘”2££L—' 2 2 2
min (7—n J2(1+ufi ). Let Qn(t) — e + u + v +

 

28 J(u + pncos 6n)2 + (v + pn sin 6n)2 +
2pn[(u -1)cos 9n + v sin 6n].

Then if t e P, n > N, and O < pn': m, we get

28

Q t) < 52 + [f(t)]? + 65 - anO 5 e + lf(t)l + 68 5

n(
(f(t)l + 51 < 1.

On the other hand, if pn Z.w, then

—_¥

v . ma

 

 

= w < 1,

so a multiplication through by épn gives

 

e\f(u + pn cos 9n)! + (v + pn sin 9n)2 < épn

But then,

 

Qn(t) < 53 + u2 + v2 + 25 JTu + Pn cos 6n)§_+ (v + pn sin 9n)2

- anO = 82 + +£(t)|2

 

-2[opn - e J(u + pn cos 6n)2 + (v + pn sin 6n)2 ]

< 82 + ff(t)|2<:5 + lf(t)| <.§1 + (1 — 51) < 1,
Thus in any event, Qn(t) < 1 on P for n > N, dn # 0.
But this is equivalent to |1 + dnl2 > (g + |f(t) + dn|)2,
or |1 + dnl > g + [f(t) + dn], or -|1 + dn| + [f(t) + dnl

< -s on P for n > N, dn # O. This is also clearly

true for all n for which dn = 0. It follows immediately

 

 

 

that
f(t) + d f t) + d I
V V _
Tiff—"61‘9“” 1+d l}—
v v

29

-.i1 + dv| + (f(t) + dv]

 

 

 

exP [ “~1 1‘ 1 < exp {- E }
|1 + d | 1 + dv
v
n f(t) + d I N f(t) + d n f(t)+d
on Pif v>N,so 1r].+ v=7r 1+d '1r 1+dv
1 v I 1 v N+1 v

 

 

 

 

n
1 _ _, -_4
< 0(1) exp {-5 NEITI + dvr ] — 0(1) as n > a) uni .

 

 

 

_. 1 n 12
formly on F. Consequently, cnk — 2wi f k+1 q 1 + dv dt

——> O as n -—> oo.

Corollary 2.6: Let f satisfy the conditions of

 

 

‘ 03(Im3J433)2
Theorem 2.1, |21| = 1, f(zl) = -1, and 2 2 < a).
1 |-1+d |
n
19 i9
If dn = pne n, let lim sup Re [(f(0) + 1)e n] < O.
n—>aa,d #0
n
0° 1
Then (f,dn,zl) is regular if g |_1 + dnl = oo.

Proof: The above proof, with only minor changes,

proves the corollary.

We remark here that inspection of the proofs above

gives immediately the following two corollaries.

Cgrollary 2.7: Let f satisfy the conditions of
Theorem 2.1, [21] = 1, f(zl) = 1, and

n 1 + pV

7; Tra— - 0(1),
v
ien len
where dn = pne . Let lim sup Re[(f(o) — 1)e ] < 0.

n—eoo,dn#o

 

30
0° 1
Then (f,dn,21) ls regular if g ‘1'1 + an) = oo.

 

There are two cases in which Theorem 2.2 and Corol-

lary 2.6 become particularly simple.

Corollary 2.9: Let f satisfy the conditions of

 

Theorem 2.1, [21] = 1, f(zl) = 1, and dn'i 0. Then
“D 1
(f,dn.z1) is regular if f T—:—a;-= oo.

COrollary 2.10: Let f satisfy the conditions of

 

Theorem 2.1, (21] = 1 f(z1) = -1, and dn': 0. Then
0° 1
f,d ,z is re lar if 2 = a).
( n 1) gu 1 tT-l + dnT

 

It is shown in [7] that the functions

1
2 + i

 

(2.11) f(z) = (1 + 12 + z”),

w(Z-1) w = eiW/a’

e .

(2.12) f(z) = w: ’

1
- 1
i¢ 1

(2.13) exp [®(z - 1) - &2(22 - 1)} , $ = e . COS ¢ = E?

H:
N
ll

all satisfy the conditions of Theorem 2.1 and have f(1) = 1.
If 21 = 1, then these functions fulfill the hypotheses of
Corollary 2.9. Moreover, if f is the function (2.13),

B = -sin ¢, and F(z) = fzv/S(z)’ then an easy but‘tedious
calculation shows that F also fulfills the conditions of
Corollary 2.9 if 21 = e-Zi¢. If f is any function obey-
ing the requirements of Theorem 2.1, then the maximum
modulus principle implies [f(0)j < 1. Then Re f(O) f‘l,

16
so if f(0)= a + ib, a - 1 < 0. Hence Re[(f(0) - 1)e n] =

31
(a - 1)cos 9n + b sin am < 0 if an is sufficiently
close to a multiple of 27. It follows that if 9n is
so restricted and 21 = 1, then the functions (2.11),
(2.12), and (2.13) also meet the requirements of Theorem

2.2.

Definition 2.14: Let f be holomorphic on the disc

00
(2| < R, R > 1, and let fn(z) = 2 ankzk for n': 0.

k-o
Then the summability method determined by the matrix

(ank) is called a Sonnenschein method.

In [7] the following theorem is proved:

.Theorem 2.15: The Sonnenschein method generated by
f is regular if and only if either f(z) = 2m for a

positive integer m , or f satisfies the conditions of

Theorem 2.1 and f(1) = 1.

The next two results are immediate consequences.

Theorem 2.16: Let f be holomorphic on the disc

 

|z| < R, R > 1, and suppose f(z) # 2m, m (a positive
integer. Then if the Sonnenschein method generated by f

is regular, so is the method (f,dn), provided

OD (Im \ldn)2 __ ien
Z T1 + d 12 < oo. lim sup Re[(f(0) - 1)e
1 n n->o:> ,dn;£o

 

1 < 0.

and C; Ti—éfEr—r== a).
1 n

32
Proof: Apply Theorem 2.2, with 21 = 1, and Theorem
2.15.

Corollary 2.17: Let f be holomorphic on the disc
|z| < R, R > 1. Then if the Sonnenschein method generated
by f is regular, so is the method (f,dn), provided

a)
1
> ——_—=
dn‘0 and}; 1+dn 00'

Proof: If f(z) = 2m for some positive integer m,
then Lemma 1.12 gives the conclusion. Otherwise, Corollary

2.9, with 21 = 1, gives the result.

CHAPTER III
COINCIDENCE OF METHODS

Lemma 3.1: The method (f,dn,zl) is Sonnenschein if

and only if dn 5 d1.

Proof: Suppose (f,dn,21) is the Sonnenschein method
generated by g. Then the matrix coefficients of the two

methods are the same, so

n f(zzl) + dk n
¥f(z;) + dk = g (z)

 

for n 3.1 on the intersection of their domains. In

 

 

 

 

 

art' 1 he 1 ha e ( ) f(zzl) + d1

1cu ar, w = we v z =

p n n g f(E1) + d1

Suppose d1 = ... = dn-1 for some n > 1. Then

gn(z) = n;1 f(221) + dk . f(zz1) + an = gn_1 2 ._f(zz1) + dn
1 f(z1) ‘l" dk f(zl) + dn f(zl) + dn'

f(zzl) + qh .
so g(z) — f(zi) + qn . But then dn — d1, so, by in-

duction, dm'E d1. On the other hand, if dn E d1, then

 

n f(zzl) + (1k f(zz ) + d
_ n _ 1 1
g f(21)+ i — g (z), where g(z) — f(517+ d1' so the

matrix coefficients of the two methods are clearly the

 

same. Moreover, since 21 is in the (open) disc of con-
vergence of f, g has radius of convergence > 1. Thus

(f,dn,21) is the Sonnenschein method generated by g.

33

34
Lemma 3.2: Let b and 21 # 0 be given. Then
every Sonnenschein method generated by a function g with
g(1) = 1 can be realized as an (f,b,zl) method for ap-

propriate f.

Proof: Define f(z) - g(z/zl) - b. Since 9 has
radius of convergence > 1, f has radius of convergence
> [21]. Then, if dk E b,

f(zzl) + dk n z

n
_n
1 f(zmdk 191 -9<z>'

 

so the (f,b,21) coefficients are those of the Sonnenschein

method.

If g(1) # 1, then the corresponding Sonnenschein
method cannot be realized as an (f,dn,21) method since

the equality of the methods would imply that

n n f(zzl) + dk
9 (z) =¥?(zl) +dk

 

for n.: 1,

whence g(1) e 1. But any Sonnenschein method can be
realized as an (

f,dn,21)* method if g(zl) = 1 for some
21 in the disc of convergence of g. In fact, the Sonnen-
schein method generated by g is easily seen to be the
I»
method (g,o,21).
If the matrix of the Sonnenschein method generated

by g is (ank) and g(1) # 1, then either lzm i ank =

lim gn(1) does not exist or it is zero. In the former
n
case the method cannot be conservative, and in the latter

35
case all constant sequences are summed to zero ([8], p. 43).
Consequently, in some sense the "interesting" Sonnenschein
methods are those for which g(1) = 1. Lemma 3.2 shows
that the class of "interesting" Sonnenschein methods is a
subclass of the set of (f,dn,21) methods. We will now
prove some results analagous to those appearing in [14]
and which may be considered extensions of those results

if attention is restricted to the "interesting" Sonnenschein

methods.

 

Definition 3.3: If sj is a term of a sequence, let
n
. n _ k n
the Operator A be defined by A sj -kEo (-1) (k)sj+k'
n 2.0.

Definition 3.4: The Hausdorff method (H,u) is de—

' ' : = n
fined by the matrix H (hnk)’ where hnk (k)A “k

for k :,n, and h 3 0 for k > n.
nk

Definition-3.5: The quasi-Hausdorff method (H*,u) is

* _ k k-n
where h - (n)A ”n

defined by the matrix H* = (hnk)' nk

for k Z.n, and h* = O for k < n.
nk

Definition 3.6: The Euler method (E,p) and the circle

 

method (T,p) are the Hausdorff and quasi—Hausdorff methods,

respectively, with ”n = pn and ”n = pn+1.

In an unpublished paper, Ramanujan has given:

Definition 3.7: Let the modified quasi—Hausdorff

 

method (Hfi,u) generated by the quasi-Hausdorff method (H*,u)

 

 

 

be given by the matrix H1 = (hzk), where
1 O 0 0
l for n=k=0 ' ‘
0
-w 0 for n=0,ki1 ._
h = so that H* = 0
“k 0 for k=0 ,n_>.1 H*
0
* > >
hn-1,k-1 for n.1,k_1 .
—. — —IH
He has also pointed out that H1(0,so,sl,...) =
H*(so,sl,sz,...), so H1 is not as artificial as it may

look; it arises naturally when considering the translativity
,problem for H*.

In [14], the following proposition is stated:

"The circle method (T,pl) is the only method, regular
or not, which is both quasi-Hausdorff and Sonnenschein.’

Bojanic [6] subsequently pointed out that the proposi-
tion is not correct as it stands, but can be made so by
replacing "circle" by "identity". In his unpublished paper
Ramanujan showed that the above proposition can be made

correct if it is altered so as to read:

(3.8) "The modified quasiéHausdorff method (H*,u) is

Sonnenschein if and only if (H*,u) is the circle method."

We remark here that his proof shows that if the Sonnen-

schein method in question is generated by g(z), then

 

(3.9) g(z) = 1 _ %i _ p)2 , where p = M0 and [1 - p| < 1.

 

37
We will now prove an analog of (3.8) for (f,dn,21)

methods.

Theorem 3.10: The modified quasi—Hausdorff method

 

(§3,u), ”n # 0, is (f,dn,21) if either
(3.11) the associated quasi—Hausdorff method (H*,u) is the
circle method (T,p) with [1 - p] <11,

or

_ CPZ = n+1
(3.12) f(z) - f(O) — 21 _ (1.- p)z , where ”n p .
[1 - p] < 1, and dn E -f(0),

and only if both (3.11) and (3.12) hold.

Proof: Suppose (H*,u) = (f,dn,21), and let the

(f,dn,21) matrix be (an)' We first make use of the fact

 

 

that cnk = hnk’ and then use the relation (f(zzl) + dn)°
= (f(zz1) + dn)(f(zzl) + dn_1)1 to get first
(f(zzl) + d )i a) k-1 k— k
(3-13) n. = 23 ( _ )A n u _ z , n 2.1
(f(zl) + dny. k=n n 1 n 1
and then
f(zzl) + d (X) _ - 1
(3°14) f(z )+ d n E (lg-ink n+ “Ln—2 2k =
1 n k=n-1
a)
k—l k-n k ‘
k=n

. _ . , ., - f(zzl) + (.11 00. k-1
Setting n - 1 1n (3.13)(g1ves f(z1) + d1

 

so if 2 = 0 we have f(O) + d1 = 0. «Let

38

OO
f(z) = % ak z . Then from (3.14) follows, for n 3,2,

 

“3 k k
(ao+dn)‘ + g akzlz . :0 (k-1)Ak-n+1 2k =
f(21) + dn k=n-1 “‘2 “’2
OO
k—l k—n k
2 (n-l)A LLn-iz ’
k=n
so equating coefficients of Zn"1 gives (a0 + dn)un_2 = 0.
whence dn = -a0 = -f(0). It follows immediately, by in-

-f(0). But then Lemma 3.1 says that

duction,that dn

(f,dn,zl) is the Sonnenschein method generated by

f(zzl) - f(O)

g(z) = f(zl) - f(O)’ Now (3.8) implies that (H*,u) is the

 

circle method, so the necessity of (3.11) has been shown

along with the necessity for dn E -f(0). Moreover, (3.9)

gives the formula g(z) =§(::%)-—ffég) = 1 _ (§z_ p)z .

ll ' p] < 1' If C = f(z1) - f(O), then f(zzl) - f(0) =

cpz _ = cpz .
1 _ (1 _ p72 , so f(z) f(O) 21 _ (1 _ p)2 . Again

- f
(3.13), ...... 2:32-16?)

 

 

 

 

 

oo
k-1 k _ p2 . . .
gilA H02 - 1 _ (1 _ p)z . But it is readily shown that
00 d3
p2 _ k—1 k _ k—1 k
- 21 1 - z - 21 A z , so
1 - (1_ p)z k=1( p) p k=1‘ 9

A #0 = A 1p and p = Ho- Definition (3.6) now implies

that u = p 1 for each n. Hence, (3.12) follows.
n

39

 

Conversely, suppose (3.11) is true. Then ”n = p , so
-* _ * _ k-1 k-n _ k-1 k-n n
hnk hn-1,k-1 - (n-l)A un—1 - (n-l)A p for n.: 1’
k :1. It follows that
00 00 a)
z h;kzk = z (:_1)Ak npnzk : z (2:1)(1_p)k-n nzk
k=n k=n k=n
Apz n n
(1_(1 _ p)z) g (2),
pz

 

where g(z) It is now immediate that

: 1 - (1 - p)z
(3*,u) is the Sonnenschein method generated by g provided
g has radius of convergence > 1, i.e., provided ]1 - p! < 1.
(3.11) insures this condition is met. Lemma 3.2 now says

that (fi*,u) is a method (f,b,zl). Finally, suppose (3.12)

 

 

 

is true. It is easily seen that c = f(zl) - f(O), so
f(zz ) + d
f(z) - fLO) - pz _ 1 m
- d thus 2 -
f(zlj - £10) 21 - (1 - p)z an g( ) f(217+ dm
_ f(zzl) - 5(0) pz

 

 

= . Since 1 - ' < 1, has
£517 - f(O) 1-(1 - p)z l Pl 9
radius of convergence > 1, so it is clear that

n f(zzl) + dm

_ n
q f(zl) + d — g (z),vhence (f,dn,zl)
m .

 

is the Sonnenschein method generated by g. We see now

that

n 00 00
n _ pz _ k-1 _ k-n n k _ k-l k-n n k
9 (z) ' (1 _ (1 _ p72) -kin(n-1)(1 P) P z -kE£n_1)A p z

 

I

for n 2.1, k 2.1, so the (f,dn,zl) matrix is that of the

modified quasi-Hausdorff method having ”n = pn+1.

40
Lemma 3.15: The only matrix which is both (f,dn,zl)

and quasi-Hausdorff is I.

Proof: From definition 3.5, the quasi-Hausdorff
CD k) k-n
Z (. A u S . If
k=n n n k

this is also an (f,dn,zl) transform, then for n = O we

transform of [sn] has the form t; =

have Akuo = 1 when k = O and Akuo = 0 when kl> 0.

It follows immediately that Mo = H1 = 1. Suppose

H0 = #1 = ... = uk_1 = 1. Then 0 = Akuo =

k . k-i . k .

_2 ('1)J(k)u- = U0 .2 (‘1)J(k) + ('1)kUk = no.2 (‘1)J(k)

3:0 3 3 3:0 3 3-0 3
- Lia-1)“ +(-1)kuk = (-1)k(uk - Mo): so L1,, = U0 = 1.

By induction, ”n E 1. Then k > n implies Ak-nun =

Ak_nuo = 0. Clearly, t; = Sn’ so t; is the identity
transform.

It is remarked_in [14] that the Euler method is the
only one which is both Sonnenschein and Hausdorff. In the

same vein is

Theorem 3.16: The Hausdorff method (H,u) is (f,dn,zl)
if either
(3.17) (H,u) is the Euler method (E,p),
or

(3.18) there is a constant di such that (f,dn,zl) =

21
21 + d; ’

(z,di, 21) and ”n = pn , where p -

and only if both (3.17 and (3.18) hold.

41
Egggj; We will first show the necessity of both (3.17)
and (3.18). Since the Hausdorff matrix is triangular, it
can be an (f,dn,zl) matrix only if f is a first degree
polynomial, say f(z) = az + 5- But then

f(zzl) + dk qzzl +_5 + dk 221 + di

f(zl) + (1k — 021 + 3‘+ dk — 21 + (1k ’

 

 

 

so (f,dn,zl) = (z,d$,zl). Now form the analogs of (3.13)

and (3.14), getting

 

(221 + d'): n
n n n-k k >
(3.19) (21 + dg—fi kEO(k)A LL}<Z 1 n _ 1i
and
221 + d. n-1 _ _ _ n _
(3.20) -E:—:—a¥'k2 (nk1)An 1 kukzk =kz (£)An kukzk , n.i 2.
n =0 =0

By letting n = k = 0, it is seen that the equality

of the matrices requires that Mo = 1. Setting n = 1

z
in (3.19) gives --L-, = ul. For n :_2, we equate co-
21 + d1
. . n . Z _ .
effiCients of z in (3.20) to get ;:-ifa:'un_1 — un’ so
n
21

 

= > - .
”n (21 + d8): for each n _.1 by induction. Suppose

d; = di for all k < n, n 3.2. Equating coefficients of

zn-1 in (3.20) gives
21 d;
* - I g = >
21 + d5 (n 1m“Ln-2 + 21 + dn L(Ln-1 nALLn—i ' n “'2'

This expression, together with the above formula for uh.

yields d8 = dg_1 = di after a simple calculation. Then

_ n _ z1
”n — p , where p —-E:—:—EI , so (3.18) follows. But now

42
definition (3.6) says that (H,u) = (E,p), so (3.17) is
true, too. Conversely, suppose (3.17) obtains, so that

”n = pn for every n 3,0 and some p. Then

n n n-k .
n n—k k n n—k 3 k
2 ( )A z = z ( ) z ( . )(-1) .z =
k=0 k L*k kao k j=o 3 ”k+3
n n kn"k n-k j k _ n n k n—k k n
2 (k)p Z ( j )(-p) z —,2 (k)p (l-p) z = (p2+ (l-p)) .
k=o j=o k=o

soihe method (H,u) is the Sonnenschein method generated by
g(z) = pz + 1-p. Lemma 3.2 now gives the desired result.

1 . . (221 + dn)1 221 + d1 n
If (3. 8) is valid, then (21 + dn)::= (jzfirjif

 

n n-k

)n km

n
.= z ( ukzk, so the method (z,di,zl) is

k 0

(pz + 1-p
also the method (H,u).

Definition 3.21: Let {gm}:D be a complex sequence
n .
with An = Z qm # 0 for n.: 0. Then the Norlund method
0

determined by the sequence [qm] is defined by the matrix

= = < =
.A (ank), where ank qn-k/xn for k _.n, and ank 0

for k > n.
'Suppose the Norlund matrix A is also an (f,dn,zl)
matrix. «Since A is triangular, we may, as in the proof

of Theorem 3.16, assume f(z) = 2. It must follow that

(221 + dn)' n qn—k k

(3.22) e .=2 2 . n11.
jzl + (inf k=o >\n

 

 

and from this,

43

221 + d n-1 q _ _ n q _
(3-23)—;‘T§£Z-%-£—12k=2—n—7\]£Zk,n32.
1 n k=o n-1 k=o n

Equating constant terms in (3.22) we get, if n = 1,

(3.24) qodl = (1121.

 

 

 

Equating coefficients of zn in (3.23) gives
zlqo = 32- so 31—:—E2-= An and
)\ (21 + d ) )\ I 21 )\
n-1 n n n-1
d q
1 + zn = 1 + An Then,
1 n—1
(3.25) dn xn_1 = qnzl , n 3,2.

Equating constant terms in (3.23), we have

 

d q q
(3.26) n “‘1 =—’l , n22.
An_1(zi + an) A

n

We need two more relations:

Qa(d1 + d2) Q1

 

 

3.27 = —— I
( ) >u(zi + d2) A2

Z
(3.28) 1 = i .

A1(zi + d2) 7\2

The second of these is obtained by equating the coef—
ficients of 22 in (3.23) with n = 2; the first is obtained
by equating coefficients of z in (3.23) with n = 2 and
using (3.24).

Substituting from (3.28) into (3.27) gives

q0(d1 + d2) = qlzl. Then, using (3.24),
we get qo(d1 + d2) = qodl , so d2 = 0. Then (3.25)

implies q2 = 0, and from (3.26) it is clear by recursion

44

that qj = 0 for j 3.2. Then from(3.25), dn = 0 for

n 3.2. Then the matrix in question reduces to
(3.29) Al
Q1 go (‘II’
-1
A _ k1 ql qo , where q1 = 0 or

q1. (10.
°- '- qlzl = godi f 0-

We have proved

Theorem 3.30: A Norlund matrix which is also an (f,dn,zl)
matrix must have the form (3.29). If q1 = 0, then the

matrix is the identity, and this case arises precisely when

Ill

f(z) = z and dn 0. The alternative case arises pre-

cisely when f(z) = 2, d = 0 for n.3 2, and dlqo = qlzl # 0.

Corollary 3.31: The identity matrix is the only one

which is both Norlund and Sonnenschein.

Proof: Let A = (ank) be a matrix which is both
Norlund and Sonnenschein, and let the generating function

of A as the latter method be g. Then g(1) = a =

1k
-1
A1 (q1 + qo) = 1, so by Lemma 3.2, A is an (f,dn,21)

OMH

matrix. Then Lemma 3.1 says dn 5 d1, whence Theorem 3.30

implies A = I.

Corollary 3.31 is PropoSition 1 of [14].

Corollaryi3.32: The identity is the only method which

is both (f,dn,zl) and (C.a).

45
Proof: (C.o) is defined for each real 0 which is
not a negative integer, and is the Norlund method deter-

mined by the sequence (q )00

n o , where qn = qn(a) =

u(o + 1)...(a + n - 1)

n' for n > 0, and qo(a) = 1. By

= u(a + 1)

2 = 0, so a = 0. But (C,0)

Theorem 3.30, q2(o)

is the identity.
Definition 3.33: Given a complex sequence [qm}:O.

n
let kn = % qm # 0 for n 3.0. Then the Riesz method

determined by [qm} is defined by the matrix A = (ank)'

_ < _
where ank — qk/kn for k _,n, and ank — 0 for k > n.

Theorem 3.34: There is no matrix which defines both a

Riesz method and an (f,dn,zl) method.

Proof: We proceed as before, Since the Riesz matrix

is triangular, if it is also an (f,dn,zl) matrix, we may

 

assume f(z) = 2. Then it must follow that
(221 + d )1 n q
—(2 + dn)' = Z XE-zk , n.i 1, and from this
1 n ° k=o n
that
221 + d n-1 q n q
(3.35) 7—47—52- 2 XLZR'Z Xhzk, n>2.
. 1 n k=o n-i k=o n

Equating constant terms in (3.35) leads to

(3.36) dnqn = kn-l 21 7f 0: n _>_'_ 2.

From this it is clear that qn # 0 for n 3.2, so

46

dn = -——————-. Plugging this value for dn into (3.35)

q q _ “'1 q q _ QO n q
(3.37) KER—B—l-zn + iL- Z (-§——£al-+ qk)zk +-X— = Z'iE-zk,
n n—1 n k=1 n-1 n k=o n
n 2.2.

Again equating coefficients, we see that qn qk_1 = 0 for
1 j_k j_n - 1, so if n3.4 there is aj 3,2 with

qj = 0, in violation of (3.36).

CHAPTER IV
INCLUSION RELATIONS

It is relatively easy to give an example of a conserVa-
tive (f,dn,zl) method which is at least as strong as the
(C,1) method. We will give such an example below after
some necessary tools have been developed. We will now

concern ourselves with results in the opposite direction.

Lemma 4.1: Let f be holomorphic at the origin and
let 21 belong to its disc of convergence. Suppose

0° k
(f(z) + dn)1 = 2 p k z , n.3 1. Then
k=o n

1 00

n
' 1 _ k
21f (21) § f(zl) + dj —(f(zl) + an): k2. k pnk 21'

 

 

Proof: Let gn(z) = (f(zzl) + dn)1 Then gé(z) -

CK) k k‘l I (I) k I _
Z kpnkzlz .£x>gn(1) = Z kpnkzl. Also, gn(z) -
k=1 k=1
n
zlf'(zzl) Z w (f(zzl) + dk) = zlf'(zzl)'(f(zzl) + dn)1
j=1 1< <
k7‘j
n 1 1

 

n

The result follows.

Corollary 4.2: Let (f,dn,zl) be regular, f(0)+dn#0,

and either
47

48
(4.3) f(zl) + dn is real and has constant sign for large n,
or
(4.4) Re[f(z1) + dn] E 0 and Im[f(z1) + dn] has constant
sign for large n.
Then if f'(zl) # 0 it follows that
00

1 k
(f(21)+ dn)T’k:1 k pnkZI # 0(1).

 

Proof: By Lemma 4.1 we need only show that

n
1 .
g f(z1)fi+ dj # 0(1). Theorem 1 of [16] asserts that if

 

f(0) + dn # 0, then a necessary condition for the regularity

oo
. 1
of (f,dn,z1) is that 31 |f(0) + d.| — oo. But then

 

Z Tf(zl) + dj[= 00, too, whence so does § f(217 + dj

 

 

1

 

The result follows.

We can now prove

Theorem 4.5: Let f(z) = 2k have radius of

o548

ak
convergence greater than p > 0, and let ak = 0 if k
is not a multiple of the integer m > 1. Suppose either
(4.3) or (4.4), f'(zl) ¢ 0, f(0) + an ¢ 0, and (f,dn,zl)
to be regular. Then the summability field of (f,dn,zl)

does not contain that of the Riesz method associated with

n
the sequence {qk}, with 7\n = z qk, if lim inf %|;k_
0 k

49

 

 

 

 

 

 

 

 

(I) 7\ z k CD 7\ k
and 2 -h' ‘—l and Z k El- both converge.
n 00 k
Proof: Let f (z) = Z ankz . The only powers of z
k=o

appearing in the expansion of fn(z) are multiples of m,
so ank = 0 unless k is a multiple of m. If (f(z) + dn)1
.0) k
= 23 p kZ , then from (2.3) it follows that p k = 0 if
k=o n n

k is not a multiple of m. But then

pnk/qk if k = Em,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P P 21 _ - =
_gk__ n,k+1 = pn k+121/qk+1 1f k+1 £m,
qk qk+1
0 otherwise,
so
1 ‘E pnk _ pn,k+1 1 |zkx I
I 1 .—
Tf<zl> + dnT'k-o qk qk+1 k
2 pnk pn,k+1zl ! A
I ' 1 —
(f(zl) + dnIT k=£m qk qk+1 k
fiib
1 nk k
. Z — 21 )\ l -
Tf(z1) + an]. k=£m qk I k
£39
A
1 pnk . k . . . 1 k
, 2 '-—~ z1 I. Since lim inf'--—— > 0,
Hal) + an k=o qk l N. k qk
there is an e > 0 and N > 0 such that gk- Z.ke for
k

 

 

1 OO
. Z

 

. k
lzlkkl:

 

k > N, so

 

pnk
qk

50

a) GD

2 k .2" k
[f(zl) 8+ d “1' k-N+1| pnkg >)(f(z]_=)€ + d n77. k=N+1 pnkz1

 

 

# 0(1) by Corollary 4.2, provided we can show that

N

1
.ka
‘(f(zl) + dn). k— 0 nkz

 

 

= 0(1) as n —9 oo.

This is easily shown to be the case. For if the (f,dn,zl)

matrix is(cnk), its presumed regularity simplies that

 

2 lcnkl :.B for every n. In particular,
lcnkl = l—(zl) + d n], _,B for all k and n, so
k
kpnkzl

:- kB for all k and n.

 

 

 

(f(zl) + dn)1

1 N
(f(El) + dn)? kio kpnkz1

 

Then

 

 

G)

1 , i Z pnk _ pn,k+121
Tszl) + dnT'. k=o

qk qk+1

 

 

k
|21 Wk] # 0(1) as

 

n -> 00, so the transformation

00 P P 2i
(4.6) (f(z y£+ d )1 Z E££'-‘—%LE:1-—
1 n k=0 k k+1

) zk x t
1 k k

 

is not conservative inasmuch as (0.1) fails. This fact is

 

00
the crux of the argument. Suppose that tn = A31 2 qksk
0 ,
7\ntn-xn-itn-l
is the Riesz transform of {sk}, so that 5n = '.
M q“
Cauchy's estimate for Ipnkl is -%n and this, together with
p
. k 1
the convergence of the series z|a—)-|-—) and

21
, | |-—)k, assures the absolute convergence of
qk +1 P

 

51

p k p 1 k
2 -£E-kkzl tk and Z —2—h:—Ikk 21 tk for every bounded
k qk k qk+1
sequence {tk]. In particular, if the Riesz method sums

(5k), then {tk] is bounded, and we can use summation by
parts to write the (f,dn,zl) transform of {sk} as

1 0° k _

(f(zl) + an): k§o pnk zl Sk ‘

 

 

 

 

 

“3 k xktk I xk-1tk-1
I 2 p zl =
00 P P 21 k
nk n,k+1
-——-— z .
(f(z_71+ d;7 k=o (qk qk+1 ) 1 xk tk

But this is the transformation (4.6) and is not conservative.
In view of the reversibility of the Riesz matrix, it is
easily seen that the transform (4.6) will sum everything
the Riesz method does if and only if the transform is con-
servative. The result follows.

As examples of functions which satisfy the conditions

of the above theorem, we give the following. In each case

21 = 1

(4.7) f(z) = 2m , m an integer greater than 1,
(4.8) f(z) = exp[zm], m an integer greater than 1,
(4.9) f(z) = cosh z,

(4.10) f(z) = log sec 2.

If f is any of these functions and [dn] is chosen so that

d 2.0 and Z

n 1 1:5; = a). then Lemma 1.12 gives the regularity

of (f,dn,zl), so all the conditions on f and [dn] are met.

52
The following is a variant of Theorem 4.5 having some—

what less elaborate hypotheses.

Theorem 4.11: Let f be holomorphic at the origin

with radius of convergence greater than p > 0. Let

 

 

 

 

 

 

 

 

 

 

 

k
21 P P 21
k n k+1
b = , [ n — ’ ] and suppose
nk (f(zl) + dn). qk qk+1
lim bnk = O for each k, lim sup 2 bnk > 0, and both
n n -> 00 k
Kk 21' kk Z1
Z'—— -— and 2) ——- converge. Then if
qk qk+1 P

lkkl -9 co, the summability field of (f,dn,zl) does not
contain that of the Riesz method associated with the

sequence {qk].

Proof: Given B > 0, there is an N > 0 such that

CO CI)
k > N => llkl :.B. SO ‘2 b klkkl': 2: bnklxkl +

01 N G) 00
Bk.§+1 bnk = kEo(lkkl-B)bnk + Bkilobnk = 0(1) + Bkiobnk.
. . _ G)

Since B is arbitrary it follows that
(D
lim sup 2 bnklkkl = oo.
n -> oo k=o
An immediate consequence of this is that the transformation
(4.6) is not conservative since (0.1) fails. The conclu-
sion of the theorem now follows exactly as in the proof of

Theorem 4.5.

53

00
Theorem 4.12: Let f(z) = Z a zk, where ak = O
o

[k

 

if k is not a multiple of m > 1. Then the summability
field of (f,dn,zl) is not contained in that of the Abel

method.

Proof: The (f,dn,zl) matrix (an) is defined by

k
_ pnk 21
cnk ~(f(zl) + any: . We have already seen that for such

 

an f, pnk = 0 unless k is a multiple of m, so the

corresponding cnk = 0, too. Consequently, (f,dn,zl)
will sum to zero the sequence {sk}, where sk = 2k,

k # gm, and s = 0, k = Em. But clearly this sequence

k

is not Abel summable.

 

Corollary 4.13: Let (f,dn,zl) be subject to the con-
ditions of Theorem 4.5 with p > |z1|. Then (f,dn,zl) is
not comparable with the Abel method or with any (C,a)

method for q.: 1.

£5992: (C,1) is the Riesz method determined by the
sequence (1]. By Theorem 4.5, the summability field of
(f,dn,zl) does not contain that of (C,1), and hence does
not contain that of (C,a) for a > 1 or that of the Abel
method. On the other hand, Theorem 4.12 says the summabil-
ity field of (f,dn,zl) is not contained in that of the Abel
method, and thus also it is not contained in that of (C,a)

for any a.

The following Lemma is well-known ([17],Lemme 4).

 

a) k
Lemma 4.14: Let 2 bk 2 a converge and let
k=o j=o
ool I oo 1: oo oo
2 b < oo. Then 2 b 2 a. = Z a. 2 b if and only
k . .
0 =0 k 0 3 3:0 3 kzj k
00 k
if lim 2 bk 2 a, = O,
Q‘>G> q q 3

Definition 4.15: The Y-transform of (5k) is the
sequence [yn}, where yo =-% so, and yn --%(sn + sn_1)

for n.: 1.

Theorem 4.16: Let f be holomorphic at the origin.
Let either (4.3) or (4.4) hold, f'(zl) # 0, and f(O) + dn # 0.
If (f(z) + dn)1 = i pnk 2k, suppose that for each n,
(--1)k pnk z? is real and does not change sign for k > N.

It follows that if (f,dn,zl) is regular, its summability

field does not contain that of the Y-method.

Proof: Inverting the Y-transform gives

k .

s = 2(--1)k z (—1)3 y.. There is a
k 0 . 3

p > [21] > 0 such that p is less than the radius of
convergence of f, and Cauchy's estimate for pnk is then
Mn/pk. Then for all bounded sequences{yj) we can apply
Lemma 4.14 and write the (f,dn,zl) transform of {5k} as

1 a) k
(f(zl) ‘9' dnT" kEO pnk 21 Sk

 

03 k
2 2 z (-1
(f(zl)—¥ an): k=o pnk 1.

 

55

2 CI) . CD

(4.17) (f(zl) + dnji jEO(-1)j(k§j(—1)k z: pnk)yj.

 

To prove the theorem, it is sufficient to show that this

transform is not conservative, and this will be done by

violating (0.1). The regularity of (f,dnfia) and (0.1)

together imply that

2 IC I = 1 u 2 Ip 2:) = 0(1) as n -¢ op.
k nk Tf(zl) + dnT. k nk _

 

Hence, we can invoke Lemma 4.14 again to write

 

 

 

 

 

 

 

 

 

 

 

 

2 cm co k k
, Z Z —1 —
Tf(zl)+dn]. j=o kaj( ) zlpnk
N 00
2 k
Tf(zi)+dnT° jzo k=j( ) Zl pnk
2 oo 00 k
. 2 Z -1 z =
|f(21)+dn]. j=N+1 k=j( ) 1 pnk
O 1 + . z 2; 21 p =
‘Tf(zl)+dnT"j:N+1 k=n nk
2 oo
0 1 . k-N =
( ) lf(zlfid I kji+1( )|21 p kl
2 N
0(1) Tf(21)+dnla kgo (k-N)l21 p k) +

2 00 'k ) k _
[f(21)+anT1kEO( -N '21 pnkI _

 

0(1) + Tf(E)EdT. C? (k‘N)|z§ P kl =
1 n ° k=o n

 

2 00 k 2N CD k
0 1 + I k - . =
( ) Tf(21)+dn]. kiol 21 pnkl [f(zl)+dn[.k‘:0|z1 pnkl

 

56

 

2 0° 1<
0(1) + Tf(zl)+d;le:;lk'zl pnkl # 0(1) by Corollary 4.2,

so (0.1) fails for the transformation (4.17) and the theorem
is proved.

Inasmuch as the Y-method is rather weak, the possibility
arises that an (f,dn,zl) method whose summability field
does not even contain that of Y might be convergencef
equivalent, i.e., that it might sum only convergent sequences.
Example 4.19 below shows that this is not the case.

The following theorem is known ([15],Theorem 2.3).

a:
Theorem 4.18: Suppose that g lf(17: dn[= oo.

 

 

CD (Ide )2
Z n
1 |f(1)+dn|2

 

< co, and that the Taylor expansion of f

about the origin has real non-negative coefficients. Then

(f,dn) is regular.

-1

Example 4.19; Let f(z) = Z3, 21 = 1, and dn = n2+1

 

 

It is trivial to show that (4.3) holds and to see that
f'(z1) # O and f(O) + d # 0. If fn(z) = Z a zk, then
n k nk

a = 1 if k = 3n and a = 0 otherwise. Then it follows
nk nk

from (2.3)that pnk = 0 if k # 32 for some 2 with

< < = ' =
0 _.£ _.n, and pnk On_k/3 otherWise, where 00 1 and
o. = Z d ... d for j > 0. From the definition of
-_.1 J
...<v.:n
J

oj and the fact that each dn < 0, it is clear that

57

sgn oj = {-1)}. Let k be any multiple of 3 for which

pnk # 0. Then either pn,k+3 = O or Pn,k+3 = On-k+3 .
3
so that sgn pn,k+3 - sgn On—k+3 = sgn On‘h._1 ' -Sgn Onih
3 3 3
= —sgn pnk° From this we have sgn[(—1)k+3 ] =

n,k+3

k+3 k+3 k
(-1) sgn pn,k+3 = {-1) {-sgn pnk) = (-l) sgn pnk =

sgn[(-1)k pnk]’ so that (-1)k does not change sign.

pnk
Finally, Theorem 4.18 assures the regularity of (f,dn), so
the hypotheses of Theorem 4.16 are met and (f,dn) is not

at least as strong a Y . Now define the sequence [sk} by

s = This sequence is

k

0, k is a multiple of 3
1, otherwise

easily seen to be summable to O by (f,dn), but it is

not convergent, and is not even Y—summable. Thus we have

Lemma 4.20: There is a regular (f,dn,zl) method which

 

is not comparable to the Y-method.

We will now give an example of a conservative (f,dn,zl)
method which is at least as strong as (C,1). The (C,1)
method is the Riesz method generated by the sequence {1].
From the proof of Theorem 4.5 we know that what we seek is

a conservative (f,dn.zl) method for which the transformation

(4.6) is conservative. Choose f(z) = z, 21 = 1, dn = n2.
. 4 L ch -. - =
Summation by parts shows t at 12((k+1)(pnk pn,k+1) i pnk’

so we have

58

1 1
(4.21) (1:5;77 i (k+1)(pnk - pnik+1) ‘ 1+dn 1 i pnk = 1.

If we assume for the moment that (z,n2) is conservative,

 

p
then lim 1+3k , must exist for each k, whence also
n n °
(4.22) lim (k+1)(pnk'pn,k+1) exists for each k.
n (1+dn).
n n

t+k2 t-1 <

On any compact set, | g 11k2|= l g( + 1+k2)l‘-

 

n ,_ a)
W (1 +'lE—£i)-: F (1 + B ) < a), where B is a bound
1 1+k2 1 1+k2

for |t-1| on the compact set. Then, if C is a circle of

radius p > 1 about the origin,

 

 

 

lpnk - pn,k+1l = l- f( 1 _ 1 )71' n t+V2 dt =
(1 + n2)E 2' C tk+1 tk+2 71TI+v2
1 n t+v2 M t- 1 1 .
2"; g :7; W dt : '2—‘f (3.1—£7} ldtl- -—]£:‘1l independently

 

of n. It follows that

1 k+1 _
(4.23) W 13014-1))ka pn k+1] : 0(1) ESE-:1 - 0(1).

The conditions (4.21), (4.22), and (4.23) are sufficient

for the transform (4.6) to be conservative. It only remains
to show that (z,n2) is conservative. We have (0.2) with

Q = 1 automatically, and because of the special choice of
f(z) and dn’ (0.1) is equivalent to (0.2) in this case.

To Show (0.3), we must show that
pnk

lgm (1 + n2):

 

l 1 l l ‘ I I fill. III I 41"

59

exists for every k. On any compact set the product

 

CD t+V2 _ if (1 + t‘]. )
1 l+v2 _ 1 1+v2

converges absolutely and uniformly. Thus if C is a

circle about the origin of radius >1, it is clear that

lim pnk a l' -l- _J;_. n t+v2 dt exists for each k
n (1+n25: gm 2ni é tk+1 K 1+v2 '

Hence, (z,n2) is conservative.
We now turn to the question of inclusion with regard
to the methods (z,dq,zl) and(f,d$,z2), that is, under what

I

conditions the latter will contain the former.

60

In [11], Meir proved the following theorem:

a) , a) . . _ ,
Let (dn}1 and (dn}1 be given With dn # 1 # dn.

oo _1 1 + dk <
Suppose § )1 + dnl = a) and 0 < 1_:—d; ._ 1 if

n.: no and 1 j_k :.n. Then (F,d$) is consistent with,
and at least as strong as, (F,dn).

By generalizing his techniques we can prove a much more
general theorem.

Define the linear operator E by Es = s 1, and de—
n—l

. 0 _ n =
fine E (sk) — 5k and E 5k E(E sk)

the (z,dn.z1) transform may be written as

n dk + 21E

= >
(4.24) tn Tlfdk + 21 so, n_o,
prOVided we define g d + 21 so = so

k
Since the (z,dn,z1) matrix is normal, it is reversible,

so there are coefficients bnm so that

n

= >
(4.25) sn § bnm tm , n._ 0.
m—o

Explicit formulas for these coefficients are in [9], p. 288.
From (4.25) we see that bnm = 0 if m > n. For conveni-
ence we also define bnm = 0 if n < 0 or m < 0.

From (4.24) and (4.25) we get

 

n n m dk + 21E
(4.26) ESQ =Sn= E, banIT'—d——_'_——Z-;So , n_>_Q.
m-o k
Then
_ n-i m d + z1E
(4.27) En 1so = s = 2 b w k so , n.: 1.
n-1 m=o n-1,m 1 dk + 21

61

Operating on (4.27) with 21E we have

 

n n—i m dk + 21E
21E so - zlsn _ § bn—1.mzlE I d” + 21 so
m-o k *
{21: b “1.1 dk + 21E
: _ - 21E 7T _.._._._____._‘ _ SO
m=1 n 1,m 1 1 dk + 21
n / d )m-i dk + z1E
= 2 b _.kZiE + W '—‘——‘ So
m=i n—i,m A m 1 dk +21
2 d m'l dk + 21E
_ b V SO
m=1 n-i,m-1 m 1 dk + 21
g b (d . m dk + z1E
= _ _ + 21) TI' —-----—— SO
m=1 n 1,m 1 m 1 dk + 21
nil b d m dk + 21E
—- I . 7T . s
m=0 n—i,m m+1 1 dk + Z1 0
n m d + 21E
= _ m-‘(d + 21) w E + so
n m dk + 21E 5
- -”__‘——— 0
Z bn-l m dm+1 Y dk + 21
m=o
n m dk + 21E
2 / 1‘ Q . 1 - >
g {bn-1,m—1\dm+2‘) bn—1,m dm+iJ Z d + 21 so, n “'1'
m—O k
It follows from this and (4.26) that
- 71 \ _ >
(4.28) bnm Z1 {bn-1,m~1(dm +21, bn‘llmchJfor m._ 0,
n 2.1.
It is clear from (4.25) that boo = 1.
Let the (f,dg,22) transform be
00 on m
t' = Z c“ s = z c' 2 b t , n.: O,
n m=0 nm m m=o nm k=o mk k
for those sequences {Sm} summed by (z,dn,z1) If the

62

order of summation can be reversed, then

CD 00 a)
... l = i . 2
(4°29) “n § ( é Cnm bmk)tk § ank tk '
-o m-k k-o
where
00
= a > >
(4.30) ank E Cnm bmk for n __O, k _.0.
m-k
For future convenience we define ank = 0 if n < O or
k < O, and c' = O for k < O.
nk

If (4.29) is a regular transformation, then (f,dfi,zz)
will sum every sequence (z,dn,zl) does, and to the same
number, that is, (f,d5,22) will be consistent with, and
at least as strong as, (z,dn,zl). We will show that if f

is a polynomial, then under suitable conditions (4.29) is

 

 

regular.
n k
Let (z + dn)i = kEo pnk z and (f(z) + d5): =
(I)
5 k
Then
n k (I)
t = , 2 p z s and t' = z p' z s
n (21 + an)a kzo nk 1 k n (f(22)+dfi)ik=o nk 2 k’

lll
...:
HI

sk E 1 gives tn , h tn.

values into (4.29) gives

so setting Plugging these

00

Z >
(4.31) kEo ank _ 1, n _.o.

In general, if f(z) = Bk z , then

Ob48

63

: (f(z)+d;1+1)i = (f(Z)+d' )2 p Z'

00 k
2 p3 z ) + d“ 2 p z
:0 nk n+1 k=o

m
+ p” )z , where q = Z 5-
m=0 nm n+1 nm nm j+k=m 3

fl = ( I >
(4'32) pn+1,m qnm + dn+1 pnm ’ n 1’ m._ 0'

Iv

, m
s‘ ‘ pnm 22 f 1
ince c = . ~ or >
nm (f(zz) + d$)£ n _, , we may use

(4.32) to derive

 

 

C5+1'm = (f(22) + da+1>i = (f(22) + d5+1)= (f(22) + da+1):

 

 

(4.33) c“ = , a a. .2 ,c' z +d' c'
n+1,m f(22)+d n+1 [ 2 j+v=m 6] nv 2 n+1 nm ]

for n 2.1, m.: 0.
From (4.30),
00 oo

= Z
m=k

k -v
c” b = 2 m , [22 Z 6.c' z; -
n+1,m mk m=k f(22)+dn+1‘ j+v=m 3 nv

 

a
n+1,k

+ dfl cl
n+1 nm]

or
1 00

m
4.34 a = , 2 z b . ' *V - -
( ) . . n+1 m=k{ 2 mkj+§=mfijcnvzz +dn+1ankl

 

64
If the matrix C' = (an) is row finite, i.e., if f
is a polynomial, then ta can be written in the form
(4.29). It is also clear that then Elankl < 00 for each

n, for suppose

m
(4.35) f(z) = § szv , 5m ¢ 0.

00
By (4.30), ank = jEk cnj bjk' But cnj = O for j > mn,
so
(4.36) ank = O for k > mn if n.: 0 and k 2.0.

From now on, we will suppose f is given by (4.35), and

m will be reserved for the degree of f. Define o =

 

 

n+1
da+1 + f(zZ). Then (4.34) gives
z-v+1 ,
an+1 k = _—{z szVk(BOCan2 V+61Cn V—122 +...+BVC no)
’ On+1 v=k
+ dn+1ank}
In View of the convention that céq = 0 for q <10 and
the assumption that Bj = O for j > m, we can write
1 z-v+1 . z-v+m
an+1 k = o { 2 22bvkmocn v 22 v+E31Cn, v-122 +""‘PBan, v-m 22 )
' n+1 v=k
+ dn+1a nk}
1 mn mn+1 mn+m
= {so 2 c b +51z2 z c' _ b +---+B 22 z c' b
On+1v-k nV Vk v=k n,v 1 vk m v=k n,v-m vk
+dn+1ank}

or

65

 

( 0 1 ( n1 j mn+j )
4.37 a = Z 6.22 2 c' .b + dI a , n.: 1,
n+1,k on+1 j=0 v=k n,v-3 vk n+1 nk
k 1.0.
Here it is necessary to introduce the notation
( ) Q
4.38 E d. . ...d. - 2 d. d. --- d. . 2.: 1,
q 11 12 12 q:i1_<_i2_:-.-:i,:0 11 12 12
and
Q
(4.39) 2 di ---di = 1 for arbitrary q and Q.
q 1 o
mn+j
We now examine .2 c' .b . When j = 0 this
V-k r1,v-j vk
mn
reduces to E cnv vk - ank' Suppose
v-k
. mn+j _.j+1 r_lj-r
(4.40) 2 c' _. - 213 z (-1) w (z1+d _ ) 9
V=k n'v j Vk r=1 v=0 k v
k+1 <
2 d. d a . for 0 i.’ —- ,
k-j+r 11 1r_1 n,k-j+r-1 J p
where
1 if Q < q,
Q <
(4.41) W (21 + d ) = 1 if v Z.k for q j.v —-Q,
v=q k-v
W (21 + dk-v) otherwise.
6.1.:va
v<k
b
(We remark here that if a 2.0, b > 0, then 2 di ---di =
a 1 v
b
2 di ---di .) Since we shall use the formula (4.28), we
1 1 v

suppose at first that k.i 1. Then

66

mn+p+1 mn+p+1

' = I '1
VEk Cn,v-p—1bvk yék Cn,v-p-1 21 [bV'1,k-1(Zl+d

- bv—1,k dk+11 =

k)

21 {(21+dk) g cn,v-p-1bv—1,k-1

mn+p+1

- 2 C' b ]
dk+1 v=k n,v-p-1 v-1,k

_1 ( )mn+p mn+p
21 { 21+, 2 C. - b _ - E C. b } .
dk v=k-1 n,v p v,k 1 dk+1v=k¥1 ,v-p vk
By the induction hypothesis (4.40), this is

p+1 p—r
z1 [zlp(zl+dk)*2 (-1)r w (z1+d
r=1 V=o

)

k-v-1

. O O Q . . a
11 1 n,k—p+r—2

p+1 p-r k+1

- r-1
21 2 ('1) ’n-(Zi+d )° 2 d. "‘d. - a }:

P+1 _lp*1-r .k

E (_1)r W (21+dk-v) ° 2 d. ---d.

- -1
Zip { 1 1 'an k+r- -2 _
1 V=° k+r-p-1 1 _ ' P

r

p+2 rp+1-r ~k+1

Z(-1) 7 (21+ )od 2 d. ...d. .a } =
r=2 v=o dk-V k+1k+r_p_1 11 1r;2 n,k+r-p-2

P

- _1 1
21p {(—1)p+ d£+1 a + w (21+d

° a
+1 nk =0 k—V) n’k_p_1

p+1 r_lpH-r k
2 (-1) w (z1+dk_v)°[ 2 d. ---d. +
r=2 v=o k+r—p—1

k+1

67

P
“9-1 _ p+1 p+1 .
21 [( 1) dk+1 ank + v:o(21 + dk-v) an,k-p-1
p+1 r_1p+1-r k+1
“T2 (‘1) W (21+ )- Z d. °--d. . a _
-p-1 P+3 r_1p+1-r k+1
21 2 (-1) w (21+ _ )- 2 d. ---d. -a _ _

so (4.40) is true for j = p+1 provided k 2.1. By in-
duction,(4.40) is true for 0 fi.j :.m if k 3,1. Now, if

k = 0, it is very easy to prove by induction that
- d1)J a for 0 :.j fi,m, so (4.40)

is valid for k 3.0 and 0.: j-ixn.

It now follows from (4.37) that

m 1 1+1 _ j-r
(4.42) 3.1.1,]. = 334 z 5433-) 2 (-1>r 1 4 (21.4%) -

 

n+1 j=o 3 21 r=1 v=o
kfi1 1
2 d. "'d. - a . + d' a for
k-jfir 11 1r_1 n,k j+r 1 On+1 n+1 nk

n.: 1,-k.: 0, under the notational conventions
(4.41) and (4.38).

Now, interchanging the order of summation in (4.42),

we write
1 m ' -r
1 m+ r_1 22 J J
218 I = Z d' a + Z (‘1) 2 B 0") 1T(21"’ _ )°
k n+1,k Ion+1| k n+1 nk r=1 j=r+1 3.21 v=o dk v
k+1
Z d. ---d. ° a . ,
k-j+r 11 1r_1 n,k-3+r-1

and, under the substitution t = j+1-r, this becomes

68

 

Elan+1,k|_ Uni-1 :2 dn+1ank +
m;1( 1)r__1m+;--J:B (qut+r-1 t7-;1(ZI+dk )
r=1 t=o t+r-1 21 v=o -v
k+1
k+§-t dil...dir-1 . an'k‘t

Again changing the order of summation inside the absolute

value signs we have

_ 1 .
3% an+1,kI - lon+1| i dn+1ank +
,m‘m+1-t_ 22 t+r-1 t- 1
tie r21 (-1> Bt+r-1(z1) V__o(zl + dk_,)
k+1
2d.-d I:
k+1-t 11 1r_1 n k t

 

 

 

 

 

r=1
m m+1-t t+r 1 t-1
2 z ( 1) B (52) (21+dk
t 1 r=1 t+r 1 21 v=0 -V)
k+1
>3 d. d f.
k+1-t 11 lr-1 n k t
T—l-rfw'fl <-1>“a i2—> " 6;? I kl +
On+1 k n r=1 r-1 21 1 n
m r-1 ‘_&)r )k+1
Z 2 —1 z + Z d. - d. a +
k t2: ) Br(z ( 1 dk 11 1r_1 I n,k-1l
m+1-t _1 22 t+r-1 t-1
I}: r21(-1)r Btu-JET) 7T (21+dk-v)

 

 

 

 

 

 

 

 

 

k+1
Z d. H-d. |a |+...+
k+1—t 11 lr-1 n -t
)m m-1
W‘Bm 21 v10 (21 + dk-v) lan,k-m|}
We recall that anv = 0 if v > mn or if v < 0, so
. m+1—t r-1 52.t+r-1 t-1
i r21 (-1) Bt+r-1(zl) ,:o(z1+dk_v)
k+1 I
z d. ...d. ola - =
k+1—t 11 1r—1 n,k t
mn+t mil-t )r-1 22 t+r-1 t- -1
Z Z -1 -- W Z +
k=t r=1 ( Bt+r‘1(21) v=o( 1 dk-V
k+1
z d. ...d. .la I =
k+1~t 11 J'r-‘1 n,k-t
'mn m+1—t t+r-1 t-1
r-1 ‘ 22 a H
kEo r21 ( 1) 5t+r~1gzl) v=o(21+dk+t-V)
ki1+t
Z d. °°-d. ° a .
k+1 11 lr-1 l nkl
Then,

1 mn m+1 r-1

l _ -1 22
Elan-n,kl :‘Ion+1| kEoildn+1 +r2£)1)r6r51'zi) dk+1

l( ‘ m( >r-1 (22)r k+2 1
21+ 2( -1 B ——- 2 d. ---d. +
dk+1) r=1 I 21 k+1 11 lr—l

t-1 m+1-t 22 t+r+1

I W (21+dk+t-v) Z (-l)r16t+r-1(z—) '
v=o r=1 1
k+1+t
2 d. ...d. l + --. +
k+1 ~11 lr—1

7O
m-1 z m
v:0(zl + dk+m-v).6m(3§o }|ank'°

Thus if we define

k t-1 m+1-t r-1 22 t+r-1
“-43) ¢t-~.- 71(21+dk+t—v)'§ ('1) Bt+r-1(-z-;) '
v-o - r—1 . .
k+1+t
2 d. ---d. , 0 j,t :.m,
k+1 11 lr-1
we have
1 m“ , k k
(4.44) Elan+1,kl :mkio{ldn+1 + ¢0l + l¢1l+...+

k k
|¢t| +...+|¢m|}|ank|

From (4.44) it is clear that if
1 k k k
(4.45) {ld' + ¢ |+|¢ |+...+|¢ l].: 1 for each k
|0n+1 I n+1 0 1 m
and for all large n, then it will follow that

(4.46) iiankl = 0(1) as n -> oo.

 

d... + ¢§ ¢E
In particular, if for all large n,‘ and -E—-
On+1 n+1

are real and all non—negative or real and all non-positive

for 1 :.t :.m, then if

k
¢t = f(zz) for each R,

"was

(4.47)

t 0

it follows from (4.44) that Z|a for all
a k

n+1,kl j'ilankl

large n, so (4.46) is true. -We remark here that from
(4.44) it is apparent that the above conditions are needed
only for 0 :.k fi.mn. *For the moment we will assume the

truth of (4.47).

71

Substituting t‘= j-r+1 in (4.42) gives

1 m 22 j
n+1,k Un+1 n+1 nk j=o j 21 t- o v=o dk v)

 

On+1 t= o j=t j :1) v-
k+1
Z d. d. . a
k-t+1 11 1j-t n,k t]

Thus if we define

 

 

k m ' 3
= ..1 3
6no 01(dn+1 + “E (q )ij(21) dk+1) and
n+1. —o
t-1
k j+t 22
<5 = ( 2 (1) B (1) Tr (21+ )
nt 0n+1 j=t v-o dk-v
k+1
2 di °° d1 ) for 1 < tim:
k-t+1 1 j-t
it follows that
m k
g .>
(4.48) an“,k £20 5“ amt]?t , n_>.1 and k_0.

Letting n = 1 in (4.48), we get

_ k . . k
azk " 510 a1k + 511‘ 31,14-1 + + 51m a1,k-m
Suppose
( ) n n k n k
4.49 a = a 75 + .2 6 - w 6. + +
n+1,k 1k v-1V° v=1 v1 v k-1 j;v+1 30
n n
k wk
2 6 w 6. .
v=1 vm v,k—m j- -v+1 jo

.l' I ..l' ‘1'! all! II. III III It. [I 1' ill. I )1 I
1"

72

in

k
d. = 1 when q < p.

where 30

":1

5:1?

From (4.48) we get

a = 5k .a + E 5k a
n+2,k n+1,o n+1,k t=1 n+1,t n+1,k-t
Using (4.49), this becomes
n+1 n n+1
k k a . k
a = a 4 W 6 .+ 2.6 a - ° w 5. + ... +
n+2,k lkflv:1 vo.,v=1 v1;v,k-1 j=v+1 30
n n+1 m
k k k
2 5 a ' W 5. + Z 5 a _
v31 vm v,k-m j=v+1 jo t=1 n+1,t n+1,k t
‘
n+1 n+1 n+1
alkvgl 6§0+ 2 651a v k— 1' w 6?0 + ... +
v=1 j=v+1
n+1 n+1
Z étm av k-mo. W 5k0
v=1 ’ j=v+1 3

By induction, (4.4?) is valid for n 2.1, k :.0.

 

 

k
8
Define 5k by 6k =' t , 1 :.t :.m. -Also, let
t nt Q
n+1
1 m
k _ k _ . _
n+1 j:
d' + f(z ) - d' - f(--za ) = f(z ) - f(-‘-z-"’- ) is
n+1 2 n+1 21 +1 ‘ 3 ' 21 dk+1

independent of n.
Suppose for each k.i 0 there is an nk such that
> < k -
for n _.nk we have 0 _.6n0< 1. vThen we claim
k

n n
(4.50) |an+1’k| :Akolz lov+1| ) .exp[-ak>1: |0v+1I } for n
' k

H.

Iv lv
o

73

 

where Ak and “k depend only on k, and o < ak.: lTk“
For every real x, 1 + x i.ex, so for k 3.0,
n n n ‘T
k _ k _ .' k <
(4-51) lvzj évol - V:.[1-(1-évo)] - 1T (1- O )_

J v=j v+1

. '1 . .
exp {-(Tkl i |ov+1| } 1f 3 > nk and

n
zlov+1|-1 - 0 for 3 > n
Suppose n > nk, 1 2.1 j.nk, and q :_0 is fixed. Then,
by (4.51).
n n
lvélfiol-lvrp‘éljorl w atolz<max1"h‘;|>.
v=i v-i v=nk+1 1.23—fink v-j °
n _1 n _1
exp {—|1k| Z |0V+1| 3 E'Qk exp {|Tkl 2|Uv+1l ].
n + 1
k
n n
_ -1 . 1 '1 q _
exp I ltklf lov+1I ) lbz-llq (§ IOv+1l )
n
Bk(§ |0v+1' ) exp {flrklg |Ov+1l 1}, where Bk
depends only on k. 0n the other hand, if 1 j.i :.n §.nk,
then
n j nk
k < k "1
F 6 _.( max w 6 ) exp { Z 0 }
lv=i vol likijhk‘wz vol Irkll I VHI

I")

n _1 n
exP{-lTk|§ l0v+1I } f"Mk exPI'1‘k|§ IOv+1

I") =B"(§|o 1'1)le {-11 £10 r1
k 1 V+1 p k 1 V+1 }'

74
and Bfi depends only on k. Finally, suppose

1.: n < i < n Then

— k.

lvziévol = 1 "'Bk(n)(§‘°v+1l ) eXp{mlfitklflgxwil ]
if -Bk(n) is sufficiently large. Let Bfi' = max Bk(n)
' 1<n<nk-1 .
Now
n n q n
k I” ‘1 "'1
lvgiévol ‘uBk (§|0V+1I ) eXP{‘ITkI§IOV+1| 1 for

1 fi.n < n and n < i.: n and B"'

k k’ k depends only on k.
Let Bk = max (Bi.~ k' Bk). It follows that there is,a
constant Bk’ independent of i, such that
n k < n -1 q n -1
(4.52) |v115v0| _. Bk(§|ov+ll ) exp{-|Tk|§|ov+ll } for

n 2.1. k 2.0. if 1 :.i Sink.

n
Now let k = 0 in (4.49) to obtain an+1,o = alovzl5go.
An application of (4.52) with q = k = 0 gives

h -1
lan+1,o|= |a10| |V5% I -'a10lBO eXP[ ITOIZ|OV+1| ). n 3.1.
If .A0 = |a10|Bo and a0 = {To|, we have (4.50) for k = 0.

Suppose (4.50) is true for columns 0, ..., k-l. From

(4.49) we have

n n n
k k {k
|a n+1, k| é.|a1kv=w15vo| +vE |6v1av,k-1j $+16jol + ... +
n n n
2|5k W 5k | = |a1k v 5k o| + T1 + ... + Tm'

v=1 vma v, k- —mj_ -v+1 JO v=1

75

In view of our convention regarding anp’ Tr = 0 if r > k.

Thus we suppose r fi.k. Let

n n nk'l n

k k k k

T = Z |5 a v 5. | 2. 2 I5 7 5. 0| +
r v=1 vr v, k- ~rj_ v+1 jo v=1 vr av.k- 'rj- v+1 j0

n k
W éjOI = SI + 52, where 32 = 0 if

n
2 l Era v, k- -r.
v=nk j=v+1
< k -1 k
nk > n. By (4'52)' 51 “'nk( max lévrav, k- il>B k<zlov+1l '
1_.v_nk
< <

n _1 n _1 k n _
exP['|?kl§|Gv+1l } =Ck(§l°v+1l ) eXPI‘Itkl§l°v+1l

We remark here that we may assume nk 3,2, and, if it is
00

further assumed that § I0 I-1

v+1 = a), that nk 18 so

n

k _
large that 2 lo I 1.: 1. ‘Now we apply the inductive

1 .

(

j+1
supposition 4.50) in conjunction with (4.51) to get

v-1 _1 k-r. v—1 - _1
$2 j'EnklévrlAk-r( Z loj+1|‘ ) exPI-ak-r § |0j+1l ] °

n k n
}‘:'Dk(§loj+1J-1) Z“l6vrl .

I-l)}, where D .= max Ak-r

l(Iv-t1 k

and Gk = Bi%(%_f lltkl). But O1<IGV+1|-1 : lTkl.|OV+1I_1

< n k“ n 1 k
52 _.eDk(ZI°3+1) ) exp [ akzloj+1l ] Z lévrl
v-nk
n _1 k-I n -1
= eDk(2]OJ+1l ) eXPEQkZ|GJ+1I 3 8r vinklov+1l
n _1 k-r+1 n _
«6 eDke (§l°3+1l ) exRL-ak ZIOJ+1l 1}

where E = max eD s
1‘ 1ir_‘m kr

of r. From (4.52),

and the inequality is independent

n k < n
Iaikvzlévol “'Fk(§]°

_1 k n _1
j+1l ) efo‘ITkl§l°j+1l ]} It now

follows immediately from the estimates for $1, $2, and

n

lalk w 5§0|, since they are independent of r, that (4.50)
v=1
is valid for the kth column and thus for every column.

As a consequence we have

(4.53) lim lan+1,kl = O for k 5,0
n-eoo

a)
provided 2 lo
1

=(I).

... I
The formulas (4.31), (4.46), and (4.53) are the
regularity conditions for the transformation (4.29).
It remains to prove (4.47). The following notation

will be used throughout the proof:

77

b
2 d. ° d E Z d - d. if r.: 1,
a 11 J'r ajiriiwi 11 1r
..<i Eb
b b
2 d d. - 1 = d ° d I Z d ° d = 1 = d 0 d I
a 11 lo 11 10 a P1 P0 P1 P0
b
2 d --~ d s 2 d --- d if r.: 1.
a p1 pr . ajplspng- p1 pr
...<prjb

By definition,

k _r-1 .m-r+1 . j-1 52 r+j-1
¢r - I (21+dk+r-v) .E {-1) Br+j‘1(21) .
v-o 3-1
k+r+1
Z d. -- d . 0 i.r i.m But
k+1 11 lj-1
r-1( ) k+r ( ) r r-v k+r
W 21+ _ = W 21 + d. - Z 21 2 d "' d ,
V=O dk+r V k+1 J .V=0 k+1 p1 pv
so
r k+r m-r+1 . r+j-1
k r-v 3—1 22
m = (.2 21 2 d --- d ) 2 (-1) a ._.(——) .
r v=o 'k+1 Pl Pv j=1 r+3 1.21
k+r+1
2 d. --- d. .
-k+1 11 lj-1

.+1
Thus ¢t is a sum of terms of the form (-1)3 T(r,v,j,P,I),

where
r+j-1
. _ r-v 22
4.54 T r,v, ,P,I - Z —— . ° ’6.
< > < 3 ) 1 (21) Br+3_1( p Pv)
(d. °°' d. )o P = (P1."'.P )I
11 lj‘l V

and .I = (i1, ---, i. ) , subject to

78
(4.55) Ofivirim. lijZm-r+1,k+1.ip1 <Pzz<

< pV :.k + r, and k + 1.: 11.: 12.: oo-

IA

i. j,k + r + 1.
j-1

-In fact, ¢: = .$_‘.(1-1)3-1 T(r,v,j,P,I), where the sum is taken

over all values of v,j,P, and I subject only to the

conditions (4.55). We will consider T = T(r,v,j,P,I)

and T' = T'(r',v',j',P',I') to be distinct unless they

have the same form, i.e., unless r'-v' = r - v, r' + j' =
..

r + j, and the sequences {pi,---,p;,.i;,---,1j,_1} and

{P1.°--.pv.il.-'-.ij_1} 'ara permutations of each other.

Clearly (-1)3-1 T(r,v,j,P,I) reduces to BIZ: if
v = 0 and j - 1, so that to prove (4.47) it is sufficient

to show that

(4.56) Z] (-1)3 T(r,v.j.P,I) = 0,
0.3.3“
(v.j>¢<o.1>
where the sum is over all r,v,j,P, and I satisfying
(4.55), except that the ordered pair (v,j) # (0,1).
(For convenience we shall use T for T(r,v,j,P,I)
where no confusion will result, and we will write ‘T 6 ¢:

to indicate that T is one of the terms comprising 5:.

'We will also assume (v,j) # (0,1).

.Lemma 4.57: Suppose T e ¢§ , where T is given by
(4.54). If j > 1 and there is a Y with 1 j_Y.: j - 1

such that i? # pu .for every n in the-range 1._ “.2 v,

k

then T e ¢r+1'

79
Proof: By (4.55), a consequence of the assumption
that j > 1 is that r < m. Hence, we may set r'= r+1,

j' = j-1, and v' = v+1. Since k + 1 j'iY fi,k + r + 1,

we may assume that there is a largest #1 such that

0.: ul.: v and pu < i , if we define p0 = k. Choose
1

Y
I: ' < < n =' I: °
pu pu 1f 1._ u._ #1 , p ”1+1 ly’ and pLL pp:1 1f
H1 + 2.: u.: v' = v + 1. Choose i; = in if 1 j,u,j,Y-1
'l=' ' < <'q_='_. ' '
and 1u 1L”1 1f Y._ u,_ j 1 j 2 It 18 ea81ly

seen that the conditions 4.55 are satisfied for v',j',r',

pi, and ii. Moreover, r' — v' = r - v, r' + j' = r + j,

and the sets [pi,...,p9,ii,...,inj._1} and {p1,...,pv,
k k

} are the same, so T' = T, T' e ¢r' =,¢

11..-.. .r+1'

1.
3-1
The construction of T' from T is illustrated by the

diagram below:

a
A
Q
o.
a
:1
a
P
o.
o.
h.

f1 pm ph‘fl pv 1\1Y:11\Y:Y+% ij:1\)
T'-: (di...d,d, d, ..d,) i....di, d1....i
p H1 p u1+1 p M1+2 pV' 1 Y'1 Y

It should be observed that the above lemma is true even
if the last hypothesis holds vacuously, i.e., if v = 0.

For convenience, define p0 = k.

emma 4.58: Let T be given by (4.54) and let

)

T e ¢ Suppose that either v = 0 or v > 0, j > 1, and

IA H:#

i.

3-1 pv. Define r" = max (r — v, pV - k - 1). Then if

80

r'.: r, in order that T 6 ¢:, it is necessary and suf-

ficient that r'.i r".

Proof: We prove the necessity first. If T 6 ¢:..
then T ='T'(r',v',j‘,P',I') and r' - v' = r - v,
r' + j' = r + j, and [pi,...,p;,, ii,...,i'j._1} must be
a permutation of (pl....,pv, i1,...,ij_1]. In particular,
if r' = r - q, then v' = v - q .and j' = j + q. 'But

vfl.: 0, so q.: v, and thus r'g: r~- v. .If v = 0, the
proof of the necessity is complete. Thus suppose v > 0,

j > 1, and ij-1 j'pv' -Since v' = v-- q, at least q

d 's must be rewritten as d.,'s. If d is so rewrit-
P ly P
u v
ten, then by (4.55), pV fi,k + r' + 1, whence r' Z.pv --k - 1.
If (i is not so rewritten, then d = d , for some n,
PV Pv Pu

so pvfik+r', or r'ipV-k>pV--k-1.

The sufficiency is trivial if v = 0, so assume v > 0.
In order to prove the sufficiency of the conditions, it is
enough to show that T can be re—expressed as T'(r',v',j',

P',I') with r',v',j',pfi, and i? in agreement with

W
(4.55). -We will so rewrite ~T. Let r' - r - q.§ r", and

set v' = v - q and j' = j + q. >Since r - q.ipr".:

r - v, clearly q fi.v, so v' 2.0. Set p; = pu if

1.: u.: v' = v - q. Let the elements of the-sequence

{pv_q+1lpv_q+2to-Oopvllll00$1j_1} be partially ordered
according to magnitude, and let (ii, i5, ..., i'j,_1) be

81

. -_ _ _ '<'= <
the resulting (q + j 1) tuple. Now, pu._ pv, pv-q —

- q j,k + r - q = k + r' for each u. By hypothesis,
r' Z.r" :,pv - k - 1, so pV j,k + r' + 1; but then

i' §.pv fi.k + r' + 1. The other conditions of (4.55) are
obviously satisfied, so T.= T' 6 ¢:,. -The construction

of T' from T is illustrated below:

T : (d d )

i.
3-1

... d d ... d )(d. ...
P1 Pv-q pv-q+1 Pv 11
1...1(...1/...)

T' : (d d , )(partially ordered by magnitude)

pi pV

Lemma 4.59: Let T = T(r*,v*,j*,P,I) be given, and
k .
suppose r* is maximal such that T e ¢r*‘ ~Then if 3* > 1,

each 1w = pu for some n.

Proof: It follows immediately from (4.55) that
r* < m. If the conclusion were false, Lemma 4.57 would

k ,
say T e 5 .thereby violating the definition of r*.

r*+1'

Corollary 4.60: Let 1T = T(r*,v*,j*,P,I) be given,
. . k
and suppose -r* is max1mal such that T e ¢r*' 'Then
T 6 ¢: if and only if r* :.r > r" = max(r* - v*,

pv* -k-1).

Proof: If j* > 1, then Lemma 4.59 implies v* > 0

. < k
and 1j*-1 _.pv*. But then :Lemma 4.58 says that T e ¢r

if and only if r Z,r". ’Now suppose j* = 1. Since we

are concerned with -T's .for which (v,j) # (0,1), j? = 1

82

implies v* > 0. Then Lemma 4.58 still applies to say

T

¢t if and only if r,: r".

Now let T = T(r,v,j,P,I) e ¢§ , and let tr be the

number of distinqt subscripts among {p1,...,pv,i1,...,i. }

5 3-1

which are :.k + r. We claim that the number of times T

k t

appears in 5 is ( r). ~Since r is fixed and T is

r V

fixed, so must v and j be fixed, i.e., if T'(r,v',j',

P',I') e ¢t and T' = T, then v' = v and j' = j.

Thus for any such T', P' = (p' ..., p5), and all the

1!

elements in this v-tuple are distinct. The number of ways

of forming such a v-tuple from tr distinct elements is

t
V

as

PI

If
r*
30

TI

( I). But T' is completely determined by P' inasmuch

there is a fixed set of subscripts from which to form
and I'.' This establishes the claim.
Suppose r* is maximal such that T(r*,v*,j*,P,I) 6 ¢:*.
j* = 1, it is obvious that tr* 3 v*; if j* > 1, then
< m and Lemma 4.57 says every iY = p for some u,
again tr* = v*. ’For each u, let pH = k + Q“. rLet

= T'(r'.V'.j'.P'.I') 6 CD1;“ where r'.: r* and 'T' ='T-

Then v'.i v* and p;, :,k + r', so we see that tr, = v*

provided pv* = k + av* fi.k + r', i.e., provided r' Z,a

. . = * _ . =
Similarly, tr' y 1 prOVlded pv*-1 k + a

v*-1._

k + r', or r' Z.a . We have—shown

(4.

- * * > > a, o = * _-
61) tr v for r __r _. v*' tr v 1 for

provided T e ¢kr for all

such r.

83

Let T(r,v,j,P,I) e 5:. Then the sign preceding T

is (—1)3-1, so all occurrences of T in ¢t have the

same sign. If T'(r-1,v',j',P',I') e ¢:_1 and T' = T,
then j' = j - 1, so the sign preceding T' is (-1)3-2,
and the sign preceding each occurrence of T in ¢:_1 is

opposite that in 5:. -To prove (4.56), then, it suffices
to show that for each fixed .T(r,v,j,P,I) for which

(v,j) # (0,1), the relation

Dru-1 t a

z k(-l) ( r ) = 0 holds, where T' =
I v I
T €¢r, r
T'=T
‘I" (r',vr,,jr,,Pr,,Ir,) = T, and the sumis overall values

of r' for which T 6 ¢:,. Equivalently, since 'Vr’

changes as jr' does, we may show

V I t I
(4.62) z <-1) r (r > = o.
k vr'
T'€¢ ,
r
T'=T

By Corollary 4.60, if r* and r“ are, respectively,

the largest and the smallest values of r' for which

T E 5k , , then r" = r* - v* or r" = p * - k - 1. Sup-
: v
pose first that r" = pv* — k.- 1 = av* - 1. -Then if
u u u -u n u k u =- u — _ _
'r (r .v .3 .P .I ) € 41... and -T T. v - V* (13* (Iva.

+ 1), so, in view of (4.61), the relation (4.62) becomes

V* * *_ *_
(4.63) 2 (-1)V(V ) + {-1)V (r av*+1(
V*-(r*-a *) V
v
v* - 1
a = O.

84
If v* = 0, then, since (v*,j*) ¢.(o,1), j* > 1. But this

Y

Thus v* > 0, and also av* 2.1. ~By (4.55), av*.fi r*, so

violates Lemma 4.59 since there are i 's but no pu's.

v* - (r* - a ) :.v*. We now need the formula

V*

b

(4.64) (a) = (b-1 b‘1

a ) (a-1)' a and b are non-negative

integers, b :,a.

If r* - av* = 0, then the left side of (4.63) reduces to
v* v*-1 .

(-1) + (-1) = 0, and (4.63) is true. Suppose (4.63)

is true for r* - a = q, and let r* - a = q + 1. Then

v* v*

using the supposition and (4.64), we write for the left
side of (4.63),
V*

v* - 1
2 <-1>V (3*) + <—1)"*'q"2 ( =
v*-q-1 ' v* - q -'2

v* ‘ '

v v* v*-q-1 v* v*-q v*-1 _
Lay“ <. > + (vs-1)] + (..-.-.)-
* v*-1 *_ _ .V* _ v*-1
(-1)V ‘q( +<-1>" q 1( +<-1)"* q( =

v*-q-1 v*-q-1 v*-q—2

* ‘v*-l *_ _ v*-1 v*-1
(-1)v -q( ) + (-1)v q 1E )+( )1 +
v*-q-1 v*-q-1 v*-q-2

( v*-1
_ v*-q =
( 1) V*_q_2 o.

By induction, (4.63) is true regardless of the value of

. Now assume that r“ = r* - v* > a - 1. Then

*-
r (1 vi-

V*
v" = 0, so the left side of (4.62) is

85
v

V* * *-
g (-1)V (V.) = (1 - 1)v = 0.

Hence, (4.62) is true, and thus also (4.56).

We have proved

 

 

 

m
Theorem 4.65: Let f(z) = 2 szv and let
0
On+1 = dfi+1 + f(zz) # 0. For 0 j,t j,m, define
t-1 m+1—t t+r-1
k r-l 22
CD = 71’ (21+ _ )9 Z (-1) B _‘("‘") -
t v=o dket v r=1 t+r 1 21
2 d. °-- di ,
k+1jir2--- 11 r‘1
<- <
_ir_1_k+t+1
-1
... = = >
where 2di1 dio 1 v:d(21 + dk-v)' Let N _,0 be
arbitrary, and for all n > N, let ' (dn+ + $2) 3.0
n+1 1
¢k
for Oikimn and :0 for litim and
n+1

0 fi.k :.mn. ~Furthermore, aSSume that for each R 2.0 there

ex1sts nk > 0 such that O._-g——- (d n+1 + ¢o) < 1 for
n+1
-1
> ' I _. . . I
all n _.nk. Finally, let §|dn+1| — oo. Then (f,dn,zz)

is consistent with, and at least as strong as, (z,dn,21).

00
We have used the fact that 2 ]on_|_1|-1 = 00 if and
1 4 .
a) -
only if 23|d3+1| 1 = oo. It should be observed that since
1

z
¢% = f(- 2%.dk+1)' the hypotheses can be cast in a slightly

different form by making this substitution.

86
We will now show that Meir's theorem is contained in
Theorem 4.65. Choose f(z) = z, m = 1, 50 = 0, $1 = 1,
and 21 = 22 = 1. Theorem 4.65 then reduces to':

Let 1 + dfi+1 # 0 and suppose

dn+1 dk+1 > 0

I
dn+1

a) for all n > N, N 2.0 arbitrary,
for Ofikin;

b) for each k 2.0 there is nk > 0 such that)

<dn+1-dk+1
0._ d' + < 1 for all n Z.nk .

+1
1+4k
+1
c) forall n>N, 1111' :0 for of.

d) z |d' n’1+1|

Then (z,d$) is consistent with, and at least as strong
(z,dn).
A moment's thought confirms that the hypotheses (a),

(b), and (c) are consequences of the single hypothesis

dn+1'-d'k+1

(e) for all n > N, N 3.0 arbitrary, let 0_ < d' +1 + 1 <

for 0 i.k :.n

But (e) is equivalent to

 

1...qu

(e') for all n > N, N 3.0 arbitrary, let 0 < 1 + d'+

for 0 :.k :.n.

Thug,substituting the (stronger) hypothesis (e') for (a),

(b), and (c), Theorem 4.65 takes the form:

87

 

Let 1 + d5+1 # 0. For n > N,«N 2.0 arbitrary, let
1 + d 00
0 (1‘4 d'+ 1.1 for 0 __k _.n. Let 2 ’dn+1l — oo.

Then (z,dr'1)1 is consistent with, and at least as strong as,
(z,dn).
This is Meir's Theorem.

szv, Bv > 0, 22 >0,

u
OMB

Corollary 4.66: Let f(z)

l-1 = d). Then

2 > Id | d < 0 d' > 0 d EPIdf
1 n ' n ' n _' ' an 1‘ n+1

(f,d$,zz) is consistent with, and at least as strong as,

(z,dn,zl).

 

 

 

 

 

Pr f- d' + 2 2V > 0 ¢k = f(- 31
-—-29-” On+1= n+1 0 Bv 2 ’ 0 21*dk+1) ’
Idk+1|) > 0. Moreover, w (21 + dk+t-v) > 0
y=0
Also, (_1)r-1 d. --- d. = Id. ---d. |, so
11 l 11 l
r-1 r—1 k
r-1 dn+1 + ¢0
(-1) Eldi --- di > 0. It follows that 0
1 r-1 n+1
k
¢ . 22
and ‘ are ositive. Finall , ——- < z , so
n+1 p Y zl|dk+1| 2
k _
d' + «v d' + f(- ).
0 .
de+1 < f( (22). Then n+1 = n+1 +1
‘ On+1 On+1z
dé+1 + f(Zz)
< ' 0 = 1. The-hypotheses of Theorem 4.65 are
n+1

satisfied, so the result follows.

88

m
Corollary 4.67: Let f(z) = z 6 zv, 5 > 0, d < 0,
0 V V n

z .
1 > o, c1;1 < o, a“ 1 f(l). Then (f,dr'1,zz)

is consistent with, and at least as strong as, (z,dn,zl).

 

. = n I >
Proof. on+1 dn+1 + f(zz) > dn+1 + f(l).. 0.
34 > “2.2. _k> > .
Moreover, zlldk+1 ._ 1, so f( Z1 dk+1) — ¢o._ f(1)_ Idnl,
dfi+1 + 4:
so 0 3.0. The remaining hypotheses of Theorem
n+1

4.65 follow as in the proof of Corollary 4.66, except for

a)
1 . . .
the condition 2 , = Q). But this is a tr1v1al con-
- 1ldn+1l
sequence of |dnl‘: f(l). The result follows.

m
Corollary 4.68: Let f(z) - 2 B zv, B > 0, d < 0,
0 V V n

f(l).: 21 > ldnl': 1. Then (f,dn,zl) is consistent with,

and at least as strong as, (z,dn,zl).

Proof: Corollary 4.67 with 22 = 21 and d5 = dn'
The above corollaries require that dn < 0. Now sup-

pose that 21 > 0, 22 > 0, 0 j'dn 2.21 :.B with 3.: 1, and

k+t+1 v—t v-t
d' 3, max |f(z)|. Then 2 di ... di 3,3 (t + 1)
|Z|:22 k+1 1 V-t

ism-1(m+1)m'1=m if ljtim. For 1_<_t_:m, let

22 j m m
at :,M{ max (5—) ] 2 B , where 2 6V = 0. Then
1jj:,m-1 1 t+1 V m+1
t-1 m+1-t t+r-1 k+t+1
k r—1 z2
¢ = w (21+ _ )° 2 (-1) B _ C——) - 2 d. ---d.
t v=o dk+t V In: t+r 1 21 k+1 11 lr-1

k+t m v-t 22 v k+t+1
= w (21+d )- 2 (-1) V(z ) 2 d1 di
k+1 v=t 1 k+1 1 v-t
k+t t m v k+t+1
z -t z
= 1r (21+dj)-[Bt(;1) + 2 (-1)V avg-1) - 2 ‘31 ...di 1
k+1 1 v=t+1 1 k+1 1 v-t
22 t k+t m 22 v—t k+t+1
343—) vr (21+d-)°[Bt- >3 Bv(;—) ~ 2 di ---di 1
1 k+1 3 v=t+1 1 k+1 1 v-t
22 t k+t 22 j m
> (-z-—-) 7r (21+d ) {at - Mt max (32—) 1 2 6V} : o.
1 k+1 1_<_j..‘_ m-l 1 t+1

2 +1 z2 V v
Moreover, Efdk+1 = CEE——)zz.i 22, so 6v(2:d ) ,2 5vzzo

 

 

21 k+1
Then
k 22 m 22 V m V . >
$0 _ f(— 21. +1) _ g BV(- E:dk+1) < § szz — f(zz) Since HLJH
k z k
' + d' + - ¢
Thus 0 < E£i£_:2 - n+1 f( zl +1) < 1 and t > 0.

_ . _.
On+1 dn+1 + f(zz) °n+1

Theorem 4.65 now yields

m
Corollary 4.69: Let f(z) = z 5V2“, 21 > o, 22 > o,
O

0.: dn.i 21 fi.B with B 2.1, d5 2. max [f(z)], and
)2): z2 A
_ _ z j m
at 3.3m 1(m+1)m 1[ max (23) ] 2 av for 1 :,t :,m,
1i.jEm-1 1 t+1
m 00 -1
where 2 B = 0. Then, if 2 (d' ) = a), it follows

m+1
that (f,d$,zz) is consistent with, and at least as strong

as, (z,dn,21)o
From Definition 3.6 it is easily seen that the Euler

method (E,p) is defined by the transformation

and this

in turn is seen to be the (z,l§2) transformation. Setting

21 = 1 and dn E léfl. in Corollary 4.69 gives

v
z z > 0 d' >
6v ’ 2 ’ n

OMB

Corollary 4.70: Let f(z) =

- m
max |f(z)|, and B Z.(m+1)m_1[ max 23] Z 8 for
< t < - V
14"!ng 1_3_§n-1 t+1
< m 00 -1
1 _.t _.m, where 2 6V = 0. Then, 1f f (dn+1) = 00 and
m+1 .

l':.p.: 1, it follows that (f,d',z ) is consistent with, and
2 n 2

at least as strong as, (z,dn,zl).
It is well-known that the (E,p) method is regular

([8], Theorem 117). Consequently, Corollary 4.70 gives

 

m
Corollary 4.71: Let f(z) = 2 5V2”, 21 > 0, an 1
o
m-1 j m
max |f(z)|, and at :.(m+1) [ max 21] 2 av for
12 l5- Z1 1_<_j_<m-1 t+1
< m 00 _q_
< ‘ = o = o
1 _.t _.m, where 2 8V 0. Then, if § dn+1 a), 1t

m+1
follows that (f,dn.zl) is regular.

10.

11.

12.

13.

14.

15.

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