A Generalization of Special Atom Spaces with Arbitrary Measure
by
Paul Alfonso, Jr.
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
December 13, 2010
Keywords: Weighted Generalized Special Atoms, Weighted Metric Space,
A ( ; ), B ( ; )
Copyright 2010 by Paul Alfonso, Jr.
Approved by
Geraldo Soares de Souza, Chair, Professor of Mathematics and Statistics
Narendra Govil, Professor of Mathematics and Statistics
Asheber Abebe, Associate Professor of Mathematics and Statistics
Abstract
A brief historical account of the development of special atom spaces is presented followed
by the introduction of two new function spaces, A ( ; ) and B ( ; ), which are general-
izations of previous special atom spaces utilizing arbitrary measures rather than Lebesgue
measure of intervals. Known de nitions relating to normed vector spaces are extended to
apply to the new function spaces of arbitrary measure. The properties of the new function
spaces are discussed including the relationship between the spaces as well as the relationship
of the spaces with well known function spaces such as Lebesgue spaces, Lp, Lip( ; ) and
 ( ; ).
Major results include H older-type inequalities for both A ( ; ) and B ( ; ). In the
case of B ( ; ), the dual of B ( ; ) is determined and a Representation Theorem for the
weighted bounded linear functionals of B ( ; ) is presented in detail. However, for A ( ; )
we mention that the dual follows the same idea of the theorem for B ( ; ). That is, that
we only need to estimate k AkA( ; ) for a  -measurable set A. Indeed we show there is
a positive constant M such that k AkA( ; )  M  (A). The duality and representation
theorems for A ( ; ) follow easily. Interpolation of Operators Theorems are presented
on sublinear operators which map B( ; 1p) into weak Lp and A( ; 1p) into weak Lp spaces.
Finally, we present the multiplication operator on A ( ; ) and B ( ; ) for  (t) = t, and
show under some conditions this operator is bounded on those spaces.
ii
Acknowledgments
I have many people to thank for their support and help with this dissertation. The
people listed in the following paragraphs are at the top of a very long list of people who
supported my e orts in some way.
First, I would like to thank God. God has blessed me in more ways than I can list and
I strive to be his servant.
I am especially thankful for my advisor, Geraldo de Souza. Without his guidance and
understanding I doubt I would have  nished my PhD. His sage advice helped me deal with
my absence from high level mathematics for so many years. I am grateful for the work of
Professor Abebe and Professor Govil, the remaining members of my committee. All of my
professors have been understanding in dealing with an Air Force pilot who still wanted to
work on a PhD in mathematics. Thank you to Professor Khodadadi for volunteering to act
as the outside reader for my dissertation.
Eddy Kwessi, Huybrechts Bindele, and Dan Brauss provided useful insight on many
occasions and were good friends during my time at Auburn.
A very special thank you to my parents, Paul and Dolores Alfonso, whom I can never
repay for all of their love and support over the years. Their constant reinforcement of the
importance of an education has driven me to achieve. I would also like to thank my sister
Sheryl, and my brother Dean as well as my extended family who have always supported my
endeavors.
Finally, and most importantly, I am thankful for my lovely wife, Kristal, and our son
Tony. Both have had to endure many nights without me as I struggled with some problem
or issue. Your love, patience, and understanding made this dissertation possible.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction and Historical Background . . . . . . . . . . . . . . . . . . . . . . 1
1.1 A Major Question in the 20th Century . . . . . . . . . . . . . . . . . . . . . 1
1.2 Special Atom Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Generalizations of Special Atom Space . . . . . . . . . . . . . . . . . . . . . 5
2 De nitions and Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.1 Basic De nitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 De nitions of A ( ; ) and B ( ; ) . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Functional and Operator type De nitions . . . . . . . . . . . . . . . . . . . . 12
2.4 De nitions of Lip( ; ),  ( ; ), Lorentz spaces and some results . . . . . . 14
3 Properties of A ( ; ) and B ( ; ) . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.1 Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.2 Completeness and Lp Inclusion . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.3 Relationships within B ( ; ) and A ( ; ) . . . . . . . . . . . . . . . . . . . 26
4 Major Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.1 H older-Type Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Duality and Representation . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
4.3 Interpolation of Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
4.4 Multiplication Operator on L(p;1) . . . . . . . . . . . . . . . . . . . . . . . 43
5 Comments on the dual of A ( ; ) . . . . . . . . . . . . . . . . . . . . . . . . . 47
5.1 Relationship between Lip( ; ) and  ( ; ) . . . . . . . . . . . . . . . . . . 47
5.2 Closing Arguments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
iv
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Appendices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
A Vector Space proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
B Veri cation of minimum in proof of Theorem 4.4 . . . . . . . . . . . . . . . . . 55
v
Chapter 1
Introduction and Historical Background
This dissertation will introduce and explore the properties and applications of two new
function spaces, denoted as A ( ; ) and B ( ; ). This dissertation begins with historical
background motivating these spaces, followed by several de nitions which streamline the
introduction of the new function spaces. We will be presenting several theorems regarding
the properties of the new function spaces as well as the relationship with other known
function spaces.
1.1 A Major Question in the 20th Century
In 1923, Frigyes Riesz introduced a new function space he named after G.H. Hardy
following a paper written by Hardy in 1915, see [32],[26]. The space is de ned as:
De nition 1.1 (Hardy?s Space Hp(D);0 <p<1) F 2 Hp(D) , F is analytic in the
complex unit disc, D and jjFjjHp = sup
0<r<1
 1
2 
Z  
  
jF(rei )jpd 
 1=p
<1.
Hardy?s Space, in turn, is related to another function space, namely Lebesgue space which
Riesz had introduced previously in 1910, see [33]. The de nition of Lebesgue Space follows:
De nition 1.2 (Lebesgue Spaces Lp;0 <p<1 ) f 2 Lp[  ; ] , f is a Lebesgue-
measurable function and kfkLp =
 Z  
  
jf(t)jpd (t)
 1=p
< 1, where  is a measure on
[  ; ].
Recall that if A =fx2[  ; ]jf(x)6= g(x)gand  (A) = 0, thenkfkLp =kgkLp. Although
both the Hardy?s Spaces and Lebesgue Spaces are de ned for p> 0, we are only concerned
with the case p 1.
1
During the  rst half of the twentieth century, a well known fact was that for 1 <p<1,
Hp(D)  = Lp[  ; ]. This follows from the fact that the Hilbert Transform of f, which is
denoted by ~f, is invariant on Lp, and k~fkLp  MkfkLp for 1 < p < 1, M a constant.
Thus, since the dual spaces of Lp were well known, the dual spaces of Hp(D) were known
for 1 < p <1. However, for p = 1, H1(D) is not equivalent to L1[  ; ] since we have
functions f in L1[  ; ] such that ~f does not belong to L1[  ; ]. Thus, the dual space of
H1(D) became a major question in harmonic analysis in the middle of the 20th century. The
search for a solution for the dual space of H1(D) led to several new areas of study, including
the origin of special atom spaces.
Charley Fe erman  rst solved the question of the dual space of H1(D) in 1971, see [24].
Fe erman solved the problem by using the space of functions of Bounded Mean Oscillation
(BMO), introduced by F. John and L. Nirenberg in 1961, see [28]. Here is the de nition of
BMO for reference:
De nition 1.3 (Functions of Bounded Mean Oscillation (BMO))
g2BMO[  ; ],g is periodic and jjgjjBMO = sup
I
1
jIj
Z
I
jg(t) gIjdt<1,
where gI = 1jIj
Z
I
g(t)dt.
Fe erman demonstrated that BMO was equivalent to the dual of H1(D) with equivalent
norms, see [24]. Following this discovery, in 1974 R.R. Coifman provided a new character-
ization of H1(D), what is now referred to as the \Atomic Decomposition of H1(D)." Coif-
man?s characterization of H1(D) introduced the concept of atoms in his de nition of ReH1
which has come to be known as the Space of Atoms. In his paper, Coifman proved that
H1(D)  = ReH1 with equivalent norms, hence the dual of ReH1 is equivalent with BMO,
see [1]. The de nition of atom follows along with the de nition of ReH1, the Space of Atoms,
as these de nitions led directly to the introduction of Special Atom Spaces.
De nition 1.4 (Atom) An atom is a function which is either a(t) = 12 or a : [  ; ]!R
satisfying:
2
 supp a I [  ; ]
 ja(t)j 1jIj;8t2[  ; ]
 
Z
I
a(t)dt = 0
where supp a is the support of the function a, and I is an interval.
De nition 1.5 (Space of Atoms ReH1) ReH1 = ff : [  ; ] ! R;periodic;f(t) =
1X
m=1
cmam(t);
1X
m=1
jcmj<1g, cm2R,jjfjjReH1 = inf
1X
m=1
jcmjwhere the in mum is taken over
all possible representations of f, and the am?s are atoms supported on intervals In [  ; ].
Notice that in De nition 1.4, the concept of atom is in general terms. The next logical
question is what speci c function satis es all the requirements to be an atom? Speci cally,
can a real-valued function be found on [  ; ] that is supported in an interval in [  ; ],
which is bounded by one over the Lebesgue measure of the interval, and when integrated
over the interval results in zero? The answer to this question was the beginning of Special
Atom Spaces.
1.2 Special Atom Space
G. S. de Souza was struck with this question when he was presented with the de nitions
of atoms and ReH1. G.S. de Souza provided the answer to this question when he de ned
Special Atoms as part of his PhD dissertation, see [15].
De nition 1.6 (Special Atoms - de Souza - 1980) Special atoms are de ned as either
b(t) = 12 or b : [  ; ]!R given by b(t) = 1jIj[ L(t)  R(t)];I = L[R; L; R are halves
of intervals I [  ; ]
Utilizing the above de nition, de Souza introduced the  rst Special Atom Space, B1, in 1980.
B1 was the beginning of a series of spaces which eventually led to the new spaces introduced
3
in this dissertation. In his 1992 work on wavelets, Yves Meyer referred to B1 as de Souza?s
Space and the name has subsequently remained, see [31]. The de nition of de Souza?s Space
is:
De nition 1.7 (Special Atom Space - de Souza?s Space B1 - 1980)
B1 = ff : [  ; ] ! R;periodic;f(t) =
1X
m=1
cmbm(t);
1X
m=1
jcmj < 1g where bm are special
atoms and cm 2R for all m2N. The norm is given by jjfjjB1 = inf
1X
m=1
jcmj, where the
in mum is taken over all possible representations of f.
Note here that the norm jjfjjB1 makes de Souza?s Space a Banach space. One useful result
after the discovery of B1 was the analytic characterization of B1. In fact, even before de
Souza was working on his space, mathematicians were working to solve another problem. In
order to formulate this problem, de ne the following function space J:
De nition 1.8 (Function Space J) F 2J,F : D!C with
kFkJ =
Z 1
0
Z  
  
jF0(rei )jd dr<1, where D is the complex unit disc and F0 is the derivative
of F.
The open problem was how to characterize the boundary value of J. That is, for F 2 J,
what is limr!1ReF(rei )? In 1983 de Souza and Gary Sampson answered the question with the
following theorem, see [8]:
Theorem 1.1 (Analytic Characterization of B1) f2B1[  ; ],f( ) = limr!1ReF(rei )
with
Z 1
0
Z  
  
jF0(rei )jd dr <1. Moreover, jjfjjB1  =
Z 1
0
Z  
  
jF0(rei )jd dr = kFkJ. That
is jjfjjB1  =kFkJ.
Following this discovery, de Souza?s space was further generalized in several iterations dis-
cussed in the next section.
4
1.3 Generalizations of Special Atom Space
The  rst generalization of de Souza?s space was with the introduction of a weight to the
bound of the special atom. The result was a generalization of B1 to Bp given in the next
de nition.
De nition 1.9 (Generalization B1 to Bp;1=2 <p<1)
f2Bp,f(t) =
1X
m=1
cmbm(t) with
1X
m=1
jcmj<1
where b1(t) = 12 , bm(t) = 1jI
mj1=p
[ Lm(t)  Rm(t)];Im = Lm[Rm; m6= 1
A norm for Bp is de ned as jjfjjBp = inf
1X
m=1
jcmj, where the in mum is taken over all
possible representations of f.
Note in the above de nition for p = 1, Bp reduces to the original B1. Bp was then generalized
to the space Cqp:
De nition 1.10 (Generalization to Cqp;1=2 <p<1;0 <q 1)
f2Cqp ,f(t) =
1X
m=1
cmbm(t) with
1X
m=1
jcmjq <1 where b1(t) = 12 ,
bm(t) = 1jI
mj1=p
[ Lm(t)  Rm(t)];Im = Lm[Rm; m 6= 1. A norm for Cqp is de ned as
jjfjjCqp = inf
1X
m=1
jcmjq where the in mum is taken over all possible representations of f.
Note that for 0 <q< 1,jjfjjCqp is not a norm in the usual sense. Again, one can easily see the
generalization that for q = 1, C1p = Bp. The spaces, Bp and Cqp, are very important spaces
in harmonic analysis since they are the boundary characterizations of spaces of analytic
functions in the disc called Besov spaces. While Bp is still a Banach space, Cqp is a complete
metric space for 0 < q < 1. Also, the duals of these spaces are Lipschitz spaces. For these
results, and a more complete summary of the above spaces, see [18],[19],[5],[16],[13],[17], and
[36].
Work continued around Bp and Cqp, resulting in weighted special atom spaces B . The
weighted special atom spaces are formed by replacing the bound in the special atoms, jIj
5
with a weight function  (jIj) where  satis es certain conditions, see [4], [3]. Note that if
 (t) = t1p then B reduces to Bp. In 2006, de Souza utilized arbitrary measure in his latest
iteration. He de ned the spaces A( ; ) and B( ; ) using a  nite measure  in place of the
original Lebesgue measure of intervals, see [21]. The de nitions of A( ; ) and B( ; ) are
included below as these spaces are integral to the new function spaces at the heart of this
dissertation.
De nition 1.11 (A( ; )) Given a  nite measure space ([  ; ];A; ), for n2N and  2
(0;1], let An, Bn, and Xn be   measurable sets inAsuch that AnSBn = Xn, AnTBn =;
and  (An) =  (Bn). De ne the space A( ; ) as:
A( ; ) =
(
f : [  ; ]!Rjf(t) =
1X
n=1
cnbn(t);
1X
n=1
jcnj<1
)
Where cn 2 R, bn(t), b1(t) = 1 ([  ; ]), and bn(t) = 1  (Xn)[ An(t)  Bn(t)]; n 6= 1. For
f2A( ; ) de ne a norm as
kfkA( ; ) = inf
1X
n=1
jcnj
where the in mum is taken over all possible representations of f.
De nition 1.12 (B( ; )) Given a  nite measure space ([  ; ];A; ), for n 2 N and
 2(0;1], let Bn be   measurable sets in A. De ne the space B( ; ) as:
B( ; ) =
(
f : [  ; ]!Rjf(t) =
1X
n=1
cnbn(t);
1X
n=1
jcnj<1
)
Where cn2R, b1(t) = 1 ([  ; ]), and bn(t) = 1  (Bn)[ Bn(t)];n6= 1. For f2B( ; ) de ne a
norm as
kfkB( ; ) = inf
1X
n=1
jcnj
6
where the in mum is taken over all possible representations of f.
Note that kfkA( ; ) and kfkB( ; ) are norms in the usual sense. As one can see, the
space A( ; ) is a generalization of previous special atom spaces using arbitrary measure
and arbitrary measurable sets rather than Lebesgue measure of intervals. The space B( ; )
is a slight variation and not a clear generalization since the \atoms"in B( ; ) consist of
the characteristic function of one  -measurable set. If one recalls the de nition of atom,
De nition 1.4, B( ; ) is not a linear combination of true atoms since the integral of bn in
the de nition of B( ; ) will not necessarily be zero. However, one interesting result is that
A( ; )  B( ; ). In fact, in 2009, de Souza and Miguel Pozo proved that A( ; ) and
B( ; ) are equivalent as Banach spaces with equivalent norms, see [6]. The spaces A( ; )
and B( ; ) became much more intriguing later that year when de Souza showed that both of
these spaces are characterizations of the Lorentz spaces L(p;1) for p> 1. Note f 2L(p;1)
if kfkL(p;1) = R2 0 f (t)t1p 1dt < 1, where f is the decreasing rearrangement of f, see
comments following De nition 2.10. Indeed, de Souza showed that for p > 1, A( ;1=p)  =
B( ;1=p) = L(p;1) with equivalent norms, see [9]. The main result in this dissertation is a
further generalization of A( ; ) and B( ; ), utilizing di erent \norms"which are de ned
as weighted metrics. The remainder of this dissertation will de ne and explore A ( ; ) and
B ( ; ), beginning with several useful de nitions, including the de nitions of A ( ; ) and
B ( ; ).
7
Chapter 2
De nitions and Comments
In order to concisely de ne our new function spaces, we will introduce several new de -
nitions in this section. Many of the following de nitions are extensions of known de nitions
and are noted as such. We will include examples and comments for clari cation as appro-
priate. For the remainder of this dissertation, we shall assume that any function denoted by
the symbol  is de ned and  nite for real numbers in its given domain.
2.1 Basic De nitions
The  rst de nition provided below de nes a class of functions which we will utilize
throughout this dissertation.
De nition 2.1 (Class C functions) We de ne C to be a class of functions
 : [0;1)![0;1) satisfying the following conditions:
 (0) = 0;  is strictly increasing and continuous (2.1)
 (  x)   ( ) (x) for some function   : R+ 7!R+ (2.2)
 (x+y) k ( (x) + (y)) for some constant k  1 (2.3)
 (x)!1 as x!1 (2.4)
In order to illustrate Class C functions we present the following Lemma.
Lemma 2.1 (Class C is not empty) For  2 (0;1] the real functions  1(t) and  2(t)
de ned by  1(t) = t and  2(t) = ln (t+ 1) on [0;1) are in the Class C functions.
8
Proof. Let  2 (0;1]. First, consider  1.  1(0) = 0 and  1 is clearly continuous. Now
 01(t) =  t  1 > 0 so  1 is strictly increasing and property (2.1) is satis ed. For property
(2.2) of Class C functions let  2R+, then  1( t) = ( t) =   t =    1(t) and we have
  1( ) =   . Property (2.3) follows directly from the inequality (t + s)  t + s , since
 1(t+s) = (t+s)  t +s =  1(t) + 1(s). This inequality is simple to prove:
(t+s) = (t+s)(t+s)  1 = t(t+s)  1 +s(t+s)  1  t t  1 +s s  1
since (t + s)  1  t  1 and (t + s)  1  s  1. Thus, (t + s)  t + s . The  nal property
is clearly true and  1 2C .
The proof for  2 is slightly more involved. Property (2.1) is clear since natural log
is strictly increasing and continuous with  2(0) = ln (1) = 0. In order to prove prop-
erty (2.2) We break  into two cases with  = 1. First consider the case  > 1. Let
g(t) =  ln(t + 1) ln( t + 1), then g0(t) =  t+1    t+1 and g0(t) = 0 occurs only at t = 0.
Since g0(t) > 0, g(t) is strictly increasing and g(0) is a minimum we conclude g(0) <g(t) for
all t> 0. Thus, we have g(0) = 0 <g(t) =  ln(t+ 1) ln(t+ 1))ln( t+ 1)  ln(t+ 1).
Second we consider the case  < 1. For t 0 we have  t + 1  ( + 1)t + 1 and from the
previous case we then have ln( t+ 1) ln(( + 1)t+ 1) ( + 1) ln(t+ 1) since  + 1 > 1.
Combining the two cases and the trivial case  = 1 we conclude ln( t+ 1) ( + 1) ln(t+ 1)
for all   0, that is   2( ) =  + 1. Finally consider  6= 1. We now have  2( t) =
ln (t + 1) (( + 1) ln( t + 1)) = ( + 1) ln (t + 1). Setting   2( ) = ( + 1) property
(2.2) is proved. Since property (2.4) is clearly true, all that remains is to prove property
(2.3). Since  2(t + s) = ln (t + s + 1), letting  = 1 we have  2(t + s) = ln(t + s + 1).
Now utilizing logarithmic properties we see  2(t) +  2(s) = ln(t + 1) + ln(s + 1) = ln((t +
1)(s + 1)) = ln(ts + t + s + 1)  ln(t + s + 1) =  2(t + s) since ts 0. Letting  6= 1,
 2(t+s) = ln (t+s+ 1) (ln(t+ 1) + ln(s+ 1))  ln (t+ 1) + ln (s+ 1) =  2(t) + 2(s)
9
by the third property proved above for  1(t) = t :2
The following de nition is an extension of a normed vector space which is applicable to
the spaces introduced in this dissertation.
De nition 2.2 (Weighted Metric Space) Let X be a vector space. X is said to be a
weighted metric space if there is a given real valued function k kX called a weighted metric
on X satisfying:
kxkX > 0 if x6= 0 (2.5)
kxkX = 0 if and only if x = 0 (2.6)
k xkX   w(j j)kxkX for all scalars  ,  w a function on R+ (2.7)
kx+ykX  kw (kxkX +kykX) for all x;y2X; kw 1, a scalar (2.8)
Note if one replaces (2.7) and (2.8) by (2.7)?: k xkX = j jkxkX and (2.8)?: kx + ykX  
kxkX +kykX for all x;y2X, then k kX is a norm in the usual sense.
Note: for convenience, we will use the typical symbol for norm throughout this disser-
tation although in many cases our representation is not actually a norm. This distinction
will be pointed out where appropriate. Wherever a metric satis es the four above restric-
tions, we will refer to this metric as a weighted metric. The next de nition de nes a useful
relationship between weighted metric spaces.
De nition 2.3 (Weighted Continuous Space Inclusion) Let X and Y be two weighted
metric spaces, one says that X is weighted continuously contained in Y if:
1. X Y and
2. There are constants M;k > 0 and a function  : [0;1) ! [0;1); (0) = 0 such that
kfkY  M (kkfkX).
10
Note if  (t) = t, then the above de nition is the usual de nition for continuously con-
tained normed spaces. For convenience, we will now de ne Weighted Generalized Special
Atoms. This de nition provides us a type of shorthand notation which we utilize throughout
the remainder of this dissertation.
De nition 2.4 (Weighted Generalized Special Atoms for arbitrary measure)
Given a  nite measure space ([  ; ];A; ), for n2N, let An, Bn, Xn, and E be   measurable
sets in [  ; ] such that AnSBn = Xn, AnTBn =;and  (An) =  (Bn). For  2(0;1], we
de ne the weighted generalized special atoms of Type I and II, bI;IIn : [  ; ]!R as follows:
bI;II1 (t) = 1 ([  ; ]); bIn(t) = 1  (X
n)
[ An(t)  Bn(t)];bIIn (t) = 1  (B
n)
[ Bn(t)]; n6= 1
where  E is the characteristic function of E.
De nition 2.5 (Completeness) A Weighted Metric Space X is said to be complete if and
only if every Cauchy sequence in X converges to an element in X.
2.2 De nitions of A ( ; ) and B ( ; )
Armed now with the above de nitions, we can now de ne the two new spaces A ( ; )
and B ( ; ), which are the foundation of this dissertation.
De nition 2.6 (A ( ; )) For  2(0;1], let (bIn)n 1 be weighted generalized special atoms
of Type I and (cn)n 1 be a sequence of real numbers,  a  nite measure on sets in a   algebra
of [  ; ] and  2C . We de ne the space A ( ; ) as:
A ( ; ) =
(
f : [  ; ]!Rjf(t) =
1X
n=1
cnbIn(t);
1X
n=1
 (jcnj) <1
)
For f2A ( ; ) we de ne a \norm" as
kfkA ( ; ) = inf
1X
n=1
 (jcnj)
11
where the in mum is taken over all possible representations of f.
De nition 2.7 (B ( ; )) For  2(0;1], let (bIIn )n 1 be weighted generalized special atoms
of Type II and (cn)n 1 be a sequence of real numbers,  a  nite measure on sets in a
  algebra of [  ; ] and  2C . We de ne the space B ( ; ) as
B ( ; ) =
(
f : [  ; ]!Rjf(t) =
1X
n=1
cnbIIn (t);
1X
n=1
 (jcnj) <1
)
For f2B ( ; ) we de ne a \norm" as
kfkB ( ; ) = inf
1X
n=1
 (jcnj)
where the in mum is taken over all possible representations of f.
The word \norm" in the previous two de nitions is in quotations since in many in-
stances,k kA ( ; ) andk kB ( ; ) will not turn out to be norms. However, we will show that
this \norm" is, in fact, at all times at least a weighted metric.
We point out here for  (t) = t, A ( ; ) and B ( ; ) reduce to the spaces A( ; )
and B( ; ), which de Souza introduced in 2006, see [9]. Thus, considering the previous
discussion, we see that A ( ; ) and B ( ; ) are indeed generalizations of all the spaces
derived from the original de Souza space. The next de nitions further generalize linear
functionals and operator norms to meet the needs of subsequent proofs.
2.3 Functional and Operator type De nitions
A few operator-type de nitions need extension for the purposes of this dissertation.
Namely, we extend the de nitions of bounded linear operators , operator norms, and dual
spaces to suit our purposes.
12
De nition 2.8 (Weighted Bounded Linear Functional) Let X be a weighted metric
space, we say that  is a weighted bounded linear functional on X if  : X!R such that:
1.  is linear
2. There are constants M;k > 0 and a continuous function  : [0;1) ! [0;1); (0) = 0
such that for all f2X, j (f)j M (kkfkX).
Note: If  (t) = t then the above de nition is the usual de nition for bounded linear func-
tionals.
De nition 2.9 (Weighted Operator Norm) Let X be either A ( ; ) or B ( ; ) and
? be a weighted bounded linear functional on X. We de ne the weighted operator norm, k?k
as follows for f2X:
k?k= sup
f6=0
j?(f)j
  1(k kfkX)
Again, note that if  (t) = t and k = 1, the weighted operator norm reduces to the usual
de nition of operator norms (or equivalent characterization of operator norms.) We now will
extend the traditional de nition of the dual space of a normed space to the weighted metric
space equivalent.
De nition 2.10 (Dual Space of a Weighted Metric Space) The space of all weighted
bounded linear functionals on a weighted metric space X is called the dual of X and is denoted
by X .
The  nal de nition of this section is a special case of a de nition given by de Souza and
Bloom in [3] relating to operators. In order to formulate this de nition we must  rst de ne
the decreasing rearrangement of a real valued, measurable function f. Let f be a real valued
measurable function on T, then for y> 0 let
m(f;y) = m(jfj;y) =jfx2T;jf(x)j>ygj
13
wherej jdenotes the Lebesgue measure on T. m(f;y) is known as the distribution function
of f. Now we can de ne the decreasing arrangement of f, denoted by f , as
f (t) = inffyjm(f;y) tg:
Armed with the previous de nition we can now de ne restricted weak type r operators.
De nition 2.11 (Restricted Weak Type r Operators) Let  be a  nite measure. We
say that an operator T is restricted weak type r if for any  -measurable set A in [  ; ]
t1r (T A) (t) M 1r(A)
where M is an absolute constant and  is the decreasing rearrangement of T A(t).
2.4 De nitions of Lip( ; ),  ( ; ), Lorentz spaces and some results
Here are the de nitions of three well known spaces for later reference. We will see that
Lip( ; ) and  ( ; ) are not only related to each other but have relationships with A ( ; )
and B ( ; ) as well as their duals. The Lorentz spaces are useful in interpolation of operator
theorems.
De nition 2.12 (Lip( ; )) For  2 (0;1] and  a  nite measure on sets of [  ; ] the
space Lip( ; ) is de ned as
Lip( ; ) =
 
g : [  ; ]!Rj 1  (B)
  
  
Z
B
g(t)d (t)
  
  <M
 
where B is a   measurable set in [  ; ]. A norm is de ned on Lip( ; ) as
jjgjjLip( ; ) = sup
B
1
  (B)
  
  
Z
B
g(t)d (t)
  
  
14
Note: the space Lip( ; ) is a generalization of the traditional Lipschitz  spaces usually
denoted by Lip . Lip is the set of all continuous functions f on [  ; ] such that for
 2 (0;1],
  
 f(x+h) f(x)h 
  
 < C where C is a constant. To see the generalization to Lip( ; ),
take  as the Lebesgue measure on [  ; ], X = [x;x + h], and f a di erentiable function,
thus we have:
1
  (X)
  
  
Z
X
f0(t)d (t)
  
  = 1
  ([x;x+h])
  
  
Z x+h
x
f0(t)d (t)
  
  = 1
h 
  
  
Z x+h
x
f0(t)d(t)
  
  =
  
  f(x+h) f(x)
h 
  
  :
If the function f above is not di erentiable, take G(x) = Rx  f(t)dt. Then G(x + h) =
Rx+h
  f(t)dt)jG(x+h) G(x)j=
Rx+h
x f(t)dt and we see:
1
  (X)
  
  
Z
X
f(t)d (t)
  
  = 1
  ([x;x+h])
  
  
Z x+h
x
f(t)d (t)
  
  = 1
h 
  
  
Z x+h
x
f(t)d(t)
  
  =
  
  G(x+h) G(x)
h 
  
  :
G.G. Lorentz originally introduced this space in 1950, see [30],[29]. The next space we
introduce,  ( ; ), is a natural generalization of the traditional second di erence Lipschitz
 space usually denoted by   .
De nition 2.13 ( ( ; )) For  2(0;1] and  a  nite measure on sets of [  ; ] the space
 ( ; ) is de ned as
 ( ; ) =
 
g : [  ; ]!Rj 1  (X)
  
  
Z
A
g(t)d (t) 
Z
B
g(t)d (t)
  
  <M
 
where A;B and X are a   measurable sets in [  ; ] such that ASB = X, ATB =;. A
norm is de ned on  ( ; ) as
jjgjj ( ; ) = sup
ASB=X;ATB=;
1
  (X)
  
  
Z
A
g(t)d (t) 
Z
B
g(t)d (t)
  
  
  is de ned as the set of all continuous functions f on [  ; ] such that for  2 (0;1], 
  f(x+h)+f(x h) f(x)
(2h) 
  
 <C where C is a constant. To see the generalization to  ( ; ), take  
15
as the Lebesgue measure on [  ; ], X = [x h;x+h], and f a di erentiable function. Let
A = [x h;x] and B = (x;x+h] then:
1
  (X)
  
  
Z
A
f0(t)d (t) 
Z
B
f0(t)d (t)
  
  = 1
  ([x h;x+h])
  
  
Z x
x h
f0(t)d (t) 
Z x+h
x
f0(t)d (t)
  
  
= 1(2h) jf(x) f(x h) f(x+h) +f(x)j=
  
  f(x+h) +f(x h) 2f(x)
(2h) 
  
  :
If the function f above is not di erentiable, one can make a similar argument as in the
Lip( ; ) case. de Souza and M. Pozo introduced the space  ( ; ) in earlier works, see [6].
The spaces Lip( ; ) and  ( ; ) endowed with their respective norms are Banach
Spaces. The proofs of these facts are simple using standard techniques. In fact,  ( ; )  =
Lip( ; ) for  2(0;1), see [6].
De nition 2.14 (Lorentz Spaces) A measurable function f belongs to the Lorentz space
L(p;q) if
kfkpq =
 q
p
Z 1
0
 
f (t)t1p
 q dt
t
 1
q
<1
for p2(0;1),q2(0;1) where f is the decreasing arrangement of f.
For q =1, the space L(p;1) is weak Lp space and L(p;p) is the usual Lebesgue space Lp.
Depending on the choices of p and q, kfkpq may not be a norm since the triangle inequality
may fail. However, under certain restrictions on p and q, kfkpq is a norm and in this case
L(p;q) is a Banach space, see [3].
16
Chapter 3
Properties of A ( ; ) and B ( ; )
In this chapter, we will examine the spaces A ( ; ) and B ( ; ) in more detail includ-
ing the relationship these spaces have to each other, other known spaces, as well relationships
within the spaces themselves. The  rst few theorems present basic properties of these spaces.
The  rst theorem is the fact that both of these new spaces are indeed vector spaces. Subse-
quent theorems present facts regarding the completeness of A ( ; ) and B ( ; ) and their
relationship to Lp spaces. The  nal subsection demonstrates the relation of the new spaces
among their own variants.
3.1 Basic Properties
The  rst theorem illustrates the simple fact that A ( ; ) and B ( ; ) are indeed vector
spaces.
Theorem 3.1 A ( ; ) and B ( ; ) are vector spaces with the usual de nitions of addition
and scalar multiplication for reals.
A complete proof of Theorem 3.1 is relatively straightforward and is in Appendix A. Proof
of closure under addition for B ( ; ) is provided below, using a technique which will be
repeated in further proofs.
Proof. Let u;v2B ( ; ) then for all n2N there are real numbers cun;cvn such that:
u =
1X
n=1
cunbIIun(t);
1X
n=1
 (jcunj) <1
v =
1X
n=1
cvnbIIvn(t);
1X
n=1
 (jcvnj) <1
17
Now consider u+v:
u+v =
1X
n=1
 c
unbIIun(t) +cvnbIIvn(t)
 
We introduce new variables and generalized special atoms of type II to re-index the above
equation:
dm =
8>
><
>>:
cum
2
; m even
cvm+1
2
; m odd
bIIm(t) =
8>
><
>>:
bIIum
2
(t); m even
bIIvm+1
2
(t); m odd
Then u + v = P1m=1 dmbIIm(t). Clearly all that remains to show is that P1m=1 (jdmj) <1.
This is clear sinceP1m=1  (jdmj) = P1n=1  (jcunj)+P1n=1  (jcvnj) <1. Thusu+v2B ( ; )
and B ( ; ) is closed under addition.2
Our next theorem states the property mentioned earlier that the \norms" of the spaces
are weighted metrics.
Theorem 3.2 (Weighted Metric) The functions k kB ( ; ) and k kA ( ; ) are weighted
metrics on B ( ; ) and A ( ; ), respectively.
We provide the proof for B ( ; ). The proof for A ( ; ) is analogous.
Proof. Proof of (2.6) (kxkX = 0 if and only if x = 0):
(() Since  2C we know  is non-negative which implieskfkB ( ; )  0. Let f2B ( ; )
such that f = 0. Then one possible representation of f is letting all coe cients cn be zero.
Hence,  (jcnj) = 0 for all cn?s so P1n=1 (jcnj) = 0. This implies kfkB ( ; ) = 0.
()) Let f2B ( ; ) such thatkfkB ( ; ) = 0. Let fk be a representation of f, then fk(t) =
P1
n=1cnkb
II
nk(t). De ne dk =
P1
n=1 (jcnkj). Now either f = 0 and the proof is complete or
there exists a sequence of representations of f,ffkg1k=1 such that dk!0 as k!1. Consider
the second case where f 6= 0 and such a sequence of representations of f exists. Then for
all n2N, jcnkj! 0 as k!1: Else, assume for some m2N, jcmkj is bounded below by
some  > 0 and  (jcmkj) > 0. Let  =  (jcmkj2 , then P1n=1 (jcnkj)  (jcmkj) > > 0 for all
k. This contradicts our original assumption thatkfkB ( ; ) = 0. Thus the coe cients in the
sequence of representations of f all converge to zero and hence f = 0.
18
Proof of (2.5) (kxkX > 0 if x6= 0):
This follows directly from (2.6) and the fact that kfkB ( ; )  0 for all f2B ( ; ).
Proof of (2.7) (k xkX   w(j j)kxkX for all scalars  ,  w real-valued function):
Let  2R, and f2B ( ; ) then f(t) = P1n=1 cnbIIn (t) and
 f(t) =  
1X
n=1
cnbIIn (t) =
1X
n=1
 cnbIIn (t)
so
k fkB ( ; ) =
  
  
 
1X
n=1
 cnbIIn
  
  
 
B ( ; )
 
1X
n=1
 (j cnj)
applying property (2.2) of  we have
1X
n=1
 (j cnj) =
1X
n=1
  (j j) (jcnj) =   (j j)
1X
n=1
 (jcnj):
Thus, taking the in mum over all representations of f in the above inequality we conclude
k fkB ( ; )    (j j)kfkB ( ; ) :
Proof of (2.8) (kx+ykX  kw (kxkX +kykX) for all x;y2X; kw 1, a scalar):
Let f;g2B ( ; ). Then (f + g)(t) = P1n=1c(f+g)nbII(f+g)n(t) where c(f+g)n 2R. Now let
 > 0. By de nition ofk kB ( ; ) there are sequences of real numbers cfn, cgn and atoms bfn,
bgn such that
f(t) =
1X
n=1
cfnbIIfn(t) and g(t) =
1X
n=1
cgnbIIgn(t)
and for the constant k given in (2.3) we have
1X
n=1
 (jcfnj) kfkB ( ; ) +  2k
 
and
1X
n=1
 (jcgnj) kgkB ( ; ) +  2k
 
:
19
Combining the above equations and multiplying by k :
k 
 1X
n=1
 (jcfnj) +
1X
n=1
 (jcgnj)
!
 k  kfkB ( ; ) +kgkB ( ; ) + :
Notice now that (f +g)(t) = P1n=1cfnbIIfn(t) +P1n=1cgnbIIgn(t) which can be re-written as
(f +g)(t) =
1X
n=1
8
>><
>>:
cfn
2
bIIfn
2
(t); if n even
cgn+1
2
bIIgn+1
2
(t); if n odd:
Therefore, taking the in mum below over all representations of f +g we have
kf+gkB ( ; ) = inf
1X
n=1
  jc(f+g)nj  
1X
n=1
8>
><
>>:
 
   
 cfn
2
  
 
 
; n even
 
   
 cgn+1
2
  
 
 
; n odd
=
1X
n=1
 (jcfnj)+
1X
n=1
 (jcgnj):
Applying property (2.3) of  and continuing above
1X
n=1
 (jcfnj)+
1X
n=1
 (jcgnj) k 
 1X
n=1
 (jcfnj) +
1X
n=1
 (jcgnj)
!
 k  kfkB ( ; ) +kgkB ( ; ) + :
Since  was arbitrary we conclude
kf +gkB ( ; )  k  kfkB ( ; ) +kgkB ( ; ) :2
The next theorem introduces the interesting result A ( ; ) B ( ; ).
Theorem 3.3 (Inclusion) For all  2C , A ( ; )  B ( ; ). Moreover, jjfjjB ( ; )  
2   12  jjfjjA ( ; ).
Proof. Let f2A ( ; ), then for all n2N there are cn2R such that
f(t) =
1X
n=1
cnbIn(t); where
1X
n=1
 (jcnj) <1
20
Re-writing f(t):
f(t) =
1X
n=1
cn
  (Xn) ( An(t)  Bn(t))
=
1X
n=1
cn An(t)
  (Xn)  
1X
n=1
cn Bn(t)
  (Xn) :
Now, since An\Bn =;and  (An) =  (Bn) we have  (Xn) =  (AnSBn) =  (An)+ (Bn) =
2 (An) = 2 (Bn). Substituting into previous equation we have:
f(t) =
1X
n=1
cn An(t)
(2 (An))  
1X
n=1
cn Bn(t)
(2 (Bn)) 
= 12 
 1X
n=1
cn An(t)
( (An))  
1X
n=1
cn Bn(t)
( (Bn)) 
!
:
We introduce two new variables to re-index the above equation:
Dm =
8>
><
>>:
Bm2 ; m even
Am+1
2
; m odd
pm =
8>
><
>>:
 cm2 ; m even
cm+1
2
; m odd
Substituting and noting that Dm are   measurable sets we have:
f(t) =
1X
m=1
pm Dm(t)
2   (Dm) =
1X
m=1
pm
2 b
II
m(t)
Finally, we have
1X
m=1
 (
  
 pm2 
  
 ) =
1X
n=1
 (
  
 cn2 
  
 ) +
1X
n=1
 (
  
 cn2 
  
 ) = 2
1X
n=1
 (
  
 cn2 
  
 )
21
Applying property (2.2) of C functions
2
1X
n=1
 (
  
 cn2 
  
 ) 2
1X
n=1
  
 1
2 
 
 (jcnj) = 2  
 1
2 
 1X
n=1
 (jcnj) <1
So, we have f(t)2B ( ; ) and A ( ; ) B ( ; ).
Taking the in mum over all representations of f of both sides of the above inequality gives
jjfjjB ( ; )  2   12  jjfjjA ( ; ):2
3.2 Completeness and Lp Inclusion
In this section, we will provide a proof that Weighted Metric Spaces are complete in
the sense of De nition 2.5 if and only if every absolutely summable series in the space
is summable. This property of Weighted Metric Spaces is very useful in several proofs
presented. The relation of the spaces A ( ; ) and B ( ; ) to the Lebesgue spaces, Lp is
then examined. This relationship with Lp spaces is vital to the last theorem presented in
this section which demonstrates the completeness of A ( ; ) and B ( ; ).
Theorem 3.4 (Completeness of Weighted Metric Spaces) Let X be a weighted met-
ric space. Then X is complete if and only if every absolutely summable series is summable.
The proof given below is an adaptation of a proof given by Royden, see [34].
Proof. ()) Let X be complete andhfiibe an absolutely summable series of elements of X.
Then P1i=1kfikX = M <1. Let  > 0 then there is an N2N such that P1i=NkfikX <  kw
where kw is the constant from De nition 2.2. Let sn = Pni=1kfikX be the partial sum of
hfii, then for n m N, we have
ksn smkX =
  
  
 
nX
i=m
fi
  
  
 
X
 kw
1X
i=m
kfikX  kw
1X
i=N
kfikX < 
22
and hence hsii is a Cauchy sequence in X and must converge to an element f in X since X
is complete. Therefore, limn!1Pni=1 fi = f in X.
(() Let hfni be a Cauchy sequence in X. Then for each integer k there is an integer nk
such that kfn fmkX < 2 k for all n;m > nk. Hence, we choose nk0s such that nk+1 > nk
and obtain the subsequence hfnki of hfni. Let g1 = fn1 and gk = fnk fnk 1, then for k> 1
we have a series hgki whose kth partial sum is fnk but kgkkX  2 k+1. Therefore, we have
1X
k=1
kgkkX  kg1kX +
1X
k=2
2 k+1 =kg1kX + 1
and hgki is absolutely summable. Thus, there exists f 2 X such that hfnki! f in X.
Finally, we show limn!1fn = f. Since hfni is a Cauchy sequence, for  > 0 there is an N2N
such that kfn fmkX   2kw for all n;m > N. Since fnk !f, there is a K2N such that
for all k K we have kfnk fkX   2kw. Take k so large that k>K and nk >N, then
kfn fkX  kfn fnk +fnk fkX  kw(kfn fnkkX +kfnk fkX) kw(  2k
w
+  2k
w
) =  :
Since  was arbitrary, we have limn!1fn = f and X is complete based on De nition 2.5 :2
The Lemma below will be used in several subsequent proofs.
Lemma 3.1 For n2N, let cn be real numbers and  2C such that P1n=1  (jcnj) <1 ,
and let k be the constant from property (2.3) of C functions, then
1X
n=1
jcnj   1
 
k 
1X
n=1
 (jcnj)
!
where   1 is the inverse of  .
23
Proof. Let n, cn,  , and k be as above in the statement of the lemma. Choose N 2 N.
Then, since  is continuous, well de ned,  nite, strictly increasing, and  (0) = 0, we have
NX
n=1
jcnj=   1
 
 
 NX
n=1
jcnj
!!
   1
 
k 
NX
n=1
 (jcnj)
!
:
Taking the limit as N!1 of the above inequality gives the desired result.2
It is interesting to note that if one chooses  (t) = tp for p2(0;1), then   1(t) = t1p and
k = 1 (see Lemma 2.1), applying Lemma 3.1 gives a well known inequality
1X
n=1
jcnj 
 1X
n=1
jcnjp
!1
p
The following theorem presents the aforementioned relationships between A ( ; ), B ( ; ),
and Lp spaces.
Theorem 3.5 (Lp Inclusion) For p 1 and  < 1p, A ( ; ) and B ( ; ) are weighted
continuously contained in Lp[  ; ].
Proof. First, consider one weighted generalized special atom of Type II (where B is a
  measurable set in [  ; ]):
jjbIIjjpLp =
Z  
  
  
   B(t)
  (B)
  
  
p
d (t) = 1 p (B)
Z
B
 B(t)pd (t)
=  (B) p (B) =  1 p (B)  1 p ([  ; ]):
Let M ;p = ( 1 p ([  ; ]))1=p, then for any weighted generalized special atom of Type II
we have jjbIIjjLp  M ;p.
Now let f2B ( ; ), then for n2N there are cn2R such that
f(t) =
1X
n=1
cnbIIn (t);
1X
n=1
 (jcnj) <1:
24
Taking the Lp norm of f we have:
jjfjjLp =
  
  
 
1X
n=1
cnbIIn
  
  
 
Lp
 
1X
n=1
jjcnbIInjjLp =
1X
n=1
jcnjkbIInkLp
 
1X
n=1
jcnjM ;p = M ;p
1X
n=1
jcnj:
Now, by Lemma 3.1 we know
1X
n=1
jcnj   1
 
k 
1X
n=1
 (jcnj)
!
:
Combining this with our previous inequality,
jjfjjLp  M ;p
1X
n=1
jcnj)jjfjjLp  M ;p  1
 
k 
1X
n=1
 (jcnj)
!
:
Finally taking the in mum of the above inequality over all representations of f we have:
jjfjjLp  M ;p  1
 
k kfkB ( ; )
 
:
In other words, B ( ; ) is weighted continuously contained in Lp[  ; ]. Now consider g2
A ( ; ). Using Theorem 3.3, g2B ( ; ) such that jjgjjB ( ; )  2   12  jjgjjA ( ; ), and
we have A ( ; ) weighted continuously contained in Lp[  ; ]: Indeed, there is a constant
M ;p such thatjjgjjLp  M ;p  1
 
k kgkB ( ; )
 
and we have the following relation between
Lp and A ( ; ):
jjgjjLp  M ;p  1
 
k kgkB ( ; )
 
 M ;p  1
 
k 2  
 1
2 
 
jjgjjA ( ; )
 
:2
We now have all of the tools needed to demonstrate that A ( ; ) and B ( ; ) are
complete weighted metric spaces. We will provide the proof for the B ( ; ) case since the
A ( ; ) case is similar.
25
Theorem 3.6 (Completeness of A ( ; ) and B ( ; )) For  2(0;1) the spaces A ( ; )
and B ( ; ) are complete.
Proof. In order to demonstrate the completeness of B ( ; ), we apply Theorem 3.4. To this
end, let hfni be an absolutely summable sequence in B ( ; ), then P1n=1kfnkB ( ; ) <1.
Let cnm 2R such that fn(t) = P1m=1cnmbIInm(t) with P1m=1  (jcnmj) <1 for each n2N.
We must show that P1n=1fn converges in B ( ; ). By de nition of kkB ( ; ), for all  > 0
there is a representation of fn such that
1X
m=1
 (jcnmj) kfnkB ( ; ) +  2n
for each n. Let f(t) = P1n=1fn(t), then f(t) = P1n=1P1m=1cnmbIInm(t) and
kfkB ( ; )  
1X
n=1
1X
m=1
 (jcnmj) 
1X
n=1
kfnkB ( ; ) +
1X
n=1
 
2n =
1X
n=1
kfnkB ( ; ) + <1:
Applying Theorem 3.4, we conclude that B ( ; ) is complete.2
3.3 Relationships within B ( ; ) and A ( ; )
The following discussion presents two theorems regarding the relation of these spaces
for di erent measures, alpha values and base functions  . Theorems for both spaces are
presented below while the proof is given for only the B ( ; ) case. The proof for the
A ( ; ) is again analogous.
Theorem 3.7 (Relationship of A spaces) Given A 1( 1; 1) and A 2( 2; 2), if  2  
 1 and there exists k1;k2 2 R+ such that  1(t)  k1 2(t), and  1(E)  k2 2(E) for all
 1; 2 measurable sets E [  ; ] then A 2( 2; 2)  A 1( 1; 1). Moreover, there exists
a constant M2R+ such that kfkA 1( 1; 1)  MkfkA 2( 2; 2).
26
Theorem 3.8 (Relationship of B spaces) Given B 1( 1; 1) and B 2( 2; 2), if  2  
 1 and there exists k1;k2 2 R+ such that  1(t)  k1 2(t), and  1(E)  k2 2(E) for all
 1; 2 measurable sets E [  ; ] then B 2( 2; 2)  B 1( 1; 1). Moreover, there exists
a constant M2R+ such that kfkB 1( 1; 1)  MkfkB 2( 2; 2)
Proof of Theorem 3.8.
Let B 1( 1; 1), B 2( 2; 2) and k1;k2 2R+ be given such that  2   1,  1(t)  k1 2(t),
and  1(E) k2 2(E). Let f2B 2( 2; 2) then
f(t) =
1X
n=1
cnbIIn (t) and
1X
n=1
 2(jcnj) <1
Substituting the de nition of weighted generalized special atom of Type II we have:
f(t) =
1X
n=1
cn Bn(t)
  22 (Bn)
=
1X
n=1
cn
   1
1 (Bn)
  22 (Bn)
  
Bn(t)
  11 (Bn):
Now, we have:
  11 (Bn)
  22 (Bn)  
k 12   12 (Bn)
  22 (Bn) = k
 1
2  
( 1  2)
2 (Bn) k
 1
2 ( 2)
( 1  2)([  ; ]):
Let M1 = k 12 ( 2)( 1  2)([  ; ]).
Utilizing conditions (2.1) and (2.2) on C :
1X
n=1
 1(
  
  cn  11 (Bn)
  22 (Bn)
  
  ) =
1X
n=1
 1(jcnj 
 1
1 (Bn)
  22 (Bn)) 
1X
n=1
 1(M1jcnj)
 
1X
n=1
  1(M1) 1(jcnj):
27
Continuing the above inequality and using the fact that  1  k1 2,
1X
n=1
  1(M1) 1(jcnj) =   1(M1)
1X
n=1
 1(jcnj)   1(M1)k1
1X
n=1
 2(jcnj):
So,
1X
n=1
 1(
  
  cn  11 (Bn)
  22 (Bn)
  
  ) <1
Thus, f 2B 1( 1; 1) and B 2( 2; 2)  B 1( 1; 1) Finally, letting M = k1  1(M1) and
taking the in mum of both sides of the above inequality over all representations of f we
conclude kfkB 1( 1; 1)  MkfkB 2( 2; 2):2
28
Chapter 4
Major Results
Armed with the basic properties of A ( ; ) and B ( ; ) we am now able to prove
deeper results. This chapter discusses the duality of the new spaces as well as interpolation
of operator theorems.
4.1 H older-Type Inequalities
To  nd the dual spaces of A ( ; ) and B ( ; ), a  rst step is to  nd a H older-type
inequality for each space. Essentially, we desire to  nd a function space X associated with
B ( ; ) such that for g2X and f2B ( ; ) we have a result similar to:
  
  
Z  
  
g(t)f(t)d (t)
  
   kgkXkfkB
 ( ; )
Similarly, we would like to  nd a function space Y to couple with A ( ; ) for a H older-type
inequality.
Consider the case for B ( ; ). As in previous proofs, we  rst consider one weighted
generalized special atom of Type II. For g in our arbitrary function space X we then have:
  
  
Z  
  
g(t)bII(t)d (t)
  
  =
  
  
Z  
  
g(t)  B(t)  (B)d (t)
  
  = 1
  (B)
  
  
Z
B
g(t)d (t)
  
  :
Thus, in order to establish our desired inequality we would like a function space X for g which
will provide the bound above. Recalling the de nition of Lip( ; ) given in De nition 2.12,
we see that Lip( ; ) is the desired candidate space to couple with B ( ; ). Similarly we  nd
that  ( ; ) given in De nition 2.13 is the desired candidate space to couple with A ( ; ).
The next two theorems and proofs validate the above choices for the desired inequalities.
29
Theorem 4.1 (H older?s-type inequality for B ( ; )) For f2B ( ; ) and g2Lip( ; )
the following inequality holds for  2(0;1):
  
  
Z  
  
g(t)f(t)d (t)
  
   jjgjjLip( ; )   1 k jjfjjB
 ( ; )
 
where k is the constant given in condition (2.3) of Class C functions.
Proof. Let f 2B ( ; ) be a weighted generalized special atom of Type II, that is f(t) =
bII(t), and let g 2 Lip( ; ). Then, by the argument in the previous paragraph and the
de nition of Lip( ; ), we have:
  
  
Z  
  
g(t)f(t)d (t)
  
  =
  
  
Z  
  
g(t)bII(t)d (t)
  
  = 1
  (B)
  
  
Z
B
g(t)d (t)
  
   jjgjjLip( ; ):
Now, let N 2 N and fN be a  nite combination of Type II weighted generalized special
atoms. Then, fN(t) = PNn=1 cnbIIn (t) and we have:
  
  
Z  
  
g(t)fN(t)d (t)
  
  =
  
  
 
Z  
  
g(t)
NX
n=1
cnbIIn (t)d (t)
  
  
 =
  
  
 
NX
n=1
cn
Z  
  
g(t)bIIn (t)d (t)
  
  
  
NX
n=1
jcnj
  
  
Z  
  
g(t)bIIn (t)d (t)
  
   
NX
n=1
jcnjjjgjjLip( ; ) =jjgjjLip( ; )
NX
n=1
jcnj:
Applying Lemma 3.1 gives the result:
  
  
Z  
  
g(t)fN(t)d (t)
  
   jjgjjLip( ; )   1
 
k 
NX
n=1
 (jcnj)
!
:
Taking the in mum of the above inequality over all representations of fN, we conclude
  
  
Z  
  
g(t)fN(t)d (t)
  
   jjgjjLip( ; )   1 k kfNkB
 ( ; )
 
: (4.1)
30
In order to extend equation (4.1) asN!1takef2B ( ; ), then there are real coe cients
cn such that f(t) = P1n=1cnbIIn (t) with P1n=1 (jcnj) 1. Now for N2N de ne fN(t) =
PN
n=1cnb
II
n (t), then fN 2B ( ; ) and it is clear that limn!1fN(t) = f(t).
Let rN = R   g(t)fN(t)d (t), then rN 2R since from equation (4.1) we know
jrNj jjgjjLip( ; )   1
 
k kfNkB ( ; )
 
:
Consider the sequence hrNi. Let M2N such that N >M, then
rN rM =
Z  
  
(fN(t) fM(t))g(t)d (t)
and
jrN rMj jjgjjLip( ; )   1
 
k kfN fMkB ( ; )
 
:
Also,
kfN fMkB ( ; ) =
  
  
 
NX
n=M+1
cnbIIn
  
  
 
B ( ; )
 
NX
n=M+1
 (jcnj)
SinceP1n=1  (jcnj) <1, we knowPNn=M+1 (jcnj)!0 asN;M!1, and hencekfN fMkB ( ; ) !
0 as N;M!1. Having  2C , we know  (0) = 0 and  is strictly increasing and it follows
that:
jrN rMj jjgjjLip( ; )   1
 
k kfN fMkB ( ; )
 
!0 as N;M!1:
Thus,hrNiis a Cauchy sequence in R and convergent in R since R is complete. In other words,
lim
N!1
rN = lim
N!1
R 
  g(t)fN(t)d (t) exists. We now demonstrate that this limit is independent
of the sequence hfNi converging to f. To this end, let hhNi be a sequence of functions in
B ( ; ) that converges to f. Then we must show:
lim
N!1
Z  
  
g(t)hN(t)d (t) = lim
N!1
Z  
  
g(t)fN(t)d (t):
31
Consider then,
lim
N!1
  
  
Z  
  
(fN(t) hN(t))g(t)d (t)
  
   lim
N!1
jjgjjLip( ; )   1
 
k kfN hNkB ( ; )
 
:
We next examine kfN hNkB ( ; ):
kfN hNkB ( ; ) =kfN f +f hNkB ( ; )  kw
 
kfN fkB ( ; ) +kf hNkB ( ; )
 
:
Since kfN fkB ( ; ) !0 and kf hNkB ( ; ) !0 as N!1, we have
lim
N!1
  
  
Z  
  
(fN(t) hN(t))g(t)d (t)
  
  !0 as N!1:
and we conclude that the limit as N!1 is independent of the sequence used to converge
to f in B ( ; ). Hence, we de ne
lim
N!1
Z  
  
fN(t)g(t)d (t) :=
Z  
  
f(t)g(t)d (t):
With this result, we can now take the limit as N!1 of equation (4.1) to obtain our  nal
result:
lim
N!1
  
  
Z  
  
fN(t)g(t)d (t)
  
  =
  
  
Z  
  
f(t)g(t)d (t)
  
   
lim
N!1
jjgjjLip( ; )   1
 
k kfNkB ( ; )
 
=jjgjjLip( ; )   1
 
k kfkB ( ; )
 
:2
Theorem 4.2 (H older?s-type inequality for A ( ; )) For f2A ( ; ) and g2 ( ; )
the following inequality holds for  2(0;1):
  
  
Z  
  
g(t)f(t)d (t)
  
   jjgjj ( ; )   1 k jjfjjA
 ( ; )
 
where k is the constant given in condition (2.3) of Class C functions.
32
Proof. Let f 2A ( ; ) be a weighted generalized special atom of Type I, that is f(t) =
bI(t), and let g 2  ( ; ). Then there are  -measurable sets X;A;B 2 [  ; ] such that
X = ASB, ATB =;,  (A) =  (B), and we have
  
  
Z  
  
g(t)f(t)d (t)
  
  =
  
  
Z  
  
g(t)bI(t)d (t)
  
  =
  
  
Z  
  
1
  (Xn)[ An(t)  Bn(t)]g(t)d (t)
  
  =
1
  (X)
  
  
Z
A
g(t)d (t) 
Z
B
g(t)d (t)
  
   jjgjj ( ; ):
Now, let N 2 N and fN be a  nite combination of Type I weighted generalized special
atoms. Then fN(t) = PNn=1cnbIn(t), and we have:
  
  
Z  
  
g(t)fN(t)d (t)
  
  =
  
  
 
Z  
  
g(t)
NX
n=1
cnbIn(t)d (t)
  
  
 =
  
  
 
NX
n=1
cn
Z  
  
g(t)bIn(t)d (t)
  
  
  
NX
n=1
jcnj
  
  
Z  
  
g(t)bIn(t)d (t)
  
   
NX
n=1
jcnjjjgjj ( ; ) =jjgjj ( ; )
NX
n=1
jcnj:
Applying Lemma 3.1 and taking the in mum over all representations of fN gives the result
  
  
Z  
  
g(t)fN(t)d (t)
  
   jjgjj ( ; )   1 k kfNkB
 ( ; )
 
: (4.2)
In a similar fashion to the previous proof, we can show that if a sequence hfNi converges
to f in A ( ; ), then lim
N!1
R 
  g(t)fN(t)d (t) exists and is independent of the choice of the
sequence used to converge to f in A ( ; ). Thus, for such a sequencehfNiand g2 ( ; )
we de ne:
lim
N!1
Z  
  
fN(t)g(t)d (t) :=
Z  
  
f(t)g(t)d (t):
Taking the limit as N!1 in equation (4.2) above provides our desired result:
  
  
Z  
  
g(t)f(t)d (t)
  
   jjgjj ( ; )   1 k jjfjjA
 ( ; )
 :2
33
The following Lemma is useful in applying the H older?s-type inequality in Theorem 4.1
in the search for the dual of B ( ; ).
Lemma 4.1 Let bII(t) be a weighted generalized special atom of Type II, then for  2(0;1)
  bII  
B ( ; )
 =  (1):
Proof. Let  2 (0;1) and p = 1 , then for f 2 B ( ; ) by Theorem 3.5, f 2 Lp with
jjfjjLp    1
 
k kfkB ( ; )
 
since M ;p = 1 in this case. Now let f be a weighted generalized
special atom of Type II, then
kfkpLp =   bII  1 L
1 
=
Z  
  
  
B(t)
  (B)
 1
 
d (t) = 1 (B)
Z
B
d (t) = 1:
By the de nition of kkB ( ; ) we have
1 =   bII  L
1 
   1
 
k   bII  B
 ( ; )
 
   1 (k  (1)):
Taking  of both sides of the above inequality we obtain
 (1) k   bII  B
 ( ; )
 k  (1)) (1)k
 
   bII  B
 ( ; )
  (1)
In other words, kbIIkB ( ; )  =  (1).2
We can now examine the duality of A ( ; ) and B ( ; ).
4.2 Duality and Representation
Perhaps the most signi cant result of this dissertation is the following theorem regarding
the dual of B ( ; ). In fact, the dual of B ( ; ) is a characterization of Lip( ; ).
34
Theorem 4.3 (Duality and Representation for B ( ; )) ? is a weighted bounded lin-
ear functional on B ( ; ) if and only if there is a unique g2Lip( ; ) so that for all f 2
B ( ; ) we have ?(f) = R   f(t)g(t)d (t). Moreover, k?k is equivalent to kgkLip( ; ). That
is there are absolute real constants c1 and c2 such that c2kgkLip( ; )  k?k c1kgkLip( ; ).
In other words, the dual space of B ( ; ) is equivalent to Lip( ; ). That is B  ( ; )  =
Lip( ; ).
Proof. Let g 2 Lip( ; ). De ne ?g : B ( ; ) ! R by ?g(f) := R   f(t)g(t)d (t) for
f 2 B ( ; ). By Theorem 4.1, ?g is a weighted bounded linear functional since j?gj 
jjgjjLip( ; )   1 k jjfjjB ( ; ) and the integral operator is linear. Now, de ne  as follows:
 : Lip( ; )!B  ( ; ); g7! (g) = ?g:
Then  is one-to-one:
Let g1;g2 2Lip( ; ) such that  (g1) =  (g2), then for f2B ( ; ) we have
Z  
  
f(t)g1(t)d (t) =
Z  
  
f(t)g2(t)d (t))
Z  
  
f(t) (g1(t) g2(t))d (t) = 0
Hence, we conclude that g1(t) = g2(t) almost everywhere which is su cient for our purposes
by the de nition of kkLip( ; ).
It remains to show that  is onto:
Let ?2B  ( ; ), then we must show that there is a unique g2Lip( ; ) such that  (g) = ?
or in other words,  = ?g. For any  -measurable set E in [  ; ], de ne  (E) := ?( E).
Note that by this de nition  is a signed measure, see [34]. We show that  <<  . That
is, we show that if  (E) = 0 then  (E) = 0. By the de nition of weighted bounded
linear functionals, De nition 2.8, there are constants M;k> 0 and a continuous real valued
function  with  (0) = 0 such that
j (E)j=j?( E)j M 
 
kk EkB ( ; )
 
:
35
Since  E(t) =   (E)  E(t)  (E), we have k EkB ( ; )   (  (E)). Now due to the continuity of
 and  if  (E)!0 then k (  (E))!0 which combined with the above inequality implies
j (E)j M 
 
kk EkB ( ; )
 
 M (k (  (E)))!0 as  (E)!0:
Therefore, if  (E) = 0 then  (E) = 0 and we have  <<  . By the Radon-Nikodym
Theorem, there is a  -measurable function g such that  (E) = REg(t)d (t). Thus,
?( E) =
Z
E
g(t)d (t) =
Z  
  
 E(t)g(t)d (t) and ?
  
E
  (E)
 
=
Z  
  
 E(t)
  (E)g(t)d (t)
by the linearity of ?. Hence, ?(bII) = R   bII(t)g(t)d (t) for any weighted generalized special
atom of Type II. Now let fN be a  nite linear combination of Type II atoms, then there
are real coe cients cn such that fN(t) = PNn=1cnbIIn (t) and we have
?(fN) = ?
 NX
n=1
cnbIIn (t)
!
=
NX
n=1
cn? bIIn (t) =
NX
n=1
cn
Z  
  
bIIn (t)g(t)d (t)
=
Z  
  
NX
n=1
cnbIIn (t)g(t)d (t) =
Z  
  
fN(t)g(t)d (t)
so
lim
N!1
?(fN) = lim
N!1
Z  
  
fN(t)g(t)d (t):
Now, let f2B ( ; ). Then f(t) = P1n=1cnbIIn (t) and similar to previous proofs, we de ne
the sequence hfNi by fN(t) = PNn=1cnbIIn (t). It is clear that hfNi!f as N !1. Since
? is a weighted bounded linear functional, we have lim
N!1
?(fN) = ?(f). So, by a similar
argument to the proof of Theorem 4.1 we conclude:
?(f) = lim
N!1
Z  
  
fN(t)g(t)d (t) =
Z  
  
f(t)g(t)d (t):
36
Thus, all that remains to show is thatg2Lip( ; ). That is, we must show 1  (E)   REg(t)d t  <
K for some constant K. Consider then
1
  (E)
  
  
Z
E
g(t)d (t)
  
  = 1
  (E)
  
  
Z  
  
 E(t)g(t)d (t)
  
  =
  
  
Z  
  
 E(t)
  (E)g(t)d (t)
  
  = ?
  
E(t)
  (E)
 
:
Again, using the de nition of weighted bounded linear functional, we can continue the above
equation with constants M;k> 0 such that
?
  
E(t)
  (E)
 
 M 
 
k
  
   E(t)
  (E)
  
  
B ( ; )
!
:
Finally, applying Lemma 4.1 we have
1
  (E)
  
  
Z
E
g(t)d (t)
  
   M 
 
k
  
   E(t)
  (E)
  
  
B ( ; )
!
 M (k2 (1)) <K
where K and k2 are positive constants. Thus, g2Lip( ; ) and ? = ?g. All that remains
to show is k?k=k?gk = kgkLip( ; ). By De nition 2.9, for g2Lip( ; ) and f 2B ( ; )
we have
k?gk= sup
f6=0
j?g(f)j
  1(k kfkX):
By Theorem 4.1
j?g(f)j jjgjjLip( ; )   1 k jjfjjB ( ; ) ) j?g(f)j  1 k
 jjfjjB ( ; )
  jjgjjLip( ; )
and taking the supremum above over all f2B ( ; ) such that f6= 0:
sup
f6=0
j?g(f)j
  1(k kfkB ( ; ))  jjgjjLip( ; ): (4.3)
37
Now let E be a  -measurable set in [  ; ] and let h(t) =
  1
 
 (1)
k 
 
  (E)  E(t). Then h(t) 2
B ( ; ) and
khkB ( ; ) =  
 
  1
  (1)
k 
  
=  (1)k
 
(see Lemma 4.1). Thus, we have the following:
j?g(h)j
  1 k jjhjjB ( ; ) =
  
  
  
Z  
  
  1
 
 (1)
k 
 
 E(t)g(t)d (t)
  (E)
  
  
   
1
  1 k jjhjjB ( ; ) =
  1
  (1)
k 
 1
  (E)
  
  
Z
E
g(t)d (t)
  
   1
  1
 k
  (1)
k 
 =   1
  (1)
k 
 1
  (E)
  
  
Z
E
g(t)d (t)
  
  
Let K1 =   1
 
 (1)
k 
 
and take the sup above over all h2B ( ; ) such that h6= 0 gives the
result
sup
h6=0
j?g(h)j
  1(k khkB ( ; ))  
K1
  (E)
  
  
Z
E
g(t)d (t)
  
  :
Taking the sup over all  -measurable sets in [  ; ] gives us k?gk  K1jjgjjLip( ; ) and
combining this equation with equation 4.3 we obtain our  nal inequality:
K1jjgjjLip( ; )  k?gk jjgjjLip( ; )
and we have k?gk =jjgjjLip( ; ).2
4.3 Interpolation of Operators
The  nal theorems we will present are interpolation of operator theorems which extend
sublinear operators from B( ; 1p) into weak Lp spaces to sublinear operators from B ( ; ) to
Lorentz spaces and similarly from A( ; 1p) into weak Lp to sublinear operators from A ( ; )
into Lorentz.
38
Theorem 4.4 (Interpolation of Operators for B ( ; )) If 1  p2 < p < p1 and T a
sublinear operator such that:
T : B( ; 1p
1
)!L(p1;1) and T : B( ; 1p
2
)!L(p2;1)
and T is both restricted weak type p1 with constant M1 and weak type p2 with constant M2
then for q 1 and t = p1(p2 p)p(p2 p1) :
T : B ( ; 1p)!L(p;q) with kTfkpq CMt1M1 t2   1
 
k kfkB ( ;1
p)
 
where C;k are constants with k the constant given in condition (2.3) of Class C functions.
Proof. Let T be a sublinear operator that meets the conditions stated in the theorem above
and let f be a special atom in B ( ; 1p), then there is a  -measurable set B in [  ; ] such
that
f(t) =  B(t)
 1p(B)
=
 
 1p1 (B)
 1p(B)
! 
 B(t)
 1p1 (B)
!
:
Therefore, f2B( ; 1p1 ) withkfkB( ; 1
p1 )
  1p1 1p(B). Similarly, f2B( ; 1p2 ) withkfkB( ; 1
p2 )
 
 1p2 1p(B). Also, from the conditions of the theorem we have:
(Tf) (t) =
 
T
 
 B
 1p(B)
!! 
(t) 1
 1p(B)
(T B) (t) M1 
1
p1 
1
p(B)
t 1p1
and similarly,
(Tf) (t) M2 
1
p2 
1
p(B)
t 1p2
Now for q 1 by De nition 2.14:
p
qkTfk
q
pq =
Z 1
0
 
(Tf) (t)t1p
 q dt
t :
39
We split this integral for some  2 (0;1) to be determined later. So we get the series of
inequalities:
p
qkTfk
q
pq =
Z  
0
 
(Tf) (t)t1p
 q dt
t +
Z 1
 
 
(Tf) (t)t1p
 q dt
t  
Z  
0
 
M1 1p1 1p(B)t1p
t 1p1
!q
dt
t +
Z 1
 
 
M2 1p2 1p(B)t1p
t 1p2
!q
dt
t  
Mq1
Z  
0
 
 1p1 1p(B)t1p 1p1
 q dt
t +M
q
2
Z 1
 
 
 1p2 1p(B)t1p 1p2
 q dt
t =
Mq1 qp1 qp(B)
Z  
0
tqp qp1 1dt+Mq2 qp2 qp(B)
Z 1
 
tqp qp2 1dt:
Continuing the algebra above:
= Mq1 
 q(p p
1)
pp1
 
(B) pp1q(p
1 p)
 q(p1 p)pp1 +Mq2 
 q(p p
2)
pp2
 
(B) pp2q(p
2 p)
 lim
 !1
pp2
q(p2 p)t
q(p2 p)
pp2
  
  
1
 
= Mq1 
 q(p p
1)
pp1
 
(B) pp1q(p
1 p)
 q(p1 p)pp1 +Mq2 
 q(p p
2)
pp2
 
(B) pp2q(p
2 p)
pp2
q(p2 p)
 
  q(p2 p)pp2
 
:
since p2 <p. So to summarize above inequalities up to this point
p
qkTfk
q
pq M
q
1 
 q(p p
1)
pp1
 
(B) pp1q(p
1 p)
 q(p1 p)pp1 +Mq2 
 q(p p
2)
pp2
 
(B) pp2q(p p
2)
 q(p2 p)pp2 :
Due to the fact that  was arbitrary we can now let  = C (B) where C is a constant to
be determined. We now de ne the function g(C) below in order to  nd a minimum for the
above bound:
g(C) = Mq1 
 q(p p
1)
pp1
 
(B) pp1q(p
1 p)
(C (B))q(p1 p)pp1 +Mq2 
 q(p p
2)
pp2
 
(B) pp2q(p p
2)
(C (B))q(p2 p)pp2 :
40
For simpli cation purposes let A1 = q(p p1)pp1 and A2 = pp2q(p p2) we continuing above:
p
qkTfk
q
pq g(C) = M
q
1 
 q(p p
1)+q(p1 p)
pp1
 
(B)A1Cq(p1 p)pp1 +Mq2 
 q(p p
2)+q(p2 p)
pp2
 
(B)A2Cq(p2 p)pp2
Notice that the exponents in the  (B) terms above are zero so
p
qkTfk
q
pq g(C) = M
q
1A1C
q(p1 p)
pp1 +Mq2A2C
q(p2 p)
pp2
It can be shown (see Appendix B) that C =
 
M1
M2
 p2p1
p2 p1 minimizes g(C). Thus, plugging this
value for C in above:
p
qkTfk
q
pq M
q
1A1
  
M1
M2
 p2p1
p2 p1
!q(p1 p)
pp1
+Mq2A2
  
M1
M2
 p2p1
p2 p1
!q(p2 p)
pp2
We note here that p2(p1 p)p(p1 p2) p1(p2 p)p(p1 p2) = 1 and if we let t = p1(p2 p)p(p2 p1) then 1 t = p2(p1 p)p(p1 p2). Thus
expanding the previous inequality and substituting t we have:
p
qkTfk
q
pq A1M
q
 
1+p2(p1 p)p(p2 p1)
 
1 M
q
 p
2(p p1)
p(p2 p1)
 
2 +A2M
q
 p
1(p2 p)
p(p2 p1)
 
1 M
q
 
1+p1(p p2)p(p2 p)
 
2 =
A1Mq(1+t 1)1 Mq(1 t)2 +A2Mqt1 Mq(1 t)2 :
Since q 1 we now have
kTfkqpq qp(A1 +A2)Mqt1 Mq(1 t)2 )kTfkpq 
 q
p(A1 +A2)
 1
q
Mt1M(1 t)2
Now let C =
 
q
p(A1 +A2)
 1
q thenkTfk
pq CMt1M
(1 t)
2 where C;M1;M2;t are all constants.
Now for 1 q p<1,kkpq is a true norm. Let h2B ( ; 1p), then there are real coe cients
cn such that h(t) = P1n=1 cnbIIn (t) with P1n=1  (jcnj) <1. Now utilizing the sublinearity of
41
T and the properties of norm:
kThkpq =
  
  
 T
1X
n=1
cnbIIn
  
  
 
pq
 
  
  
 
1X
n=1
cnTbIIn
  
  
 
pq
 
1X
n=1
  c
nTbIIn
  
pq =
1X
n=1
jcnj  TbIIn   pq 
1X
n=1
jcnjCMt1M(1 t)2 = CMt1M(1 t)2
1X
n=1
jcnj CMt1M(1 t)2   1
 
k 
1X
n=1
 (jcnj)
!
by Lemma 3.1. Taking the in mum above over all representations of f we have
kThkpq CMt1M(1 t)2   1
 
k kfkB ( ;1
p)
 
:
and we conclude T : B ( ; 1p)!L(p;q) as desired.2 The next theorem is the interpolation
theorem for A ( ; ).
Theorem 4.5 (Interpolation of Operators for A ( ; )) If 1  p2 < p < p1 and T a
sublinear operator such that:
T : A( ; 1p
1
)!L(p1;1) and T : A( ; 1p
2
)!L(p2;1)
and T is both restricted weak type p1 with constant M1 and weak type p2 with constant M2
then for q 1 and t = p1(p2 p)p(p2 p1) :
T : A ( ; 1p)!L(p;q) with kTfkpq CpMt1M1 t2   1
 
k kfkB ( ;1
p)
 
where Cp;k are constants with k the constant given in condition (2.3) of Class C functions.
Proof. Let T be a sublinear operator that meets the conditions stated in the theorem above
and let f be a special atom in A ( ; 1p), then there are  -measurable sets X;A;B in [  ; ]
such that ASB = X, ATB =;, and  (A) =  (B) with
f(t) = 1
 1p(X)
( A(t)  B(t))
42
Then since A( ; 1pi) B( ; 1pi) for i = 1;2 we have
kTfkpq =
  
  
 T
 
1
 1p(X)
( A(t)  B(t))
!  
  
 
pq
 
  
  
 T
 
 A(t)
 1p(X)
!  
  
 
pq
+
  
  
 T
 
 B(t)
 1p(X)
!  
  
 
pq
=
  
  
 T
 
 1p(A)
 1p(X)
  A(t)
 1p(A)
!  
  
 
pq
+
  
  
 T
 
 1p(B)
 1p(X)
  B(t)
 1p(B)
!  
  
 
pq
 
  (A)
 (X)
 1
p
  
  
 T
 
 A(t)
 1p(A)
!  
  
 
pq
+
  (B)
 (X)
 1
p
  
  
 T
 
 B(t)
 1p(B)
!  
  
 
pq
:
Now since  (X) = 2 (A) = 2 (B) by Theorem 4.4 we have:
kTfkpq 
 1
2
 1
p
CMt1M1 t2 +
 1
2
 1
p
CMt1M1 t2 = CpMt1M1 t2
where Cp = 21 1pC. The remainder of the proof is identical to the proof of Theorem 4.4.2
We now take a slight detour to discuss an interesting side result.
4.4 Multiplication Operator on L(p;1)
One interesting result of the research for this dissertation involves a special case of the
Lorentz Spaces (see De nition 2.14), with q = 1. Recall in Chapter 1 that de Souza showed
that the Lorentz space L(p;1) is equivalent to B( ; 1p) for p > 1 with equivalent norms,
see [9]. We use this fact to characterize all functions g so that the multiplication operator
Tg de ned by Tgf := g f maps L(p;1) into L(p;1) and is bounded.
We  rst review the motivation for this result. Given the linear multiplication operator
Tg de ned above, we would like to characterize all functions g so that for f 2 L(p;1),
kTgfkL(p;1)  CkfkL(p;1) where C is a constant. To this end, recall that for f2L(p;1), then
there are constants cn and  -measurable sets Bn in [  ; ] such that f(t) = P1n=1cn Bn(t)
withP1n=1jcnj 1p(Bn) <1since L(p;1) is equivalent to B( ; 1p). Thus we consider a generic
43
 -measurable set B in [  ; ]. We note  rst that for a function g we have:
 B(t)g(t) =
8
>><
>>:
g(t); if t2B
0; if t is not in B
and thus ( B(t)g(t)) =
8
>><
>>:
g (t); if t2[0; (B)]
0; otherwise
where  denotes the decreasing rearrangement of a function. As a result, for Tg B =  B g
we have:
Z  
  
(Tg B(t)) t1p 1d (t) =
Z  
  
( B(t)g(t)) t1p 1d (t) =
Z  (B)
0
(g(t)) t1p 1d (t) =
 1p(B) 1
 1p(B)
Z  (B)
0
(g(t)) t1p 1d (t):
The above equation leads to our next de nition which provides the means to solve the
characterization required.
De nition 4.1 (Xp) Given a  nite measure space ([  ; ];A; ), for B 2A, de ne the
space Xp as:
Xp =
(
g : [  ; ]!Rj 1
 1p(B)
Z  (B)
0
(g(t)) t1p 1d (t) <M
)
where M is an absolute constant. For g2Xp de ne a \norm" as
kgkXp = sup
 (B)6=0
1
 1p(B)
Z  (B)
0
(g(t)) t1p 1d (t):
Utilizing this new space we now present our theorem.
Theorem 4.6 (Characterization of Multiplication Operator on L(p;1))
For 1 < p < 1, Tg : L(p;1) ! L(p;1) de ned by Tgf := g f is bounded if and only if
g2Xp. Moreover, kTgk =kgkXp, and Xp L(p;1) with kgkL(p;1)  (2 )1pkgkXp.
44
Proof. ()) Assume Tg is bounded so there is a constant C such that for f2L(p;1),
kTgfkL(p;1)  CkfkL(p;1). Let B2A and f(t) =  B(t). Then
Z  
  
( B(t)g(t)) t1p 1d (t) C
Z  
  
( B(t)) t1p 1d (t) = C
Z  
  
 [0; (B)](t)t1p 1d (t) =
C
Z  (B)
0
t1p 1d (t) = pC 1p(B):
So Z
 
  
( B(t)g(t)) t1p 1d (t) =
Z  (B)
0
g(t) t1p 1d (t) pC 1p(B):
Thus,
1
 1p(B)
Z  (B)
0
g(t) t1p 1d (t) pC
and taking the supremum above we conclude kgkXp <pC and thus g2Xp.
(() Now let g2Xp. First consider  B(t) where B2A. Then
kTg BkL(p;1) =
Z  
  
(Tg B(t)) t1p 1d (t) =
Z  
  
( B(t)g(t)) t1p 1d (t) =
Z  (B)
0
g (t)t1p 1d (t)
=  1p(B) 1
 1p(B)
Z  (B)
0
g (t)t1p 1d (t)  1p(B) sup
 (B)6=0
1
 1p(B)
Z  (B)
0
g (t)t1p 1d (t):
We conclude
kTg BkL(p;1)  kgkXp 1p(B):
Now letf2L(p;1) then there are constantcn and setsBn inAsuch thatf(t) = P1n=1cn Bn(t)
with P1n=1jcnj 1p(Bn) <1, then
kTgfkL(p;1) =kTg
1X
n=1
cn Bn(t)kL(p;1)  
1X
n=1
jcnjkTg Bn(t)kL(p;1)  
1X
n=1
jcnjkgkXp 1p(B) =
kgkXp
1X
n=1
jcnj 1p(B) kgkXpkfkB( ;1
p)
45
where we obtained the last inequality above by taking the in mum over all representations of
f in B( ; 1p). Thus since B( ; 1p) is equivalent to L(p;1) with equivalent norms we conclude
that Tg is bounded. Note that since  [  ; ](t) 2 L(p;1) and k [  ; ](t)kL(p;1) = (2 )1p,
then b(t) = 1
(2 ) 1p
 [  ; ](t) 2 L(p;1) with kbkL(p;1) = 1. Thus for g 2 Xp, Tg [  ; ] =
 [  ; ](t) g(t) = g(t) 2 L(p;1) and Xp  L(p;1). If we take f(t) = 1 on [  ; ] then
kfkL(p;1) = (2 )1p and kgfkL(p;1) = kgkL(p;1)  (2 )1pkgkXp. Thus, Xp  L(p;1) with
kgkL(p;1)  (2 )1pkgkXp2.
46
Chapter 5
Comments on the dual of A ( ; )
This chapter examines the possible dual of A ( ; ). As we have seen in Theorem 4.3,
Lip( ; ) is equivalent to the dual of B ( ; ). While we have not proved that the dual of
A ( ; ) is  ( ; ), we will show that this is most likely the case due to several factors. We
start by formalizing the relationship between Lip( ; ) and  ( ; ).
5.1 Relationship between Lip( ; ) and  ( ; )
The following theorem demonstrates the connection between Lip( ; ) and  ( ; ) and
is due to de Souza and Pozo, see [6].
Theorem 5.1 (Equivalence of Lip( ; ) and  ( ; )) For  2(0;1) the spaces Lip( ; )
and  ( ; ) are equivalent as Banach Spaces. That is, there are positive constants M;N such
that
MkgkLip( ; ) <kgk ( ; ) <NkgkLip( ; ):
In order to prove Theorem 5.1 we provide another Theorem also by de Souza and Pozo,
see [6].
Theorem 5.2 (de Souza and Pozo) Let ( ;A; ) be a  nite positive, nontrivial measure
space and  2(0;1) with f :  !R such that f2L1( ;A; ), then
sup
C;D2A; (C4D)6=0
  R
Cf(t)d (t) 
R
Df(t)d (t)
  
  (C4D)
 = sup
A2A; (A)6=0
  R
Af(t)d (t)
  
  (A)
 = sup
A2A; (A)6=0
R
Ajf(t)jd (t)
  (A) :
We now prove Theorem 5.1. Although one can use Theorem 5.2 for the complete proof, we
will provide the argument for one side of the inequality in Theorem 5.1.
47
Proof. Let g 2 Lip( ; ) and A;B;X be  -measurable sets such that ATB = ; and
X = ASB. Consider then:
1
  (X)
  
  
Z
A
g(t)d (t) 
Z
B
g(t)d (t)
  
   1
  (X)
   
  
Z
A
g(t)d (t)
  
  +
  
  
Z
B
g(t)d (t)
  
  
 
=
  (A)
  (X)
 1
  (A)
  
  
Z
A
g(t)d (t)
  
  
 
+  
 (B)
  (X)
 1
  (B)
  
  
Z
B
g(t)d (t)
  
  
 
 2kgkLip( ; ):
Taking the sup of the above inequality over all  -measurable sets A;B;X such that ATB =
; and X = ASB gives the result
kgk ( ; ) < 2kgkLip( ; ):
For the other side of the inequality, apply Theorem 5.2 with CTD =;.2
5.2 Closing Arguments
Given the H older?s-type inequality for A ( ; ) in Theorem 4.2, the dual of A ( ; )
is in fact  ( ; ). In order to see that we need an estimate for k AkA( ; ) where A is a
 -measurable set in [  ; ]. Indeed using Theorem 1 by F.F. Bonsall along with the Closed
Range Theorem for quasi Banach spaces one can show that k AkA( ; )  M  (A), see [37].
Thus, along with the fact that  ( ; ) and Lip( ; ) are equivalent as Banach spaces, it
seems reasonable that the spaces A ( ; ) and B ( ; ) are equivalent (recall the dual of
B ( ; ) is Lip( ; )). Recall also that de Souza showed that the spaces A( ; ) and B( ; )
are equivalent so the extension to A ( ; ) and B ( ; ) as equivalent spaces is logical.
48
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51
Appendices
52
Appendix A
Vector Space proof
This appendix provides the remainder of the proof of Theorem 3.1 which states A ( ; )
and B ( ; ) are vector spaces. We will provide the proof for B ( ; ). The proof in the
A ( ; ) is analogous. Closure under addition was proved following the statement of Theo-
rem 3.1.
Proof. Let u;v;w 2 B ( ; ) and a;b be real scalars. Then by de nition, u;v;w
are all linear combinations of weighted general atoms of Type II. Thus, associative and
commutative properties are true by properties of summations. Also, the identity element for
addition is zero which is in B ( ; ) by setting all coe cients to zero. Clearly av2B ( ; ).
Since v2B ( ; ), there are real coe cients cvn such that
v =
1X
n=1
cvnbIIvn(t);
1X
n=1
 (jcvnj) <1:
Hence,
av =
1X
n=1
acvnbIIvn(t) and
1X
n=1
 (jacvnj) 
1X
n=1
  (jaj) (jcvnj) =   (jaj)
1X
n=1
 (jcvnj) <1
by property (2.2) of C functions and therefore av2B ( ; ). The distributive properties
are clear due to properties of summations: for any scalars a;b, we have a(v +w) = av +aw
and (a + b)v = av + bv since we are dealing with the summation operator which is linear.
Now the identity element for scalar multiplication is 1. Again, it is clear that 12B ( ; ),
since
 ([  ; ])bII(t)1 =  ([  ; ]) ([  ; ]) = 1:
Finally, we show that for v2B ( ; ), there exists a w2B ( ; ) such that v+w = 0. Let
v2B ( ; ) then similar to above we have real coe cients cvn such that
v =
1X
n=1
cvnbIIvn(t);
1X
n=1
 (jcvnj) <1:
Now, let
w =
1X
n=1
 cvnbIIvn(t)
53
then w2B ( ; ) and
v +w =
1X
n=1
cvnbIIvn(t) +
1X
n=1
 cvnbIIvn(t) =
1X
n=1
(cvn cvn)bIIvn(t) = 0:
Therefore, we conclude that B ( ; ) is indeed a vector space over R.2
54
Appendix B
Veri cation of minimum in proof of Theorem 4.4
This work will prove the assertion in the proof of Theorem 4.4 that C =
 
M1
M2
 p1p2
p2 p1 is
a minimum of g(C). Recall that
g(C) = Mq1A1Cq(p1 p)pp1 +Mq2A2Cq(p2 p)pp2
with
A1 = pp1q(p
1 p)
and A2 = pp2q(p p
2)
:
Note that g(C) can be re-written as
g(C) = Mq1A1C 1A1 +Mq2A2C 1A2:
Taking the derivative of g(C)
g0(C) = Mq1A1 1A
1
C 1A1 1 +Mq2A2
 
 1A
2
 
C 1A2 1:
Setting g0(C) = 0 leads to
Mq1C 1A1 1 = Mq2C 1A2 1 )
 M
1
M2
 q
= C 1A2 1C1 1A1 = C 1A2 1A1:
Consider  1A2  1A1 :
 1A
2
 1A
1
= q(p2 p)pp
2
+ q(p p1)pp
1
= q(p1(p2 p) +p2(p p1))pp
1p2
= q(p2 p1)p
1p2
:
Substituting above  
M1
M2
 q
= Cq(p2 p1)p1p2 )C =
 M
1
M2
 p1p2
p2 p1
:
It remains to show that this value for C is indeed a minimum. We now take the second
derivative of g(C):
g00(C) = Mq1
 1
A1  1
 
C 1A1 2 Mq2
 
 1A
2
 1
 
C 1A2 2:
55
Now if g00(C) > 0 at C =
 
M1
M2
 p1p2
p2 p1 as desired we would have
Mq1
 1
A1  1
 
C 1A1 2 >Mq2
 
 1A
2
 1
 
C 1A2 2 )
 M
1
M2
 q 1
A1  1
 
>
 
 1A
2
 1
 
C 1A2 2+2 1A1 =
 
 1A
2
 1
 
C 1A2 1A1 =
 
 1A
2
 1
  M
1
M2
 q
:
So we must show 1A1  1 > 1A2  1 or 1A1 > 1A2 . If we show this inequality then one can
work backwards through the above inequalities showing g00(C) > 0 for C =
 
M1
M2
 p1p2
p2 p1 which
guarantees this value of C is the minimum of g(C). Now p1 pp1 > p2 pp2 since p2 <p<p1, which
implies that q(p1 p)pp1 > q(p2 p)pp2 which is equivalent to 1A1 > 1A2 . Therefore C =
 
M1
M2
 p1p2
p2 p1 is
the minimum of g(C).2
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