Abstract
Title of thesis: PRECISE ESTIMATES FOR WEIGHT
FUNCTIONS SATISFYING A WEIGHTED
FOURIER TRANSFORM INEQUALITY
Hatshepsitu S. H. Tull, Master of Science, 2007
Thesis directed by: Professor Raymond L. Johnson
Department of Mathematics
This paper shows that for a given weighted Fourier transform inequality,
certain weight functions will satisfy it. The work done in my paper is a
continuation of similar ideas found in Yuki Yayama?s thesis. She proved
that a nonessentially increasing weight function w with a finite number of
zeros can satisfy a given weighted Fourier transform inequality. Her proof
includes estimations of distribution functions, the sine and the arcsine
functions both near zero. My paper provides another proof by using
precise values of distribution functions certain approximations used
only when necessary.
1
PRECISE ESTIMATES FOR WEIGHT FUNCTIONS
SATISFYING A WEIGHTED FOURIER TRANSFORM
INEQUALITY
by
Hatshepsitu S. H. Tull
Thesis submitted to the Faculty of the Graduate School of the
University of Maryland, College Park, in partial fulfillment
of the requirements for the degree of
Master of Science
2007
Advisory Committee:
Professor Raymond L. Johnson, Chair
Professor John J. Benedetto
Professor Manoussos G. Grillakis
1
Acknowledgements
I would like to thank everyone who has supported me throughout my grad-
uate endeavor. Several faculty members at Spelman College, Morehouse
College and the University of Maryland, College Park provided me with in-
valuable wisdom and insight. I am particularly grateful to my family and
friends for their support and encouragement. I am honored and fortunate
to have such a tremendous and diverse network of people who have always
believed in me and in my goal of successfully completing my graduate degree.
ii
Table of Contents
Acknowledgements....................................................ii
Table of Contents.....................................................iii
List of Figures.........................................................iv
1 Introduction ......................................................... 1
1.1 Discussion of theorem for essentially increasing weight functions w ... 1
1.2 Essentially increasing weight functions w with a finite number of zeros 1
2 Definitions ...........................................................2
2.1 Distribution function ................................................ 2
2.2 Nonincreasing rearrangement ........................................ 2
2.3 Weight class F?2,2 .....................................................2
3 Main Problem ...................................................... 3
3.1 The case for a weight function w1 with one zero for 0 < x <? .......3
3.2 The case for a weight function wn with n zeros for 0 < x <? ...... 11
4 Summary ...........................................................14
References ............................................................16
iii
List of Figures
Figure 1 ............................................................... 15
Figure 2 ............................................................... 15
Figure 3 ............................................................... 15
iv
1 Introduction
1.1 Discussion of theorem for essentially increasing, even weight
functions w
The weighted Fourier transform inequality studied in this paper is:
integraldisplay ?
??
|hatwidef(x)|2w
parenleftbigg1
x
parenrightbigg
dx?C
integraldisplay ?
??
|f(x)|2w(x)dx, (1)
wherew is a non-negative weight function, C is a constant, and hatwidef the Fourier
transform defined by
hatwidef(x) =
integraldisplay ?
??
e??xyf(y)dy.
The conditions of weight functions that satisfy (1) have been discussed in
[1] and [2]. It was shown in [2] that if w(x) is an even weight function,
nondecreasing on (0,?), we have (1) if and only if w(x) ?A2, where A2 is a
Muckenhoupt weight class and consists of all non-negative locally integrable
functions w(x), such that for all intervals (a,b),
parenleftbiggintegraldisplay b
a
w(x)dx
parenrightbigg1/2parenleftbiggintegraldisplay b
a
w(x)?1dx
parenrightbigg1/2
?C(b?a) (2)
holds. This theorem is still valid if one replacesw(x) by even weight functions
which are essentially increasing on (0,?) and have no zeros. This was shown
in Yayama [1]. A weight w(x) is an essentially increasing function on (0,?)
if there exists an increasing functionU(x) and positive constantsC1,C2, such
that C1 ? w(x)U(x) ? C2. In her thesis, Yayama proved that a weight function
w(x) with n zeros which is not essentially increasing can satisfy (1). She
proved this by using estimations of the distribution function and also using
the fact that sint?t and arcsint?t for t? 0. This paper will discuss the
computation of the precise value of the distribution function and make the
approximations only when necessary, thereby providing another proof of the
result.
1.2 Essentially increasing weight functions w with a finite num-
ber of zeros
We show that ifw is essentially increasing and there exists at least onez such
that w(z) = 0, the weight w cannot satisfy (1). We suppose that w(x) is an
essentially increasing function with w(z) = 0. Then there is an increasing
function U(x) and positive constants C1,C2, such that C1 ? w(x)U(x) ?C2.
1
Then U(x) = 0 for x?z and so w(x) = 0 for x?z. If we also suppose that
w belongs to A2, then the inequality (1) must hold. If we choose
f(x) =
?
??
??
1 |x| ? z2
0 |x|> z2,
then integraldisplay ?
??
|f(x)|2w(x)dx = 0
Since |hatwidef(x)|2 > 0 on 12z ?x? 1z, we have
integraldisplay ?
??
|hatwidef(x)|2w
parenleftbigg1
x
parenrightbigg
dx> 0.
Hence, w(x) cannot satisfy (1) [1].
2 Definitions
Consider the following definitions from [2] and [3].
Definition 2.1
Let E be any subset of R. If a function f belongs to Lp(E), the distribution
function Df of a function f is defined by Df(s) = m{x ? E : |f(x)| > s},
where m is the Lebesgue measure.
Definition 2.2
The nonincreasing rearrangementf? of a measurable function f is defined on
a measure space by f?(t)=inf{s : Df(s) ? t}, where Df is the distribution
function defined above.
Definition 2.3
The weight class F?2,2 , is the collection of all pairs of non-negative, locally
integrable functions (u,v) on R such that
sup
s>0
parenleftbiggintegraldisplay 1/s
0
u?(t)dt
parenrightbigg1/2parenleftbiggintegraldisplay s
0
parenleftbigg1
v
parenrightbigg?
(t)dt
parenrightbigg1/2
<?. (3)
We write (u,v) ? F?2,2. Note that if (u,v) ? F?2,2, then u satisfies inequality
(1). Here, we interpret ??0 to be 0.
2
It is shown in [3] that, if (u,v)?F?2,2, then
integraldisplay ?
??
|hatwidef(x)|2u(x)dx?C
integraldisplay ?
??
|f(x)|2v(x)dx (4)
holds for some constant C, for every f(x) such that the right hand side is
finite.
Remarks
1. Both Df and f? are nonincreasing functions on the positive real axis.
2. Clearly, if f is a nonincreasing function, f = Df.
3 Main Problem
We will examine even, nonessentially increasing weight functions with a finite
number of zeros by giving concrete examples, and then determine if they can
satisfy (1).
3.1 The case for a weight function w1 with one zero for 0 <x<?
First, consider the case for n = 1 :
w1(x) =
braceleftBigg |sinx|a |x| ? 3pi
21 |x|> 3pi
2
,
where w1 has two zeros, at x = 0 and x = pi. Then,
?w1(x) = w1
parenleftbigg1
x
parenrightbigg
=
braceleftBigg |sin 1
x|
a |x| ? 2
3pi1 |x|< 2
3pi
,
and
1
w1(x) =
braceleftBigg 1
|sinx|a |x| ?
3pi
2
1 |x|> 3pi2 .
The graphs of w1(x), ?w1(x) and 1w1(x) are labeled Figure 1, Figure 2 and
Figure 3 respectively in the List of Figures. Because w1, 1w1 and ?w1 are all
even functions, then only consider the positive real axis for each of these
functions. If A(x) ?B(x), then the following is true:
DA(s) ?DB(s) and A?(t) ?B?(t). (5)
3
The distribution function of 1w1(x) is
D 1
w1
(s) = 2
bracketleftbigg
m
parenleftbigg
0 ?x? 3pi2 : 1|sinx|a >s
parenrightbigg
+m
parenleftbigg
x> 3pi2 : 1 >s
parenrightbiggbracketrightbigg
?p1(s),
where
p1(s) =
?
?
?
6arcsin
parenleftbigg
1
s1/a
parenrightbigg
, s? 1
? 0 ?s< 1.
By definition of the nonincreasing rearrangement function,
parenleftbigg 1
w1
parenrightbigg?
(t) = inf
braceleftbigg
s : D 1
w1
(s) ?t
bracerightbigg
.
Since D 1
w1
(s) ?p1(s), then by (5)
inf
braceleftbigg
s : D 1
w1
(s) ?t
bracerightbigg
? inf{s : p1(s) ?t}. (6)
Therefore, parenleftbigg
1
w1
parenrightbigg?
(t) ? inf{s : p1(s) ?t}.
Recall that if s ? 1, then s1/a ? 1 and 1s1/a ? 1 also. This implies that
arcsin
parenleftbigg
1
s1/a
parenrightbigg
approaches arcsin(1) = pi2. Moreover, as s ? 1, then p1(s) =
6arcsin
parenleftbigg
1
s1/a
parenrightbigg
?? 6
parenleftbigg
pi
2
parenrightbigg
= 3pi. Consider
6arcsin
parenleftbigg 1
s1/a
parenrightbigg
= t
arcsin
parenleftbigg 1
s1/a
parenrightbigg
= t6
1
s1/a = sin
parenleftbiggt
6
parenrightbigg
s1/a = csc
parenleftbiggt
6
parenrightbigg
s =
parenleftbigg
csc t6
parenrightbigga
.
Hence, parenleftbigg
1
w1
parenrightbigg?
(t) ?P?1 (t)
4
where
P?1 (t) =
?
?
?
parenleftbigg
csc t6
parenrightbigga
, 0 ?t< 3pi
1, t? 3pi
The distribution function of ?w1(x) is
D?w1(s) = m{x : ?w1(x) >s}.
To simplify D?w1(s), consider the following background calculations. Let
|sin 1x|a = s. For x> 0, then we have that
parenleftbigg
sin 1x
parenrightbigga
= s
sin 1x = s1/a.
The points x1,x2, and x3 where
parenleftbigg
sin 1x
parenrightbigga
= s occur when
2
3pi ?
1
x1 ?
1
pi,
1
pi ?
1
x2 ?
2
pi,
2
pi ?
1
x3 ? 0.
That is, when
pi ?x1 ? 3pi2 , pi2 ?x2 ?pi, 0 ?x3 ? pi2.
After further calculations,
x1 = 1pi+ arcsins1/a, x2 = 1pi?arcsins1/a, x3 = 1arcsins1/a.
For ease of calculations, let T = arcsins1/a. From these values, we get that
D?w1(s) = 2
parenleftBigg
(x1 ?0) + (x3 ?x2)
parenrightBigg
= 2
parenleftBigg 1
pi+T +
1
T ?
1
pi?T
parenrightBigg
= 2
parenleftBiggT(pi?T) + (pi2 ?T2)?T(pi+T)
T(pi2 ?T2)
parenrightBigg
5
= 2
parenleftBiggpiT ?T2 +pi2 ?T2 ?Tpi?T2
T(pi2 ?T2)
parenrightBigg
= 2
parenleftBigg pi2 ?3T2
T(pi2 ?T2)
parenrightBigg
.
At this point,
D?w1(s) =
?
????
???
???
????
0, s> 1 (7)
4
3pi, s = 1 (8)
2(pi2?3T2)
T(pi2?T2), 0 <s< 1 (9)
The value for (7) is due to the fact that |sinz| ? 1. Now we will approximate
(9) for s close to 0. Note that as s? 1, then s1/a ? 1 and T = arcsins1/a ?
pi
2. Let q(T) =
2(pi2?3T2)
T(pi2?T2). Hence,
q(T) ? 2(pi
2 ?3(pi
2)
2)
pi
2(pi
2 ?(pi
2)
2) =
2(14)pi2
pi
2(
3
4)pi
2 =
4
3pi.
As s ? 0+, then s1/a ? 0+, T ? 0, q(T) ? 2T ? +?. For t ? 0, then by
Taylor series we have for f(t) = arcsint, fprime(t) = 1?1?t2 and fprime(0) = 1. The
Taylor expansion is
f(t) = f(0) +fprime(0)t+fprimeprime(0)t
2
2! +... = t+...,
and hence arcsint?t. Then for 0 <s<?, then T = arcsins1/a ?s1/a. Now
the expression for q(T) can be simplified to
q(T) = 2T
parenleftBiggpi2 ?3T2
pi2 ?T2
parenrightBigg
? 2T
because
parenleftBigg
pi2?3T2
pi2?T2
parenrightBigg
? 1. Hence, T ? s1/a which implies that q(T) ? 2s1/a for
0 <s< 1. Since D?w1(s), then q(T) is decreasing.
We can now rewrite D?w1(s) as D?w1(s) ?Q1(s) where
Q1(s) =
??
???
???
???
????
0, s> 1 (10)
4
3pi, s = 1 (11)
2
s1/a, 0 <s< 1 (12)
6
Now we can write the nonincreasing rearrangements ?w?1(t) and Q?1(t) as
?w?1(t) = inf{s|D?w1(s) ?t} and Q?1(t) = inf{s|Q1(s) ?t}
and by (6) we have that ?w?1(t) ?Q?1(t) where
Q?1(t) =
braceleftBigg 1, 0 ?t? 2
(2t)a, 2 <t<?.
Now consider
(
integraldisplay 1/s
0
?w?1(t)dt)1/2(
integraldisplay s
0
(1/w1)?(t)dt)1/2,
where 0 < s. The supremum of this product must be bounded in order
to have (?w1,w1) ? F?2,2 and moreover that ?w1 satisfies inequality (1). To
demonstrate that this is indeed true, we only need to show that
(
integraldisplay 1/s
0
?w?1(t)dt)1/2(
integraldisplay s
0
(1/w1)?(t)dt)1/2
is bounded. Let
integraldisplay 1/s
0
?w?1(t) = C1 and
integraldisplay s
0
parenleftbigg 1
w1
parenrightbigg?
(t) = D1.
Now compute C1D1 for 0 <s and determine if C1D1 is bounded. Since
?w?1(t) ?Q?1(t) and ( 1w1)?(t) ?P?1 (t), then
C1D1 ?
parenleftbiggintegraldisplay 1/s
0
Q?1(t)dt
parenrightbiggparenleftbiggintegraldisplay s
0
P?1 (t)dt
parenrightbigg
.
Because of the definition of (?w1,w1) belonging to F?2,2, it is enough to show
that for any s> 0,
parenleftbiggintegraldisplay 1/s
0
Q?1(t)dt
parenrightbiggparenleftbiggintegraldisplay s
0
P?1 (t)dt
parenrightbigg
is bounded for all fixed a where 0 <a< 1.
Based on Q?1 and P?1 , there are four cases of s to consider. Further work
will illustrate that these four cases can be combined into three cases.
Case 1 : 1s ? 2 or s? 12. Then
integraldisplay 1/s
0
Q?1(t)dt =
integraldisplay 1/s
0
1dt = 1s.
7
Case 2 : 1s > 2 or s< 12. Then
integraldisplay 1/s
0
Q?1(t)dt = integraltext20 Q?1(t)dt+integraltext1/s2 Q?1(t)dt
= integraltext20 1dt+integraltext1/s2 (2s)adt
= 2 + 2a(integraltext1/s2 t?adt)
= 2 + 2a
bracketleftbigg
1
?a+1t
?a+1
bracketrightbigg1
s
2
= 2 + 2a
parenleftbigg
t1?a
1?a
parenrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
1
s
2
= 2 + 2a
parenleftbigg
(1s)1?a?21?a
1?a
parenrightbigg
= 2(1?a)1?a + 2asa?1?21?a
= 2asa?1?2a1?a
Case 3 : s< 3pi. Then
integraldisplay s
0
P?1 (t)dt =
integraldisplay s
0
parenleftbigg
csc t6
parenrightbigga
dt.
In general, since one cannot exactly evaluate integraltext(cscx)adx where 0 < a < 1
in closed form, then we must perform the following calculations to obtain an
estimate for integraltext(cscx)adx.
Consider the function ?(x) = x?tanx, where ?(0) = 0. Then
?prime(x) = 1?sec2x< 0 for x> 0.
Consider another function ?(x) = tanx?x, where ?(0) = 0. Then
?prime(x) = sec2x?1 > 0.
Note also that ?(x) ? 0 for all x> 0.
If 0 ? x ? pi2, then x ? tanx. Now consider the function F(x) = sinxx .
Then
Fprime(x) = cosx[x?tanx]x2 .
8
On the interval
parenleftbigg
0, pi2
parenrightbigg
, Fprime(x) < 0. Hence, for 0 <x< pi2, then
2
pi ?
sinx
x ? 1,
but letting x? pi2 or x? 0, then the ineqality is true for 0 ?x? pi2.
Now we have all the information needed to determine a lower and an up-
per bound for integraltexts0 (csc(t6))adt. Recall that for 0 <x? pi2, we have
2
pi <
sinx
x ? 1,
1
x <
1
sinx ?
pi
2x,
1
x < cscx?
pi
2x.
For 0 <t? 3pi or 0 ? t6 ? pi2,
6
t ? csc
t
6 ?
6pi
2t,
parenleftbigg6
t
parenrightbigga
?
parenleftbigg
csc t6
parenrightbigga
?
parenleftbigg6pi
2t
parenrightbigga
.
Now integraldisplay s
0
parenleftbigg6
t
parenrightbigga
dt?
integraldisplay s
0
parenleftbigg
csc t6
parenrightbigga
dt?
integraldisplay s
0
parenleftbigg3pi
t
parenrightbigga
dt,
6as1?a
1?a ?
integraldisplay s
0
parenleftbigg
csc t6
parenrightbigga
dt? (3pi)
as1?a
1?a ,
that is for s? 3pi,
6as1?a
1?a ?
integraldisplay s
0
P?1 (t)dt? (3pi)
as1?a
1?a .
9
Case 4 : s? 3pi. Then
integraldisplay s
0
P?1 (t)dt = integraltext3pi0 P?1 (t)dt+integraltexts3piP?1 (t)dt
= integraltext3pi0
parenleftbigg
csc t6
parenrightbigga
dt+integraltexts3pi 1dt
From the previous four cases, one can conclude that there are actually three
cases (I, II and III) of s to consider. In each of these cases, let
parenleftbiggintegraldisplay 1/s
0
Q?1(t)dt
parenrightbiggparenleftbiggintegraldisplay s
0
P?1 (t)dt
parenrightbigg
= H.
We will disregard the lower bounds for Cases I, II and III because the integrals
in each of these cases are positive.
Case I : 0 <s< 12. Then
H ?
parenleftbigg
2asa?1?2a
1?a
parenrightbiggparenleftbigg
(3pi)as1?a
1?a
parenrightbigg
= 2a(3pi)a?2a(3pi)as1?a(1?a)2
? 2a(3pi)a(1?a)2
= (6pi)a(1?a)2
= Z1
Recall that C1D1 ? H. Since 0 < a < 1, then H ? Z1 where Z1 ? R+ and
hence H is bounded above.
Case II : 12 ?s< 3pi. Here,
H ? 1s
parenleftbigg
(3pi)as1?a
1?a
parenrightbigg
= (3pi)a(1?a)sa
? (3pi)a(1?a)(1
2)
a
= (6pi)a(1?a)
= Z2
10
Recall that C1D1 ? H. Since 0 < a < 1, then H ? Z2 where Z2 ? R+ and
hence H is bounded above.
Case III : s? 3pi.
H ? 1s
parenleftbigg
(3pi)a(3pi)1?a
1?a +s?3pi
parenrightbigg
= 1s
parenleftbigg
(3pi)a(3pi)(3pi)?a
1?a +s?3pi
parenrightbigg
= 1s
parenleftbigg
3pi
1?a +s?3pi
parenrightbigg
= 3pi(1?a)s + 1? 3pis
? 3pi(1?a)s + 1
? 3pi(1?a)(3pi) + 1
= 11?a + 1
= Z3
Recall that C1D1 ? H. Since 0 < a < 1, then H ? Z3 where Z3 ? R+
and hence H is bounded above. Now, C1D1 ? max(Z1,Z2,Z3). This means
C1D1 is bounded above which implies that ( ?w1,w1) ?F?2,2. Thus, ?w1 satisfies
inequality (1).
3.2 The case for a weight function wn with n zeros for 0 <x<?
Consider the case for any natural number n where
wn(x) =
braceleftBigg |sinx|a |x| ? (2n+1)pi
2
1 |x|> (2n+1)pi2
Then,
?wn(x) = wn
parenleftbigg1
x
parenrightbigg
=
braceleftBigg |sin 1
x|
a |x| ? 2
(2n+1)pi
1 |x|< 2(2n+1)pi,
and
1
wn(x) =
??
?
1
|sinx|a |x| ?
(2n+1)pi
2
1 |x|> (2n+1)pi2 .
11
As in the case for n = 1, only consider the positive real axis for each function
since they are all even. Recall also from this case that if A(x) ?B(x), then
the following is true:
DA(s) ?DB(s) and A?(t) ?B?(t).
We will use similar reasoning to conclude that (?wn,wn) belongs to F?2,2 for
every natural number n and hence ?wn satisfies inequality (1).
The distribution function of 1wn(x) is
D1/wn(s) = 2
bracketleftbigg
m
parenleftbigg
0 ?x? (2n+ 1)pi2 : 1|sinx|a >s
parenrightbigg
+m
parenleftbigg
x> (2n+ 1)pi2 : 1 >s
parenrightbiggbracketrightbigg
?pn(s),
where
pn(s) =
??
?
2(2n+ 1)arcsin
parenleftbigg
1
s1/a
parenrightbigg
, s? 1
? 0 ?s< 1.
By definition of the nonincreasing rearrangement function,
parenleftbigg 1
wn
parenrightbigg?
(t) = inf
braceleftbigg
s : D 1
wn
(s) ?t
bracerightbigg
.
Using similar reasoning as in the case n = 2,
parenleftbigg 1
wn
parenrightbigg?
(t) ? inf{s : pn(s) ?t}.
Recall that if s? 1, 1s1/a ? 1, arcsin
parenleftbigg
1
s1/a
parenrightbigg
approaches pi2 and
pn(s) = 2(2n+ 1)arcsin
parenleftbigg 1
s1/a
parenrightbigg
?? 2(2n+ 1)
parenleftbiggpi
2
parenrightbigg
= (2n+ 1)pi.
Consider
2(2n+ 1)arcsin
parenleftbigg 1
s1/a
parenrightbigg
= t
arcsin
parenleftbigg 1
s1/a
parenrightbigg
= t2(2n+ 1)
1
s1/a = sin
parenleftbigg t
2(2n+ 1)
parenrightbigg
12
s1/a = csc
parenleftbigg t
2(2n+ 1)
parenrightbigg
s =
parenleftbigg
csc t2(2n+ 1)
parenrightbigga
.
Hence, parenleftbigg
1
wn
parenrightbigg?
(t) ?P?n(t)
where
P?n(t) =
?
?
?
parenleftbigg
csc t2(2n+1)
parenrightbigga
, 0 ?t< (2n+ 1)pi
1, t? (2n+ 1)pi
As n increases, the domain of wn(1x) decreases. Hence, wn(1x) ?w1(1x) for all
natural numbers n. Moreover by (5),
?w?n(t) ? ?w?1(t).
Since ?w?1(t) ?Q?1(t), then ?w?n(t) ?Q?1(t). Let
integraldisplay 1/s
0
?w?n(t) = Cn and
integraldisplay s
0
parenleftbigg 1
wn
parenrightbigg?
(t) = Dn.
Then we have
CnDn ?
parenleftbiggintegraldisplay 1/s
0
Q?1(t)dt
parenrightbiggparenleftbiggintegraldisplay s
0
P?n(t)dt
parenrightbigg
.
Using analogous computations as those found in Section 3.1 show that
(?wn,wn) ?F?2,2 and thus, ?wn satisfies inequality (1).
13
4 Summary
Yayama?s thesis used estimations of the distribution function and approxi-
mations for sint and arcsint to prove that a nonessentially increasing weight
function w(x) with n zeros can satisfy the inqequality (1). This paper pro-
vides more precise estimates of the distribution function and uses approxi-
mations as a last step.
14
 
-6 -4 -2 0 2 4 6 
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
x
w
1
(x
)
Figure 1
 
List of Figures 
 
-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
x
ti
l
d
e
(
w
1
(x
))
Figure 2
 
-6 -4 -2 0 2 4 6
0
1
2
3
4
5
6
7
8
9
10
x
1/
w
1
(x
)
Figure 3
 
15  
References
[1] Yuki Yayama, Examples Of The Weight Functions Which Are Not
Essentially Increasing But Satisfy The Weighted Fourier Transform
Inequality, 1-22.
[2] J. J. Benedetto, H. P. Heinig and R. L. Johnson, Fourier inequalities
with Ap weights, General Inequalities V, ed. W. Walter, Birkhauser
Verlag, Basel, 217-231.
[3] Elias M. Stein and Guido Weiss, Introduction to Fourier Analysis on
Euclidean Spaces, 57 and 188-192.
[4] J. J. Benedetto, H. P. Heinig and R. L. Johnson, Weighted Hardy Spaces
and the Laplace Transform II, Math.Nachr. 132 (1987), 29-55.
16