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This is to certify that the

thesis entitled

ON THE SYMMETRIC DERIVATIVE

presented by

Lee Matthew Larson

has been accepted towards fulfillment
of the requirements for

Ph. D. . Mathematics
degree in

 

 

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fly Major professor

April 18, 1981
Date

 

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ON THE SYMMETRIC DERIVATIVE

BY

Lee Matthew Larson

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
DOCTOR OF PHILOSOPHY

Department of Mathematics

1981

ABSTRACT
ON THE SYMMETRIC DERIVATIVE
BY

Lee Matthew Larson

A class of functions, 0*, is defined and is shown to
contain all known symmetrically differentiable functions.
It is proved that if f60*. then f is in the first Baire
class. Using this result, it is shown that there is
associated with each f€o* another function, g, which retains
the symmetric differentiation properties of f while at the
same time "maximizing" many of the more desirable properties
of f such as differentiability, continuity and upper semi-
continuity. Such a function. g, is in Baire class one and
is uniquely determined up to its values on a set with
countable closure. We call 9 the "nice copy" of {2

Using the properties of the nice copy, many of the
standard theorems of ordinary differentiation can be refor-
mulated in terms of the symmetric derivative. In particular,
analogues of the mean value theorem and the Darboux property
are presented. The methods also give simplified proofs of
several well-known theorems. These results are then applied
to develop an abstract Zahorski class structure for
symmetric derivatives.

In addition, several structure theorems for completely

arbitrary symmetric derivatives are proved.

ACKNOWLEDGEMENTS

I would like to thank my parents, without whom
I would not be, my wife, Ruth, who encouraged
my dream-quest of unknown end and Professor

C. E. Weil, whose patience was astounding and

whose advice was invaluable.

ii

Introduction

Chapter
Section
Section

Section

Chapter
Section

Section

Chapter
Section
Section

Section

Chapter
Section
Section
Section

Section

Bibliography

I:

1.1:

II:
2.1:

2.2:

III:
3.1:
3.2:

3.3:

IV:

4.1:
4.2:
4.3:
4.4:

TABLE OF CONTENTS

0 O O O O O O O O O O O O O O O O O O O O 1
Notation. Definitions and Basic Theorems

Notation O O O O O O O O O O I O O 0 O O O 5
Some Preliminary Function Theory . . . . . 12
A Covering Theorem . . . . . . . . . . . . 18
The Class of Arbitrary Symmetric Derivatives
Comparison with Baire Class One . . . . . 21
Arbitrary Symmetric Derivatives . . . . 24
The Structure of Functions in 0*
Nice Copies of Functions in 0* 28
Nice Copies of Functions in o . . . . . . 39
Monotonicity and Mean Value Theorems . . . 49

Symmetric Derivatives and the Zahorski Classes

The Abstract Zahorski Classes . . . . . . 58
Symmetric Derivatives and the Class m2 . . 64
Symmetric Derivatives and the Class m3 . 67
Symmetric Derivatives and the Class M; . . 71

. . . . . . . . . . . . . . . . . . . . 79

iii

INTRODUCTION

If f is a real-valued function defined on IR” then
the symmetric derivative of f at x (often called the first
Schwarz derivative of f) is

s _1im f baht-f Lx-h)
f m" h-o 2h '

The symmetric derivative arises naturally in studies of the
pointwise convergence of Fourier and Taylor series as well
as other areas of harmonic analysis. In this work, however,
we do not consider these applications of symmetric differ-
entiation, but rather, we investigate the symmetric deriv-
ative viewed as a generalization of the ordinary derivative.
Specifically, our goal is to expose similarities between the
well-known structure of ordinary derivatives and the struc-
ture of symmetric derivatives.

We begin in Chapter I by presenting much of the
terminology used throughout this work and by stating the
fundamental theorems needed in the succeeding chapters. In
particular, we define a class of symmetrically differentiable
functions, 0*, which is the "domain" for most of the later
theorems. It is shown that 0* contains all measurable,
symmetrically differentiable functions and therefore all
known symmetrically differentiable functions, since the

question of the measurability of such functions remains

1

unresolved. Chapter I is concluded with the proof of a
'partitioning theorem which was first stated in a slightly
weaker form by B. S. Thomson [26].

One of the most useful theorems available for the
study of ordinary derivatives, due to Zahorski [29]. is
that any ordinary derivative belongs to the first class of
Baire (81). It was proved by Filipczak [7] that the sym-
metric derivative of an approximately continuous function
is in 81. The main theorem of Chapter II is that this
result can be extended to the more general case of 0*. In
Chapter II, we also examine the question of whether there
are any symmetrically differentiable functions which are
not contained in 0*. While no answer to this question is
reached, several results are obtained which strongly suggest
that if any such function, f. exists, then fsis in $1.

It is well-known that if f is a finite-valued ordinary
derivative, then any primitive function for f is determined
up to an additive constant. That this is not the case with
a symmetric derivative can be seen by considering the
following two functions. Let f(x)=o everywhere and let.

x-2 for x=t1. il/Z.

9(X)=
0 otherwise
Then it is easy to see that f8(x)=gs(x)=0 everywhere, but
f(x)-g(x) is not constant. Because of this lack of a unique
primitive. many of the standard theorems of ordinary differ-

entiation are either false or much harder to prove with the

symmetric derivative.

A solution to this uniqueness problem is presented in
Chapter III with the introduction of the "nice copy" of a
function in 0*. The nice copy of f 60* is a.function, g.
which in some sense "maximizes" several of the desirable
properties of f such as differentiability and continuity,
while at the same time retaining the symmetric differen—
tiability properties of f. In particular, it is shown that
there is a set. A, with countable closure, such that gs(x)
agrees with f3(x) on A“: and further, that g is uniquely
determined and upper semicontinuous on 54:.

The existence of the nice copy for any f in 0* leads
at once to the existence of a "nice primitive" which is
uniquely determined up to an additive constant and its
values on a set, A, with countable closure. This nice
primitive solves the uniqueness problem presented above, and
thus gives us a means of establishing many of the classical
theorems of ordinary differentiation in terms of the symmet-
ric derivative. For example, the quasi—mean value theorems
of Aull [l] and Evans [5] and the monotonicity theorems of
Weil [28] and Evans [5] can be generalized to 0*. Another
consequence of the methods employed in Chapter III is a
simplification of a proof due to Charzynski [4] showing
that the set of discontinuities of any f ('0* such that f3
is finite-valued must be countable with no dense in itself
subset.

Finally, in Chapter IV, we extend the results of

Zahorski [29] on the associated sets of derivatives to

symmetric derivatives. In so doing, the results of Kundu
[16] are considerably strengthened. In particular. it is
shown that if f8 is the symmetric derivative of an f 6,0*
such that f8 has the Darboux property. then £86 7/121. Kundu's
theorems, in the cases of‘m3 and m4, are proved without his
assumptions that f is continuous and f8 has the Darboux
property. Examples are given to show that certain of the
proved Zahorski class containments are proper with symmetric

derivatives.

CHAPTER I
NOTATION. DEFINITIONS AND BASIC THEOREMS

Section 1.1: Notation

In this section we introduce most of the basic defin-
itions and notation which will be used in later chapters.
Several of the "classical" theorems concerning symmetric
derivatives are also stated to motivate some of the new
concepts.

Throughout this work the real numbers will be denoted
by IR and the extended reals, {-0, a], will be denoted by
111*. 2 will stand for the integers and 2+ will represent
the positive integers.

If ACJR is Lebesgue measurable, then the measure of
A will be denoted by [A]. In fact, the only measure we shall
have occasion to use is Lebesgue measure, so terms such as
"measurable,“ "almost everywhere," etc., should be inter-
preted accordingly. Ac will stand for the complement of A.

Let f be a real-valued function defined on an open
interval, I. If x 6 I, we define the upper (lower) symmet-
ric derivative of f at x to he

?s (x)=1im suphdof (x+h)2-;lf jx-h)

s =lim inf f( +h -f -h
(_f (x) h~o x )2h (x ) ) '
5

6
When fs(x)=§§(x). whether finite or infinite, we call their
common value the symmetric derivative of f at x and denote
it by fs(x). If fs(x) exists at every point of the domain
of f. then f is said to be symmetrically differentiable.

D+f(x°), D_f(x°). etc. stand for the Dini derivatives
of f at x,: f+(x°). f'(x.) and f'(xO) denote the ordinary
right, left and bilateral derivatives of f at x0, respec—
tively. If both of the sums, D+f(x)+D-f(x) and D+f(x)+D_f(x),
make sense, then it is easy to see that

-:(D+f(x)+D-f(x)) s f‘(x) s fact) s %(D+f(x)+D_f(x)).
Therefore, if both f+(x) and f‘(x) exist, then so does fs(x).
and fs(x)=%(f+(x)+f-(x)) . Further, if f'(x) exists, finite
or infinite, then fs(x)=f'(x). Thus, the symmetric deriv-
ative is a generalization of the ordinary derivative. To
see that it is a strict generalization, consider f(x)=[x]
for which £3(0)=o, but f'(O) does not exist.

f is said to be symmetrically continuous at xo‘if

lim (f(x°+h)-f(x°4h))=0.
h~0
As usual, a function which is symmetrically continuous at
each point of its domain is called symmetrically continuous.
It is clear that if fs(x.) exists and is finite, then f is
symmetrically continuous at x0.

It easily follows from the definitions that if f is
continuous at x,. then f is also symmetrically continuous
at x.. That the converse is not true can be seen from the
function f(x)=cos% which is symmetrically continuous (even

symmetrically differentiable) at x=O. but certainly not

continuous there. Therefore, just as symmetric differen-
tiability is an extension of ordinary differentiability, so
is symmetric continuity an extension of ordinary continuity.
We shall denote, for any function, f.
D(f)=[x: f'(x) exists and is finite]
and
C(f)=[x: f is continuous at x ].
The following proposition will prove useful in later
chapters and will be employed without constant reference to

this section. A proof of it may be found in [12].

Proposition 1.1. Let I be an interval and f a function

defined on I. Then C(f) is a G6 set.

Let A CR and x be a limit point of A. If there exists
at least one sequence from A increasing to x, we define
A-lim sup f(t)=lim sup [f(t): t€(x-6. x)nA]

trx- 6~0
and

A-lim inf f(t)=lim inf [f(t): t€(x-6, x)flAQ,.
tdx- 6-0

If both of the above limits agree. their common value will

be denoted by A-lim f(t). The right-hand limits through

tex-
A are defined analogously. The meanings of A-lim sup f(t).
t-x
A-lim f(t). etc.. are now Obvious. If, in the above def-

trx
initions, A=nl, then it is omitted from the above expressions

to conform to standard notation.
Suppose A CR and x0619. . We denote the reflection of
A through x° by Rx (A). For example, 31([O,4))=(-2,2].
0

Further suppose that I is an open interval and f is a

8
function defined on I such that C(f) is dense on I. Let

x061 and [biz i€Z+] be a sequence of positive numbers dec-
reasing to 0 such that (xo-éi,x°+6i)CI for all iEZ+.
Choose any kEZt Since any 65 set which is dense in an
interval is residual in that interval, we see that C(f) is

residual in both (xo-é . x0) and (x0, x°+6k). It is then

k

clear that R.X (C(f)n(x,-5 . x,)) is residual in (x,. xo+6kl
O

k
and thus
C(f)f]R. (C(f)f](x°-5 , x°))7=’¢.
xo k
Choose xk to be an element of the above intersection and

let ykéflx°(xk). This procedure can be followed for each

kEZ+ to generate two sequences, [xiz i€z+] and [y1 i€Z+],

which satisfy

(1) Rxo(xi)=yi for all i€Z+,
(2) limn‘cxne-limn‘cyn=xo
and

(3) [xi: i€Z+]U[yi: iEZ+}CC(f).

Given a function, f, we say two sequences, [xi: i€Z+]
and [yi: i€Z+}, satisfying (1)-(3) converge C(f)-symmetric-
ally to x,. From the above considerations, the following

proposition is clear.

Proposition 1.2. Let f be a function defined on an open
interval, I, such that C(f) is dense on I. Then each xoél

has a pair of C(f)-symmetric sequences converging to it.

With f and I as in the proposition, we define fsc(x°)

to be

9
f (yn) -f (xn)

n-OO —x
yn n

 

lim

if the limit exists and is the same for all C(f)-symmetric

sequences, [xn: nEZ+} and [ynz n€Z+}, converging to x,.

For example, if f3(x°) exists and C(f) is dense in a

neighborhood of x0. then £s°(x°) exists and equals £3(x°).
The following theorems, which motivated several of

the definitions given above, are fundamental to the results

in Chapters II and III.

Theorem 1.3. (Fried [10]) Suppose the set of points at
which f is symmetrically continuous is residual on an open

interval, I. Then C(f) is also residual on I.

Theorem 1.4. (Preiss [21]) If f is symmetrically contin-

uous on an interval, I, then it is continuous a. e. on I.

Theorem 1.5. (Khintchine [13]) Let f be a measurable
function defined on an open interval, I. Then f has a
finite ordinary derivative at almost all points for which

_f_8 (X) >-°.

Suppose f is a function defined on an open interval,
I: SUCh that fs(x) is finite everywhere on I. Then f is
symmetrically continuous on I and theorems 1.3 and 1.4
both imply that C(f) is dense. Further, if f is measurable
on I and symmetrically differentiable (infinite values
allowed), then by considering f and -£ we see from theorem
1.5 that f has a finite ordinary derivative a. e. and thus

C(f) is again dense. The common thread which seems to bind

10
all three of these theorems is that if f is a reasonably
behaved function which is symmetrically differentiable,
then C(f) is dense. This observation motivates the

following definition.

Definition. Let I be an open interval. Define 0*(I) to
be the class of all functions, f, such that C(f) is dense
on I and fs(x) exists, finite or infinite, everywhere on I.
Define 0(I) to be the class of all functions, f60*(I),

such that fs(x) is finite at each x61.

Analogously, we denote by A*(I) the class of all
functions, f, such that f'(x) exists everywhere on I and by
A(I) the class of all functions, f€A*(I), such that f'(x)

is finite everywhere on I.

Using the above notation, we write

0*s(I)=[fs: f€0*(I)} and A*'(I)=[f‘: f€A*(I)}.
05(1) and A'(I) are defined similarly.

In order to make the notation slightly less cumber-
some, we denote 0*(1R), A*(R), etc., as just 0*, A*, etc..
Most propositions will be stated with this simplification,
it being clear in all such cases that the restriction of
the statement to an arbitrary open interval is valid. As
a further notational convenience, if I and u happen to be
classes of functions, we denote Inn by In.

The following corollaries are easy consequences of

the definitions and the three theorems.

ll
Corollary 1.6. Let f be a measurable function such that
fs(x), finite or infinite, exists everyWhere. Then
(a) [[x: |£s(x)1=e}]=o
(b) f'(x) exists and is finite a. e.

(c) f€0* .

Proof. By applying theorem 1.5 to f and -f, we see that
th: £3(x)=e}]=o and [[x: fs(x)=—~}1=o and (a) follows.
(a) and another application of theorem 1.5 yield (b).
Since f'(x) exists and is finite a. e.. it follows that f

is continuous a. e. so C(f) is dense and f60*.

Corollary 1.7. £60 if and only if fs(x) exists and is

finite everywhere. In this case, f is measurable.

Proof. If £60, then fs(x) exists and is finite everywhere
by the definition of 0. On the other hand, if fs(x) is
finite everywhere, then f is symmetrically continuous and
theorem 1.3 implies that f is continuous a. e.. Thus, C(f)
is dense and £60. We note that any function which is con-

tinuous a.e. is measurable.

Putting these results together, we see that if i is
the class of all measurable symmetrically differentiable
functions, then ACOC£C0*. The question of whether there
are any symmetrically.differentiable functions which are
not measurable is unanswered and will be examined further
in Chapter II. Using a theorem of Zahorski [29] that any
f€A* is discontinuous on at most a countable set, we see

that ACA“ C£C0*.

12

Section 1.2: Some Preliminary Function Theory

Let f be a function defined in an open interval, ICIR,
taking on values in IR*. f is said to be in Baire class
one if there exists a sequence, [fn: n€Z+], of functions
continuous on I such that f(x)=limnd.fn(x) for each x61.
The class of Baire one functions defined on I is denoted
by 251(1) . If I=1R , we write 81(I)=31.
Following is the fundamental theorem characterizing

81. Proofs of it can be found in Goffman [12] or Natensen

[20].

Theorem 1.8. The following statements are equivalent:
(a) £681:

(b) For all sent, the sets [x: f(x)$a] and [x: f(x)2a]
are 35 sets:

(c) For all aent, the sets [x: f(x)<a] and [x:.f(x)>a}
are FG sets:

(d) If P is a perfect subset of hi, then the restriction

of f to P has a point of continuity.

The primary importance of the class of Baire one func-
tions lies in the fact that it contains all ordinary deriv-
atives. If f is a function such that f'(x) exists and is
finite everywhere, then evidently f is continuous, and it
is easy to see that f'éfil. If infinite derivatives are
allowed, the situation is not as clear because f need not
be continuous. For proofs that f'EEl, even in this case,

see Zahorski [29] or corollary 2.6 of this work. In fact,

13

in Chapter II, we will prove that 0*SC31.

A function, f, defined on an interval, I, is said to
have the Darboux (intermediate value) property if whenever
x and y are in I and o is any number between f(x) and f(y),
then there exists a number, 2, between x and y such that
f(z)=o. We shall denote the class of all functions defined
on I which have the Darboux property by 3(1). As usual,
.DUR) is just written as .0.

We shall rarely have use of fi’by itself, but rather,
will make much use of the properties of the class 381.
There are more than a dozen known ways of characterizing

£31. The following theorem contains the ones we will need.

Theorem 1.9. Let féfil. The following are equivalent:

(a) £6831:

(b) For each xénl there is a sequence, [yn: n€Z+],
increasing to x, and a sequence,[zn: n€Z+}, decreasing to
x, such that llmnd°f(yn)=llmnflof(zn)=f(x):

(c) For each x6111 ,

f(x)€[lim inft _f(t), lim supt_x_f(t)] n

"X

n [lim inf x+f(t) , lim suptfl+f(t) ] .

t-o
For a proof of theorem 1.9, as well as many other of
the characterizations of 3%1, see Bruckner [3, p. 9].
It is well-known that the derivative of a continuous
function has the Darboux property. That the same is not
true of the symmetric derivative can be seen from the

function f(x)=lxl, where fs(x)=1 for x>O, fS(O)=O and

14
fs(x)=-1 when x<O. The Darboux property is clearly violated
at x=O. In section 3.2 and in Chapter IV we will explore
some consequences of this lack of the Darboux property for
symmetric derivatives.
Before proceeding much further, the following prop-
osition is prdbably worth noting to avoid some possible

misconceptions.

Proposition 1.10. 81 is closed under addition and multip-
lication by constants. fifil is closed under multiplication

by constants, but not addition.

Proof. The assertion for 81 follows easily from the defin-
ition of $1. That fifil is closed under constant multiplic-

ation can be seen from theorem 1.9(b).

Let

sin %- x¥0
f(x)= 1 x=O

‘2'

and

sinjib x¥0

9(X)= 1
'5' F0.

It is easy to see from theorem 1.8(d) that f and g are in
81. Now, use theorem 1.9(c) to show that f and g are in

3% But,

1 .
o x740
f (X)+g (x)= 1 x30

so f+g££$l.

One of the main uses we will have for £81 is contained

within the following theorem.

15
Theorem 1.11. (Weil [28]) Let rebel such that £s(x)>-

everywhere and §§(x)20 a. e.. Then f is nondecreasing.

Evans [5] extended this theorem from 381 to the
larger class of all measurable functions, f, satisfying
(1) lim inft‘xf(t)$f(x)slim supt‘xf(t)
at each x. (Note the similarity to theorem 1.9(c).) In
fact, he showed that this class is the largest class of
measurable functions for which a statement like theorem
1.11 is true.

For our purposes, we will need a slightly more

general version of theorem 1.11.

Theorem 1.12. Let f be a function satisfying (1) such that
C(f) is dense, §§(x)>-an everywhere and §§(x)20 a. e.. Then

f is nondecreasing.

Proof. We pattern our proof after that of Weil [28].

First, let f§(x)>0 everywhere. Suppose f is not
nondecreasing. Then there exist a0 and b0 with a°<b° such
that f(a°)>f(b°). Choose any o€(f(b°), f(a°)) and define
Ea=[x€[a°, b0]: f(x)sa] and Ea=[x€[a°, b0]: f(x)2a]. Suppose
that neither Ea nor Ea contains an interval. Then, if
x€(a°, b°)nC(f), it is clear that f(x)=o. According to
proposition 1.2, if x°€(a°, b0) we can choose C(f)-symmetric
sequences, [xn: n21] and [yn: n21], converging to x0. Then

_f_s(x°) slim f(xnl’fwn) =lim ‘3‘“ =0

n“ 11-90
x - -
n yn xn yn

 

which contradicts our assumption that §§(x°)>0. Thus,

16
either Ea or Ea contains an interval.

Suppose, for example, that Ed contains an interval.
(If E“ contains an interval, the argument is similar.) We
can then choose an interval (c,d)CEa such that
(2) c= inf [x: (x,d)C:Ea] .
c>a°, for otherwise (2), (1) and a<f(a°) would imply that
§§(ao)=-. Since §?(c)>0, it follows that there is a 6
with O<6<d-c such that f(c-h)<f(c+h)sd whenever O<h<6,
which implies that (c-o,c)CEa. Thus, 1im supt‘cf(t)$o,
so that by (1), f(c)‘a and cEEa. From this, it follows
that (c-6,d)CEG, which contradicts (2). This contradiction
shows that the points a0 and b0 cannot exist and f must
therefore be nondecreasing.

Next, suppose §§(x)20 everywhere. Let c>O and define
f€(x)=f(x)+cx. Then £:(x)2e>0 everywhere and f6 satisfies
(1), so according to the above argument, fe is nondecreasing.
Since fe is nondecreasing for every c>O, we can take the
limit as c~O to see that f is nondecreasing.

Finally, let f be as in the statement of the theorem.
In Zahorski [29], it is shown that for any set, A, such
that [A]=O, there exists a continuous and nondecreasing
function, g, which is differentiable everywhere and for
which g'(x)=~ whenever x€A. Let e>O and A=[x: §§(x)<0].
Since ]A]=O, there is a function, g, as above. Define
fe(x)=f(x)+cg(x). Then, clearly, §:(x)20 everywhere and
fe(X) satisfies (1), so fe is nondecreasing. Again,
letting e~O shows that f is nondecreasing, and the theorem

follows.

17
Following Evans [5], we define the class m-l to
consist of all functions, f, satisfying (1) such that C(f)

is dense. Then from theorem 1.12, the following is clear.

Corollary 1.13. Let f€M_l0* such that fs(x)>- everywhere

and fs(x)20 a. e.. Then f is nondecreasing.

A function, f, defined on an open interval, I, is

upper semicontinuous at xEI if

1km suptdxf(t)sf(x).
f is lower semicontinuous at x if -f is upper semicontinuous
at x. If f is upper (lower) semicontinuous at each point of
its domain, then it is said to be upper (lower) semicontin-
uous.

Note that this definition appears at first glance to
be slightly different than that in some common books because
of the way we defined the upper and lower limits of a
function. The following theorem shows that our definition

is the same.

Theorem 1.14. Let f be a function defined on an open
interval, I. The following statements are equivalent:
(a) f is upper semicontinuous:

(b) For each xEI, f(x)21im supt‘xf(t);

(c) For each sent, [x: f(x)2a} is closed relative to I:

(d) For each a632, [x: f(x)<a] is open.

Proof. (a) is obviously equivalent to (b) and (c) is
Obviously equivalent to (d). Let aénl and A=[x: f(x)2a].

Suppose that (b) is true. xEA iff there is a sequence,

18

[xn: n21]:A, such that llmnfiaxn=x. Then, (b) implies that
f(x)2a so that x€A. Therefore, A is closed and (c) is
true. Suppose (c) is true and f(x)<1im suptfixf(t). Then
there is a sequence, [xn: n21]CI, such that limn‘axn=x and
limnd¢f(xn)= 11m supt_xf(t). Let f(x)<a<11m suptdxf(t)
and A=[x: f(x )Za]. A is closed. There exists an NEZ+
such that for all nZN, xnéA. Since A is closed, this

clearly implies that x6A. This is impossible, and the con-

tradiction shows that (b) must be true.

Note that theorem 1.14(c) and theorem 1.8(b) imply
that if f is upper semicontinuous, then £681. Analagous
results can clearly be established for a lower semicontin-

uous function, f.

Section 1.3: A Covering Theorem

Suppose that (a, b) is an open interval such that
there is a 5(x)>O associated with each xE(a, b). For each
x, we define

Sx=[[x-h, x+h] : O<h<6(x)]
and

c=Ux€(a, b)sx’
Gris called a full symmetric cover for (a, b).

The above definition is due to Thomson [26], who
used it to prove a slightly weaker version of the following

theorem.

19
Theorem 1.15. Let (a, b) be an interval and c=(a+b)/2.
Suppose G»is a full symmetric cover for (a, b). Then
there is a set, DC(O, (b-a)/2) such that D has countable
closure and G-contains a partition of [c-x, c+x] for
every xéDc. Further, each of these partitions can be
chosen to contain an element of Sc. (Where Sc is as in

the above definition.)

Proof. To simplify notation, we assume that c=O and that
6.13 a full symmetric cover of the interval (-b, b). With
the assumptions that 6(x)=6(-x) and 6(lx])<lxl for x¥O, we
lose no generality because cvis at worst made smaller.
Define .

D=[x€(O, b): cvcontains no partition of [-x, x]]
and let

a: sup [x€(O, b): Dh(O, x) is countable].

We must show that a=b. Suppose this is false; i. e.,
suppose o<b. First, note that a>6(O). By the definition
of a, for every c>O, Dh(a-c, a] is countable and 5h(c,a+e)
is uncountable. But, if xEDcn(a-6(o), a), then C-contains
a partition of [80(Rd(x)), Rd(x)], from which it follows
that ENG, d+6(o))CRa(Dn(o-o(o) , 01)) , which is countable.
This contradiction shows that o=b.

Now, let x°€(0, b) and let

-x°=ao<a1< ...<an=x°

be a partition of [-x°, x0] from an There is a k€[O,1,...,n]

such that akso and ak+120; i. e., O€[ak, ak+1163. Because

20
6(x)<'lx[ when X510, it is clear that whenever xaKO and IESX.

then O£I. This implies that [ak. ak+1]€SO'

CHAPTER II

THE CLASS OF.ARBITRARY SYMMETRIC DERIVATIVES

Section 1.1: Comparison with Baire Class One

It is an easy matter to prove that a finite-valued
ordinary derivative is a member of 81. The situation would
appear to be more complex in the case of the symmetric
derivative because the primitive function may have many
discontinuities. Nevertheless, we have the following
theorem, which will prove very useful in the succeeding

chapters.

Theorem 1 .1 . 0*3c 81 .

Proof. According to theorem 1.8(b) it suffices to show
that for any aenl, [x: f3(x)2a] and [x: fs(x)5a] are both
G6 sets. To do this, define g(x)=f(x)-ax so that C(f)=C(g),
g is symmetrically differentiable everywhere with
93(x)=rs<x>-a

and
<1) (x: fs(x)=a]=(x: 93(x)=0}.

Choose a 6>O. Using proposition 1.2, we see that

for each xéni, there is an h€(O, 6) such that x+h and x-h

are both elements of C(g). From this observation, it

21

22
makes sense to define
A=[x: sup0<h<bg(x+h)-g(x-h)>0, where x+hEC(g) and x-h€C(g)].
If x€A, then there are h, o and 6, each positive, satise

fying the following inequalities:

(2) x-h, x+h€C(g)n(x-o, x+5);

(3) g(x+h)-9(x4h)>2a:

'(4) lX+h-yl<e implies 19(X+h)-9(y)l<a:
(5) lxth-yl<e implies l9(x4h)-g(y)l<a:
(6) e+h<6.

Choose any x06}! such that O<Ix-x°]<e and choose B>O with
[x-x,]+B<c. Let

(7) Rl=(x°4h-8, x°4h+B) and R2=(x°+h-B, x°+h+8).

Since Ix-x,[+B<e, it is clear that

(8) R1C(x4h-c, x-h+e) and R2C(x+h-e, x+h+e).

Since C(g) is a dense Gbset, it is residual. From
this, it follows that Rx,(c(g)an) and C(g)nR2 are both
residual in R2 and cannot be disjoint. So, there is an
h'>O such that x,-‘h'€C(g)flRl and x°+h'€C(g)nR2. From (7)
and (8) it follows that h'<h+B<h+e<6. From (4) and (5) it
follows that

]g(x+h)-g(x°+h)]<o and |g(x-h)-g(x,-h')l<d.
Combining these two inequalities with (3), it is seen that

'g<x.+h') -g<x.-h'>>o
so that xoeA. Since the only requirement on xo was that
[x-x°]<c, it follows that (x-c, x+c):A. Thus, for each
xEA, there is an e>O such that (x—e, x+e)CA. Therefore,

A is open.

23

Similarly, if we define, for each n€Z+, the set An

to be
[x: sup 19(x+h).g(x.h)>—%?-where x+h€C(g) and x-hEC(g)].
0<h<"
n

then An is alsO'open;
It is clear that
s O
(x: 9 (x)20}=fln___,1 An'

so the set in (l) is a G6 set.

The set [x: fs(x)3a] can also be shown to be a G6 set

by considering -f instead of f.

Using theorem 1.3 with corollary 1.6 and theorem 2.1,

we arrive at the following corollaries.

Corollary 2.2. Let f be a symmetrically differentiable
function such that fs(x) is finite on a residual set. Then
3
f 631 0
Proof. This follows easily from theorem 1.3 and the obser—
. . s
vatlon that if f (x) is finite, then f is symmetrically

continuous at x.

Corollaryp2.3. If f is measurable and symmetrically differ-

entiable, then fsehl.
Proof. By corollary 1.6, f€0*.

The above statements represent improvements over the
strongest previously known similar result, due to Filipczak
[7], who proved that if f is approximately continuous and
symmetrically differentiable, then £8681.
It is not known whether every symmetric derivative

24
is in Baire class one. Consideration of corollaries 2.2
and 2.3 shows that such a function would have to be very
badly behaved: at the very least, it would have to be
nonmeasurable and have an infinite symmetric derivative on
a second category set. The question of the existence of a
nonmeasurable symmetric derivative was posed as long ago
as 1928 by Sierpinski [24] and still remains open. We

explore these questions further in the next section.

Section 2.2: Arbitrary Symmetric Derivatives

As noted above, it is perhaps possible that an arbit-
rary symmetric derivative could be a very badly behaved
function. However, a few statements concerning the behavior

of such a function can be established.

Theorem 2.4. Let f be a symmetrically differentiable
function and -¢Sa<Bs~. Suppose A=[x: fs(x)$a], B=[x: fs(x)2B]
and I is an interval such that ICAUB. Then both A and B

cannot be dense in I.

Proof. It may be assumed without loss of generality that
o<O<B, for otherwise, we just consider g(x)=f(x)-ax, where
d<a<B, as in the proof of theorem 2.1.

Suppose both A and B are dense in I. Since ICAUB,
at least one of the sets, A or B, must be of the second
category in I: suppose B is of the second category. Define
for n62+

Bn=[x€I: f(x+h)-f(x-h)>0 for O<h<%-].

25
Since B<:LE=1 Bn and B is of the second category, there is
an n°€Z+ and an open interval J<:I such that Bno is dense
in'J.
Because A is dense in J also, we may choose an a 6A OJ

and two sequences, [xn] and [yn], from Bn With [xn]

increasing to a and [yn} decreasing to a such that yn—xn<%
O
for each n€Z+. Then
f(a+(yn-xn))-f(a-(yn-Xn))=f(yh+(a-xn))-f(xn-(yn-a))
=f(ynf(a-xn))-f(yn-(a-xn))+f(yfir(a-xn))-f(xn-(yn-a))
=(f(yn+(a-xn))-f(yn-(a-xn))+(f(xn+(yn-a))-f:xn-(yn-a)))>O
1 .
because yn xn<no and xn<a<yn imply that yn a<no and
a-x-(l- With both x and y elements of B .
n 110 n n no
Thus, if h =y -x , we see that h decreases to O
n n n n
and f(a+hn)-f(a-hn)>0 for each n. This implies that
fs(a)20, because f is symmetrically differentiable at a.
But, fs(a)sa<0 because aEA. This contradiction shows the

supposition to be false, so both A and B cannot be dense

in I.

Theorem 2.4 shows that the associated sets for an
arbitrary symmetric derivative behave much like the assoc-
iated sets for a function in Baire class one. In view of
theorem 2.1, this is hardly surprizing.

Theorem 2.5 rules out another form of pathological

behavior for symmetric derivatives.

Theorem 2.5. Let f be any function. Then [x: lfs(x)]=°]

contains no interval.

 

 

26

Proof. Suppose, to the contrary, that there is an interval
Ic[x: lf3(x)]=°]. According to theorem 2.4, both of the
sets, A=[x: f8(x)=-¢] and B=[x: fs(x)=¢], cannot be dense
in I. So assume there are a,B€I such that a<B and (a, B)CB.

For each xE(o, B) and each p>O, there is a 6(x,p)>O
such that if O<h<6(x,p), we have [x-h, x+h]C(o, B) and
(9) f(x+h)-f(x-h)>2hp.

For each n€Z+, define
(10) Jh=[[x-h, x+h]: x€(a, B) and O<h<6(x,n)}.
Each Jh is a full symmetric cover of (a, B), so by theorem
1.15, there is a set Dnc((a+B)/2, B) with [Dn[=O such that
J; contains a partition of [(c+B)/2-x, (d+B)/2-tx] for

every x+(d+s)/2e((o+s)/2, B)-Dn. Let

(11) D== Un=1Dn
and
(12) E=((a+B)/2.B)-D.

Since [Dl=0, E#¢, so we may choose an x+(d+B)/2€E.

Let

2:2.
2

and choose any n€Z+. By (11) and (12), there is a partition

(13) d==f( +x) -f(3§£-x)
of [(o+B)/2-x, (o+B)/2+ x] in J5. Denote the intervals

in the partition by [oi. Bi]. i=1.....m. Then, using (13),
(10) and (9), we see

(14) d-Ei=lf(Bi) f(ai) >ZIE=1(Bi-oi)n 2xn.

Since n is arbitrary and x>O, (14) clearly leads to a
contradiction of (13). Thus, B can contain no interval.

Similarly, A can contain no interval.

27

I

*
Corollary 2.6. A :81.

Proof, According to theorem 2.5, the set on which f' is
finite is dense. Whenever f'(x) is finite, f is continuous

at x, so C(f) is dense. Now apply theorem 2.1.

Corollary 2.6 was apparently first proved by Zahorski

[29, p. 14].

CHAPTER III

THE STRUCTURE OF FUNCTIONS IN 0*

Section 3.1: Nice Copies of Functions in 0*

The primary goal of this section is to prove the

following theorem.

Theorem 3.1. Let f60*. Then there are two sets, A1 and

 

A2. each with countable closure, and two functions, 91 and
g2, each in Baire class one satisfying:

(a) g:c(x)=fs(x) everywhere, i=l,2;

A“

(b) g:(x)=fs(x) everywhere on i' i=1,2;

(c) 91 (92) is upper (lower) semicontinuous on A: (A3);
(d) C(f)CC(gi) and f(x)=gi(x) for each xEC(f), i=1,2;
(e) D(f)CD(gi) and f'(x)=gi(x) for each x€D(f), i=l,2:

(f) If I is a component of Ag, then gi€m_1(I), i=l,2.

The proof is accomplished with the aid of the
following series of lemmas, some of which are interesting

in their own right.

Lemma 3.2. Let I be an open interval, C a dense subset of

 

I and f any function defined on I. Define

28

29

u(x)=C-lim suptdxf(t) and.£(x)=C-lim.inft‘xf(t).
Then u is upper semicontinuous and t is lower semicontinuous.
Proof. Let den: and A=[x€I: u(x)2a]. If A=¢, then A is
closed. Otherwise, we may choose a sequence, [xnz n€Z+]CA
such that 1imnflaxnex. From the definition of u, for each

+ I C I 1 1
n62 , there is a tnGC satisfying [tn-xn]<n and f(tn)+E>a.
Then, clearly 1imhfiatn=x and
. . . 1 _

C-lim.suptdxf(t)zlim.supn~°f(tn)zlim|supnqa(a—E)-a
so u(x)2a and xeA. Thus A is closed and it follows that
u is upper semicontinuous.

The proof that z is lower semicontinuous follows by

noting that -L is u for -f.

Qemma 3.3. Let f, C, u and 1 be as in lemma 3.2. Then
(a) C(f)CC(H) (C(f)CC(£)) and f(X)=H(x) (f(X)=¢(X)) for
each x€C(f).

(b) D(f)CD(H) (D(f)<'-'-'D(£)) and f'(x)=u'(X) (f'(X)=£'(X))
for each x€D(f).

Proof. We may suppose without loss of generality that
OEC(f) and f(O)=O. Then, given an e>O, there is a 5>0
satisfying [f(h)]<c whenever [h]<6. Fix an h such that
[h]<6 and choose a sequence, [xn: n€Z+]CC, such that [xn[<6
for each n and limnq°f(xn)=u(h). It is clear that u(O)=O,
so that

[u (h) -u (0) .l=limn_,,lf (xn) 1‘s
and it follows that OEC(u) and u(O)=O. Therefore (a)

follows.

30
Now, suppose O€D(f). We may assume without loss of
generality that f(O)=O=f'(O). (Otherwise we just add an
appropriate linear function to f.) Then, given an e>O,
there is a 6>O such that when ]h[<5, [f(h)]<e]h]. Fix an
h with O<]h]<6 and a sequence, [xn: n€Z+]:C, as above.
Then
lu(h) [=1i.mn

ralf<xn)(slimn.,lxn!e=1h(c.

Because c may be chosen arbitrarily small, we see that
O€D(u) and u'(O)=O=f'(O). Therefore, D(f)CD(u) and
f'(x)=u'(x) for each x€D(f).

The assertions for 1 follow by noting that -L is u

for -f.

Lemma 3.4. Let f€0* and define u and L as above with
C=C(f). If u (1) is finite in a neighborhood of x0. then

us(x°) (13(x°)) exists and equals fs(x°).

Proof. By translating f, we may assume without loss of
generality that x°=O and there is an K>O such that (“(x)]<¢
whenever [x]<n.

First, suppose fS(O)=O. Then, given c>O, there is a
66(O,n) such that when O<t<6,
(1) lf(t>-f(-t)l<2te.
Fix a t6(0,6) and choose a sequence, (Sn: n€Z+]CC(f), such

‘0

that limn sn=t, O<sn<6 and

(2) llmnflaf(8n)=C(f)-llm sudetf(x)=u(t).
Since (an: n€Z+]CC(f), for each nEZ+ there is a pn>0 such

1
that when [sn-x]<pn, then [f(sn)-f(x)]<n. pn may also be

31

chosen small enough so that snfpn<6 and limhqapn=0.

For each n€Z+, let

Gn=RO(C(f)n(sn,sfi+pn)).

C(f) being residual in n! clearly implies that Gn is
residual in (-sn-pn,-sn), so C(f)nGn#¢ for each n€Z+. Form
a sequence, (tn: n€Z+], by choosing tn€(sn,sn+pn) such that
Ro(tn)€C(f)nGn. It follows then that limn”.tnet, O<tn<5
for each n and [f(sn)-f(tn)]<%u Using (2), we see that
(3) limnflf(tn)=u(t) .
From the definition of u, (3) and then (1), we see that
(4) u(-t)21im supndaf(-tn)zlim supnqo(f(tn)-2tnc)=u(t)-2te.
Similarly, it follows that
(5) I-l(t)2u(-t)-2te.
(4) and (5) imply
(6) ]u(t)-u(-t)]szte.
Since 5 may be chosen arbitrarily small, we conclude from
(6) that us(O)=O=fS(O).

If fs(O)=oEI{, we consider the above argument applied
to g(x)=f(x)-ax to see that u3(0)=c.

Now suppose fs(O)=¢. Then, given dent, we can choose
a 66(O,x) such that whenever O<t<6,
(7) f(t)-f(-t)>2tc.
Fix a t€(O,5) and choose a sequence, (Sn: n€Z+]CC(f)n(-6,0)
such that limnfiasn=-t and

1imnn°f(sn)=C(f)-lim supx‘tf(-x)=u(-t).

In the same manner as above, we may choose a new sequence,

(tn: n€Z+]CC(f)n(-6,0) such that 30([tn: n62+])CC(f),

32

1imnd°tn=-t and

(8) 1imn”.f(tn)=u(-t).

Then, using (8) and (7),

u(t)-u(-t)=C(f)-lim.supxatf(x)-C(f)-lim.sudetf(-x)
=C(f)-1im supx‘tf(x)-limnqaf(tn)

(9) 21in supn‘;f(-tn)-limn‘°f(tn)

21in supnd¢(f(-tn)-f(tn))
>lim supnd¢2]tn]o=2to.

By choosing a arbitrarily large in (9), we see that uS(O)30.
The case when fs(O)=-° succumbs to a similar argument.
Therefore, us(O)=fs(O).

The assertion that 15(x°)=fs(x°) follows by noting

that -L is u for -f.

It should perhaps be noted that some condition such

as requiring u and J to be finite in a neighborhood of x0

is necessary. To see this, for n€Z+, let

 

. -n -n -1
El? 4 ”(€32 ) x6(2-n_4-n'2-n+4-n)_[z-n}
0 otherwise
and

fn(x)= Zn=lfn(x)+Z:=lfn(-x) .

Then £60. but u(x)=°=-£(x) whenever x€[t2-n: nEZ+], so the
difference quotients

u(h)-u(-h) £(hl:£(-h)
2h and 2h

are undefined whenever h=2-n for some n€Z+. Thus, us(o)

and 13(0) are also undefined.

33
Lemma 3.5. Let f€0* with
A=[x: lim.suptdxf(t)=¢] and B=[x: lim.inft‘xf(t)=-¢}.

Then both A and B are countable closed sets.

Proof. we prove the lemma in the case of A. The assertion
for B then follows by considering -f.

Using c=rz in lemma 3.2, we see that A is closed and
because C(f) is dense, A.must be nowhere dense. Being
closed, A may be written as A=PUN, where P is perfect and
N is countable. Suppose P¥¢ and let (d,B) be a component
of PC. (a or B could be infinite.) Since P#¢, a or B must
be finite. Suppose B is finite. Then, since P is closed,
BéP. P being perfect and (d,B)CPc, we see that for each
O>Oo [5.3+6)OP is uncountable. Since An(o,B)CN is countable,
we may choose a sequence, (Bu: n€Z+]CP, such that Bn
decreases to B and RB({Bn: n€Z+])nA=¢. Using the facts
that BnGA for each n and

lim supt#RB(Bn)f(t)€[-°o“)
we may choose a tn>B for each n62+ such that [tn-Bnl<%

and f(tn)-f(Ra(tn))>n. Clearly, limnflatn=B and

f(tn)mf(fic1(tn)) lim inf n

1im inf _____=.
n”‘ 2(tn-B) 2 n‘°2(tn-B)

 

so fs(B)=“. Similarly, if a>-, then fs(o)=-.

Now, we note that
c o
P - Un=1(an.Bn)

where (on,Bn) is a component of Pc for each n. Clearly,

then since P is a nowhere dense set in hi, the sets

34
(on: n€Z+] and (Bu: n€Z+] are both dense in P. This implies,
from the above, that the sets
I+=[x: fs(x)=°] and I-={x: fs(x)=-°]
are disjoint dense subsets of P.

P, being closed, is a G set and according to theorem

6

2.1, I+and I- are 66 sets, so POI+ and POI- are dense Gd

subsets of P. As such, both PnI+ and PnI' are residual in
, + _

P. Since P is a Baire space, POI OI ¥¢, which contradicts

the fact that I+nl'=¢.

Therefore, we conclude that P=¢ and A=N, a countable

set.

Enough machinery has now been developéd‘to accomplish

our primary goal.

Proof. (Theorem 3.1) Let f€0*, u and 1 be as in lemma 3.4
and A and B be as in lemma 3.5. Define

Al={x: lu(x)]=°l and A2=ix= l£(X>l=°}

and let
u(X) xEAi £(x) xeag
91(X)= and 92(x)=
f (x) x€A1 f(x) xéA.2

Since A1UA2CAUB, the countable closure of both A1 and
A2 follows from lemma 3.5. (b) follows from lemma 3.4. (c)
follows from lemma 3.2. (d) and (e) follow from lemma 3.3.
(a) follows from the definitions of u and L and (d).

The rest of the theorem will be proved in the case

i=1, the proof in the case i=2 being similar.

Choose an aénl. The upper semicontinuity of 91 on

35

A: and theorem 1.14(c) imply that for each n€Z+,
- f 0 l
En-[xeAl . gl(x)za+I-1-]

-c

is closed relative to A1 and so is an F0 set relative to

IR . Since

—c - O
F1=[x€A1 . g1 (x)>a]— Un=lEn'
we see that F1 is also an F0 set relative to It. It is
clear that
(10) F2=[x€A1: gl(x)>a]

is an F0 set, because A1

[x6112 : 91 (x) >a]=F1UF2

is countable. Thus,

is also an F0 set.
Similarly, the upper semicontinuity of g1 implies
that
- "C,

F3-[x6A1. 91(X)<a}
is open in A: and so is open in El. Thus, F3 is an F0 set
relative to JR . As above,

F4=[XEA13 91(X)<a}
is also an F0 set relative to it. Therefore,
(11) {xemz gl(x)<a]=F3UF4
is also an F0 set relative to fit. Since a was chosen
arbitrarily, (10),(ll) and theorem 1.8(c) imply that 91631.
Using the definition of g1 and (d), we see that if

x655, then
lim 1nft-xgl(t)‘ C(f)-1im lnfthgl(t)=
(12) =C(f)-lim inft”xf(t)sC(f)-lim eupt‘xf(t)=

=91(X)=C(f)-11m suPtdxgl(t)‘llm supt~x91(t)'

Comparing the above relations, we obtain (f).

36
An examination of the statement of theorem 3.1 shows

that we can slightly improve on the semicontinuity relations

 

in (c) and (f). In particular, (c) implies
(l3) gl(x)21im suptdxgl(t)
whenever xeAi, and

(14) 92(x)51im inftdxg2(t)
whenever XGAS. (f) implies that

(15) 91(X)‘llm supt‘xgl(t)
whenever xEA: and

(16) 92(x)2lim lnftuxgz(t)

for all xeAg. Combining (13)-(16), we see
Corollaryg3.6. Let f, A1, A2, g1 and 92 be as in theorem
3.1. Then

(17) gl(x)=lim supt‘xgl(t)

for all x65? and

(18) 92(x)=lim inftdxgz(t)

for all x653.

Definition. Let f, A1, A , g1 and 92 be as in theorem 3.1.

2
We call 91 (92) the upper (lower) semicontinuous nice copy
of f. A1 (A2) will be called the upper (lower) essential

set for f.

In the following, we shall adopt the convention that
if g is said to be the nice copy of f, it will be the upper
semicontinuous nice copy unless it is specifically noted
otherwise. Similarly, an essential set will be an upper

essential set unless contrary mention is made. These

37
conventions will cause no problems because of the similarities
in the behavior of the upper and lower semicontinuous nice

copies of a function.
Corollary 3.7. If f€0*, then ][x: [fs(xX=¢]]=O.

Proof. Let g be the nice copy of f and A the essential
set for f. Then, from theorem 3.1. 9681 and is thus

measurable. Now apply corollary 1.6(a).

The following theorem answers some natural questions

concerning the uniqueness of the nice copy of a function.

Theorem 3.8. Let f and g be functions in 0* and suppose
that Dcfil is any dense set. If f(x)=g(x) for every x6D,
then the essential sets for f and g are equal and the nice
copies of f and g are equal up to their values on the

essential set.

Proof. Suppose xOEIR and
(21) C(f)-1im supt f(t)=o
4x0
where a may be infinite. Then, there is a sequence,
(tn: n€Z+]CC(f)
such that limnfia 'tn =x° and

(22) lmn«f(tn) =01.

Since (tn: n€Z+]CC(f), for each n€Z+, there is a 6nE(O,%)

such that when [tn-y]<6n.

1
(23) (f(tn) -f(y) (<3.
Because 960*, C(g) is dense, so for each nEZ+, there is a
6
u €C(g)n(t -—Eut ). Then, for each n€Z+, there is an
n 6 r1 2 n

n
nn€(0,1f) such that when ly-un]<nn.

38
(24) 1 () ( >l<i
9 Y '9 un n'
Finally, D being dense, we may choose, for each n€Z+, a
vn€(un-nn,un)nD. Thus, clearly
limndcun=limhdavn=xo,
and using (24), the fact that f(vn)=g(vn) for each n€Z+,
(22) and (21) we see
- ° 2 ' 2
C(g) lim supt‘x°g(t) lim.supnqog(un)
. l .
211m.supnq¢(g(vn)-E)zlim supnd°f(vn)2
. l__ .
211m supn~¢f(tn)-a-C(f)-lim.suptdx°f(t).
The reverse inequality can be shown by interchanging f and
g in the above argument. Therefore, at each XOEEI,
C(f)-1im suptdx°f(t)=c(g)-lim supt*x°g(t).
From this. the theorem easily follows using the definition

of the essential set and the nice copy.

As a consequence of this theorem, we see that the
nice copy of f is in some definite ways the “best" repres-
entation of f. For instance, suppose that in a nontechnical
manner, we consider a "copy" of f to be any function which
has the same symmetric differentiation properties as f and
agrees with f on a "large" set. Then.g of theorem 3.8
is a copy of f. The theorem says that both f and 9 have
the same nice copies. Moreover, from (d) and (e) of
theorem 3.1, we see that if h is a nice copy of f and 9,
then C(g)CC(h), C(f)CC(h), D(g)CD(h) and D(f)CD(h). This
can be interpreted to mean that of all the copies of f, its
nice copy is the "most continuous" and the "most different-

iable." Using similar arguments, the nice copy can be shown

39

to "maximize" many other such properties such as upper
semicontinuity, monotonicity and higher order different-
iability. Thus, it does seem appropriate to call it the

"nice copy" of f.

Section 3.2: Nice Copies of Functions in 0

If we restrict our considerations in theorem 3.1 to
functions in 0, some of the properties of the nice copy
can be considerably sharpened. Before we state and prove
this new version of theorem 3.1, we first establish the
following interesting lemma, which will be a useful tool

in several of the following proofs.

Lemma 3.9. Let f and g be elements of 0 such that
fs(x)=gs(x) everywhere. Then there exists a constant, cent,
such that the set

M=tx: f(x)#g(x)+c}
is countable and has no nonempty subset which is dense in

itself.

Proof. Let r>O. It suffices to show that the set Mn(-r,r)
satisfies the lemma.

To do this, we define Y(x)=f(x)-g(x). Since fs and g5
are finite everywhere, Ys(x)=0 everywhere, so by corollary
1.7, Y60 and C(Y) is dense. We may suppose without loss of
generality that O€C(Y) and Y(O)=O. We will show that c=O
satisfies the lemma.

Let e>O. Because OEC(Y), there is a 61>O such that

When O$]h]<61. [Y(h)]<§u Since Ys(x)=0 everywhere, for

40
each xEIl, there is a 6(x)6(0,61) such that when O<h<6(x),
(25) [‘1’(x+h) -Y(x-h) I <1?-
Using this 6(x), we form a full symmetric cover for fit by
defining
Jx=[[x-h,x+h]: O<h<6(x)] and .(hU xGJR Jx'
Then, by theorem 1.15, there is a set, DC(O,r), with
countable closure, such that J contains a partition of
[-x,x] for every xEDcn(-r,r), and further, each such partit-
ion contains an element of Jb. Choose any x°€(0,r)nDc and
let
(26) -x°=o1<d2< ...<ak_1<0<ok< ...<an=xo
be a partition of [-x.,x°] from J. ‘Using (25) and the facts
that ok<61 and Y(O)=O, we see that
tux.)(=1r(x.)-~r(0)1sz’.‘;11r(d)r<dl+l)l+1r(ok)-r(on<
fizi=kmi+lfli )+€<—n+ E:Zxr 2< e.
A similar argument shows that ]Y(-x°)]<c.

Therefore, for each x6(-r,r)-(DURO(D)), we see that
]Y(x)]<c. Since D has countable closure, this implies that
[x€(-r,r): ]Y(x)]zc]
also has countable closure. If for each n€Z+, we define
En=[x: ]Y(x)]2%], then, for each n€Z+, En is countable.

Since

Mn(-r,r)=LJ;=lEn,

it follows that Mfl(-r,r) is countable.

Define S=[x€fil: MflRx(M)#¢}. Thus, if x65, then there
are u and v in M such that x=nga Since M is countable,
there are at most a countable number of such pairs, so 5

41
must be countable. Let S=[sn: nGZI].

Now, suppose M.has a subset, N#¢, which is dense in
itself. We inductively choose two sequences, [xnz n€Z+]
and (en: n€Z+]c(O.1), as follows: First, choose any xléN
, choose 6 such that

1 l
O<cl<lx1-sl] and el<]Y(xl)]. Suppose x2. ...,xn_1 and

such that x1#sl. Then, using this x

cl. '°"€n-l have been chosen. Select anN such that xn#sn,
xnalxi for i<n and [xn-xn_l]<cn_1. (This selection is
possible because N is dense in itself.) cn6(0,%) is chosen

such that [xn-en.xn+en]C[xn_1-cn_l,xn_1+en_l], en<lxn-sn]
and en<lY(xn)]-

From this selection procedure, it is clear that there
is an xEEl such that lim. x =x. For each n€Z+,

ndo n
lxn-Xl<en<lxn-snl.

so xaisn for any n. Therefore, xfls, and by the definition
of S, it must be true that Y(Rx(xn))=0 for each n. Since

Ys(x)=0, we see .
]Y(xn)-Y(Rx(xn))l ]Y(xn)] en

. _ . . _ 1
0:1an ZW-xn] _ 11mn--«=I'2 Ix-xn[ > llmn-«IDZer1 - 2

 

This contradiction shows that M has no nonempty dense in

itself subset.

A function, f, is said to be symmetric if for every
x6}! there exists a 6(x)>O such that f(x+h)=f(x-h) whenever
O<h<6(x). A set, A, is a symmetric set, if its character-
istic function is symmetric.

It is clear from the definition that if f is symmetric,

then fs(x)=O everywhere. We may apply the above lemma to

42
simplify the proofs of theorems due to Ruzsa [22] and

M. Foran [9].

Corollary 3.10. If f is a symmetric function, then there
is a cénzsuch that the set, M=[x: f(x)#c} has countable

closure.

Proof. From lemma 3.9, it follows that there exists a

cent such that the set, M=[x: f(x)#c], contains no nonempty
dense in itself subset. It is easily seen from this that
M.must be nowhere dense. So, without loss of generality,

we may assume there exists a 61>0 such that f(x)=c when
O<x<61. Since f is symmetric, for each xéni, there is a
6(x)€(0,61) such that when OSh<6(x), then f(x+h)=f(x4h).
Using this 6(x), we form a full symmetric cover, J, for fit

as in the proof of lemma 3.9. The rest of the proof proceeds
as in the first half of the proof of the lemma, where it

can be shown that MCDURO(D), with D the set of theorem 1.15.

Corollary 3.11. If A is a symmetric set, then either A or
Ac has countable closure.
Proof. This is immediate from the definition of a symmetric

set and corollary 3.10.

We are now ready to state the restricted version of

theorem 3.1.

Theorem 3.12. Let £60 with Al (A2) the upper (lower)
essential set for f and g1 (92) the upper (lower) semicon-
tinuous nice copy of f. Then Al and A2 are symmetric sets

and gland 92 satisfy:

43
(a) g:(x)=fs(x) everywhere, i=l,2:
(b) g1(gz) is upper (lower) semicontinuous on A: (A3);
(c) C(f)CC(gi) and f(x)=gi(x) for each x€C(f), i=l,2;
(d) D(f)CD(gi) and f'(x)=gi(x) for each x€D(f), i=l,2;
(e) If I is a component of Ag, then giéfifll(1), i=l,2.

Further, gi is determined up to an additive constant and

its values on Ai by (a), (b) and (e), i=l,2.

Proof. (b), (c) and (d) follow from theorem 3.1. The rest
of the theorem will be proved in the case i=1. The case
when i=2 then follows by considering -f.

To prove that A1 is symmetric, let xoéfil. Since
fs(x°) is finite, f is symmetrically continuous at x0. so
there is a 6>O such that if O<h<6. then
(27) (f(x,+h)-f(x,eh)]<l.

Suppose yoeAln(x°-6,x°+6) and that C(f)-11m suptqyof(t)=°.
Then we may choose a sequence, [yn]C(x°-6,x°+6)flC(f) such
that limnqayn=y° and limnd¢f(yn)=o. For each n€Z+, there
exists a 5n€(0,%) such that (yn-6n, yn+5n)C(x°-6,x°+5) and
[y-yn]<6n implies that
(28) lf<y>-f(yn>l<1.
Using the fact that C(f) is residual, we may then choose,
for each n, a new

zn68x0((yn-6n,yn+6n)nC(f))flC(f).
Then, using (27) and (28)

C(f)-11m supt“R (y°)f(t)zlim supnfiaf(zn)2
x0

211m supnq¢(f(flxo(zn))-l)21im supnda(f(yn)-2)=o

44
so that Rx°(y°)€Al. A similar argument shows that if
C(f)-1im supt‘y°f(t)=-".
then Rx°(y°)EA1 once again. Therefore,
axe(Aln(x°-6,xo+6))=Aln(x°-6,x°+6).
Since x. was chosen arbitrarily, it follows that A1 is a
symmetric set.

Again, let xoenl. ‘We may, without losing generality,
assume fs(x°)=0. Then, for each e>O there is a 6>0 such
that whenever O<h<6.

(29) [f(x,+h)-f(x°rh)l<2he

and because A1 is symmetric, we may choose 6 small enough
so that

(30) Rx°(Aln(x°-6,x°+6))=(x,-6,x,+6)nA1.

Fix an hE(O.6). If x°+h€A then by (30), the definition of

1.
91 and (29),

(31) [91(x°+h)-gl(x°-h)l=lf(x°+h)—f(x°4h)[<2he.

If x°+h¢A1, we can use the same argument as in the proof of
lemma 3.4 to show that

(32) [gl(x°+h)-gl(x,-h)l<2he.

If we let a go to O, we see from (31) and (32) that gs(x°)=0
and (a) follows.

Let I be a component of A: and xoel. Since 9160, 91
is symmetrically continuous at x0. According to corollary
3.6.

gl(xo)=lim suptdxogl(t).
Combining these two facts, we see that

(33) 91(x,)=1im Supt”x +91(t)
O

45

and
(34) gl(x°)=lim supth°_gl(t)
so that
(35) gl(x°)€[lhm inftdx°_g1(t),lim supth°_gl(t)]n

n[lim inftdx°+gl(t) .lim suptdxfigflt) l -
According to theorem 3.1, 91681. Now, we apply theorem
1.9(c) with (35) to see that gléfifil(l).

NoW, let h be a function satisfying (a), (b) and (e)

l
and let I and x0 be as above. Since hl€£$1(I), it follows

from theorem.l.9(b) that

(36) h1(x°)slim.suptdx°+hl(t)
and
(37) h1(x°)slim.supt‘x04hl(t).

Using the upper semicontinuity and the symmetric continuity

of h at x.. we see that

1
(38) h1(x°)21im suptdx04hl(t)
and
(39) h1(x°)21im supt~x°+h1(t).
Combining (36)-(39) yields
(40) hl(x°)=lim supt‘x°+h1(t)
and .
(41) hl(x°)=lim supth°_hl(t) .

Define ?(x)=g1(x)-h1(x). Through the addition of an
appropriate constant (the constant of the theorem), we may
assume without loss of generality that Y(x°)=0.

According to (40) and (33), we may choose sequences,

46

[xn: n€Z+] and [ynz n€Z+], increasing to x0 and satisfying

(42) limndohl (xn)=h1(x°)
and
(43) lirttn_,.,,gl (yn)=gl (x0) .

Then, from (42) and (34),
. = - - - s

1im inftexo-Y(t) lim infth°_(gl(t) h1(t))

(44)Slim infn‘°(gl(xn)4h1(xn))=lim infndagl(xn)-h1(x°)$
Sgl(x.)-h1(x,)=Y(x°).
Then, from (41) and (43),
. = - - 2

1im.suptdx°_Y(t) lim.supth°_(gl(t) h1(t))

(45) 21.1m 8npn_..(91(yn)-h1(yn))‘=91(x,)-lim inf1,1_,,,,,h]_(yn))2
2gl(x°)4hl(x°)=Y(x°).
Combining (44) and (45) yields
. . s .
(46) lim inftfix°_Y(t)$Y(x°) lim.suptdx°_Y(t).
In a similar fashion, it follows that
(47) lim inftdx°+Y(t)$?(x°)Slim supt~x°+Y(t).
(46) and (47) imply
(48) Y(x°)€[lim Infth°_Y(t).llm suptdx°_Y(t)]n
n[lim inftdx°+Y(t),lim.suptdx°+Y(t)].
Since both h1 and g1 are in 81(1), we see that Y681(I). Now
apply theorem 1.9(c) and (48) to conclude that Y€fi$1(l).
Clearly, from its definition, Ys(x)=0 everywhere. By
theorem 1.11, Y is seen to be constant on I. Since Y(x°)=0,
Y(x)=0 for each xEI.
Using the same argument as above, if J is any other

component of Ag, then there is a C(J)€ni such that

Y(x)=C(J) for each xEJ. From lemma 3.9 and the fact that

47
Y is identically O on I, it easily follows that C(J)=0 for

each component, J, of Ag. Therefore gl(x)=hl(x) on X:-

Again, the semicontinuity-type conditions in theorem
3.12(b) and (e) can be improved. Specifically, (33) and

(34) imply the following corollary.

Corollary¥3.13. Let f, 91. 92. A1 and A2 be as in theorem
3.12. Then

91(xo)=lim suptrx°_gl(t)=lim suptdxfiglm

for all XOEAE and

gz(y°)=lim inft

for all yoéAg.

°_gz(t)=lim lnftdyo+92(t)

”Y

The following propositions are meant to explore the

relationship between an arbitrary £60 and its nice copy.

Corollary 3.14. Let £60 and let g be the nice copy of f.
Then the set [x: f(x)#g(x)] is countable and has no dense

in itself subset.

Proof. This is clear from lemma 3.9 and theorem 3.12(a)

and (c).

Corollary 3.15. Let f and g be elements of 0 such that

f8(x)=gs(x) everywhere. Then the essential sets for f and
g are equal. If A is this essential set and f1 and 91 are
nice copies of f and g, respectively, then there is a cent

such that f1(x)=gl(x)+c on Ac.

Proof. By lemma 3.9, there is a cent such that f(x)=g(x)+c

n. e.. Now, apply theorem 3.8 to f and g(x)+c.

48
Theorem 3.16. If f60. then C(f)c is countable and has no

subset which is dense in itself;

Proof. Let A and B be as in lemma 3.5 with C=AUB. Then,

according to lemma 3.5, C is closed and countable, so

c°=U

n=l In
‘ where the In are pairwise disjoint open intervals. (Some
of them may be empty.) Let u and i be as in lemma 3.4.
Then, by that lemma, us(x)=zs(x)=fs(x) on Cc and
u(x)=£(x)=f(x) on C(f). By lemma 3.10,

Dl=[x6Cc: £(x)¥f(x)] and D2=[x6Cc: u(x)#f(x)}
both have no subset which is dense in itself. It is clear
that if x6C(f)c, then either f(x)#u(x) or f(x)#t(x). From
this, it follows that C(f)c=CUD1UD2. Since each set in this
union has no subset which is dense in itself, it follows

that C(f)c has no subset which is dense in itself.

Theorem 3.16 was first proved by Charzynski [4],
building upon methods developed by Mazurkiewicz [17] and
Sierpinski [24]. The proof given here is much easier.
Szpilrajn [25] showed that if a set, B, has no dense in
itself subset, then B=C(f)c for some f60, thus characterizing
C(f) for an arbitrary f60.

Viewing theorem 3.16 as an extension of theorem 1.4,
one might suspect that a similar extension of theorem 1.5
is also true. This was shown to be false by J. Foran [8],
who constructed a continuous £60 such that D(f)c is

uncountable.

49
Eggollary 3.17. Let £603. Then there is a unique symmetric
set, A, and a function, F6031, satisfying:
(a) Fs(x)=f(x) everywhere:
(b) F is upper semicontinuous on.Ac;
(c) If I is a component of Ac, then F63$1(I):
(d) F is unique up to an additive constant and its values

on A.

Proof. Since £603, there is a g60 such that gs(x)=f(x)
everywhere. Let F be the nice copy of g and A be the
essential set for g. The uniqueness of A and (d) follow

from corollary 3.14. Theorem 3.11 yields (a), (b) and (c).

Definition. Using the notation of corollary 3.17, we shall

call F a nice primitive for f and A the singular set for f.

Corollary 3.17 cannot be extended to 0*3 because there
is no guarantee of a unique primitive when f is allowed to
attain infinite values. A discussion of the problems

involved in this case can be found in Bruckner [3, p.80].

Section 3.3: Monotonicity and Mean Value Theorems

In this section, we will present present some standard
theorems of ordinary differentiation in terms of the symmetric

derivative.

Theorem 3.18. Let f60* such that fs(x)20 a. e. and fs(x)
is never -°. Then the nice copy of f is nondecreasing

and the essential set for f is empty.

50
Proof. Let A be the essential set for f and g be the nice
copy of f. Suppose x is an isolated point of A and (x,B)
is a component of Ac. According to theorem 3.1(f),
g€m_1(x,B). By supposition, gs(t)20 a. e. on (x,B) and
gs(t) is never -¢ on (x,B), so theorem 1.12 implies that
g is nondecreasing on (x,B). From this and theorem 3.1(d)
it is clear that
(49) C(f)-lim suptfix+g(t)<°.
Similarly, if (o,x) is a component of Ac, we can see that
g is nondecreasing on (0.x) and
(SO) C(f)-1im suptdx_g(t)>-.

Since x6A, either C(f)-1im suptdxg(t)=9, or

C(f)-lim.suptdxg(t)=-. Assume the former. Using (49)

and (50), we see

(51) C(f) -1im suptdx+g(t)=a<°
and
(52) C(f)-1im suptdx_g(t)=°.

According to (52), we may choose a sequence, [xn: n6z+]CC(f)
such that xn increases to x and limnd¢g(xn)=°. Since
C(f)CC(g) by theorem 3.1(c), for each n6Z+ there is a
1 . .
6n6(0,n) such that Iy-xn]<6n implies that Ig(y)-g(xn)]<l.
C(f) being residual in r1 implies that for each n, there
. . lim 6 =0, 2
is a zn6C(f)flRx((xn-6n,xn)flc(f)). Since n~° n n
decreases to x. From the choice of 2n and (49),
' — — — — = ' — ‘

1im supn‘.(f(x+(zn x)) f(x (zn x)) 11m supn~¢(f(zn) f(Rx(zn))

Slim supnd~f(zn)-lim infnflcf(Rx(zn))<

<a-limndm(g(xn)+l)=-,

51
from which we see that fs(x)=-, a contradiction. Therefore,
(53) C(f)-1im supth_g(t)<°.
Similarly, it can be shown that
(54) C(f)-lim.suptdx+g(t)>-°.
(49). (50). (53) and (54) imply that

[C(f)-1im supt_,xg(t) l<°

which contradicts the choice of x6A. Therefore A has no
isolated points.

A having no isolated points implies that A has no
isolated points. A closed set with no isolated points is
perfect, and all nonempty perfect sets are uncountable.
Since A'is countable by theorem 3.1, we conclude A=¢.

Thus, AC=IR , so by theorem 3.1(f) , g67R_1. Now, apply

theorem 1.12 to see that g is nondecreasing on r1.

Corollary 3.19. Let f60s such that f(x)20 a. e.. Then

any nice primitive for f is continuous and nondecreasing.

Proof. Let F be a nice primitive for f. Since F60, F

' clearly satisfies the conditions of corollary 3.18. Hence,
we may conclude that the essential set for F is empty and
F is nondecreasing. According to theorem 3.12(f), F6351.
Any monotone function satisfying the Darboux condition

must be continuous.

Corollary 3.19 is also true in the more general case

of the parametric derivative, as shown in [6].

Corollary 3.20. Let £60S such that f is bounded above or

below a. e.. Then any nice primitive for f is continuous.

52
.Proof. Suppose there is an MEI: such that f(x)2M a. e..
Then g(x)=f(x)-M20 a. e. and 9608. Apply corollary 3.19

to g. If f is bounded above, then consider -f.

Corollary 3.21. Let f and g be elements of 05 such that

f(x)=g(x) a. e.. Then f(x)=g(x) everywhere.

Proof. Let h(x)=f(x)-g(x). It is clear that h60S and
h(x)=0 a. e.. By corollary 3.19, any nice primitive of
h is constant. Therefore, h(x)=0 everyWhere and f(x)=g(x)

everywhere.

In the above corollary it is necessary that f(x)=g(x)
a. e.. If the two functions are not equal a. e., the
conclusion is not even true for the ordinary derivative.

A proof of this may be found in Bruckner [3, p. 202].

When considering symmetrically differentiable
functions the ordinary mean value theorem is not true
because symmetric derivatives need not satisfy the Darboux
condition. However, some replacements in the same spirit

as the mean value theorem can be established.

Theorem 3.22. Let f60* and 0,B6C(f) with o<B. Then there
are nonempty G6 sets, A and B, both contained in (d,B),

such that

 

s f(B)-f(q) s
f (a) S 341 ‘f (b)
for all a6A and b6B.

Proof. Let F be the nice copy of f. Theorem 3.1(c) implies

that F(a)=f(0) and F(B)=f(B). Through the addition of an

53

appropriate linear term, we may suppose
(55) F(a)=F(B) .
Define

A=[x6(o,B): fs(x)‘O] and B=[x6(o,B): fs(x)20].
According to theorem 2.1, A and B are G6 sets. Suppose
B=¢. Then fs(x)>0 for each x6(a,B) and applying corollary
3.18, we see that F is strictly increasing on (0,B).
Since 0,B6C(f), we see from theorem 3.1(c) that F is
strictly increasing on [6,5]. This implies that F(G)<F(B),
whcih contradicts (55). Therefore, B¥¢. A similar

contradiction is reached if we assume.A=¢.

Corollary 3.23. Let f,d,B,A and B be as in theorem.3.22.

If fs(x)>- for all x€(a,B), then ]A(>o.

Proof. If we proceed as in the proof of theorem 3.22 and
assune that [A]=O, we arrive at a contradiction in the same

way via theorem 3.18.

Corollary 3.24. Let £60 with a,B, A and B as in theorem

3.22. Then both A and B have positive measure.
Proof. Apply corollary 3.23 to f and -f.

Propositions of the type of theorem 3.22, corollary
3.23 and corollary 3.24 are often called quasi-mean value
theorems. Theorem 3.22 was apparently first proved by Aull
[l] for continuous functions. It was later extended by
Evans [5] and Kundu [14] to functions satisfying certain
monotonicity conditions.

As mentioned above, theorem 3.20 cannot take on the

54
form of the usual mean value theorem because fs may not
satisfy the Darboux condition. For example, let f(x)=[x],
=-l and B=2. Then

f(B)-£(Q1_=l,
B-a 3

and fSUR) =[-1,0,l] . However, if £863, then we do arrive

at the usual statement of the mean value theorem.

Corollary 3.25. Let f60* such that f3 has the Darboux
property. Suppose o,B6C(f) such that o<B. Then there is
a ye(d,s) such that f(B)-f(o)=fs(y)(B-c1).

Proof. This is immediate from theorem 3.22.

Even though fS need not satisfy the Darboux condition,
there is a weaker "Darboux-like" condition which it must
satisfy at every point.

Theorem 3.26. Let £608. Then, for each x6fii,

f(x+h)+f(x-h), f(x+h)+f(x-h)
2 .

(56) lim inf 2

h-O Sf(x).$1im suph”O

Proof. We may assume, without losing generality, that x=0
in (56). Suppose the right-hand inequality in (56) is
false. Through the addition of an appropriate constant,
we may assume that there is an 0631 such that

f(h)+f(4h).

(57) :f(0)>cx>0>lim.suph_.O 2

It is clear that f60S implies that f(--x)60s and
g (x) =f (x)-I2-f ( -x) 608 .
(57) may be rewritten as

(58) 9(0)>a>0>lim sushd09(h).

55
(58) implies that there is a 6>0 such that g(h)<0 whenever
O<[h]<6. By corollary 3.20, there is a primitive, G, for
9 such that G is continuous and decreasing on (-6,6). This
implies that g(0)=GS(0)$O, which contradicts (58). Thus,
we conclude that (57) is impossible and the rightehand
inequality in (56) is true.

The leftehand inequality is established similarly.

Corollaryg3.27. Let f60*s and F be a nice copy of a

primitive for f. Then C(f)CD(F).

Proof. Let x6C(f) and e>O. Then there is a 6>0 such
that [f(x)-f(y)]<§-whenever [y1s6. Let -6<h<6. By corollary
3.20, F is continuous at x and x+h. Theorem 3.22 can then

be applied to see that
f(x)-efP(x+h%'F(x) s f(x)+e.

 

Letting 8‘0, we see that F'(x) exists and equals f(x).

Corollary 3.27 was first proved by Aull [l] for con-
tinuous F and then extended by Evans [5] to measurable F
in.m_l.

Corollary 3.28. Let f60*s and F the nice copy of a primitive

for f. Then F'(x) exists and is finite on a residual set

of full measure.

Proof. Since £681, C(f) is residual (see [20]). Corollary
3.27 then implies that D(f) is residual. By theorem 3.1,

F681, so F is measurable. Now apply corollary 1.6(b).

Corollary 3.28 extends results of Evans [5] and

56

Muckhopadhyay [18] to 0*.

Corollary 3.29. Let f60*s such that f is continuous.

Then fEA'.

Proof. This is immediate from corollary 3.28, or even

the fundamental theorem of calculus.

Finally, we conclude this section with an extension
of a result due to Kundu [16] which will be useful in

Chapter IV.

Lemma 3.30. Let £603 and M6El such that f(x)$M a. e.. If
F is a nice primitive for f and a<b, then

F(b)-F(a):SM]{x: f(x)>0](](a,b)].

Proof. According to corollary 3.20, F is continuous. For
each nez+ and each xerz, define
l
(59) Fn(x) n(F (xi-H) -F (x) ) .
Fn is continuous for all n62+ and if x6D(f), then
(60) limnflFn(x)=f(x) .

According to corollary 3.28, (60) is true a. e.. Now
1im inf fbr (x)dx=lim inf nj‘b(r(x+l)-r(x))dx=
1'1"O a n 1 11"“. a n
b+-
=1im inf n(j' nF(x)dx-j“"l~(x)dx)=
nee +l_ ‘ a
_ n a+£
(61) =llm infn4.n(fb nF(x)dx-j‘a nF(x)dx)2

b+l- 1

al—
211m infnflcnj‘b nF(x)dx--lim supndcnja nF(x)dx=

=F (b) -F (a)
by the fundamental theorem of calculus.

Using theorem 3.22 with (59) it follows that

57
Fn(x)$M for all n and x. Define, for each n6Z+,
cn(x)=max(Fn(X) . 0) .
Then, Gn is continuous for each n with O‘Gn(x)SM and from
(60), limndaGn(x)=max(f(x), 0) a. e.. Applying the
dominated convergence theorem, we see

° ° b x)dx 5 lim b
lim inf ”[ F ( a G (x)dx:
(62)

nor“. -. "
f

ah,“ f(x,>o}f(x)dx ‘MI (x: f(x)>o)n(a.b)]

A combination of (61) and (62) yields the lemma.

 

CHAPTER IV

SYMMETRIC DERIVATIVES AND THE ZAHORSKI CLASSES

Section 4.1: The Abstract Zahorski Classes

In 1950, Z. Zahorski [29] began a classification of
derivatives based upon the structure of their associated
sets. In the course of this work, he defined a descending
sequence of subclasses of £51 which he called mi. i=0, ...,5.
If we represent the classes of functions which are, res-
pectively, approximately continuous, bounded in A' and
both bounded and approximately continuous by an bA' and
bdfi then Zahorski's conclusions can be represented schem-
atically.

”(013ml 3 7”2 3’ 7”32’ 7724. D 77" ‘7
(l) U U U U U
.681 23A“ :A' 3166' 31:4

5:

Kundu [16], in 1976, defined abstract Zahorski
classes and succeeded in demonstrating a similar structure
for continuous functions, f, such that £863. In the
following sections, we will extend Kundu's theorems to

larger subclasses of 0*8.

Definitions . Let ACIR .

MO(A) is the collection of all F0 sets, F, such that

58

59

for all xeAnF, x is a bilateral limit point of F.

M1(A) is the collection of all FO sets, F, such
that for all x6AflF, x is a bilateral condensation point
of F.

M2(A) is the family of all F0 sets, F, such that for
all x6AflF and all 6>0, [(x-6,x)nF[>O and l(x,x+6)]>0.

M3(A) is the collection of all F0 sets, F, such that
if xeAflF and (In: n6Z+] is any sequence of closed inter-
vals converging to x (i. e., any neighborhood of x contains
all but a finite number of the In) such that InnF=¢ for
all n, then

. I 1
lim 1 n =0.
n“ d(x,In)

M4(A) is the collection of all FC sets, F, such that
there is a sequence of closed sets, [Fn: n6Z+], and a

sequence of numbers, (nu: n6Z+]C(O,1), such that F= n=1Fn
and for every c>0 and any x6AflFn there is an c(x,c)>0 such
that for any two real numbers, h and hl, satisfying hh1>0,
h<ch1 and ]h+hl]<c(x,c), the relation

FnJ
J >nn

is true, where J is the interval with endpoints x+h and
x+h+h1.

M5(A) is the collection of all F0 sets, F, such that
for all x6AnF, x is a density point of F.

For i=0,1, ...,5, define the abstract Zahorski class,
mi(A), to be the collection of all functions, f, such that

for any aEIZ, [x: f(x))a] and [x: f(x)<a] are both in Mi(A).

60
Finally, define the class, Z(A), to be the collection
of all functions, f681, such that for each x6A, each e>O
and each sequence, (In: n6z+], of closed intervals conver-
ging to x such that for each n, f(y)2f(x) on In, or f(y)Sf(x)
on In.

liyEIn= if (y) -f (x) (26H
n-m lIn]+d (onn)

 

lim =0.

If A=nl in any of the above definitions, we omit the
reference to A: e. g., MiUR) =Mi' 7711f!!!) =7/(i and Z(JR)=Z.

It follows easily from the definitions that for any
ACJR , Mi+1 (A)C-'Mi(A), i=0, ,4. Therefore, ”(1+1 (A)(:77(i (A),

i=0, ...,4. The following lemma is less obvious.
Lemma 4.1. Let ACJR . Then Z(A)C?Iz3 (A) .

Proof. Let f62(A) and a6nl. It must be shown that the
sets [x: f(x))a] and [x: f(x)<a] are in M3(A). We will
prove that [x: f(x))a]6M3(A). The proof of the other
inclusion is similar.

Let B=[x: f(x))a]. By the definition of Z(A), we
see that f681, so theorem 1.8(c) implies that B is an F0
set. If AnB=¢, it follows vacuously that B6M3(A). So,
suppose that AfiB¥¢ and choose an x6AflB. Let e=f(x)-a and
choose a sequence of closed intervals, (In: n6z+], conver-
ging to x such that InnB=¢ for each n62+. Then, since
f62(A) and f(y)Sf(x) for each yEIn and each n62+,

Hyexn= (f(y)-f(x)(2e1(_

0=lim _
ndo IIn]+d(x,In)

 

61

=11!“an ‘In‘ = limn d(onn)

]In]+d (X.In) 1T
from which it easily follows that
‘In‘

__——_=Oe

1im

The classes Mi and.mi, i=0, ...,5 are due to
Zahorski [29]. The generalized Zahorski classes, Mi(A)
and mi(A), for i=0, ...,4, are apparently due to Kundu

[16], although he states several of them in a different,

 

but equivalent way. M5(A) and m5(A) are extensions of
Zahorski's original classes in the spirit of Kundu's "m
generalizations. The class, Z, was defined by Weil [27],

who proved that ACZCM3 and that the containment is strict.

A function, f, is said to be nonangular iff

(2) D+f(x)2D_f(x)
and
(3) D-f(x)2D+f(x)

for all x631. f is said to be angular at x if either (2)
or (3) fails to be true. (This definition is due to Garg
[11].)

A typical example of a function which is not nonang-
ular is f(x)=[x], which is angular at x=0 because

D-f(0)=-l<l=D+f(O)

violates (3). It is clear from the definition that any
f6A* is nonangular. In general, it is an easy consequence
of the proof of the Denjoy-Saks-Young theorem [23] that

the set of points at which any function is angular is at

most countable.

62
Our interest in nonangular functions is motivated
by the fact that much of the rest of this chapter concerns
functions in 30*3. Using nonangularity, we can state a

sufficient condition for f8 to be in 30*3 when f60*.

Theorem 4.2. Let f60* such that f is nonangular and

symmetrically continuous. Then £86361.

Proof. According to theorem 1.1, fsem , so by theorem

1
1.9(c), it suffices to show that for any x6fii,

s . . s . s
f (x)6[lim inftdx_f (t),lim.suptdx_f (t)](3

(4)

. . s . s
(1 [lim inf f (t) ,11m suptflx+f (t) ] .

t-x+
Suppose (4) is false. Then, for example, we may

suppose that there is an x6ni such that

(5) f8(x)<lim inf t_,x+f"’(t) .

Then there must be an e>0 and a 6>0 such that fs(t)>fs(x)+6

when t6(x,x+€). Through the addition of an appropriate

linear term to f, we may assume

(6) fs(x)<o<fs(t)

whenever t6(x,x+e). (6) implies, via theorem 3.18, that

the nice copy of f is strictly increasing on (x,x+c).

Suppose

(7) f(x))lim f(t) ,

tdx+
where the limit on the right exists because of the monoton-
icity of f on (x,x+c). Then the symmetric continuity<if f
implies that

(8) f(x))lim suptdx_f(t).

From (7) and (8), we see that D+f(x)--° and D_f(x)=°,

which is a violation of (2). Therefore, f(x)$limt +f(t).
~x

63

In a similar manner it can be shown that f(x)Zlimtflx+f(t).
Thus, f(x)=limtdx+f(t). Because f is increasing on (x,x+c),
we can now see that
(9) D+f(x)20.

Since fs(x)<0 and f(t)>f(x) on (x,x+e), it is easy
to see that there is an n>0 such that f(t)>f(x) on (x-n,x).
Therefore,
(10) D‘f(x)so.
(3). (9) and (10) imply that D+f(x)=o‘f(x)=o.

 

D'f(x)=0 implies that there is a sequence, [xn:n6z+},

increasing to x such that

f (x) -f (xn)

1' =0.
J’mn-oan x-xn

 

D+f(x)=0 implies that

. . f(ax(xn))-f(x)
lim infn“ x-xn -20.

 

Now, consider,

f<ax(xn))-f(xn)
«09 2 (x-xn)

. f(ax(xn))-f<x) f(x)-f(xn)
=llmn~¢( 2(x-x ) '1 2(x-x ) -

n n

f (Rx (xn) ) -f (x) . f (X) -f (xn)

ndo 2(x-xn) -+1imn‘° 2(x-xn)

 

s .
f (x)-limn

 

 

 

 

=1im inf 20.

This is a contradiction of (6), so we are forced to con-
clude that (5) never occurs.
The impossibility of the other assumptions which

violate (4) is established similarly.

The converse of this theorem is false. To see this,

consider

64

x .‘l

.-3-SinX x>0
f(x)=' . 0 x=0

x

. l
3 Sinx+x x<0.
It is clear that f has a finite derivative on (-,O)U(O,°),
so there is no problem with angularity or the Darboux prop-
erty on either of these intervals. It is also easy to

show that fs(0)=£u so f60. Since f is continuous and

g; 8111;" - --]’—-2 cos}; x>0
3x
f'(x)=
%—sin!’--—1§cos}-+ l x<0,
3x x

theorem 1.9(c) can be used to see that £6361. But,
D+f(0)=%-and D_f(0)=%-shows that (2)is violated and f is

angular at 0.

Section 4.2: Symmetric Derivatives and the Class m2

The main purpose of this section is to prove the

following theorem.

Theorem 4.3. £0*sdm2.
To prove this theorem, we use the following lemma,

which should be viewed in light of theorem 3.18.

Lemma 4.4. Let f6£0*s with F a primitive for f. If f(x)20

a. e., then any nice copy of F is nondecreasing.

Proof. Without loss of generality, we may assume that F
is the nice copy of itself. It then suffices to show that

F is nondecreasing. To do this, according to theorem 3.18,

65
we must show that
A=[x: f(x)=-°]

is empty.

Define

B=[x: f(x)$-l] and C=[x: f(x)=-2].

Since f6£0*s, theorem 2.1 implies that £6361 from which it
follows, using theorem 1.8(b), that A, B and C are G6 sets.
We claim that A is relatively dense in B. To see this,
suppose it is not. Then there is an open interval, I,
such that IflA=¢ and IDB¥¢. An application of theorem 3.18
shows that F is nondecreasing on I, so IOB=¢, a contradic-
tion. Therefore, A is relatively dense in B.

We now claim that C is also relatively dense in B.
To see this, let x63. Since A is relatively dense in B,
we may choose a sequence, [xnz n6Z+]CA, such that

lim x

= ° . , 2 .
nae n x. By assumption, the set (x. f(x) 0] is dense

in It, so we may choose for each n6Z+, a yn such that
[xn-ynl<%»and f(yn)20. Since £633 for each n62+, there
is a 2 between x and y such that f(z )=-2. Because
n n n n
1 . . _
[zn—xn]<]yn-xn]<n. we see that limndazn=limh~¢xn—x.
Therefore, the claim has been established.

But, B,being a G set, is a Baire space. Any dense

6
G6 subset of a Baire space is residual in that space. So,
A and C must be disjoint residual subsets of a Baire space,
which is impossible. This contradiction forces us to

conclude that A=¢, and the lemma follows.

Proof. (Theorem 4.3) Let f6.DU*S and A=[x: f(x)>0].

66
Suppose there exists an x6A and an c>0 such that
(11) l(x,x+e)nA]=O.
Then f(x)$O a. e. on (x,x+e). By the lemma, it follows
that f(t)$0 everyWhere on (x,x+e). Now, theorem 1.9(b)
implies that f(x)‘0 so that x6A. This contradiction
shows that (11) is false for every x6A and every e>O.
In a similar manner, it can be shown that
[(x-e,x)fiA]>O
for each x6A and each e>0. '
Therefore, A6M2.
Through the addition of an appropriate linear term
to a primitive of f, it may be shown that
[x: f(x))a]6M2
for any a6[-¢,°). By considering -f, we see that
[x: f(x)<a]6M2
for any a6(-,°].

Therefore, f67n2 and the theorem follows.

Corollary 4.5. If f60* such that f is nonangular and

symmetrically continuous, then £86m2.
Proof. This follows at once from theorems 4.2 and 4.3.
Corollary 4.6. 36*'Cm2.

In particular, the derivative of any continuous
function is in‘mz. That theorem 4.3 cannot be improved
to include m3 instead of m2, even for the ordinary derivat-
ive of a continuous function, was shown by Zahorski [29].

The Darboux condition is also necessary because mzCfifil.

67
Theorem 4.3 improves a result of Kundu [16], who
proved that if f60* is continuous, (x: \fs(x)]=°] is count-
able and £865; then fSEmZ. Corollary 4.6 was first shown

by Zahorski [29].

Section 4.3: Symmetric Derivatives and the Class m3

Comparing theorem 4.2 and (1), one might be tempted
to conclude that fiUSCMB. Unfortunately, the situation is
a bit more complicated, as can be seen from the following

example.

Example. There is a continuous and nonangular f60 such

that fseasl and [fs(x)]53 for every x, but fs£m3.

To construct such a function, for n6Z+, let

-n-l

In=(3 .3-n]. For each x6In. define

 
  
 

 

 

 

 

2 l 2 2
3n+1 3n+1 3n+1 32n+1
2n+l
rn(X)' 3 2 (x‘ §+1)2+ £11‘- 2i+1 §+l" 2§+1 X‘ nil+ 2:+1
3 3 3 3 3 3 3
2 2 2 1
-2x-— ——+——<x$—
3n 3n+1 32n+l 3n

 

Using these functions, we define a function, f, with

domain fit by:

(-1)nrn(x) x61n
l
rn(x)= 0 x=0 or (xl>§
rn(-x)+x x6RO(In)

It is an easy calculation to show that f is differentiable

on (-°,0)U(0,°) with [f'(x)]SB whenever x#0. It is also

68
evident from the symmetry of its definition that fs(0)=%.
Therefore, f60.

For each nEZ, rn(x) attains its maximum value on In

 

at 2 . From this, we see that
3n+1
+ 2n+l 2 . 1
D “0) = llmn-oO—2_—r2n(—2h:—I)=lmn~¢(l "23) = 1

3 3
and similarly, D+f(0)=-1. From this and the definition

of f, it follows that D-f(0)=0 and D_f(0)=-2. Thus, f
satisfies (2) and (3) at x=0. f satisfies (2) and (3)
everywhere else because f'(X) exists when x¥0. Therefore,
f is nonangular.

Since each Dini derivative of f is finite everywhere,
f is continuous. Using the facts that f is continuous and

nonangular, theorem 4.2 is applied to show that £86661.

 

 

Let A=[x: fs(x)>0]. Since fs(0)=%u we see that 06A.
From the definition of rn(x), it is clear that rfi(x)$0
_._jL_.Ja -
whenever x6Jn—[3n+l,3n]. Observe that J2nnA-¢ for all
n62 and [Jn: n6Z+] converges to O. In addition,
.. .(lJin', . :21:
n ' 2n n 2-3

Therefore, ffm3.

Notice that this example also invalidates the next
natural assumption from (1), that a bounded symmetric
derivative with the Darboux property is in MA.

The following theorem somewhat clarifies the situation.

Theorem 4.7. Let f60*s with F any primitive for f. Then

f6Z(D(F)).

69

Proof. According to theorems 3.1(d) and 3.8, there is no
generality lost in assuming that F is the nice copy of a
primitive of f, because D(F) is, at worst, made larger.
Let x6D(F). Since f(x)6nl, through the addition of a
linear function, it may be assumed that f(x)=0=F(x).
Because x6D(F)CC(F), there is a 6>0 such that [F(t)]<1
whenever lt-x]<6. Let [a,b]C(x,x+6) be such that f(t)20
on [a,b] and let c>0. Define

A=[t6[a,b]: f(t)zc].
We claim that

(12) e\A|slimt_.b_F(t)-1im F(t) .

t-Oa-I-

To see this, first note that the limits in (12) make
sense because theorem 3.18 guarantees that F is nondecreasing
on (a,b). Define

_ .1 _
Fn(t)-n(F(t+n) F(t)) .
It is clear that if y6D(F), then limndaFn(y)=f(y), so that
by corollary 3.28, Fn converges to f a. e.. Choose any
[c,d]C(a,b) such that [c,d]CC(F). Then, since f is non-
negative and measurable on [c.d],
d . .
c][x6[c,d] . f(x)=c] [if c lim infndan(t)dts

d+l c+l

. . - d . n . n
slim Infnwchn(t)dt$11mn_.°fd nF(t)dt-limnwa nF(t)dt=

=F (d) -F (c)
because F is continuous at c and d. Now, choose two seq-
uences, (ch: n6Z+] and [dn: n6Z+], contained in C(F),such
that cn decreases to a and dn increases to b. Then

c]A]=limndac][x6[cn.dn]= f(X)2€}l‘

7O

slim supn“ (F (dn) -r (cn) ) slimtdbj (t) -lim p (t)

trad-
which is (12) .

Since x6D(F) and f(x)=0, we see that given an n>0
there is a C6(0,6) such that O<t-x<§ implies that
[F (t) -F (x) [fit-x] en .
Therefore, if [a,b]C(x,x+g), then

[limtflb_F (t) -lim F (t) l=

t-va-I-

s]1imt_.b_F (t) -F (x)+F (x) -lirrlt_.,,+

(mu-rm) 1s

F(t)):

S]1imt_.b_ (r (t) -r (x) ) 1+ llimtsar

stunt-0b—

en(2(a-x)+ (b-a) ) ‘2€n(d(x. [a.b])+] [a.bl l) .

enlt-x]+limtda+en]t-x]=en[b—x]+en]a-x]=

Combining this with (12), we see that
(13) lAlSZn(d(x,[a.b])+l[a.bll)

In a similar manner, (13) can be established if
f(t)Sf(x) for all t6[a,b] or if [a,b]c(x-6,x). Since n

can be chosen arbitrarily, the theorem follows.

The following corollaries are immediate from theorem

4.7 and lemma 4.1.

Corollary 4.8. Let f and F be as in theorem 4.7. Then

f6m3(D(F)).
Corollary 4.9. A'C'I7l3.

Theorem 4.7 is a generalization of Weil's original
theorem [27], which was that A'CZ. Corollary 4.8 is an
improvement on Kundu's theorem [16] which required f to
have a continuous primitive and to be in 303. Corollary

4.9 is Zahorski's original result [29].

71

Section 4.4: Symmetric Derivatives and the Class m;
Theorem 4.7 and Zahorski's original results motivate
the following theorem.

Theorem 4.10. Let £60s such that f is bounded. If F is

any primitive for f, then f6m4(D(F)). In
The proof is immediate from the following lemma.

Lemma 4.11. Let f60*S such that f is bounded above and let

 

F be a primitive for f. If 061R, then [x: f(x)>o]6M4(D(F)) . J

Proof. We may suppose that F is the nice copy of a primitive
for f because this at worst makes D(F) larger. Let o=0 and
define B=[x: f(x))o]. According to theorems 2.1 and 1.8(c),
E is an F0 set. Let x°6EnD(F) and f(x°)=a. Using the fact
that x°6D(F), we may write, for h sufficiently close to 0
F(x°+h)=F(x°)+ah+hn(h)
where limhyon(h)=o. If hh1>0 and Ih+h11 is sufficiently
small, then

F(x°+h+hl)-F(x°+h) h(n(h+h1)-n(h))

 

 

(l4) bl = a hl + n (h+h1) .
Choose c>O with O<fil<c. For h small enough, say h<e(x°,c),
1

(14) implies that

a
(D(h) km-

Then, ]h+h1]<e(x°,c) implies

a F(x,+h+hl)-F(xo+h)
(15) O<§< h
1

 

72
Now, let M>l be an upper bound for f. Then lemma
3.28 implies
(16) F(x°+h+h1)-F(x°+h)$Ml{x: f(x°)>0]fl(x°+h,x°+h+hl)l.
It then follows from (15) and (16) that
[[x: f(x))0]n(x,+h,x°+h+h1)] a

>c-:>O.
hl 2M

Theorem 2.1 implies that [x: f(x)>%] is an F0 set for

(17)

 

+ +
each n62 , so we may choose, for each n62 , a sequence of

: m6Z+], such that

closed sets, [En m

[X: f(X) >-} = UélEn m

It is clear that

E= -nU___1[X= f(x)>-}= Un___1Um== 1En,m

and since x063, there are integers, n and m, such that

 

x0613n m' Then a)% and from (17) we see
[[x: f(x)>0]fl(x°+h,x+h+h )l
l :> 1 O
hl 2nM> °
Therefore, if we choose nn =2nM6(0,2), the definition of

M4(D(F)) is satisfied.
Corollary 4.12. bACm4.

Theorem 4.10 improves on a result of Kundu [16] which
was that if f60 is continuous such that £86.80s is bounded,
then £86m2(D(f)). The corollary is due to Zahorski [29].

From (1), we see that baCbA'Cm4. It is easy to see
that bAflcbfios and the example in section 4.3 shows that
b.80s is not contained in m3. .Therefore bA' is properly

contained in bfios. The next two examples show that even

73

the containments bA'C2b037/74 and A'dc0sd are proper.

Example. There is a bounded symmetric derivative, f6m .
which is not a derivative.

Let In=[2-n,2_n+1] for n62+. Partition each In into
2n equal subintervals, 1:, k=l, ...,2n.. If we write

[o.B]=I:, for some k and n, then we may define

4 B-0
0_ _B(x-o) x6[o,o+—Zf0

—4-(x- BEE) x6 [c+E-;—c,o+-u—fl4iz-]

9:(X)= B.“
a—‘f-Bm-s) xe (d+-3—$’i—'°—l.el
o xetdplc

Using these functions, we define

a n
Zn. 423:. (g: (no—gk n(-x))+x(_,,' O) (x) xao
f1(x)=

x=0

NIH

We must show that flem4. Let 06H! and define
F=[X61R : fl(x)>a] .
If 02%» then it is clear from the continuity of each 9:

that F is open and consequently FEM4. So, we suppose that

G<%. Then we may write
F= u" m=:1[x6]R f (mam-137?“)
Using the definition of f1' we see that for each n6Z+, the
set

rn={xem: f1(x)2d+1‘:———9-}

consists of the set [0] and a sequence of disjoint closed

74

intervals converging bilaterally to zero. Thus, Fn is
closed for each n, which implies that F is an E0 set. We
choose the sequence,[Fn], to be the sequence of sets in
the definition of M4.

Now, let A=[x>0: f1(x)z%]. In the same manner as
above, we see that A is closed and also that ACFn for each
nEZ+.

n-3}. Pick

Choose a c>O and let k°=min[n6z+: c<2
e(0,c)6(0,24k°) and let h and hl be positive numbers such

-n°+1) for some

that h<ch1 and h+h1<c(0,c). Then h€[2-n°,2
n°>k°.

We claim that there exist integers, k and n, such
that I:c(h,h+hl). To see this, suppose not. Then (h,h+hl)
can intersect at most two if the intervals (1:: n6Z+,lSk$2n],
because otherwise it must contain one of them. Using the

-n°+l

fact that h<2 , we see

2no 1 _ , —2n -2n +3
hlslrno 1+[Ino_1]—s 2 °<2 o ,

This implies that

h 2-n, k -3
- >—-——__ )2 °
h1 2 n°+3

>c
by the choice of k0. This violates the assumption that
h<ch1. Therefore we are forced to conclude that the claim
is true.

Let [J1, ...,Jm} be the set of all intervals contained
in (1:: n62+,l‘ks2n] such that Jin(h,h+h1)#¢ and such that

the Ji are arranged in increasing distance from zero. It

is clear that m23, for otherwise (h,h+hl) contains no I:-

75

, h+h

Thus, h6J 6Jm and JiCKh,h+hl) for 2£iSm-l. From the

l 1
definition 9:, we see that for Ziism-l,

.. ,1;
[AfiJifl(h.h+hl) l-lAnJi] 4]Ji]

so that

(18) [AflUi_2Ji ]=-]"']Um:2J Ji 1.

From the definition of the 1:, it follows that [J1]‘]J2]

and [Jm]<4]Jm_1] so that

 

 

(19) (011.: I (J l+IU’§;1J I+IJ “anew“;2 Jil.
(18) and (19) imply that
1 m-lJ
(20) Izs.rI(h.h+hl)I2 [AfiUi;2J i] -IU;.2J i‘__l_
h 24‘
1 ‘Ui: l Ji‘ 26‘L‘Ji=2J i‘

Choose n such that 0<n<§%u Then (19) and (20) imply

an(h.h+hl) ( [Afl(h.h+h1) ]
1 1

are chosen to be negative such that h<ch

 

If h and h

l 1

and h+hl>-c(0,c), we note that RO(Afl(O,°))CA to establish

(Fn(h+hl,h) l [An(h+hl,h) ]
___.— 2 2

"hi ‘hl

(30(An(o,~) )r)(h+h1.h) I [An(-h,-h-h1]

2 = > T] .
‘hl ‘hl

Therefore, in the definition of the class M , if we let

 

 

 

nn=n for each n6Z+, then the definition is satisfied at O
with the set F. If x6F such that x¥0, than there is a
neighborhood (x-p,x+p)CF. From this it is evident that
the criteria of the definition of M are satisfied at x.

4

Therefore, F6M4.

76
If B=[x: f1(x)<d}, then it is similarly established
that EGMA. Therefore, fIEMA.
Now let

F(x)=j"c§fl(t)dt.

Since fl is continuous on (-°,0)U(O,°), F is differentiable
with F'(x)=fl(x) whenever x#0. Also

F (x) -F (-x)
0 2x _

S .
F (0)=limx‘
_ - _L X -X _
4me 2x (j‘ ofl(t)dt-j‘ o f1(t)dt)—
=1im —1-j"‘(f (t)-f (-t))dt=
eribc-C) 1 1
_ . _l_ a: l
-1:I.Inx_.0 ZXJ‘ O dt=§=fl(0) .
Thus, F60 with f1(x)=Fs(x) everywhere and fEboSm4.
Define the function

a 2n k k
Zh=02k=l(-9n(x)-gn(-x))+x(_,'o) (x) x740

1 _
2 x—O

f2(X)=

Then, in the same manner as above, it can be established

3
that f2€bo Mk. But,

0 x>0
fl(x)+f2(x)= 1 x=O
2 x<0

which violates the Darboux condition at x=O. Thus, either

fl or f2 cannot be a derivative. Let f be either f1 or 152

such that f is not a derivative.

ExamEle. There exists an féms which is a symmetric deriv-

ative, but not a derivative.

77
For n62+ define In=[2-n, 2-n+2—2n] and let gn be a

nonnegative continuous function supported in In such that

-n-l
g -2
IIn
Then, let 9(X)=Z%:19n(x) and
9(x) X>O
f (x)= O " x=0
-g(-x) x<O.

f is continuous on (-°,O)U(O,¢). We must show that f is
approximately continuous at 0. To do this, let c>0,

k=max{n62+: Z'nzc} and N={x€fii: f(x)=0}.

Then
_ a -i -2i_
‘NM-c, c)122c-21U:=k1n‘-2c-221=k2 +2 -
=2c-2(2 k+1+§2‘2k+2)22c-2‘k.

From this it is clear that

1Nfl(-c, c)l___l

limc-OO 2c
Therefore, 0 is a density point of N. Since f(O)=O, we see
that f is approximately continuous at 0.
Let F(x)=f :f(t)dt. Then F' (x)=f(x) whenever xalo

because of the continuity of f. Also,

fgfItIdt-j’zgh f(t)dt

=1 ~o 2h

 

I29 (t) dt-J' ‘3, g (t) dt
11“h-oo 2h

 

=0.

Therefore, F60 and FS=onSd.

78

To see that fiA', we first note that

+ . n 2-n _ n 9 -m-1
D F(O)Zlimn_‘°2 f0 f(t)dt—2 23ng >l>f(0)

so F is not differentiable at x=0 and therefore FKA. Since
F is absolutely continuous, it must be the nice copy of
itself. Suppose GEA is an ordinary primitive for f. Then
Fs(x)-Gs(x)=0 everywhere and by corollary 3.19 there must
be a cent such that F(x)=G(x)+c. But, this implies that
FGA, which is a contradiction. Therefore f has no ordinary

primitive and is therefore not an ordinary derivative.

l.

10.

11.

12.

13.

14.

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