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

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) 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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