ON A GENERALIZATION 0F HAA‘R SERIES

Thesis for the Degree of Ph. D.
MICHIGAN STATE UNIVERSITY
MELVIN ANDREW NYMAN
1972

'.‘. ..........

   
 

LIBRARY

Michigan State
niversity

    

g“ n A ‘5.

"-2

4!

This is to certify that the

were
‘ '4‘“? 4W;
"i"

thesis entitled
ON A GENERALIZATION OF HAAR SERIES
presented by
Melvin A. Nyman
has been accepted towards fulfillment

of the requirements for

Eh.D. degreein Mathematics

 

#614 5/! 61/2; lg -

Mnjor professor

 

 

ABSTRACT

ON A GENERALIZATION OF HAAR SERIES

BY

Melvin Andrew Nyman

For each natural number n let Dn be a

set {tn,0'tn,l"°°'tn,rn}' where

n,O ,r

O = t < tn,l<°'°<tn n = l.

m
Assume that Dl C'D2 C... and that D = L] Dn is
n=l

dense in [0,1]. For each natural number n let 3h

be the system of all intervals [t t .] (j = l,...,rn).

n'j-l' 11,]
A function f is called regular if f is of bounded

___ fJx+) H? (x-)
2

variation on [0,1], f(x) for x e (0.1),

f(0) = f(0+) and f(1) = f(1-). For each n 2_l let
8n be the vector space of all regular functions on
[0,1] which are constant on the interior of each interval

from 35. Set T1 = 81 and

1
Tn = [f e snzfgfg = o for all g e gn_l} (n = 2,3,...).

A K-series with respect to {Du} is any series of the
m

form where fn 6 Tn (n = 1,2,...). If

f .
n=l n

X0,xl,... are the Haar functions and a0,al,a2,... are

m
numbers, then n§13n_lxn_1 is a K-series (the corresponding

Melvin Andrew Nyman

sequence of sets IDn} are easily constructed). An

analogous assertion holds for the Rademacher functions.

The Fourier—K-series of a function which is
Perron integrable on [0,1] is defined. It is proved
that if f is Perron integrable and sn(f) is the nEh-
partial sum of the Perron-Fourier-K-series for f,
then I; sn(f) = f f for every interval J of
3h (n = l,2....).J From this it may be shown that the
Perron-Fourier-K-series of a Perron integrable function
converges almost everywhere to the function. It is
also proved that the Fourier-K-series of a Lebesgue
integrable function converges to the function at every
Lebesgue point of the function. It is shown that if
f G Lp[0,l] (l g_p < m), then the Fourier-K-series
of an arbitrary finite Borel measure on [0,1] is

defined and the behavior investigated.

A uniqueness theorem for K—series which

generalizes those known for Haar series is proved.

Let f be a finite function on D. Derivates
of f 'with respect to {Dn} are defined. Using the
properties of such derivates, a class of K-series is
found such that no member of the class has an infinite

sum on a set of positive Lebesgue measure.

Melvin Andrew Nyman

Let f be a bounded function on [0,1]. Let

co

' fn be a K-series with respect to {Dn}. Conditions
n=l

on the sequence {Dn} are found so that if

E fn(x) = f(x)

n=1

for every x 6 [0,1], then Z) fn is the Fourier-K-series
n=1

for f. Examples are given to show that this need not

always be the case.

ON A GENERALIZATION OF HAAR SERIES

BY

Melvin Andrew Nyman

A THESIS

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

DOCTOR OF PHILOSOPHY
Department of Mathematics

1972

TO ARLENE

ii

ACKNOWLEDGMENTS

I wish to thank Professor Jan Marik, under
whose guidance this thesis was prepared, for his
help and encouragement. I am sure that without it

this would never have been completed.

iii

TABLE OF CONTENTS

CHAPTER I
Introduction . . . . . . . . . . . . . .
CHAPTER II
Definitions and Some Theorems on Fourier-
K-Series o o o o o o o o o o o o o 0

CHAPTER III
A Uniqueness Theorem for K-Series . . .

CHAPTER IV
Existence of Infinite Sums for K—Series
on Sets of Positive Measure . . . . .

CHAPTER V
Remarks on Everywhere Convergence . . .

Bibliography 0 O O O O O O O O O I O O O O 0

iv

CHAPTER I

INTRODUCTION

Consider the Haar, Walsh and Rademacher systems
of orthonormal step functions. The first two of
these are complete in L2[O,l] and the last is not,
Walsh's system being the completion of the Rademacher
system. The similarities and dissimilarities of
these systems with one another and with the trigonometric

system have been investigated by several mathematicians.

The basic elements of the theory of Haar-Fourier
series as well as a complete discussion of the relation-
ship between the Haar, Rademacher and Walsh systems may
be found in the book by Alexitis [1]. Fine [4],[5]

has developed much of the theory of Welsh series.

In this paper the notion of K-series is defined
and is shown to include series with respect to the
systems of Haar and Rademacher. Since the partial sums
of order 2n (n = 0.1.2,...) of a Walsh series may be
written as the Znth_ partial sum of a Haar series
(see [1], pg.62) the convergence theory of K-series

also includes these convergence questions.

Wade [14],[15] has given conditions under which
a Haar series is the Haar—Fourier series of a Lebesgue
integrable function. In Chapter III, we are able to
prove more general theorems for K-series and deduce

stronger versions of Wade's theorems as corollaries.

In Chapter IV, the question of whether a K-
series may have an infinite sum on a set of positive
measure is considered. Talalyan and Arutyunyan [13]
have proved that this is impossible for Haar series.
Skvorcov [12] deduces the same theorem by another
method. Using the method of Skvorcov we can find a
class of K-series for which the question has a negative
answer. By applying directly a theorem of Gundy [7],
we can find sufficient conditions for the answer to

be affirmative.

Faber [3] has given an example of a Haar
series which converges to zero at all but one point
of [0,1]. Skvorcov I10] has shown that if a Haar
series converges everywhere on [0,1] to a bounded
function then it is the Haar-Fourier series for the
function. These two results suggest the analogous
questions for K-series. Results in this direction are

given in Chapter V.

Since the Perron integral is used extensively
in this paper, we will include the definition and
some of the most important properties. For a more
detailed discussion of the Perron method of integration

see [9].

Definition. Let f be a function (not
necessarily finite) defined on an interval [a,b].

The function U is a majorant for f if

 

a) U(a) = 0

b) DU(X) = lim inf U(Y)_U(X) > -m for all x 6 [a,b]

c) {2U(X) 2_f(x) for all x 6 [a,b].

 

The function V is a minorant for f if

 

a) V(a) = 0

 

b) DV(X) = lim sup V(Y)-V(§) < +m for all x 6 [a,b]
y-R Y’X

c) DV(X) 3 f(x) for all x 6 [a,b].

It can be shown that

(l) sup[V(b):V is a minorant for f} g_inf[U(b):U is a
majorant for f}.
If f has at least one majorant, at least one minorant
and equality holds in (1), then f is said to be Perron
integrable on [a,b] and the common value is denoted by
(13)];b f. The following are some of the most useful
a

properties of the Perron integral.

Let f be Perron integrable on [a,b].
a

Set (P)] f = 0. Then for any x 4 [a,b]
a

f is Perron integrable on [a,x] and on

[x,b]. Furthermore (P)fx f + (PM;b f = (P)‘[;b f
a x

a

for every x 6 [a, b] and the function
(P)I: f is continuous on [a, b]. Any
function of the form F(x) = C+(P)f: f is

called an indefinite Perron integral of f.

Let f be Perron integrable on [a,b] and
k be a finite constant. Then the function

is Perron integrable on [a,b] and

(PM: kf = k(P)j'b f.
a

Let f and g be Perron integrable functions
on [a,b] such that fhg is defined on

all of [a,b]. Then f+g is Perron integrable

on [a,b] and (P)‘rb (f+g) = (P)‘rb f + (P)fb g.
a a a

Let f be Perron integrable on [a,b], U
a majorant for f, V a minorant for f
and F(x) = (P)f: f. Then U-F and F-V

are non-decreasing functions on [a,b].

If f is Lebesgue integrable on [a,b],

then is Perron integrable on [a,b]

and (11]: f = (L)]JD f.
a

6. If the function F possesses a finite
derivative F’ everywhere on [a,b],

then F' is Perron integrable and

Nb) - F(a) = (1>)]'b F’.
a

7. Every Perron integrable function is
measurable and is almost everywhere
finite and equal to the derivative of

its indefinite integral.

8. Let f and g be functions on [a,b]
such that f=g a.e. Assume f is Perron
integrable on [a,b]. Then 9 is

Perron integrable and (NJ;b f = (HIb g.
a a

9. Let f be Perron integrable on [a,b] and
9 have finite variation on [a,b]. Then

the product fg is Perron integrable.

When there is no danger of confusion we will write jb f
a

for (P) fb f.
a

Let g be a finite function on a set A. If
a,b 6 A, a < b and I = [a,b], denote g(b) - g(a)
by g(I). If E is a set of real numbers, then [E]
will denote the outer Lebesgue measure of E: in case
E is measurable, then, of course, \E] is the measure
of E. Which of these is meant will be clear from the

context.

CHAPTER II
DEFINITIONS AND SOME THEOREMS
0N FOURIER-K—SERIES

We start by defining the notion of a K-series°
This gives simultaneous generalization of series in
the Haar and Rademacher orthonormal systems. Mbreover,
the concept of K-series includes as a special case
the partial sums of order 2n for series in the
walsh orthonormal system. We shall also define the
Fourier-K-series of an integrable function and of a
measure and investigate the convergence properties of

such series.

1. Definition. Let Dn be a finite set

[tn'0,tn'l,...,tn'rn] where o = tn'o < tn'1\...\tn'rn= 1
(n = 1,2,...). Set D = L) D . Assume Dl C D C...
n=l n 2

and that D is dense in [0,1]. For each n b 0 let
3h denote the system of all intervals of the form

[t (j = l,2,...,rn). Set .9: 31 an.

. t .
nil-1' n03] n
we say a function g is regular on [0,1] if

g is of bounded variation and 9(0) = g(0+), g(l) = g(1-),

_ g(X+)+g(X-1 for x 6 (0,1).

g(x) — 2 For each n > 0

denote by 8n the space of all regular functions on

[0,1] 'which are constant on the interior of each
interval of fin. Notice that Sn has dimension r .

Define T and

1 = 81

1
Tn = [f e Sn3Jg fg = o for all g e gn_l} (n = 2,3,...).

1,2,...)

(n

A series of the form Z) fn where fn E T

n=l n

will be called a K-series with respect to the sequence

{Dn]n=1°

2. Lemma. Let f be a Perron integrable

 

function on [0,1]. Let T be a finite dimensional
vector space of regular functions on [0,1]. Then there
1
is a unique function g E T such that I (f-g)t = 0
O

for every t 6 T.

Proof. Since every non-zero regular function
is non-zero on a set of positive measure, we may choose

a basis v1,...,vn for T such that

l . .
I Vivj = éij (1,3 = 1,...,n).
0
Set
1 I
[31 = J10 fVi (1 = 1,...,n).

n
Then 9 = Z: fiivi is the required function. Suppose
i=1

1

h e T fulfills ] (f-h)t = o for all t e T. This
0

is equivalent to

1 1
jg fvi = [5 hvi (i = 1,...,n).

n
Since h e T we have h = Z} Vivi. Thus

i=1
1 l n n l
fv. = I (Z Y.v.)v. = Z W]. v.v. = Y. (i = 1,...,n).
JNo 1 o j=1 3 3 l j=1 3 o 1 3 1
Hence :9. D

3. Definition. Let T,f and g be as in

 

lemma 2. Then g is called the orthogonal projection

 

of f to T.

 

4. Definition. Let f be a Perron integrable
function on [0,1]. For each n 2_l let fn be the
orthogonal projection of f to Tn' The K-series

on

is the Perron-Fourier-K—series for f, hereafter

 

fn
n=l
denoted by PFK—series for f. In case f is Lebesgue

 

integrable we say that Z: fn is the LFK-series for f.
n=1

5. Lemma. Let S be a finite dimensional vector
space of regular functions on [0,1]. Let T be a linear

subspaCe of S. Let

1
V = [f e 8:] fg = 0 for all g e T].
0

Let h be Perron integrable on [0,1]. Let $1 and
$2 denote the orthogonal projections of h to T and
V respectively. Then the orthogonal projection of h

to S equals $1 + $2.

Proof. Choose an arbitrary u G S. Let t
be the orthogonal projection of u to T and let

v = u - t. Obviously v E V so that

H

1 1
[0(h-Il-IZIu = ]‘ (h-wl-wz) (t+v) = J‘OOI-wfit

O #10

1 l
¢t+ (h-¢)v- wv=o. a
2 IO 2 IO 1

6. Lemma. Let f be Perron integrable on

 

m
[0,1] and Z: f be the PFK-series for f. Then

n k=1 k
23 fk is the orthogonal projection of f to Sn (n = 1,2,...).
k=l
Proof. By induction. The assertion is obvious
n
for n=1, since 8 = T . Assume f is the
1 l k=1 k

orthogonal prOJection of f to 8h. Because Dn CDn+1
we have Sn C Sn+l' Therefore we may apply lemma 5

With S = Sn+1, T = 8n, V = Tn to see that the orthogonal

n
projection of f to 8n+1 is Z) fn + fn+1. D
k=1

7. Theorem. Let f be Perron integrable on

 

[0,1]. Let g be the orthogonal projection of f to

gn. Then

I g = I f for every J 6 3h (n = 1,2,...).
J J

Proof. Fix an n. Let J ,...,J be an
-—-——- l rn

enumeration of the intervals of Ah. Let vi be a

regular function such that vi = l on int Ji and

vi = 0 off Ji' 0bv1ously vi 6 8n (1 = 1,...,rn).

10

Since 9 is the orthogonal projection of f to 3n,

we have

1
[J (f-g) =[O(f-<JIV-1 = o (i = 1,...,rn). [:1
i

on n
Given any K-series Z) fk' define 5n = Z) f
k=l k=l

(n = 1,2,...). In case 2: fk is the PFK-series for
k=l

n
a Perron integrable function f write sn(f) = Z} k'
k=1

k

We will write sn(f,x) for (sn(f))(x).

We now present theorems on the behavior of the

K—Fourier series of Perron integrable functions.

8. Theorem. Let f be Perron integrable on

[0,1]. Let Z) fn be the PFK-series for f. Then
n=1

1 .
Sn(pr) = TET'IJf for X 6 Int J, J 6 3h (n = 1,2,...).

Proof. This is a direct consequence of lemma 6

and theorem 7. D

9. Theorem. Let g be Perron integrable on

[0,1]. Let 23fn be the PFK-series for 9. Then
n=1

Z fn(x) = g(X)
n=1

for every x such that

fine”: 9) = 90:).

11

Proof. Let G be an indefinite Perron

integral of 9. By theorem 8 we have

_ 1 .
sn(g,x) — TET [J g for x 6 int J, J 6 3h

and
sn(g,x) = EEET [J g + EfiET [L g for X 6 Dn'

x a common endpoint of J,L e fih. Therefore, if

G'lx) = g(x) ‘we have g(x) = lim sn(g,x) since D
[1400

is dense in [0,1]. D

10. Corollary. Let g and Z: fn be as
a: n=1
in theorem 9. Then Z: fn(x) converges to g(x)
n=1
almost everywhere on [0,1].

Proof. Follows directly from theorem 9 and

the fact that for any Perron integrable function N
d
gym]: t) = tux) a.e.

ll. Corollary. Let g and Z} fn be as in
n=1

theorem 9. Then 2) fn(x) converges to g(x) for
n=1

every x which is a point of continuity of g. Further-
more, if g is Lebesgue integrable, then 2: fn(X) = g(x)

n=1
at every Lebesgue point x of g.

12. Lemma. Let f be Lebesgue integrable on
[0,1]. Let Z) fn be the LFK-series for f. Assume f

n=1
is continuous on an interval I C [0,1]. Then for any

12

compact interval I1 contained in the interior of 1,

co
Z) fn converges uniformly to f on 11' In case
n=1

I = [0,1], then 2‘ fn converges uniformly to f
n=1
on [0,1].

Proof. Let I be a compact subinterval of

1

int I. Let 12 be a compact interval such that

1 C int 12 and 12 C int I. Let c > 0. By the

uniform continuity of f on I

I

2 there is a 6 > 0

such that ]f(x)-f(y)] < e for every pair x,y 5 I2

with |x-y| < 6. Because of the density of D and
because Dn C Dn+1 (n = 1,2,...) there is an integer
N0 so large that if n > N
[J] < 5.

0 and J 6 fih, then

Now there is an N1 such that if n > N1,
J 6 3h and J n Il # ¢. then J c 12. Let
n > max(No,N1) and let t e 11. If t t on, then

there is a J 6 fih such that t 6 int J and by theorem 8

(1) [sn(f.t)-f(t) I = T~17T [f [f(u)-f(t)]dU\
J

g_T%T-]J[f(u)-f(t)\du < a.

If t 6 Dn n (0,1), then there are two intervals
J1,J2 6 3h for which t is a common endpoint and by

theorem 8

13

(2) [sn(f.t)-f(t) I _<_ 27%?) [J |f(u)-f(t) [an
1

l s e
+ |f(u)-f(t)]du < - + — = e.
2]J2] J‘Jz 2 2
Assume that I = [0,1]. Take I1 = 12 = [0,1]

in the first part of the proof. Then there is an N

0"
L"
(D

(1.

such that if n > N and J 6 3%, then [J] <
n > N and t f'Dn' Apply theorem 8 as above to

conclude that for J e fig and t 6 int J
, 1
l ) s (f,t)—f(t) / f(u)—f(t) du < .
( in ‘3- ‘J‘ J1)" \ C

If t 6 (0,1) 0 Dn' then applying theorem 8 as we did

to obtain (2) we get

(2 4') ‘sn(f, t) -f(t) ‘ S FWI |f(u) -f(t) [du
J
1

l /
+ 313:)- sz‘f(u) -f(t) ‘du '\ e.

In case t=0 or t=l, theorem 8 gives (1') where t

is an endpoint of J. D

The inequality which is derived in theorem 16
is the direct analogue of one which is know for the

Haar-Fourier series of an Lp function (1 < p < co)

[6, pg. 72].

We begin with some lemmas.

14

13. Lemma. Let p 2.1. Let f be a
Lebesgue integrable function on [a,b]. Let Y ‘be a

number such that [b Y = [b f. Then
a a

fIvIng’IfIP.
a a
Proof. We have

Mao-a) = I:le

IJbYI = Ilbfl _<. (Tb 1)1-1/p(jb|flp)l/p
a a a a
(b-aIP'l/pUb mp) VP.
3

Taking pth power of both sides of this inequality we

see \Y[p(b-a)p g (b-a)p-1jb\f[p. Thus
a
IbIYIp = IYIptb-a) _<_ flap. r:
a a

14. Proposition. Let Z: fn be a K-series.
n=1

For each x 6 [0,1] and each “.2 1 set

s;(x) = max ‘s.(x) [.
igjgn 3
Then for all A > 0

>JII---|

[[x:s;(x) > All 3, I Isn(t)[dt

Ix:s;(x)>x]

1
f6 [sn(t))dt.

V\
yll—I

15

Proof. Take A > 0 and a natural number n.

*
Let A = [xzsn(x) > A} and Aj = [x:|sl(x)‘ g_x,...,
‘sj_1(x)] g_x,[sj(x)] > A] (j = 1,...,n). Obviously

n
A.nA.=¢ for i;£j and A= U ..
J 1 j=1 3

Let 35 be the algebra generated by the

elements of £3 (j = 1,2,...). Now sj 6 8j and

hence is constant on interiors of intervals of £6. It
follows that Aj is composed of interiors of intervals

of £3 together with some points of Dj' Thus

Aj 6.7]. (j=l,2,...,n). If jgn and Jefij,

then [3 SJ. = I; Sn' so that

J‘Jisj‘ = \IJsj‘SIJ13n‘°'
Since Aj 6 35 we have

[1")st _<_ [Ajsn] (j = 1,...,n).
J 3

Hence

IISnI =23
A

3:1 Jr‘s“ 2 3°51 IA.‘Sj‘ 2 3'31 X‘Aj‘

J J
= MA].

This completes the proof. D

15. Lemma. Let 1 < p < m. Let f E Lp[0,l]

Q
and let :3 fj be the LFK-series for f. Let s; be
j=1

16

as in proposition 14. Then

1*p p1 p/ p1 p
[0(sn) 3 (SET) jg|sn(f)\ ;,(§EI9 ]6\f] (n = 1,2,...).

Proof. Let A > 0 and n 2,1. Define
1 if 5 > A
¢(§I)\) = o
o if 0 g g g x
Then the inequality
A)[x:s;(x) > M] _<_ ]* [sn(x) [ax
{Klsn(XI>x}

of pr0position 14 becomes
1 * l *
AIONWSDUE): A) dX g foidsn (X), X) ‘sn (x) ‘dx,
Thus

m - m _ 1
[Exp 1([:I(s;(X).x)dXIdx g.f 1? 2(j I(S;(X).A)[sn(X)]dX)dx.
O 0

Since we may interchange the order of integration, we

have
3*(x)

n

(1) I: %(8;(x))pdx - xp-1d1)dx

I
O I-‘
A
‘fi

0

I
I

(J’pr‘lw (8:. (x) . 1) d1) dx

0 1&0 H

y\

00 _2 *
(Iéxp t(sn(XI.x)|sn(x)\dx)dx

*
s (x)
-2
(Ion AP ‘sn(x)]dk)dx

I
OLTP

H

 

1 * -1
= I; p_1<sn<x))P [sn(X)|dX-

 

17

By Halder's inequality we have
(2) f:(s;(x))p—l\sn(x)[dx
'3 []:([s;(x)]p'1)p/(P‘1)dx]l‘l/P[]:\sn(x)[de11/P.
From (1) and (2) we get
— Pfo (sn *(x))Pax 3p ——-1-C1)1(] [5* (x)]pdx)1 1/P(]“:]sn(x) [pdx)1/p.
Hence
([:(s;(x))de)l/P / —P—1(j:[sn(x> [pdx)1p./

Taking pth powers we have the first inequality of our

assertion.

The second inequality follows from theorem 8

and lemma 13. U

16. Theorem. Let 1 < p < m; f E Lp[0,l].

Let Z) fj be the LFK—series for f. Define
j=1

3*(x) = sup‘sn(f,x)\ for all x 6 [0,1].
n

Then

1 1
[0(s*(x))pdx g (BET)PJ‘O|f(x) [pdx.

18

*
Proof. Let sn be as in 14 and 15 (n = 1,2,...).

* *
Then it is easily seen that sn r s . Hence we have

1 * 1 *
(s )p = lim (5 )p
‘170 nam I0 n

by the Lebesgue Monotone Convergence Theorem. This

combined with lemma 15 proves our assertion. U

Next we show that the LFK-series for a function

in Lp[0,l] (l g_p < 0°) converges to f in Lp norm.

17. Theorem. Let 1 g_p < m. Assume
f E Lp[0,1]. Then the LFK-series for f converges

to f in Lp norm.

Proof. Let c > 0. There is a continuous

function 9 such that Hf-gup < g. By lemma 12,
sn(g) 4»g uniformly on [0,1]. Thus there is an
integer N such that [sn(g,x)-g(x)| < § for n 2.N.

1
Therefore “sn(g)-ng = (J‘0]sn(g)-g]p)1/p < § for n 2_N.

Combining theorem 8 and lemma 13 we have

“SnIf-glllg IlIsntf-gI [P = 2'3] [sn(f-g) [P
0 6.8 J

n

s JganIf-ng = filf-glp

= Hf-gup (n = 1,2,...).
P

From this and the triangle inequality we get

l9

nsn<f>-fnp _<. usn<f-g> up + IISnIgI-gllp + ug-fnp
g_2Hf-ng + Hsn(g)-9Hp < e for n 2_N.

The following two sections contain the definition
of the formal integral of a K—series and one of the

elementary properties.

18. Definition. Let Z: fn be a K-series.
co n=1
The sum of the series ijx fn will be denoted by F(x)
n=1 0
at points of convergence.

19. Lemma. Let :3 f. be a K-series. Let n
j=1
be a natural number; let x 6 Dn' Then F(x) = [X Sn'
0

Proof. By orthogonality we have [X s. = IX 5
01 on
for 1.2 n. Thus

fgfj=§3~rxfj rsn=F(x).L’j

j=1 j=1 O 0

Next a short investigation of Fourier-Stieltjes-
K-series for measures is undertaken. By a measure we will
mean a finite, signed measure defined on the Borel sets

in [0,1].

20. Definition. For any J = [a,B] E 3’ define
* * *
J by J = [0.6) if [:5 < l and J = J if [3:1.
Let u. be a measure. For each n 2_l define sn e 3n

'1'
by I 5n = U(J ) for each J 6 Ah. Set f1 = 31 and
J

20

f = s -s Take

n n n-l for n 2_2. 0bv1ously f1 6 T

1.
n‘z 2 and let I]_,...,Irn-l be an enumeration of fih-l'
Define vj to be regular and

Vo = (j =1,2'ooo'r )0

1 on int Ij
J

0 off Ij n-l

Then [vl,...,vr } is a ba51s for Sn-l' F1x a j.
n-l

Then Ij = JlU...U JP where Ji 6 fih (1 = 1,...,p)

*

*
and Ji 0 J1 = 0 for i # 1. (Note: it may happen

that p=l). Now applying the definitions of Vj' fn,

sn and s it follows that

n-l

l
fofnvj = I fn = I (Sn-Sn-l) =.§: I Sn _ I Sn-l

I. I. =1 J. I.
J J 1 1 j
p * * * *
= 23 uni) - MI.) = MI.) - U(I.) = 0.
i=1 3 J J
m
Therefore f E T . Hence 2) f is a K-series with
n n n=1 n

respect to [Dn}.

Q
The K-series Z: fn is the Fourier-Stieltjes-
n=1
K-series for u, hereafter denoted by FSK-series for H.

 

g1. Definition. Let [Dnl and [fih] be as
in definition 1. Let x 6 [0,1) and n 2_1. Define
dn(x) and Bn(x) by an(x) g_x < Bn(x) and
[an(x),8n(x)] 6 Ah. For x 6 (0,1], define dg(x)

and 85(x) by d’(x) <x_<_B’(x) and
[a;(x).6n’(x)] E fin (n = 1.2....).

Set an(1) = Bn(l) = B;(l) = 1 and

21

d;(0) = 85(0) = a (0) = 0 (n = 1,2,...).

For any x 6 [0,1] set Jn(x) = [an(x)ofin(x)] and

J;(x)

[ag(x),Bg(x)], where [0,0] = [0} and

[1,1] = [1].

- _ 4’
Note that if x E Dn' then Jn(x) — Jn(x)
(n = 1,2,...). Furthermore, if x 6 [0,1] n Dn then

(1n(X) = 813x) = x (n = 1,2,...).

22. Proposition. Let u, be a measure. Let

Z) fn be the FSK-series for U. Then
n=1
lim 5n = u([y]) for each y 6 [0,1)
n-w J (y)
n
and

U({1I)

lim sn
nda J;(1)

Proof. Let y 6 [0,1). Then it follows directly

from the definition that for n so large that Bn(y) \ 1

]‘ sn = u[an(y).Bn(y)).
Jn(y)

The first assertion follows from this since

F)[an(y).Bn(y)) = {y} and [U[0,1]| < a. similarly
n=1

lim 5 = U([1))- C3
naw £%:UJ n

22

23. PrOposition. Let n be a natural number

m

and x 6 Dn' Let u 'be a measure and Z: fj its
j=1

FSK-series. Let F be as in definition 18. Then

L{0,x) = F(x), if x < l and u[0,l] = F(l).

Proof. Let k be such that x = tn k. Then
k tn j k
.. = I =
F00 — JZSn jglft Sn .§1u[tn,j-l' tnpj)

n,j-l 3-

= Ll[0,x) if x < l,

and
rn-l
tnlj ._
F(1)= rgift sn —.Ei U[tn'j_ 1' j) + u[t n'rn_1,1]
j- n,j- _1 J
= u[0,l].

24. Proposition. Let x E D. Assume H. is
m
a non-atomic measure. Let 23 f be the FSK-series

n=1
for 11. Then F(x) = u[0,x).

Proof. Choose a natural number n. Then

fa=fӴ+fw

O O

F(an(x)) + [x s

an(x)

n

u[0,an(x)) + Jx s ,

an(X) n

by proposition 23. Now

lim u[0,an(x)) = u( U [0,0Ln(X))) = U[0.X)-
n-Oco n=1

Mereover,

 

n(X)
”2100931) = ‘Bn (X)"°‘n (,0 Man (XLBn (X))\ 3 Ma n(X).I:3n(X))[

 

23

. x-an (x)
Since ‘Bn(x)'an(x)‘ < 1. But

 

u[an(XI.Bn(X)) -+ u([x}) = O.
This proves the assertion. D

We will next show how the singular portion of

the measure may be recovered.

The following theorem is well known (see for

example [8], pg.154).

gs. jgheorem. Let u be a Borel measure on
[0,1]. Set g(x) = u[0,x) for x 6 (0,1] and

9(0) = 11(0). Then

a) g is differentiable a.e. with respect
to Lebesgue measure,

b) g' is Lebesgue integrable on [0,1],

c) there is a measure Us such that
U(E) = HS(E) + f g’(x)dx for every
Borel set E ang us is singular with

respect to Lebesgue measure.

(D

_g. Theorem. Let. L1 be a measure and Z} fn

its FSK-series. Let g be as in theorem 25. T%::
(1) g'(x) = 55 fn(x) except for a set of

Lebesgu2=measure zero, and if u is

non-atomic,

24
(2) I1 (0.x) = Z rfn - Jx Z fn for all x 6 [0,1].
S ~ n=1 0 0 n=1

Proof. Let x F (0,1) be such that g'(x)

exists. Then we have

 

 

, . g(Bn(X))-9(an(X)) . u[an(x).Bn(X))
g (X) " it: Bn(x)-an(X) 3,113: Bn(x)-anIXI
= :1: sn(x+).
But sn(x+) = sn(x) for x 6 Dn' Thus g'(x) = lim sn(x)

11-00

for almost all x with respect to Lebesgue measure.
This proves (l).
o
If u is non-atomic, we have u[0,x) = ZZIX fn(t)
for all x 6 [0,1] by proposition 23 and 24. 2:mbgning

this With (1) and part (c) of theorem 25 we have (2). C]

CHAPTER III

A UNIQUENESS THEOREM FOR K-SERIES

In this chapter we will show that a Haar
series is a K-series and prove a uniqueness theorem
for K-series which generalizes those known for Haar

series. we will start by defining the Haar functions.

Set xo(x) = l for all x 6 [0,1]. Set
1 if x E [0,%)
X1(X) = 0 if x = 5
-1 if x e (%,1].
Given any integer “.2 1, write it uniquely as
n = 2m+k, where 0 g_k < 2m. Define Xn to be regular

and

m k 2k+l
./2 for x E (:5 ’33:]?

 

_ m 2k+l k+l

xn(x) — -\/2~ for x 6 (2m+l' 2!“)
It k+1

0 for x £ [35» -;Efl.

Note that this gives X1 as defined above if we take
a
n=1. Any series of the form 23 anXh, where an are

n=0
real numbers, is called a Haar series.

25

26

In papers on Haar series the following condition
is often imposed, [ll], [14], [15].

. co
1. Definition. A Haar series 23 anxn satisfies
n=0
condition GH if for every x 6 [0,1]

a

“k
1im.-——T—T-= O
k-ow xnk X

where Ink} is the sequence of integers m for which

xm(X) a! 0-

The corresponding conditions for K-series are

the following:

CD
2. Definition. A K-series Z: fn satisfies
n=1

condition G if

lim I f = 0 and lim I f = 0
n a n
n-oao Jn (x) n-ooo Jn (x)
for every x 6 [0,1], where Jn(x) and J;(x) are as

in section 21 of Chapter II.

0
The K—series Z} fn satisfies condition H if
n=1

lim 3n = 0 and lim s
11-»: Jn(x) n-ooo Jr:(x) n

for every x 6 [0,1]. (Note that Jh(x) = J;(x) if x 5 Dn.)

27

The next Objective is to show that a Haar series
is a K—series and that for Haar series conditions GH, G
and H are equivalent. Define a sequence of finite sets
[An] as follows. Set A1 = {0,1}, A2 = {0,%,1]. For

n 2;2 write n-l = 2m+k where 0.3 k < 2m as in

the definition of the Haar functions. Set

1 2 2k+1 k+1 k+2

A = {O'———-—-—'-————'...' p ’ poo-,1}.
n 2m+l 2m+l 2m+1 2m 2m

 

. _ l 1 _ l l 3
For 1nstance A3 - [0.3.5.1]. A4 - { IZOEIZII]I

0
_ _ l 1 3 l 3
5 - {O'é'Z'i'Z'JJ' A6 — [O,§,1,§,§,3,1], etc.

m
3. Proposition. Let Z) a xn be a Haar
w n=O n
series. Then 23 anxh is a K-series with respect to
n=0
(An). Moreover, for Haar series conditions GH and G

are equivalent.

Proof. Define fn+1 = anxn 0°(n = 0,1,2,...).
Then f E T for each n. Hence Z) f is a K—series
n n n=1 n
‘with respect to {An}.

Let Z: a xn be a Haar series. Define
n=0 n
fj (j = 1,2,...) as above. Let y 6 [0,1] and let
{nk}:;1 be the sequence of integers i such that

xi(y) # 0. Fix a k > 1 and write nk = 23+m where

0 < m < 23. If y g A , then
_. nk

28

 

I in
Jnk+1(Y) k

_ 1 ’5'
(l) ' 2j+1“"‘nk‘V6

] f
I n+1
Jnk+1(Y) k

[ank i

2 finkhfi‘ '

+1

 

 

 

If y E An‘\\[0,1}, then y is an endpoint of the
k
support of xn since Xn (y) #'0. Therefore, of the
k k

two integrals

 

 

 

I f and I f
I +1 n +1
J (y) nk J (y) k
nk+1 nk+1
one is zero and the other has absolute value
a
2j+1 =ank(y) I
If y=0 or 1, then one of the two integrals
I f and IJ, fn +1
(Y) n +1 (Y) R
nk+l k Jnk+ly
is zero and the other has absolute value
Ian
1 j _. k.

we will show that there is an N such that

(4) I; ( )fn+l = IJ’ ( )fn+l = 0 for n > N and
n+1 Y n+1

n #nk (k = 1,2,...).

29

There are two cases.

Case 1. Suppose y g A =

3
n # nk (k = 1,2,...). Then y A An+1' Consequently

fiC:8

A.. Let
3

— ’ '
Jh+l(y) — Jh+l(y). Furthermore xn is zero on
int Jn+l(y) Since xh(y) = 0 and xn is constant on
the 1nter1or of Jn+1(y). Thus

JJ ( )fn+l = £Ja ( )fn+l = 0'
n+1 y n+1 Y

So in this case (4) holds for any n # nk (k = l,2,3,...).

Case 2. Assume y F A. Then y 6 AP for
some p. Let n be an integer such that xn(y) = 0.
Then either y is the midpoint of the support of yh,
or y is outside the support of xh. Suppose y is
the midpoint of the support of xn. Then y 6 Ah+l and
y f Ai for i g.n. Therefore, since Ai C Ai+1
(i = 1,2,...), we have n+1 g.p. From this consideration
we see that if n 2_p and xh(y) = 0, then y is
outside the support of Xn' Let n 2_p and n # nk
(k = 1,2,...). Since the support of xh is the union
of two adjacent closed intervals with endpoints in An+l
and since y is outside the support of Xn' the support

0 J
of xn does not overlap With J (y) and J +1(y).

n+1 n
Hence
I f = I f =
n+1 a n+1
Jn+1(Y’ Jn+1(y)

for n # nk (k = 1,2,...). This establishes (4).

30

m
Now assume that Z: anxh satisfies condition
n=C)
GH. Let s > 0. There is a K such that

< e for k 2_K.

 

 

Combining this with (l), (2) and (3) we have

C

2- and <

<

I fn +1 2

f
n +1
Jnk+l(y) R

J’ (y) k
nk+l

 

 

 

for k 2 K.

Therefore by (4) we see that

If nt <

and l] f t <
Jn(y) Jn(y)

n

Nlm
Run

for all n > max(nk+l,N+l). Thus 23 fn satisfies
n=1
condition G.

C!)
Conversely, suppose satisfies condition
[1:

G. Let c > 0. Then there is an N such that

f
1 n

[f f ] < 5 and [I f [ < e for n > N.
n 4 .r n 4

Jn (y) Jn (y)

Choose K such that nK > N. Then from (1), (2) and (3)

it follows that

ank

2:337?

< e for k 2_K.

 

 

Therefore 23 anxn satisfies condition GH. f]
n=0

The equivalence of conditions G and H for Haar

series follows from a more general assertion.

31

4. Definition. We say [Dnl satisfies
condition Q if there exists q 6 (0,1) with the
following property: if J F fih' L E fih+l, L CtJ, then

either L=J or [L] < q‘J‘ (n = 1,2,...).

 

5. Theorem. Let [Dn} satisfy condition

00' Assume Z) fn is a K-series with respect to [Dn].
an n=1
Then 2) fn satisfies condition G if and only if it
n=1

satisfies condition H.

Proof. Assume Z: fn satisfies condition H.
n=1
Let y 6 [0,1]. Then

n

lim I 3n = lim 2) fj = 0 implies
new Jn(y) nam j=1 Jn(y)
11m] f = 0.
114:» J (y) n
n
Similarly
lim 5n = 0 implies lim fn = 0.
new Jg(y) .naw Jgty)

Q
Therefore 2) fn satisfies condition G.
n=1

Conversely, suppose fn satisfies condition
n=1
G. Let x 6 [0,1]. Set
6. = sup 1' )fk] (j = 1,2,...).

3 is kac)‘
Let q be the number in definition 4 and let

m-l
bm — qu +...+Bm_1q+8m (m — 1,2,...).

32

Since Bj+l g_Bj for 3 2.1, ‘we have
b < 8 ( m"1+ +q+1) < i;- for all m 2 1
m _ 1 q 0.. _ l-q 0

An easy induction shows that bm+1 qu + Bm+1

(m = 1,2,...). Therefore

lim sup bm = q lim sup bm + lim 8m = q lim sup bm.
mam mam mam mam

Since lim sup bm < m it follows that lim sup bm = lim bm = 0.
mew mew mam

For each j'Z 1 let cj be the number of distinct

intervals occurring in J1(x),...,Jj(x). For each n 2.1

we have
(1) I [f 1 - ‘Jh(X)‘ I If t < Cn-cj] If I
Jn(X) j l"Tj (X) I Jj (x) j _ q Jj (X) j

(j = 1,...,n)

by condition 00. If Jj(x) = Jj+1(x) for any j.

then fj+1 = 0 on int Jj+1(x)' because of the orthogonality.
Combining this fact with (1) we have

C _-
(2) f +...+ f _<_ q n f
IJn(x)‘ 1‘ IJflbc)‘ n‘ IJ1(X)‘ 1‘
+ sin-'31“ If I+ +q°“' “‘11 If I
JZIXI 2 Jh_1( ) ”'1
c —c c —1 c 2
+ q n n I ‘fn‘ S.q n 61 + q n 82 +...
Jn(X)
+ q6 + B = b
cn-l cn cn

since Jj+1(x) #’Jj(x) implies cj = cj+l.

33

Now

”Jn(x)sn‘ = [and ‘S“‘ '<‘ IJn(X) ‘fl‘ +°”

+ f
IJnm \

n‘ (n = 1,2,...).

Thus it follows from (2) that

(3) S _<_b (n=l,2,...).
\IJn(x) a

From the density of D it follows that lim on = m.
n-Ooo

Hence, given a > 0, there is an N such that n 2_N

implies bc < c. From this and (3) it follows that

n
s < e for n'z N.
lIJn(x) nl

Similarly, we may show lim 5n = 0.
11-000 Jr: (X)

m
Therefore 23 fn satisfies condition H. [3
n=1

It should be remarked that assumption of
condition Qo is not used in proving the "if" part of

the assertion.

Combining theorem 5 and prOposition 3 we see
that for Haar series conditions GH, G and H are equivalent.
In view of the following prOposition, condition H seems
to be the more natural condition to impose for the study

of K-series.

34

6. Proposition. Let g be a Perron integrable
function on [0.1]. Then the PFKHseries for g satisfies

condition H.

Proof. Let G be an indefinite Perron integral

of 9. Then

s (g) = 9 = G(f3 (XH - G(a (XH
IJ (x) n J (x) n n
n n
and
s (g) = = G(B’(X)) - G(a’(X))
fJ'(X) n J‘J’(X) n n
n n
(n = 1,2,...).
Since D is dense, lim Bn(x) = lim an(x) = x and
n-an n-ooo
lim fi’tx) = lim a’(x) = x. Therefore lim I s (g) = 0
11400 n40 n n-wo Jn(x) n
and lim I s (g) = O by the continuity of G. D

; n
n-m Jn (X) -

It should be remarked that prOposition 6 remains
valid if the Perron integral is replaced by any method
of integration for which the indefinite integral is
continuous.

m

7. Lemma. Let Z) fn be a K—series, F as

in definition 13 of Chapter II. Then the following

pr0perties hold:

35

(i) for X’éDn

F(Bn(x))-F(an(x))
Sn(X) = Bn(xjian(xj (n = 1,2,...).

 

(ii) for x e Dn 0 (0.1)

S (X) = l F(X)-F(an(x)) + F(Bn(x))-F(X)
n 2 3n(X)-x

 

 

]

x-a;(x)
(iii) if an fulfills condition H, then

lim[F(En(x)) — F(an(X))] = O for all x e (0.1)
n4m

and
lim F(ag(x)) = lim F(Bn(x)) = F(X) for x 6 D.

n4m nqm

Proof. For any x 6 [0,1]

(x) (x) a (X)
(1) f3“ 8n = fin sn - I n s = F(Bn(X))-F(an(X))

an (x) O n
by lemma 19 of Chapter II. For X 5 Dn we have

Bn(X)

s = s (X)[S (x) - a (X)
an(x) n n n ]

n
since sn is constant on (an(x),bn(x)). This proves (i).

If x 6 Dn n (0.1), then an(x) = sn(x+) + sn(x-)
by definition. Now sn(x-) = sn(t) for any t E (ag(x),x)
and sn(x+) = Sn(T) for any T 6 (x,Bn(x)). Now we
apply part (i) in order to compute sn(x+) and sn(x-).

This proves (ii).

36

Assume Z) fn satisfies condition H. ‘From
n=1
(1) we observe that

(X)
(2) 1F(Bn(X)) -F(an(X))| = If“ Sn"
a (x)
Thus F(Bn(x))-F(an(x)) 4 O as n a m.

If x E Dm for some m, then dn(x) = x for

n 2_m. Therefore for n 2.m (2) becomes

Bn(X)
\F(Bn(X))-F(X))‘ = \f sn|.
x
Similarly,
‘F(x)-F(ag(x))| = \fx Sni for n 2Dm.
' 050:)

This completes the proof of (iii). U

m
8. Proposition. Let Z)fn be a K—series.
n=1

Suppose F(x) = O for every x E D. Then sn = O

(n = 1,233,000) 0

Proof. This is an immediate consequence of
lemma 7, (ii) and the regularity, for we have 3n = O

on the interior of any interval of fih (n = 1,2,...). D

9. Pr0position. Let f be a Perron integrable
function on [0.1]. Let 23f be the PFKfseries for f.

n=1n
IX fn exists as
l 0

M8

Then for each x e [0.1]. F(X) =
n

a finite number and

F(x) = (p) fat)“.
0

37

Proof. Let x e D . Then x = t for some
'—-—- m m,k

k. Hence by lemma 19 of Chapter II

R t .
F(X) =ij:3 S
j=1 t

m 21W f
t
m,j- -l

m,j—l j:

(P)fx f(t)dt.
0

Let x 6 [0,1] \~D. Fix an n. Then by the

first paragraph

an(X)
F(an(x)) = (P)f§ f(t)dt.

Thus
a (x)
fxsn (t)dt _ (O “ Sn + fx 8 = F(an (x))
an(X) n
a (x)
+ s — (P) n f + .
ix (X) n IO Ix (x) n
Now
(X
H" 211:} =\ <::P>I
an(x)Sn
since sn is constant on (an (x),Bn (x)). By the
fin (X)
continuity of the integral we have (P)j: 4 O and

(1 (X) n(X)
(P)I6n f 4 (P)j:f (as n 4 m). Therefore

lim jxsn(t)dt = (p)jxf. a
O

naw O

10. Definition. Let a,b E D, a < b. Let

 

/
8a b be the class of all finite real-valued functions
I

on D 0 [a,b] which fulfill the following properties:

38

P1: If x 6 [a,b) n D and l = lim f(Bn(x)),
n-Om
finite or infinite, exists, then f(x) 2 l.
P2: If x e (a.b] n D and u: 11‘“ “0‘13””

1"l-+OlD

finite or infinite, exists, then f(x) 3 u.

11. Remark. Let a,b E D, f E 8a b' and let
I

g be continuous on D n [a,b]. Then f+g E 8a b'
I

Proof. Let x 6 [a,b) H D and let the limit

y = 1im[f(Bn(x))+g(fin(x))] exist. Then also the limit
n 400

6 = lim f(Bn(x)) exists and Y = 6+g(x) 3 f(x) + g(x).
1140::

So f+g' fulfills p1.
Similarly we have property P2 for f+g. E

12. Lemma. Let a,b E D, a < b. Assume
f 6 8 ‘b' J 6 3'. J c:[a,b] and that f(J) 2’0.
a, n
Then there exists an m > n and intervals Jk E 3k
such that J = Jn D Jn+1 3...: Jm, f(Jk) 2.0

(k = n,...,m) and Jm C int Jh.

Proof. Denote J = [x,y]. Among all the intervals

of 3' contained in J there must be at least one,

n+1
say L, such that f(L) 2_O. In this way we may construct
intervals Li 6 £3 such that J = Ln 3 Ln+1 D...,f(Li) 4.0
(i 2_n). Let p be the smallest integer greater than

n for which Lp # J. The existence of such a p

follows from the density of D.

_'

 

39

Suppose the assertion is false. Then we have

either x 6 Li or y 6 Li for any i 2.n.

1) Assume that x E Lp. Then x 6 Li for all
i 2_n, and, therefore, Li = [x,Bi(x)] for all 1.

Choose an i > p. Denote the set Di n Li by

—l
(To....,TS}, where x = $0 <...< TS = Bi_1(x). Clearly

71 = 61(x). If s=1, then f(Bi(x)) = f(Bi_l(x)).

If s > 1, then we must have f(Tj_1) > f(Tj) for

j = 2,...,s, for otherwise the assertion would be

true with m=L. Jk = Lk (k = n,...,i-l), Ji 2 [Tj_l,Tj];
thus f(Bi(x)) = f(Tl) > f(TS) = f(Bi_1(x)). We

have therefore f(Bi(x)) Z f(Bi_1(x)) 2_f(x) for any

i > p. Because of the density of D there are

infinitely many numbers i > p for which 3 > 1 and,

therefore, f(Bi(x)) > f(Bi_l(X)). It follows that

lim f(Bi(x)) > f(x) which contradicts P

. 1'
14m

2) If y E Lp then we prove analogously

that lim f(a;(y)) < f(y) which contradicts P2. This
14m
proves our assertion. C

13. Lemma. Let a,b E D, a < b. Assume that
[a,b] 6 £5 for some j. Let f E 3a b and f(a) g f(b).
I
Then there exists an x 6 (a,b)‘\ D such that

f(an(X)) < f(Bn(x)) (n = j.j+1....).

40

Proof. Take n=j, J = [a,b] in 12 and
conStruCt the Intervals Jj'ooo’Jmo set p0 = 3, pl = m.
NOw apply 12 with n=m and J=Jm. getting intervals

J D J 3...: J such that J E , f J ) 0

pl p1+l p2 k 3k ( k 2-

(k = p ,...,p ) and J c int J . Continuing in this
1 2 p2 p1

fashion we construct sequences of intervals

Jj 3 Jj+1 D... and integers pO < p1 <... such that
Jk 6.8 , f(Jk) 20 (k = j,j+l,...)
and J c int J (i = 1,2,...).
pi i-l
Since each Jk is closed and D is dense in
[0,1] there is a point xO such that F( JR = {x0}.

k=J
It is obVious that Jk = [ak(xo),Bk(xo)] for each R

and that xo 6 int Jk

remarked that f(Jk) 2_O (k = j,j+1,...), the proof

for each k, so xo E D. Having

is complete. [3

The following lemma is a generalization of a

lemma in [2], due in its present form, to J. Marik.

14. Lemma. Let a,b E D, a < b. Assume
f 6 3a,b' Let C be a countable set in (a,b). Let

lim inf f(Jn(x)) 3.0 for x 6 (a,b)\\ D. Let

1 n3” f(Jh(x)) b)\\ )
im inf < O for x 6 a, CUD . Th f
nqm <1Jh(x)‘ _’ ( ( en

 

is non-increasing on D n [a,b].

41

nggf. Suppose not. Then there exists
y,z G D n [a,b] such that y < z and f(y) < f(z).
There is a j such that y,z 6 Dj' Then we can find
J = [v,w] 6 35 for which f(J) >10. Choose an e > 0
such that f(v) + 3c < f(w). We may assume C is an
infinite subset of (v,w)‘\ D. Let [c1,c2....} be

an enumeration of C. Define functions

(X) = ‘23 --€-- sgn(x-c ).
m n=1 2n n
R(x) = CMX) + ex. and
¢(x) = f(x) - R(x).

Now R(w) - R(v) gR(l) - R(O) = cp(l) + e - cp(0)
= 36, so ¢(v) < $6M). Since m is continuous on

[0,1] \\C, R is continuous on D. Thus W e 3v‘”.
I

For each x E (O,l)\\(CUD) we have

((Bn(x))-¢(an(x)) f(Bn(X))-f(an(x))

lim inf lim inf[ -
M” Bn(x) - anaa S W, Bum - anal ‘1

3,-6

since m is non-decreasing.

 

 

If x E C, then

lim inf[w(Bn(x))-¢(an(x))] = lim inf[f(Bn(X))—f(an(X))]
n-oao n-m

— lim[ cp(Bn(X) )-cp(an(x) )]
ham

3 1im[cp(an(X))-cp(6n(x))] < o.

11400

42

Therefore, for each x e (0,1)\\D there is an n such

that ((Bn(x)) < ¢(an(x)). This is a contradiction with

lemma 13. [j

15. Theorem. Let, Z} fn be a K—series
n=1
fulfilling condition H. Let a,b 6 D, a < b. Let
C be a countable set in (a,b). Let f be a Perron

integrable function on (a,b). Assume

(1) lim sup 3 (x) 2_f(x) a.e. on (a,b),
n
11-000
(2) 1im sup sn(x) > -m and 1im inf sn(x) < J...
n-bm n-N'D

for x 6 (a,b)\ (CUD).

Then
(3) Z} fn(x) exists and is finite a.e. on (a,b),
n=1
(4) Z) fn(x) is Perron integrable on (a,b), and
n=1
(5) F(x) - (P)fx Z) fn is constant on D n [a,b].
0 n=1

Proof. we may assume lim sup sn(x) 2_f(x)
nan

everywhere on (a,b). Let G be a minorant of f.

Define H on D n [a,b] by H(x) = G(x) - F(x).

Set ((x) = (P)‘fx f for x e [a,b]. Then W
a
is continuous and G-w is non-increasing. Thus

G(x+) g_G(x) for any x 6 [a,b). If x 6 [a,b) n D,

 

43

then 1im F(Bn(x)) = F(x) by lemma 7, part (iiiL so

n-Ooo
that lim H(6n(x)) g H(x). Similarly lim H(aéXx)) 2 H(x)
n-ND nag:

for any x 6 (a,b] O D. Hence H E 8a b'
I
Furthermore, for each x 6 (a,b)

lim inf[H(Bn(x))-H(an(x))] = lim inf[G(Bn(x))-Gflan(x))
11-an 11-000

- F(Bn(X)) + F(an(X))] S. 0

since DG(x) # +0° and lim[F(Bn(x))-F(an(x))] = O by
[1400
lemma 7.

For each x 6 (a,b)\\(CUD) ‘we have by lemma 7,

 

 

part (i),
. . H (Bn (X) ) -H (01.1n (X)) . _ G (Bn (X) ) -G (an (X) )
llfiqinf Bn (X? - unfit) = 11:1;“fi 6,100 - an(XT ' Sn(X)]

g_fiG(x) - lim sup sn(x) 3,0
nam

since DG(x) < +m, lim sup sn(x) > -m, and
nam

DG(x) 3 f(x) g_lim sup sn(x). Thus we may apply lemma l4
n-Oco

to conclude that H is non-increasing on D n [a,b].
Since D is dense in [0,1], H may be extended to a
non-increasing function H on all of [a,b]. Since H
is monotone, H5(x) exists and is finite almost everywhere

on (a,b). Furthermore

44

(of fi’zmb) - H(a)
a

N

which implies H' is Lebesgue integrable on (a,b).

Recall ¢(x) = (P)£:f. Since w-G is non-
decreasing there is a set A of measure zero such that
for any x 6 (a,b)\.A, w-G has a finite derivative.
Furthermore, the function 5(w-G) is Lebesgue integrable
on (a,b). Since ('(x) exists and is almost everywhere
f(x), there is a set B of measure zero such that
G'Cx) = (’(x) - (w-G)’(x) exists as a finite number
for every x 6 (a,b)\\B. Thus 5G is Perron integrable
on (a,b) since 5) is Perron integrable and 5(w-G)
is Lebesgue integrable on (a,b). For any x e (a,b)\Dn

we have
G(ES,r1 (X) )-G (an (x) ) H (Bn (x) ) -H (an (x) )
(6) Sn(X) = BnGcT-anbc) ' BnTx) -an(X) 7

 

 

(n = liZ'B’ooo).
Since H'(x) and G'(x) exist and are finite for almost

every x 6 (a,b), it follows from (6) that lim sn(x)
n-m

exists and is finite for almost every x 6 (a,b).

Furthermore the function qfix) = lim sup sn(x) being
nam

equal almost everywhere to the difference of two Perron
integrable functions, is a Perron integrable function on

(a,b). This proves (3) and (4).

45

To prove (5) construct sequences of minorants

[Un}n=l and majorants [Ln]n=1 for m such that
(7) lim Un(x) = lim Ln(x) = (p)fxn
n-Oco 11-000 0

for every x 6 (a,b).

Since we may take f=m in the first part of
the proof, we see that the function Un - F fulfills

the conditions of lemma 14 (n = 1,2,...).

Part (iii) of lemma 7 and the fact that Ln is
a majorant (QLn(x) ¥ -m for all x) imply that the
function K = F - L belongs to 5 and that
n a,b

1im inf[K(B.(x))-K(a.(x))] = O (n = 1,2,...).
11-00:) J 3

Moreover, for each x e (a,b)‘\JCUD)

Klfij(x))-K(aj(x))

lim inf
j” 53°62) " ajG‘)

 

Ln(Bj(X))—Ln(aj(x))

= lim inf[s.(x) - _
j4m J Bj(X) aj(x)Y

 

3 lim inf sj(x) - QLn(x) g 0 (n = 1,2,...)
3-...

because Ln is a majorant for m, 2Ln(x) > -m,
1im inf s.(x) < +m, and lim inf s.(x) g m(x) g_QL (x).
j-Om j-Oco J n
Therefore for each n > O, F - Ln and Un - F

fulfill the conditions of lemma 14. Hence F - Ln

46

and Un - F are non-increasing on D n [a,b] for
each n. From this and (7) it follows that
F(x) - fxm and Ixm - F(x) are non-increasing on

a a
D 0 [a,b].

Whence, F(x) — Ix lim 5. is constant on
a jam
[a,b] n D. D

16. Theorem. Let Z} fn be a K-series satisfying
n=1
condition H. Let C be a countable set in (0,1),

f be a Perron integrable function on (0,1). Assume

lim sup Sn(X)-Z f(x) a.e. on (0,1),
nqm

lim sup sn(X) > -m and lim inf sn(x) < +m
n-Mn n46)

for all x E (0,1)‘\(CUD). Then 23 fn converges a.e.
n=1 a:
to a Perron integrable function m and Z? fn is the
n=1
PFxbseries for m. Moreover

{12:11: fn = (NJ: cp for all x 6 [0,1].

PrOOf. Take a=0, b=l in theorem 15. Then

by 15, Z} fn
n=1

integrable function m. Furthermore F(x)-(HIx m is

O
0, so F(x) = (”Ix co
0

converges a.e. on [0,1] to a Perron

constant on D. But F(O)

for x e D.

47

Let 2) 9n be the PFK-series for m. By
n=1

pr0position 9. 23 Ix 9n = (P)fx m for all x 6 [0,1].
n=1 0 0

Therefore, proposition 8 applied to the K—series

co

n=1 (fn-gn) gives fn = gn (n = 1,2,...). This

proves the assertion. D

to
17. _Theorem. Let Z} fn be a K-series
n=1
satisfying condition H. Let g be a Perron integrable

function on [0,1]. Let C be a countable subset of
[0,1]. Let {an,k} and {bn,k} be two non-negative.
limit preserving matrices such that {k : an k # O}
I
and [k : bn k # O} are finite (n = 1,2,...). Define
m ' Q
— ’ —
on -kEl anksk and on —kEl bnksk' Let {nj} and
{n5} ‘be two increasing sequences of natural numbers.

Assume

lim on = g in measure,
j*w j

I

lim sup on,(x) > -m for x E (O,l)\.C, and

j-w° 3

lim inf Og.(x) < +m for x 6 (O,l)\(C.

3
Then 2} fn(x) = g(x) for almost every x in [0,1]
n=1
and Z3fn is the PFK-series for g.
n=1

Proof. There are integers jl < j2 <... such

that putting mt = n. , we have lim 0m (x) = g(x)
3! Law I

48

for almost all x. Since the matrices are non-negative
it follows that lim sup sn 2_lim sup 0 , lim sup 5 .2
n4m nam r1 ndm 1‘

lim sup 0’, and lim inf s g_lim inf 0;. Because of
D400 n 400 n 400

this 1im sup 5 (x) 2.1im sup on(x) 2_lim cm (x) = g(x)
n 400 n 11 4m 400 I,

almost everywhere. For x 6 (O,l)\\C

lim sup Sn(x)-2 lim sup o£(x) 2.1im sup o',(x) > -m
nam nqw jam j

and

lim inf sn(x) 3 lim inf o;(x) 3 lim inf o’,(x) < +m.
ndm n4» jam j

Q

Therefore.theorem 16 may be applied to see that Z) fn
n=1

converges a.e. to a Perron integrable function m for

which it is the PFK-series. But 1im sn(x) = qKX) a.e.
nqm
implies 1im 0n(x) = ¢(x) a.e. since {an k} is limit
ham ' -

preserving. Therefore g=¢ a.e. by the uniqueness of

the limits. This completes the proof. (3

Generalizations of the theorems of Wade [14],
[15] for Haar and Walsh series may be obtained as
corollaries to pr0position 3 and theorems 5,16 and 17.
In each of these corollaries C will denote an arbitrary
countable set in [0,1] and A the set of dyadic

rationals in [0,1]. If 23 akxk is a Haar series,
k=0
n-l

then we W111 denote Rigakxk by 5n (n = 1,2,...).

49

an,k = bn,k = O for n # k and

an,n = bn,n = l (n,k = 1,2,...) in theorem 17, then

If we put

we have the following assertion.

O

18. Corollary. Let Z} a xn be a Haar series
n=O n
satisfying condition GH. Let g, [nj]”and [n;} ‘be as
n. n.
in theorem 17. Set 2 3 = mj and 2 3 = m5. Assume

1im Sm = g in measure,

j+~ 3

lim sup Sm. > -w on (O,1)\.C and
jam J

lim inf sm’ < +o on (0.1)\\C .
3*” 1

Then Z} anXh(x) = g(x) for almost all x in [0,1]

n=0
Q
and Z) anxn is the Perron-Haar-Fourier series for 9.
n=0
Let W ,w ,... be the walsh functions and
O 1 n_1
ao.a1,... be real numbers. Denote jg; ajwj by 3n

(n = 1,2,...), as we did above for Haar series. It is
well known (see for example [1] or [14]) that there are
real numbers YO'Y1"" such that

Zn-l 2n

23 a.) (X) = Z;

1
yixi(x) for x Q'A

and n 2_O. Wade [14] has shown that if 1im a = O,

k-Om k

m

0
then 2‘ Yixi satisfies condition GH. wade [14] also
'=O

has shown.that if ZiYixi is the Haar-Fourier series
130

m
of a function 9, then 23 a is the Walsh-Fourier

j=o jwj
series for g. Putting C U A for C in conallary 18

‘we get the following corollary to theorem 1?.

50

19. Corollary. Let Z: a
n=0
series. Let 9, [nj} 'and [n3.’] be as in theorem 17.
n. n.
Set 2 J = mj and 2 J = mj’. Assume

(I be a Walsh
n n

lim sm = g in measure,
3-.” 3'

lim sup Sm' > -oo on (0.1) \ (CUA).

j-w J

1im inf sma < +oo on (O, 1) \ (CUA), and
3'4” 3'

lim a = O.

k-wo

m
Then X anwn is the Perron-Walsh-Fourier series for
n=0

9 and lim 5 n (x) = g(x) for almost all x 6 [0,1].
ndw 2

CHAPTER IV
EXISTENCE OF INFINITE SUMS FOR K-SERIES
ON SETS OF POSITIVE MEASURE
In a 1965 paper, A.A. Talalan and F.G. Arutyunyan
[13] have shown that a Haar or Walsh series cannot
have an infinite smnn on a set of positive measure.
This problem has been considered by R.F. Gundy [7]
and V.A. Skvorcov [12]. In this chapter we will
show that under suitable restrictions on the sequence
{Dn}, a K-series with respect to {Dn] can not
have an infinite sum on a set of positive measure.

The method is similar to that used by Skvorcov.

1. Definition. Let g be a finite function
on D. Recall that if [x,y] = J 6 3, we write

g(J) = g(y) - g(x). For x (’D, define the symmetric

 

D:derivates of g at x to be the limit points of

 

the sequence

g(Jh(X))

‘Jh(x)‘ °

In particular.define
g(Jh(X))

S g(x) = lim inf
—D n” IJn(x) |

51

52

 

and
_ ) g(Jh(X))
S g(x = lim sup .
D mm r
23 Lemma. Let Z: fn be a K-series. Let
n=1

 

x 6 [0,1] \ D. Then the set of limit points of the
sequence {sn(x)}::=l is the same as the set of
symmetric D-derivates of F. In particular

lim inf sn(x) = §DF(x), lim sup sn(x) = SDF(x) and

D40 n4»
Q
at points of convergence 2: fn(x) = SDF(x).
n=1

Proof. This is an immediate consequence of

lemma 7, part (ii) of Chapter III.

3. Lemma. Let G be a finite function on D.

 

Let a > 0. Let E C:[O,l] \ D be a set of positive
outer measure. Assume §DG(x) < a for every x e E.
Then for each 6 > 0 there is an integer n.2 l and

an interval Q C fih such that

|Q| < e: |E n 0‘ > (l-e)|Q( and
6(0) < a|Q‘.

Proof. Let x be a point of outer density
for E. There is a 6 > 0 such that ‘E n J‘ > (l-€)|J\

for all intervals J such that x E J and ‘J‘ < 6.

53

By the hypothesis we have

lim inf
n-Ooo

Choose an i such that

G(Ji(X))
(Ji(x)‘ \ a
Take Q = Ji(x) . c3

4. Definition.
satisfies condition Q1

with the pr0perty: if

G (Jn (x))

< a.
Jn x

and (Ji(x)\ < min (6,6).

We say the sequence {Dn}

if there is a q 6 (0,1)

J 6 3h, L 6 fih+l and L C J,

then +§+ > q (n = 1,2,...).

5. Lemma. Let

[Dn] satisfy condition 01:

let q be the corresponding number from definition 4.

Let G be a finite function on D. Let E C'[O,1],

Q E I. Let 5 > O and b be arbitrary numbers.

Assume that

(l) ‘E n 01 > (1-€)|Q(

(2) G(J) > O for every J 6 3' such that

JCQ and JnE#¢

(3) SDG(x) >.b for all x 6 E.

Then 6(0) >b(1 if) 101.

l

54

Proof. We may assume E c Int Q‘\.D and that

all points of E are points of outer density. Since

 

1 1 G(Jn (x)) _ ) b
' = S G
H 12.21“" E1001 D (X >

for each x E B, we may associate with each x e E

and n > O a J 6 fi' such that

(2) G(J) >b‘J‘, x 6 J and ‘J| < n.

The collection of all intervals J 6 fi' such that
(3) J OE7-’¢, JCQ and G(J) >b|J|

is therefore a Vitali covering for E. Set 61 = €‘Q|.
Applying the Vitali covering theorem, we can find
a finite number of non-overlapping intervals

Jl’...’JN E .8: J1 c Q SO that

(4) G(Ji) >b(Ji|

(5) Ji nnslgx (i=1,2,...,N), and

N
(6) (E \iil Ji‘ < 61'

But (E) > (l-e)‘Q| by assumption and
N N
‘E\.Ul J“ 2 (El - Z \Ji‘.
1:

i=1
Combining this with (6) we see that

N N
(7) (u Ji] = 2:131! > |E| - e1 >- (1-€)1Q\- em
i=1 i=1

= (l-ze)\Q|.

55

Choose K so large that none of the intervals J1....,JN
belongs to 3h for every n 2DK. Let N ‘be the

system of all intervals L such that (i) L C’Q,

(ii) L 6 3k for some k 3.x, (iii) either L is

one of J1....,JN or L n E #'¢' and L has the
property that of the intervals belonging to fik+l whose

union is L, at least one doesn't overlap with any

of the intervals J1,...,JN.

We first show that Q = LJL. For if not, then
LEN

we could find an interval Lo 6 fik ‘which is contained

in Q but not contained in any interval of u. Now

Jj 6 M for j = 1,...,N by the definition of u.
Suppose Lo overlaps with Jj for some j. Then

since Dk c Dk+1 for each k, either Lo C'Jj or

Jj c Lo' From the choice of K it follows that Jj c L0
is impossible. So Lo c.Jj. But L0 is not contained
in any interval from u. Hence Lo overlaps with

none of the intervals J1,...,JN. Therefore, if

Lo n E #'¢, then Lo fulfills (i), (ii) and (iii)

in the definition of H. But Lo g u. Hence Lo 0 E = ¢.

Consider the interval L1 6 fiK-l which
contains Lo' ‘We have L1 c:Q because 0 6 3, so if
Ll n E ¥’¢. then L1 satisfies conditions (i). (ii)

and (iii) of the definition of fl. But this implies

 

56

L CU L, acontradiction. So L nE=¢. If

° L691 1

L1 = Q, the construction terminates. In case L1 # Q
proceed as follows. We claim L1 overlaps with no
interval belonging to u. To see this recall that

Dk c Dk+l for every k. Therefore, if L1 overlaps
with any interval I of U, either I :ILl or

Ll : I. If L1 c: I, then L1 c: U L, which implies

L6”
L0 clJ ‘L, a contradiction. On the other hand, if
LG”
I c'Ll, then L1 n E # ¢' since I n E #’¢' by the

definition of T, again a contradiction. Hence L1
overlaps with none of the intervals from M. In

particular, L1 overlaps with none of J1....,JN.

Now conSider L2 6 fiK-Z such that L1 c L2. Clearly

L2 fulfills (i) and (ii) in the definition of 2L

If L2 n E 51¢. then L fulfills (iii) in the

2

definition of ll. But this implies L c L c L c U L,
o l 2 Let!
a contradiction. So L2 0 E = ¢. In this way we

could construct a finite ascending sequence of intervals
disjoint from E and belonging respectively to 3 ,

3K

sequence is Q. Therefore Q n E = ¢, contradiction.

-1""'Jb, where Q 6 3h. The last term in this

Hence Q = L) L.
Lefl

For any pair of intervals in. 21 which overlap,
one is contained in the other. Thus we may replace the
system a iby a system ”1 c N such that Q = L) L and

LEfll

no two intervals of ”1 overlap.

57

Set A = L) Ji' Then by (4) we have

Jie’fi

(8) Z G(Ji) > Z b‘Ji| =b\A\.
Jieul J em

i 1
Set u2=[Le9.Il:L9!Ji,lgi_<_N}. Then

 

(9) Q\A= UL

and L 0 E #'¢ for each L 6 £5, so that by (2)

(10) o < Z G(L).

Leflé
If L 6 $12, then L ;! Ji (1 = 1,...,N) by the construction
of ”5. Since #2 c M any L e u: satisfies (iii)
in the definition of u. Therefore, if L 6 3:, say

(1) such that L(1)

‘- 3j+1

overlaps with none of J1....,Jn because of (iii). By

L 6 35, there is an L

condition 01' (L‘ < é\L(I)\ for each such pair L
(l)

and L . Therefore

\Q\A(= z ‘L“<l z ‘L(1)‘.
Lei!2 q Lcsl2
However, the union of all such L(1) is contained in
N
(Q\UJi). Thus
i=1
N
1 1) /1 ,.1
(11) (Q \A‘ <§\U L( Ii§IQ\,U1Jil<~-§2€|Q\

Leflé 1:

in view of (7). That is

(12) m > (1 - £31m.

58

NOW

6(0) = Z G(Ji) + Z G(L) > Z G(Ji)
JiEQIl Leflz Jiefill

by (10). This combined with (8) and (12) gives

G(Q) L‘>b\A‘ >b(1 - 2:13))0‘. a

6. Theorem. Let (Du) satisfy condition Q1.

Let G be a finite function on D. Let
A = (x e [0,1]:§DG(X) > -m or SDG(X) / +m].

Then §DG(X) = SDG(X) and is finite a.e. on A.

Proof. Let A1 = (x : SDG(X) ; §DG(X) / -w}.
Suppose ‘Al‘ ) 0. There is a number a \ O and a set
B CiAl such that \B‘ > o and SDG(x) - §DG(X) > a
for all x e B. Let q be the number of definition 4.

Choose 3 > 0 such that

 

(1) 2€<a(l-2€).
q
For each integer n set
Bn = {x e B : n6 < §DG(X) g_(n+l)€}.

Since ‘8‘ > 0, there exists a p such that ‘Bp‘ > O.
For each x 6 Ep there is an integer m(x)

such that for m 2 m(x)

G(Jm(X))

(2) p5 W .

59

Now define

E = (x e Bp : m(x) g_m] (m = 1,2,...).

m
Then B = L) E , so there is an m such that
p m=l m 0
‘E | > O. Denote E by E and set 6 = min (J).
m m
o o Jéfih
0

We have for all x F E
(3) O < _S_D[G(x)-pex] = §DG(x) - p6 g e < 26

in E c:B and e S G x l e on B . Further,
s ce p P < -D ( ) g (P+ ) p

we have for all x 6 E

(4) §D[G(x)-pex] > §D[G(x)-pex] - _S_D[G(x)-pex]
= §DG(x) - §_DG(x) > a.

Also

(5) G(Jm(x)) - pe‘Jm(X)‘ > o for all x e E

and m 2_mo.

Applying lemma 3 and (3) we can find an

interval Q E 3' such that

(6) (Q) < min(e,6)
(7) ‘E 0 Q) > (1 - min(e,5))‘Q‘ and
(8) 6(0) — pe‘Q‘ < 2e(Q|.

If JEfim, JflE¥¢ and JCQ, then (6)
implies that m 2_mo. Thus from (5), (4) and (7) we

see that lemma 5 may be applied to the function G(x)-pox.

60

Hence

(9) G(Q) - palm > a(l - 2519))“-
From (8) and (9) we see that
2e(Q| > a(l - 3&5) (0|.

which implies 23 > a(l - %f). This is a contradiction
with (1). Therefore ‘A1‘ = 0. Similarly we show

{x : §DG(x) < SDG(x)<'+m} 'has measure zero.

It remains to show that M = [x : SDG(x) = +m]
has measure zero. For each x 6 M there is a natural

number m(x) such that m 2_m(x) implies

 

G(Jm(x))
> O.
ij (X? T
Set
Mj = {x e M : m(x) g.j] (j = 1,2,...).
Assume that [M‘ > 0. Then since M.= L) M. there is
j=1

an n such that ‘Mh‘ > 0. Let y E Mh be a pOint

of outer density and choose m > n such that
|J (y) flM|\(1-l)‘J (m
m n ’ 2 m '

Set P = Jm(y). From the definition of Mn it follows
that if x E’Mn and j 2_n, then G(Jj(x)) > 0.
Therefore, if J E I) J C P and J n Mn # Q , then

G(J) > 0.

Let x 6 Mn. Since x E M. we have

SDG(x) = SDG(x) >|b for every real number b. Therefore

61

conditions (1), (2) and (3) of lemma 5 are fulfilled

with e = , from'which it follows that

NIH

G(P) > b(l -(l"1)‘P‘

for every real number b. But this is impossible.
Hence M has measure zero. Similarly, we show

[x : SDG(x) = —m} has measure zero. D

7. _Theorem. Let {Dn} satisfy condition Ql.
Let E; fn be a K-series with respect to [Dn}' Let
E bg—a set of positive measure. Assume that for each
x 6 E, either

lim sup 3 (x) < +m
nan

or

1im inf Sn(x) > -m.
naw

m
Then 23 fn converges to a finite number almost everywhere
n=1
on E.

Proof. This follows immediately from lemma 2

and theorem 6. D

8. Corollary. Let {Dn} satisfy condition 01'

 

Then a K-series with respect to {Dn} cannot diverge

to either +m or -m on a set of positive measure.

62

9. Definition. We say {Dn} satisfies
condition Q2 if for each natural number n there
0 o \ .
is an interval J 6 fih such that Qn+l Dn c.J
Note that if Dn has n+1 points (n = 1,2,...),

then {Dn}n=1 satisfies condition Q2.

R.L. Gundy [7] has considered the problem of
representing arbitrary measurable functions on a
probability space by means of series with respect
to complete orthonormal sequences of step functions. If
we take [0,1] with Lebesgue measure as the probability

*
space and construct an H -system [u )m

accordin
n n=1 9

to Gundy's definition, then there is a sequence

{Dn]:;1 fulfilling condition 02 such that any series
Q

2: a u is a K-series with respect to {D }. Let

n=1 n n n

{Uh} be such a sequence. Then applying theorem 2.2

of [7] we see that there exists a K-series with respect

to {Dn] with an infinite limit on a set of positive

measure if and only if {Dn} does not satisfy

condition 01'

This leads us to conjecture that the assumption

of condition Q1 is essential in theorem 7.

CHAPTER V

REMARKS ON EVERYWHERE CONVERGENCE

V.A. Skvorcov [10] has shown that if a Haar
series converges pointwise everywhere on [0,1] to
a bounded function 9, then the given Haar series
is the Haar-Fourier series for g. On the other
hand, L. Faber [3] has constructed a nontrivial
Haar series which converges to zero at all but one
point in [0,1]. In this chapter the corresponding

questions for K-series are considered.

First an example of a nontrivial K-series
which converges to zero everywhere on [0,1] is
constructed. The partial sums of this example are a
subsequence of the partial sums in Faber's Haar

series example.

1. ‘Example. The example is constructed by

defining the partial sums. Let {Dn}:el be given by

e .3.- '= n = '
Dn [ n . j O,l,...,2} (n 1,2,...). Define Sn

2
to be regular and

63

64

 

(
n-l l 1. l
_ n-l l 1 _1_
Sn - < "'2 0n (5: i + n)
2
0 ..f[;-la,;._lfi]
k 2 2

(n = 1,2,...). Set fl = 31’ fn = sn-sn_1 for n 2.2.

Take an n 2 2. The elements of fin-l are of the form

Jj = [fififilv “35$? (j = 1.. -v2n-1’-

Let vj be regular and satisfy

 

l on int J.
V. = { j (j = 1'2’ooo'2n-1)0
3 0 off J.
3
Then {Vl""'V2n-l] is a baSis f0: 8n-1' In order to
show fn 6 Tn we need only show I; fnvj = I; fn = 0
. n-l 3
for J = 1300002 0 BUt
n-2 l 1 l. l l l 1
2 on (- " __ 0‘ "' "‘9 U (-,- + —)
2 2n 1 2 2n 2 2 2n
_ n—2 l .2; l l _L_]. l
fn - '2 0“ (a ' n'i) U (5 + n'2 + n—l)
2 2 2
1 l l. l
0 off [5 - 33:1w5 + n-1]'
Therefore, I f = 0 (j = 1,...,2n-1).
J. n

3
m
Since fn E T (n = 1,2,...) the series Z} fn

n
n=1
is a K-series with respect to [Dn}.

By the regularity sn(%) = O for all n. Fix

xa!

Nil-4

: there is a j such that sn(x) = O for n > j.

65

Q
Therefore fn(x) = O for every x. Notice that
n=1

_ l
I 1 Sn — 2
for all n, so 23 fn does not satisfy condition H.

Example 1 shows that for general K-series
there can be no analogue to Skvorcov's Haar series
theorem. In order to obtain a class of K-series for
which there is an analogue to Skvorcov's theorem we

will impose the following conditions.

Recall from Chapter IV that the sequence [Dn]

satisfies condition Q2 if for any n, there is an

interval J 6 fih such that Dn+l\\ Dn C.I.

2. Definition. we say {Du} satisfies
condition Q3 if there is a number K ‘with the property
that for each pair of intervals J1,J2 G fih+l with
J1 UJ2 CL Ebn we have

)Jl)

T3?

< K (n = 1,2,...).

3. Lemma. Let {Dn} satisfy condition Q2.

Let Z? fn be a K-series with respect to {DD}. Let
n=1
P = [a,b] 6 5P, a1 6 int P, P1 = [a1,b] E 3H and

sq 2_sp on int P1. Assume that for each n > p and

each J G fih ‘with the property J c int P we have

66

SD g,sp on int J. Then there are numbers

q = m1 < m2 < ... and a1 < a2 <...< b ‘such that

P. = [a.

J "‘3' J
on int Pj' Moreover (mj) may be chosen so that

= fm.+2 =...= fm 1 = O on int P. (j = 1,2,...).

f
j j+l- 3

.+1
"‘3

Proof. Because of condition 02 and the orthogonality

there exists L 6 £5 such that the support of fn+1 is

contained in L (n = 1,2,...).

If fn = O on int P for all n > q, the

l
assertion is trivial. Otherwise proceed as follows.

Let n be the smallest integer i > q such that fi

is not identically zero on int P1. By orthogonality

and condition Q2 there is at least one interval I e fih
such that I c P1, I # P1 and fn > O on int I. From
our assumptions it follows that we cannot have I c int P,
because sn > Sn-l =...= sq 2_sp on int I. Let

I = [a,fi]. Then a < a1 g_a < B g_b. If B < b, then

I Clint P. Therefore fifib and a1 < a. Set m = n,

2

a2 = a, and P2 = I.

Now assume that m1,...,mj and P1,...,Pj

have been constructed. Repeating the argument above

1 and

P. i l , . . .
J_1 n p ace of P we construct m3+1 and Pj+l D

with m.j in place of q, Pj in place of P

 

67

4. Theorem. Let f be a bounded function on

[0,1]. Let [Dn} satisfy conditions Q2 and Q3. Let
Z: fn be a K—series with respect to {Dn}. Assume
ngl a

23 f (x) = f(x) for all x 6 [0,1]. Then 2: En
n=1 n n=1

is the LFK-series for f.

Proof. Let \f(x)\ g_c for all x G [0,1].

We will show that (sn(x)\ g_c for all x and n.

n
0

Suppose not. Then there is an no and Jo 6 3
such that Isn [ > C on int J6. For the sake of
o
definiteness, assume sn > C on int JO.
0
It will be shown that there exists a sequence of

intervals {Jj} and an increasing sequence of integers

{nj}' with the properties

(1) Jj+l g int Jj,
(2) J]. 6 .Bn'.
3
(3) Sn. .; Sn. 2 C on int Jj+1 (j = 0,1,2,...).
3+1 3

The sequences are constructed by induction. We have

n and Jo. Assume that no,n1,...,nk and J

O JIOOO'J

0’ 1
have been constructed. Suppose that for every n > n

k

k
and every I 6 3h 'w1th I c int Jk we have sn < snk
on int 1. (We wish to deduce a contradiction.)

68

Obviously there exists an index i > nk such that fi
is not identically zero on int Jk. Let m be the
smallest such i. By property Q2 and orthogonality

there is an I 6 3m such that I c:Jk, I # Jk and

fm 2 O on int I. Since sm > Sm-l =...= snk on

int 1, the interval I must, by the assumption, have

a common endpoint with Jk' For definiteness, assume

the common endpoint of J and I is the right,

k
denoted by yo

Now apply lemma 3 with p = nk, P = Jk' <;=m

and P1 = I to find integers m = m < m <... and

l ‘ 2
intervals Pj 6 3h (j = 1,2,...), all having y
as a common right endpoint, such that fm g_o off Pj'
i
f >0 01’! int Pol f =f =ooo=f =0
m. —- .+1 .+2 . -l
3 J m] m] m3+1
on int Pj (j = 1,2,...).
Define M. = -min f (t) (j = 1,2,...).
3 t mj
Suppose for j 2_2
4 M. > f - .
( ) 3 _, ml(y )

Let P; denote a member of fifi upon whidh —Mj is

J
attained by fm . Then we have

(5) MjIP5‘ g fmj(Y-)\Pj| (j = 2,3,...).

Because of condition Q2 we have fm (y-) = me (y)

3 3

69

(j = 1,2,...). This fact combined with (4), (S) and

condition Q3 gives

IP-I ,
'(y) P, < me. (y)K
J J 3

(6) o < 2fml(y) g M]. g 2fm

for j 2.2, where K is a constant independent of j.

But the convergence of Z: fn(Y) implies 1im fm (y)
n=1 j-Ooo j
a contradiction with (6). Therefore, there exists j \

for which Mj < fm1(Y-).

By condition Q2 and orthogonality the support of

fm is contained in Pj-l (j = 2,3,...). Therefore
3'
f c P. (j = 2,3,...). Now P3 is disjoint from
J 3-1 J
and to the left of y so pj“ c int Jk (j = 2,3,...) .
Let j1 > 1 be such that ‘fm. (t)| < fm (y-) for
31 1
. ; . 4)
every t 6 int Pj . Then fm + fm_ > O on int Pj .
l l 31 1
Since P? c P. for j > 2, f > O on int P3
3 3-1 " mj" 31

(j = l,2,...,jl-l). Combining this fact with the choice

of (mj} we have

Sm = sm -1 + fm +...+ fm + fm > sm _1
31 l l jl-l jl l
= 3n on int Pf ,
k 3l

a contradiction.

0.

70

Therefore there is an n > nk and J 6 fih

such that J c int JR and s 2.3 on int J. Set
n nk
nk+l = n and Jk+l = J. This completes the construction.

From the density of D and from (1), (2), and

(3) we have F) JR = [z] and lim sn (2) >.C 2_f(z),
k=l kdm k
a contradiction. Therefore (sn(x)| g_c for all

x 6 [0,1] and all n.

Fix an integer m 2.1 and J 6 Sb. Set

9 = characteristic function of J. Then
I1 9 sn = f sn for n 2_m.
O J

By the Lebesgue Dominated Convergence Theorem we have

lim 5 = lim 9 s = (1 gf = f.
nee J n ham 0 n o IJ
Because of the orthogonality
I s = I s for n > m.
n m '—

Combining these last two facts with theorem 8 of Chapter

m
II proves that Z: fn is the LFK series for f. [3
n=1

In the remainder of this chapter it will be
assumed that Dn has n+1 points. Note that this is

the case for Haar series, that condition Q2 is automatically

3-: ..

71

fulfilled and that conditions 00' Q1 and Q3 are
equivalent under this assumption. Furthermore if Dn
has n+1 points, then 3h has n elements and

3h \\fi _1 has two elements (n = 2,3,...).

5. Definition. Let I; \~fih—l have two

elements, say J and K. Set

= nmax(+—+-, (~19 (n = 2, 3,...).

Under the assumption that Dh has n+1 points,

 

condition Q3 is equivalent to boundedness of the
sequence {qn}. In the next example a sequence [Dn}

and a non-trivial K-series 2: f with respect to
[1:1 00
{Dn} are constructed such that Z: fn(x) = O for every
n=1
x 6 [0,1]. This example satisfies all the hypotheses

of theorem 4 except for the boundedness of (q }.
n

 

6. Example. Let 2 = A1 < A2 <..., lim An = w,
n-boo
. . l 1. 1 1 l l
llm(A -A ) = 0. Define J = [- - ——v— , L = [-,- + —-]
nqm n+1 n n 2 An 2 n 2 2 An
(n = 1,2,...). It is easy to see that there is a sequence

{Dn} fulfilling the requirements of definition 1 of
Chapter II and a sequence of natural numbers 2 = R1 < k2 g...

with the properties

(1) Dn has n+1 points

72

(2) Jn'Ln G fikn

(3) Jn+1 E fikn+l

(4) Ln+1 6 3k +2'
n
Set 51 = O. For each n
to be regular and
A on
n
.n
O on
Let skn+l be regular and
An+1 on
SR +1 = -An on
O on
Let skn+2 be regular and
An+1 on
Skn+2 = ”An+1 °n
O on
Set s = s =...= s
kn+1 kn+1 l kn+3
fl = s, and fm = Sm-Sm-l (m =

is identically zero for m ¢ kn+l

.2 1 define sk
n
int J
n
int L
n

[001] \ (Jn U Ln) .

int Jn+1

int L
n

[0,1] \ (Jn U Ln) .

+1

int Jn+1
int Ln+1

[O'l]\\(Jn+l U Ln+l)°

= 3 Define

kn+2

2,3,...). Then fm

and m fi'kn+2

(n = 1,2,...). Therefore fm E Tm for m # kn+l and
11,...,Ik be an enumeration of the intervals in fik .
n n

73

to be regular and

{:1
V . =
3 0

Then

Define vj

on

on
(J = lI"°Ikn)'

8k .

n

{V1, ...,Vk}
n

In order to show f

k +1 5 T
n

Show

Ilf
v. —
O kn+l j

for j = 1'°°°'kn' Obviously f

int 1., if I. # J . Since
3 J n

0. Hence fkn+1

we have I fk +1 =
Jn n

Similarly let P1.---.Pkn+1

of the intervals of 3k +1. Define
n
and

on int

1 .
u , =
3 0 off Pj

Then {u1,...,ukn+l] is a baSis for

[0,1]\1.

k +1
n

int 1.
J

J

is a basis for

we need only

 

- O on
:1: S y
I; kn+l
n
6 Tk +1’
n

be an enumeration

uj to be regular

P.

3 (j = 1,...,kn+l).

. To show
8kn+l

fk +2 6 Tk +2 we need only show
n n
(1 f I f o
u. = =

O kn+2 j Pj kn+2
for j = 1,...,kn+2. ObViously fkn+2 = O on int Pj
if Pj # Ln' Since I; skn+l = l = IL skn+2. we have

n n

f = 0.
IL
n

74

Consequently, fm 6 Tm for every m. So

Z} fm is a K-series with respect to [Dm]. Let
n=1
x = %. Then there is an integer m such that sn(x) = O
l 1
for n 2.m. For x — 5 we have skn+1(§) — An+l - An
and
l l l
s (—) = s (-) =...= s (-) = O
kn+2 2 kn+3 2 kn+1 2 r:
(n = 1,2,...). Therefore 1im sm(%) = 0.
co Ill-00° FL
Hence m2; fm(x) = 0 for every x, but fkn+l and

 

fkn+2 are non-zero functions (n = 1,2,...).

The example constructed in section 6 has several
interesting prOperties which are the subject of the

next remarks.

7. Remark. Let {An}. {kn}. {Ln}, (Jh}, be

as in example 6. Set

Then q =q = =—_:—_ (n: 1,2,...).
kn+l kn+2 ‘Kn+1‘ An+l An

Moreover, there does not exist an integer N such that
qk +1 g_n for n > N. To see that this is the case,
n

suppose that there were such an N > 1. Then for any

A
n > N ‘we have % g_—%il-— 1. That is
n

75

l /
(1) Ar“1 + H) 5-An+l

There is an d > O for which
(2) An 20.11 (n = l'ooopN+l)o

Let n be an integer such that An 2_dn. If n g_N+l
then (2) holds. If n > N, then by (l) we have
.An+l 2_d(n+l). So that (2) holds in any case. From

(1) and the fact that lim(A
A [1460

that lim 79-: O, a contradiction.
n-boo

n+1-An) = 0 it follows

8. Remark. We have seen in section 7 that no
matter how the sequence {An} is chosen it is impossible

for qk +1 to be eventually bounded above by n.
n

HOwever, choose 8 > 1 and d 6 (8-1,1). Define
An = 2n“ (n = 1,2,...). Then
-1 n+1 a
qk+1=<T’ -1-
n
Now
(Efl)O‘-l=(l+-l-)a-l=9b,
n n n n

a

where bn = (l + j?)a-1 and O < an < 1. Therefore
1im b = 1, so there exists an N such that b :>——-

n n Dd

new

1 \ 9 IL.
+1 ’ n [_(1.

n

for n 2.N. For n > N we have then q;

Thus qk +1 < n8 for n 2'N.
n

9. Remark. Let 1 < bl < b2 <..., iii'bn = m.

It is clear that the sequence {kn} in example 6 may

 

76

be chosen so that qkn+1 < bkn+l and qkn+2 < bkn+2:
at the same time we may choose the sets Dn so that
qm = l for m # kn+1, kn+2 (n = 1,2,...). Then

/
qm \ bm for any m.

In View of example 6 and these remarks, it
appears that the assumption of condition Q3 in theorem 4
cannot be substantially weakened. A natural question
is whether the imposition of conditions on the rate
at which the lengths of the smallest and largest intervals
in the nEE- partition tend to zero might imply the

conclusion of theorem 4.

10. Definition. Let {Du} be given. Denote

v = max \J‘ and = min ‘J‘ (n = 1,2,...).
n Jean H“ Jean

11. Lemma. Assume that there exist numbers
d,B such that nun > a and nvn < B (n = 1,2,...).
Then {Dn} satisfies condition Q3.

Proof. Take n > 1 and J,L 6 3h. Then

[JV/
(L '3 BE \ d

IUD

 

- D

 

lg. Corollary (to theorem 4). Let [Dn}
satisfy the assumptions of lemma ll. Assume Dn has
n+1 points for each n‘2 1. Let Z: fn be a K—series
n=1

 

77

with respect to {Dn] which converges everywhere on

[0,1] to a bounded function f. Then Z? fn is the
n=1
LFK-series for f.

The next example shows that if we only assume
{nun} is bounded away from zero the conclusion of

theorem 4 need not hold.

13. Example. Let 0 < d < 1. Set

A = 2nd

n (n = 1,2,...). Define sets E C E C E C...

O l 2
1
as follows. Set Eo — {0,1}, El — {0,5,1},

+ —-l}. Let m 2.2 and let finite
sets EO c E1 c...c Em be defined. Define

(1) __ . l (2) ._ . 1
Em —[t€Em.t<§}, Em —{t€Em.t>§]. Let
(1)

[t1 <...< tr} be an enumeration of Em and
/ / - (2)
{tr+l \ tr+2 \...< ts} be an enumeration of Em .
l) / .
Let &£ = {[tj—l'tj] : 2 ;_j g.r] and

déz) = [[tj-l'tj] : r+2 g.j g_s}. Denote 5m = dél) U 6é2).

Let Fm be the set of midpoints of intervals from 6m.

Now define

 

 

E -A1 p%+ l ]o
m+l

NIH

=E UF U[
m

m+l m

Now we prove Em c [0,1] (m = 0,1,2,...). The
assertion is clear for m = 0,1,2. Assume n‘g 2 and

En CT[O,1]. Then certainly Fn c [0,1]. Since

 

78

l
n+1

 

 

. / 1
An+1 > 2 it follows that O \ 2 - < l.

A

Therefore E C [0,1]. This establishes the assertion.

n+1

1_1_l_
Next we prove that max{t 6 En . t < 5] — 5 — An

(n = 2,3,...). This is obvious for n=2. Assume n > 2

l _ 1
and max{t C En t < 2} — A . Thus

NIH NIH

l
max{t P E U F . t < 5} +1

 

n
- i. Thus max{t 6 E :
A n
n
1

 

since 2 -
2 A

 

> % - ig— and
n+1 n

q
I a

1 }.
n+1

 

 

A This proves the

assertion. Similarly it may be shown that

%] - 1 +-JL (n = 2,3,...).

min{t G En : t > — 2 An

From these two facts it follows that

 

 

{l_ 1 1 l
2

p -' "I'
A 2 An+1

} n E = ¢ (n = 2,3,...)
n+1 n

because An+1 > An (n = 1,2,...).

Now we may show that En has 2n+1 points
(n = 0,1,2,...). This is clear for m = 0,1,2. Suppose
n 2,2 and En has 2n+l points. It follows from
the definition of Fn that Fn 0 En = Q. Moreover

n .
Fn has 2 -2 pOints. We have seen above that

l l l l
I“ " I - 'I'
2 An+1 2 A

 

 

)nE =¢.
n+1 n

79

Therefore
1 l 1. l
En+1 _ En U Fn U [2 - An+1'2 + An+1]

has (2n+l) + (Zn-2) + 2 points. This establishes

the assertion.

Set D = E (n = 0,1,2,...). In order to
2n n

define Dk for any natural number k, choose an n.

Let J1....,J n be an enumeration of the elements of
2
fi'n' Further assume that the indices are so chosen that
2

the right endpoint of Ji is the left endpoint of Ji+

 

 

l
(i = 1,...,2n-1). Set D = E U {midpoint of J 1.
n n l
2 +1
and D = D U [midpoint of J.) for 1 < j < 2n-
n . n . j -—
2 +3 2 +j-l
Define
l l
D = D U {- + }
2n+2n 1 2n+2n 1-1 2 An+1
and
l 1
D = D U (- + }.
2n+2“‘1+1 2n+2n’1 2 An+1
Define D n = D n U {midpoint of J.) for
- . J
2 +3 2 +j-1

2n—1+1 < j 3 2n.

Since (Egiia < 2 (n = 2,3,...) and An = 2nd

 

 

(n = 1,2,...), we have
X2.._=—-—2a>——-—————l a=A1 (n=2,3,ooo)o
n 2n 2(n-l) n-l
Therefore
1 l l
(l) A >A A (n—2,3,...).

 

l

80

Now we prove that

 

l l 1 ,
(2) -2-(A - A ) -\ for n > 4.

_1__
n-l n An A

To see this rewrite the inequality as

3—2(1 + 1 'a > (1 - 1)"O‘.
n n

Now using the Taylor expansion for (1+x)-a we see

that there are sequences {cn} and {dn} satisfying

 

 

/ l l
O \ Cn < H: "' a < dn < O!
l d a G(G+1)
(l + H — l - E + 2 a+2, and
Zn (1+cn)
l -a _ a djg+l) _
(l _ H) — 1 + F1 + 2 (1+1 (n —' 2,3,...).
2n (1+dn)
Hence
(3) (1+l)‘“/1—9+a—L——La+1 .
n \ n 2n2

Now let n > 4. Since n.2 5 and O < d < l, we have

 

\ l a+2 4 d+2 4 3
an / " g: (1+dn) > (g) > (3') I

that is

1 125

(1+d )o+2 < 64 < 2

n
so that
(4) (l - %)'“ < l + g + 9195l1-.

n

From (3) and (4) it follows that

 

81

3-2(l+%)-a (l-é'a>3-2+29-‘--91Cltl—)--l

 

 

 

n
n
_ a _ d(d+l) _ d _ 2d d+l)
B 2 - E 2
n n
d
= —§-ln - 2d(d+l)] > O
n
This establishes (2).
_ 1 1.1 _ 11 .1
Set Jn - [5 ‘ A '5]' Ln — [5'5 + A 1'
n n
l 1. l l l 1. l l
K _ [‘ " """"I" ’ J: I - [- + ,- + ---]
n+1 2 An 2 An+l n+1 2 An+1 2 An
(n = 1,2,...). Now we prove
_ . 1 1 .1.
(5) U n — mln[§ un—l’A _ A] (n — 2,3,...).
2 2 n-l n
Obviously u. = l. By (1) we have 1 - 5L-< JL :
2 2 2 A A
2 2
therefore
. l l l l l
u =min{_-——,——)=————.
4 2 1112A2 2 A2
Sinc u +-l; - u and J / SL- we have u / l u
e 4 A — 2 ‘4\A p 4\2 2

2 2
which is (5) for n=2. Now suppose that (5) holds for
some n'g 2. Let J c fi’n be such that (J) = U’n‘
2

2
Because of (l)

. l 1 l 1
2n 2 2n 1 An-l An An

Therefore J is not Jn or Ln. Now in constructing

D2n+l from Dzn, each interval of fién\‘[Jn'Ln} is
diVided in half. Thus for each I 6 fién+l\‘{Jn+l'Kn+l'
l
Ln+1'In+l] we have (I) 2 5 (J). By (1) we have

(Ln+l1 = 1Jn+11 1Kn+11 = 1In+11°

 

82

 

 

 

 

 

Thus
1 l l l
u = min[- (J(,(K (} - minf- U r—— - 1
2n+1 2 n+1 2 2n An An+1
This proves (5).
L t — u Now w ro that U - Zs—nu
e U32 — . e p ve 2n —
for n 2.5. The relation is clear for n=5. Suppose
that for some n 2'5 u.n = 25_nu. From (5) it
2
follows that u < 1' --l;. Thus
n —-A A
2 n-l n
l l l 1
2 2n 2 An-l An
Combining this with (2) gives
1 . l l
2 2n An An+1
From (5) and (6) it follows that
. l l l
u =m1n[-H.—- ]=-u-
Therefore u l = l(ZS-nu) = 25—(n+l)u. This proves
211+ 2
the assertion.
1

Let n > 4 and 2n < j < 2n+ . Then

. . n
u U n’ so that juj.2 ju n+1 > 2 u n+l'

2 2 2 2
Thus juj > l6u. There is a number c > 0 such

(A

n+1-3 ”j

that mum > 2c for m = 1,...,32. Thus
. l . \
juj > l6u — 5 32u32 > c. Therefore k+k,” c for

every k 2_1.

 

83

Set 51 = 0. Define $2 to be regular and
A1 on int Jl
32 = .
-Al on int L1.
Let 53 be regular and
A2 on int J2
53 = -Al on int L1
0 off J2 U Ll'
For n‘Z 2 set pn = 3-2n_2+l and define s to
n
be regular and
An on int Jn
spn = —An on int Ln
0 off J U L .
n n
Set 5 = s =...= s . Let s be
pn pn+l pn+l-2 n+1-l
regular and
f .
An+1 on int Jh+l
s - -A on int L
pn+l—1 n n
0 off Jn+1 U Ln.
Then s E S and s 1 E Q _1 since J ,L
pn pn n+1 pn+1 “ n
n+1 n n+1 1
Take fl = 51 and fn = Sn-Sn—l for n 2.2.

6 fl
pn

Then it follows in exactly the same way as in example 6

 

84

that f E T for each n > 1 and that Z: f is
n n —- n=1 n

a non-trivial K-series with respect to [Dn} which
converges to zero at every point in [0,1]. However,
as was observed above, there is c > 0 such that

nun > c for every n.

 

BIBLIOGRAPHY

 

ll.

BIBLIOGRAPHY

G. Alexitis, Convergence Prdblems in Orthogonal
Series, Permagon Press, New York, 1961.

R.B. Crittenden and V.L. Shapiro, Sets of Uniqueness
on the Group 2w, Annals of Mathematics, 81 f1

L. Faber, fiber die Orthogonal funktionen des Herrn -
Haar, Jahresber. Dtsch. Math. Ver., 19 (1910), "3,

 

N.J. Fine, On the Walsh Functions, Trans. Amer. Math.

, Fourier-Stieljes Series of Walsh Functions,
Trans. Amer. Math. Soc., 86 (1957), 246-255.

 

A. Garsia, Topics in Almost Everywhere Convergence,
Markham Publishing Co., Chicago, 1970.

R.L. Gundy, Martingale Theory and Pointwise Convergence
of Certain Orthonormal Series, Trans. Amer. Math.
Soc., 124, number 2 (August 1966), 228-248.

W. Rudin, Real and Complex Analysis, MbGraw-Hill,
New York, 1966.

S. Saks, Theory_of the Integral, Dover, New York, 1964.

 

V.L. Skvorcov, A Theorem of Cantor Type for the Haar
System, Vestnik Moscow Series I Mathematics
and Mechanics, l (1964), number 5, 3-6.

, The Calculation of the Coefficients of
an Everywhere Convergent Haar Series, Mathematics
of the U.S.S.R. - Sbornik, 4 (1968), number 3,
349-360.

85

12.

13.

14.

15.

86

, Differentiation with Respect to Nets and
Haar Series, Mathematical Notes of the Academy
of Sciences of the U.S.S.R., 4 (1968), number 1,
509-513.

 

A.A. Talalyan and F.G. Arutyunyan, Convergence to +m
in the Haar system, Math. Sbornik, 66 (1965),

W.R. Wade, A Uniqueness Theorem for Haar and Walsh
Series, Trans. Amer. Math. Soc., 141 (1969),
187-194.

, Uniqueness Theory for Cesaro Summable Haar
Series, Duke Math Journal, 38 (1971), number 2,
221—227.