TRANSACTIONS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 259, Number 1, May 1980
ON GENERALIZED HARMONIC ANALYSIS
BY
KA-SING LAUI AND JONATHAN K. LEE2
Motivatedby Wiener'swork on generalizedharmonicanalysis, we
considerthe Marcinkiewiczspace 6XP(R)of functionsof boundedupperaveragep
power and the space St(R) of functions of bounded upper p variation. By
identifyingfunctionswhose differencehas normzero, we show that St(R), 1 <p
< oo, is a Banach space. The proof dependson the resultthat each equivalence
class in cVP(R)
containsa representativein LP(R).This result,in turn,is based on
Masani'sworkon helixesin Banachspaces.
Wiener defined an integrated Fourier transformationand proved that this
transformationis an isometry from the nonlinear subspace '5liV(R)of %R2(R)
consistingof functionsof bounded averagequadraticpower, into the nonlinear
subspaceGW(R)of V(IR) consistingof functionsof boundedquadraticvariation.
By usingtwo generalizedTauberiantheorems,we provethat Wiener'stransformation W is actually an isomorphismfrom .)2(R) onto 'V2(R).We also show by
counterexamplesthat W is not an isometryon the closed subspacegeneratedby
6V2(R).
ABsTRAcT.
1. Introduction.The purpose of this paper is to find out how Wiener's generalized
harmonic analysis [18] fits into the framework of contemporary functional analysis.
For a complex valued Borel measurable function f on R such that
limT. oo(2 T)-f T'If(x)12 dx exists, Wiener [18] defined the integrated Fourier
transformation g = W(f) of f as
g(u)=27
(f-x +
f(x)e.
dx+ff(x)
-ix
dx)
We call W the Wiener transformation.By using a deep Tauberian theorem, he then
proved that the mean square modulus of the above function f equals the quadratic
variation of its transformation g, i.e.
lim
T--),oo
1f|TT
If(x)12dx
=
rMn
h--)o
? | g(u + h)-g(u-h)I2du.
2h J__
(1.1)
Now, for allf E L 2 (R),let
Ilfi
=
Ilf
2=
liM
21
T
f(X)12 d)/
(1.2)
Receivedby the editorsNovember10, 1978.
AMS (MOS) subjectclassifications
(1970).Primary42A68; Secondary42A32,46B99.
Key wordsandphrases.Banachspaces,generalizedharmonicanalysis,helixes,Marcinkiewiczspaces,
Wienertransformation.
Tauberiantheorem,upperp-variation,
'Partiallysupportedby the Facultyof Artsand Scienceresearchgrantof the Universityof Pittsburgh.
2Partiallysupportedby NSF GrantsGP7808and GP7809.
( 1980 American Mathematical Society
0002-9947/80/0000-0204/$06.75
75
76
K.-S. LAU AND J. K. LEE
and
9V3(R) = {ff E- L2(R), Ilfil < o0).
Note that the set of functions VlS(R)for which the limit on the left-handside of
(1.1) exists and is finite is a nonlinearsubspaceof the linearspace %2(R).
Next, for all Borelmeasurableg on R, let
IIgll=
= lim (2
gIINII
I
JI
~~~~~~~~~1/2
00
g(u + h) - g(u
-
h)12du
(1.3)
and
= { g: g is BorelmeasurableandlIgII < oo.
2212(R)
Then the set of functionsfor which the limit on the right-handside of (1.1) exists
and is finite is a nonlinearsubspaceof the linearspace 21(R).
Similarly,we can define the classesof functions91P(R)and 'V\(R).Both 91P(R)
and SVt(R)are normedlinear spaces when two functionsin any one of the spaces
whose differenceshave norm zero are identified.Marcinkiewicz[13] and independently Bohrand F0lner [3] showedthat 6XP(R)is complete,but the questionof the
has been open.
completenessof cV?(R)
In ?3, we show that ?I(R) is completefor 1 <p < oo. For this, we find that all
usual methodsof provingcompleteness(cf. e.g. [5], [12])fail. We have to appealto
the theory of helixes in a Banachspace X, i.e. continuousfunctionsx(.) on R to X
such that for all a, b, t E R, Ut{xb-xa} = xb+t-xa+t, where {Ut}eR is a
stronglycontinuousgroup of isometries[8], [14]. Using resultsfrom the theory of
helixes, we are able to show that each equivalenceclass in St'(R), 1 <p < oo,
containsa functionin LP(R).This enablesus to get hold of a limit for any Cauchy
sequencein 'V?(R),1 < p < oo.
The case p = 1 has been consideredby Nelson recently [17]; he showed that
Y(R) is isometricisomorphicto the space of countablyadditive,Borel measures
on R with finite variation.Hence S(R) is also complete.
Equation(1.1) shows that the Wiener transformationW is an isometryon the
nonlinear subspace qPS3(R)of <JT2(R). In ?5, we show that W is an isomorphism
from DO2(R)onto Y2(R)with
llWil
where h(x)
=
00h(x)dx)
(2sin2x)/
TX2, X
1/2
and IIW-ll =((max xh(x))
> 0, and h(x)
=
-1'2
/
(1.4)
sup,>. h(t), x > 0 (i.e., h is the
smallestdecreasingfunctionwhichdominatesh). The proof dependson two special
types of Tauberiantheoremswhichwe will developin ?4 (Theorems4.5 and 4.6).
It follows from(1.4) that
11
(fInh (x) dx)>
(fn
2
)
ANALYSIS
HARMONIC
ON GENERALIZED
77
and
(
2sin2x -1/2
2m)
IIW-1li =max
1.05 and jjW'-jI
jz
(Numerically, WI
WI
> 1.
1.49.) In view of (1.1), it is natural to ask
the closed linear subspacegeneratedby 'll9(R),
whetherW restrictedon <VW2(R)>,
is an isometry.We answerthis questionnegativelyand therebydisprovea conjecture by Masani [161.
Finally, we observe that the Wiener transformationW is also a bounded linear
into cVI"
(R), 1 <p < 2, i/p + l/pr = 1.
operatorfrom MPR(R)
and of
ACKNOWLEDGEMENT.
The proof of the completenessof the space cV)T(R)
the fact that W is an isomorphismon 902(R) onto 512(R),the definitionof h and
the proof of the result
wi(fl
h(x)dx)
11Wll < (max xh(x))'2
and
were given by Lee in his IndianaUniversitydoctoraldissertationin 1971 (unpublished) (cf. [101,[111).The equality(1.4) and the Tauberianresultsare due to Lau.
Both authors would like to express their gratitude to Professor Masani for his
supervisionand commentson this work. Their thanks are also due to the referee
for many helpfulsuggestionsin simplifyingthe paper.
2. The space 6)1P(R). Throughout,we assume thatf is a complexvalued, Borel
measurablefunctionon R. Let w be a positiveBorel measurablefunction.We will
use LP(R,w(x) dx) to denote the Banachspace of functionsf such that
/ o
dx
lIf! = (f|_f(x)IPw(x)
I/p
< oo.
Let MP(R), 1 < p < o, denote the set of locallyp-integrablefunctionsf such that
IIfll =
fI
T
( _T
<T<2oo
sup
-T
'/1'
If(x)IPdx)
< 0.
Let IP(R)be the subspaceof f in MP(R)such that
T
I
Em
T- )oo
4Tf -T lf(x)lP dx = 0.
Let )P(R) be the Marcinkiewicz space defined as in the introduction and let
6SP(R) be the set of f in GYP(R) such that
lim
T-oo
2T
|
PT
If(x)IP dx
exists.
2.1. Let 1 < p < oo and let a > 0. Then
PROPOSITION
MP(R) S LP(R, 1
K.-S. LAU AND J. K. LEE
78
PROOF.
fT
For any T > 1,
If(x)IXl
1+
IX
dx
a
=
1+1
fT
1+
I+
d
IA
d(
I+a
')
IxI+a '1T
(1
+a)
XT
+ (I + a)|(
(1
J
2T
I + Tl+'
1
CT
I
f(y)P
2T
+2(1 + a)f
x
+ Xl+a)2
1+
I(Y)I P ay) dx
y
a
(+
f(Y)IP
dy
dx.
This implies
dx < k(IlfllMpy
aIX+a
where
=
dx
O
+
a)
~1
k~2+ ~~~aj1+
Xl+a'
Hence
MP(R)
Y
1 +
C LP(R,
The strict inclusion follows from the fact that f(x) = (log x)'/1PMIl,x)(x) is in
LP(R, 1/(1 + IxIl+a)) but is not in MP(R). EO
2.2. Let 1 < p < oo. Then
PROPOSITION
(i) MP(R) is a Banach space,
(ii) G1j(R) is isometric isomorphic to the quotient space MP(R)/IP(R) under the
natural identification.
PROOF.We leave the simple proof of (i) to the reader.To prove (ii), we identify
functionsin DIP(R) whose differenceshave zero norm.We will still usef to denote
the equivalenceclass of f in 1?(R). The map T: MP(R)--* 6XP(R)with T(J) = f is
clearlya contraction.It is also a sujection, for if f is in YP(R),we let
f'11(x)
=
f
f(x),
= lf )'
x>
X
1,
Lx:< 1.
Also note
= 0. Hence T(f') = f' = f in DY1?(R).
that T 1(O) = IP(R). This induces a bijection f: MP(R)/IP(R)
G1P(R) and 11f11
Thenf' E=MP(R)and Ilf - f'Il
..*
< 1. To show that siis an isometry,we need only show that
inf
g ERIP
(R)
Tlf+gjjmp
<
(
Tm
T--*oo
If(x)tP
T
dx)
79
HARMONIC ANALYSIS
ON GENERALIZED
But this followsdirectlyfrom
sup (T
-T
1 <IT
-
IIf(X)
fX[-a, a(X)lp
sup(fT
If(xIP dx)
I
2
a<T
dx
Va>l1.
Fl
T)
is not a linearsubspacein
In the following,we will give an examplethat 6YSP(R)
I1MP(R):Let En = [(2n)!, (2n + 1)!], n > O,and letf = Xu E '
=
g(x)
XuEuI(X)-2R
XO?
O,
x <O,
and h(x)={2'
O,
xV
x <O.
O,
It is clear thatf = g + h,
1I TT
-I jgljdx = T--*oo
lim -I
lim
T--+oo 2 TiT
l2T I
)P
-
-
dx=
and
lim
T--
Hence g, h E
ISUP(R).
2T
jihiP dx=-T
Observe also that
1
lim
n-*o2(2n)'
(2n)! tfl
lim
-=
n--*o
(2n)!
< li
n- *
2(2n)!
2(2n)!0
Ifl
jj
J
(2n -1)!
2(2n)!
=0
and
nlm
m2(2n 1+ 1)!'
(2n + )!
(2n +)!
2(2+
ffl-40=
n-im
>
(m
n--*oo
This shows thatf V
1SP(R)
2(2n + 1)!
(2nl+ 1)!
2n + 1)!-(2n)!
2(2n + 1)!
and ~SP(R) is not a linear subspace of
fIP
= 1
2
M1P(R).
(R) with
We remarkthat in [9], we prove that for 1 <p < oo, each f E 6WSP
such f,
of
set
The
the
of
an
extreme
is
=
unit sphere S(6TP(R)).
point
If II 1
=
For
of
(R))
1,
S(9.
extreme
points S(G1PP(R)). p
however,does not exhaust all
does not containany extremepoints. The nonlinearsubspace61P(R) has also been
studiedby Masaniin [15]wherehe introducedvectorgraphtheoryand conditional
and its subspaces,the readermay
Banach spaces. For other propertiesof YTP(R)
referto [1], [3] and [9].
3. The spaces \f(R) and their completeness. For each h E R, we define
Th and Ah
as
(Trf)(x) = f(x + h)
and AJ =
(Th -
I)f
be defined as in
where x E R andf is a Borel measurable function on R. Let C\>l'(R)
the introduction; it follows directly from the definition that forf E 2P(R),
K.-S. LAU AND J. K. LEE
80
1
_
\l /p
X
(h f-o
hlim
h0O+ 2
llf=
=
lim (jI(
h
(hh - T-h)f(x)IP
l)fx)LP dx)
oo
_h0+
dx)
By identifyingfunctionswhose differenceshave zero norm,it is easy to prove that
('il(R),
I * 11)is a normed linear space. In the following, we will show that each
f E ??f(R),1 <p < oo, is equivalentto a g E LP(R), i.e., Ilf - gI = 0. This fact
will be usefulin provingthe completenessof 'VP(R),1 <p < oo (Theorem3.6) and
the surjectivityof the Wiener transformationfrom <2(R) onto cV2(R)(Theorem
5.2).
Let A be a subset in a Banach space X; we will use <A> to denote the closed
linear subspacegeneratedby A. Let x(.) be a continuousfunction on R to X; we
call Sx = <{xb - xa: a, b E RI> the chordal subspace of the curve x(.). The
functionx(.) is a helix in X if thereexists a stronglycontinuousgroupof isometries
{ Uj)tE. on Sx onto Sx such that, for any t in R,
Ut(xb -
Xa) = Xb+t
-Xa+t;
{ Ut}teRis called the shift groupof the helix x(
3.1 (MASANI[14]). Let x(.) be a helix in X with shift group {Ut}PER
THEOREM
Then
rm
-
e-(xo
ax=
x,) dt
(Bochner integral) exists and is in Sxd Moreover,
xb
Ut dt)(ax)
Ua
xa=(Ub
Va, b ER.
We call ax the average vector of the helix x(.)*
3.2. Let 1 < p < oo and let f E cVJ(R).If x, = TJ - f, then x(.) is a helix
LEMmA
in LD(R) with shift group {it) }tR and the average vector is given by
Tf) dt E Sx C LP(R).
e-t(f-
PROOF.Sincef E 'V"(R),xt = (T
-
l)f E LP(R) and
00
him
J_(Th
-
I)f(X)I" dX = 0.
It follows thatfor any t E R,
lir
fJ
Ixt+h(X) - xt(X)IPdX = J
h-O
-
-
fI(t+h
fI(rh
-
;)f(X)IP dA
I)f(X)IP dX = 0.
81
ANALYSIS
HARMONIC
ONGENERALIZED
Hence x(.): R -* LP(R) is continuous. By definition, we can show that, for any a, b
and t E R, T(xb - xa) = xb+t - xa+t, This implies that {;ItER is a strongly
continuous group of isometries from Sx (C LP(R)) onto S. and x(.) is a helix in
5 LP(R)
LP(R)with shift group {Tlt}ER.Finally,by Theorem3.1, thereis a g E
such that
00
g =J
xt) dt
et(xo-
00
=
e-t(f-
TJ) dt.
3.3. Let f E cV (R), 1 < p < oo. Then there exists a g E LP(R) such
THEOREM
that lf- gI1j, = 0.
PROOF.Let g
= fo'et(xo
xt) dt
-
be as in Lemma3.2. For any h > 0, we have
(Theorem3.1)
(Th
-
)
T-h)(g
=
dt
Ttg
and ([8, p. 82])
|| t'llp dt < 2hll|gllp.
Tt dt| < |
l tg
Hence
If- glv=
,:izm
I/p (
1/P
_ (1
h- Th))
-
f )II
= 0. E
The theorem is not true for p = 1 as V (R) is isometric isomorphic to the space
of countablyadditiveBorelmeasureson R with finite variation[171.
In the following,we will consider the completenessof Stl(R), 1 <p < oo. The
of V (R).
casep = 1 followsdirectlyfrom the above isometriccharacterization
Let B(LP(R))denote the space of boundedlinear operatorson LP(R).For any
a, n in R witha < b, we let
la,b
=
b- a
{
t
dt,
the integral being a Riemann integral in the strong operator topology in B(LP(R))
([8, pp. 62-67]). We note the followingfacts:
Vs E R,
Vf E LP(R),
TsIa,b
lM
Io, h(f)
(3.1)
Ia,bTs
=f
in LP(R),
(3.2)
(3.3)
Vh > O,
Va, b E R, a < b
< 1,
1IIO,hll
and Vf E LP(R),
'a,b(f)
e
6DA,
(3.4)
where 6DAis the domain of the infinitesimalgeneratorA of the translationgroup
on LP(R) ([8, p. 307]). It is well known that A is the restrictionof the
{Th),heR
differential operator on 6DAand 6DA = {f E LP(R): f is absolutely continuous and
f E LP(R)}. Let g E 6DA;then
82
K.-S. LAUANDJ. K. LEE
lim
-
((Th
h \*O2hJ
/oh-
lim
=
T-09p
(2h)"'1'
2h
(Th -
IP
= 0 * IIAgllp= 0.
Therefore,we have
1 < p < oo, let f
LP(R). Then IlIfIK,= 0.
PROPOSITION3.4. For
f'
f
E
LP(R) be absolutely continuous and
E
Let a < b; then (3.4) and Proposition3.4 imply that IIIa,b(f)IIcv = 0 for all
LP(R), 1 <p < 00. From this, we immediately draw the following conclusion:
E
LEMMA 3.5. Let 1 <p < oo, let f,fl,
bI, . . ., bk E R with a, < b, Then
. ..
E LP(R)
. fk
|tf 2 Ia bn
(fn)||
and let a,, ... .
ak,
llfllcvp
THEOREM3.6. For 1 < p < oo, the normed linear space cV)(R) is complete.
PROOF.
For convenience,we let Io b = Ib. Let {fn} be a Cauchy sequence in
cVJl(R);it sufficesto show that {fnj has a convergentsubsequence.Withoutloss of
generality,we assumethat
llfn+l -fnill?
1/2 n+l.
<
Also, by Theorem 3.3, we may assume that fn e LP(R). For n > 1, select a
decreasingsequenceof positivenumbers{hn) in (0, 1) such that limn,, hn = 0 and
for 0 <h
<h,
(1/2h)"/Ill(Th
-
-f)I1p
T-h)(fn+l
<
(3.5)
1/2
Define el = 1 and en, n > 2 satisfying
{en)j\O
asn-oo,
(3.6)
11(Ien- l)fijjP < (2hn)'/P/2n,
II(IeI-(1)(fm-fi)jIp < (2hn)"//2,
m = n, n-1
(3.7)
and I=1,
. . ., m-1.
(3.8)
((3.7) and (3.8) follow from(3.2).)For any positiveintegersj < k,
k
k
||
,(Ien+,
Ij,fn|
<;
)(fn
1j(l(r,,+
+ l('e - 1)(fn
+ 11
(I,", ,-)filip
< 1/2i-4
fi)llpD
- f)llp
+ 11(I - O)filip)
(by (3.7) and (3.8)).
ON GENERALIZEDHARMONIC ANALYSIS
83
Therefore, the sequence
k
Ie{
+
I(fl)
k
00o
Iew
)(fn)}
(Ie-+l
n=l
kml
is Cauchy in L-(R) and converges to, say, f E LP(R), i.e.,
f
h1(f1)+
=
2
(Ie"
Ie,)fn in LP(R).
-
(3.9)
Note that for each g e LP(R), by (3.2) and the telescoping of the terms,
g = Ie(g) + 2 (Ie +,-Ie,)g
in LP(R).
(3.10)
We will estimate the term IIf- fkII:
If
-
I
( 2h
f+
T
PTh
lh-O(2h2)
h
I
-
fk) +
-h)(Iei(f-
(h
(I,+
-
I.,)(fn -fk))
(by (3.9), (3.10))
00
1\i/p
( 22hh )
i
2
l(Th-T-h)
~
h--+O
n-k+1
(I +, - I, )(fn
-fk)
p
(by Lemma3.5)
For abbreviation,for any positive integersr < 1 and h > 0, let
A
( )I/P
=
(Trh -
T
+1 -If
I)(fn
fk))
Fix h e (0, hk) and let q > k be the unique integer such that hq < h < hq
1 > q, we have Ah,k,l < Ah,k,q+ Ah,qi, Observethat
Ah,k,q
=
~~q
(1) I /P I/P
(h
2h )
( 2h)
-
)
-T-
n-k+1
(I
()
-
~
n
I
2
f-+)
jk+1
(fj
For
i-1
~~~q
I/P
||(Th
-
*
T-h)
(Ie+-
Ie)(f1 - fj|
p
(changingorderof summationand addingup telescopingterms)
h
(~~~s
I\/p
< 22h( )
q+1
2
j-k+1
+
)
T-h)(fj - j-l)lp
II(h - T-h)(fj -f-1)ilp
(by (3.3))
K.-S. LAU AND J. K. LEE
84
and
I/
nq+
Ah,q,l
A^,,
2
I
n<2
JI
1(Th -T _ h)(Ie,,+ l -Ie,
I
1
I
2n
< 1/2 k-2
l/p
l!(Ie+ -I,)(fn
2hn
I
h
fk)li
fk)llp
(n n-
l/p
(11( e,,+
-
fk)IIp
1)(fn
(by (3.3))
)ll
-fk
+
ll(Ie,,
-
fkn
) 11P)
(by (3.8)).
We have shown that for any integer k, for any h E (0, hk), and for large integer I
(i.e., 1 > q as defined previously), Ah,k,l < 1/2k-3. This implies that
<
Ilf -fkll'\
1/2 k-3
and hence {fk) converges to f in Vcw(R). f
We conclude this section by considering a related space, SV(C), C = [0, 2X],
which consists of those Borel measurable functions on R with period 27T and
_ O(h
f71
)1/P
By identifying functions whose differences have zero norm, it is easy to show that
Stl(C) is a normed linear space. If we consider functions on C as 2X-periodic
functions on R, we can prove the following (compare this to Theorem 3.3):
THEOREM 3.7. For
REMARK.
<p
1 < p < oo, CVP(C)C LP(C).
In [6], Hardy and Littlewood-proved that 'fP(c)
n L'(C) C LP(C),
1
< 00.
The proof depends on two results due to Carroll [4] and Boas [2]: Let ALP(C)
denote the set of functions f (not necessary measurable) on C such that A,f = (Th
- 1)f is in LP(C). Carroll proved that if f E ALP(C), 1 < p < oo, thenf admits a
decompositionf = g + H + S where g E LP(C),
Iff
H(h) =f
A,f(x)
dx,
which is additive on C, and AhS(x) = 0 for almost all x. Moreover, Boas proved
that if the above S is measurable, then S is constant a.e.
PROOFOF THEOREM3.7. Let f E 'VPf(C).Then A,f E LP(C). Let g, H and S be
defined as above. Since f is measurable and A,f is integrable on C, Tonelli's
theorem applied to (AJr)+ and (AJf- shows that H is measurable. In addition, the
expression for H shows that H is additive and periodic on R, so that it is identically
zero. This implies that S is measurable and hence S is a constant a.e. Therefore
Ca.e.andf ELP(R). O
f=g+
By using Theorem 3.7 and the same argument as in Theorem 3.6, we obtain
THEOREM3.8. For 1 < p < o0, 'VP (C) is a Banach space.
ON GENERALIZED
HARMONIC
ANALYSIS
85
4. Tauberian theorems. In proving the identity (1.1), Wiener introduced a fairly
general form of Tauberian theorem which applies to functions in QS(R). In this
section, we will consider two similar types of theorems which apply to functions in
ct1?(R).
Let g be a function of bounded variation on [a, b]. For all x E (a, b], let
ttg(a, x] = g(x + ) - g(a + ). Then it is well known that jig has a unique countably
additive, regular extension to the a-algebra of Borel subsets of [a, b]. The following
integration by parts holds.
LEMMA
4.1. Let f, g be measurablefunctions on [a, b] such that f is integrableand g
is of boundedvariation. Then
fbf(x)g(x)
(L
dx =
f(x) dx)g(b)
(
-
f(t) dt) dLg(x).
PROOF.The result follows by applying the Fubini theorem to the right-hand side
of the identity
fbf(x)g(x)
dx
dt&g(t)+ g(a +)
Lf(x)(f
=
dx.
O
Let S + denote the set of positive Borel measurable functions on [0, oo] such that
T
sup
f(x) dx < 1.
1
T
<1
For any T, a > 0 and for any f E S +, the substitution x = t/ T shows that
f(Tx) dx = a aT
J
f(t) dt) < a.
PROPOSITION
4.2. Let h be a positive decreasing integrablefunction on [0, oo). Then
(i) fo?f(Tx)h(x) dx < f 'h(x) dx for allf E S +, T > 1,
(ii) limnaO f a0f(Tx)h(x) dx = 0 uniformlyfor allf E S +, T > 1.
PROOF. (i) Note that because h is a decreasing function, the corresponding
is negative.Hence for any / > 0, f E S +,
measureILh
f
f(Tx)h(x)
dx = (f
f(Tx)
<(fh(f))-f
-
dx)h(1)
|
f
f(Tt) dt) diL(x)
h(x)dx.
xd,u.=f
Letting , -B oo, we obtain (i). To prove (ii), we observe that by (i), for each
f E S+, T > 0, f0f(Tx)h(x) dx < oo; hence
urnmf
t00
aa+o f(Tx)h(x)
dx
=
0.
In order to obtain the uniform convergence forf E S + and T > 1, we let
ha
(x)
h (a),
hx,if
if x
a,
x >a.
K.-S. LAU AND J. K. LEE
86
By applying(i) to ha, we have forf E S +, T > 1,
f( Tx)h(x) dx <
f
f( Tx)ha(x)dx
<
h. (X) dx = ah(a) + f
f
h(x) dx.
Note that becauseh is decreasing,
a
h(x)dx -0
S
h(a)
asa-*oo.
/2
Also,
f
h(x)dx-*0
asa-oo.
This impliesthat
lim
0
f
f(Tx)h(x)dx = 0 uniforily forallf
S+, T > 1. 0
LEMMA4.3. Let h be a positive continuous function on [0, oo) and let h(x)=
sup,>xh(t). Suppose that h E L'[0, oo) and suppose there exist disjoint intervals
(ai, bi), i = 1, . .. , k, in [0, oo) such that for each x E (a,, b,), h(x) < h(b,). Let
=
(h(bi)(bi
?
a)
-
h(x) dx).
-
Then
00
f(Tx)h(x)
lim
sup
T-*
feS+
PROOF. It suffices to
00
h(x) dx +
dx >
n.
0O
show that for any 0 < e < q, there exists an f E S + such
that
lim
f(Tx)h(x)
f
dx >
h(x) dx + (q - e).
f
(4.1)
We will considerthe case k = I only. The case k > I follows from the same idea
of proof. We write a, = a and b, = b and without loss of generalityassume that
a > 0. Otherwise,let {adj 0 and let
r=
h(b)(b
-
n)
h(x) dx;
-f
then {,n}
q. We can prove(4.1) for (d,n,b) and qn.
Since h is continuous, for e > 0, we can find 0 < el, 0 < 8 < e/8 such that
x-
bI<
8 implies that jh(x)
'q
-
e
<
(h(b)
h(b)l < e1 and
-
-
b+(
e1)(b +
2~~~~~~~~~~
'5-
a)
-f
h (x) dx.
(4.2)
87
ON GENERALIZED HARMONICANALYSIS
Since h is decreasingand integrable,Proposition4.2(ii) impliesthat there exists an
a > b + 8 such that
p|
p(Tx)h(x) dx < |
p(Tx)h(x) dx < 8
epE S +, T > 1.
In particular,
t00
h(x) dx
f
< 6.
Let { ) be a sequenceof positivenumberssuch that I n6,,8 < 8 and let M be the
upperboundof h. Let T1 > 1 and let
f1(x)
Suppose we have chosen Tn
i,
tx
1' fn-
( o
*
0 <(x < aT1,
aT1 < x.
Select Tnsuch that
Tn> max{(-Tn-
T}-I
1
and define
0,
O<x<aTn-,
1,
aTn_I
0,
fn(x)=
aTn
(x<aTn,
< X < bTn9
b+ 8 -a,
bTn< x < (b + 8)T
1,
(b + 8) Tn <Sx <aTn,
aTn
0,
<x.
It is easy to
=
Note that the functions {fn})have disjointsupports.Letfffn.
show that
T
f(x) dx = 1
for T E[1, oo)\ U (aTn, (b + 8)Tn)
and
T
+T
f(x) dx < 1 for T E
0
00
U
(aT,, (b + 8)Tn).
~~~~~n-2
Hence f E S+. We will show that
lim
n--+oo
f
f(Tnx)h(x) dx >
f
h(x) dx +
(r-
).
K.-S. LAU AND J. K. LEE
88
Forn > 1,
dx
f(T.x)h(x)
<
f. (T"x)h(x) dx
-
Jf f(Tnx)h(x)
i=l
< "l
M
~~~i-n+lO
dx + 8
fJ(Tnx)h(x)
n-I
- TaT, + 8 < i
<
o
X
dx + 0 + 8
fa>i(x)
n-I
J fn(Tnx)h(x) dx + 8
-
ff(Tnx)h(x) dx +
O
f
<
dx
1
+ 8 < 28
and
a
fn(Tnx)h(x) dx
a
>f
>
n
fh(x) dx-M
dx +
h(x) dx
b+
8
h(x) dx
+
+ (b + 8-a)(h(b)-el)
(?1-j)
h(x) dx - n +
h(x)
+
h(x) dx
)-a
b
aT,,- I/ Tn
>
(b +
b+8
h(x)dx
(by (4.2))
(28 +
-
Combiningthe above two estimationswe have, for n> 1,
f
f(Tnx)h(x) dx >
f
h(x) dx +
(7 -E).
E
PROPOSITION4.4. Let h be a positive continuousfunction on [0, co). Let h(x) =
supt>xh(x) and assume that hi is integrable. Then
sup
f E=S+
f(Tx)h(x) dx) =f
lm
T-o
h(x) dx.
?
?
PROOF.
Sinceh is decreasing,by Proposition4.2(i),
sup (lim
f ES +
Too
f(Tx)h(x)
f
?
dx) < sup (im
f =S +
< f
f(Tx)h(x)
dx)
Too ?o
h(x) dx.
We will prove the reverseinequality.That h is integrableyields an a > 0 such that
f 'h(x) dx < E. Since h is continuous,so is h; hence the set {x E (0, a): h(x) >
h(x)} is the union of a (finite or infinite) sequence of disjoint open intervals
89
HARMONIC
ANALYSIS
ON GENERALIZED
{(a,, b,)} in (0, a). Let 1 < r < oo be the numberof such intervals.Note that h is
constanton each (a,, bl). It follows that
dx
fh(x)
h(x) dx +
= f
>
(h(bi)(b
ai) - f
-
h(x) dx)
Lemma4.3 impliesthat for any integerk, 0 < k < r,
sup
f es+
00
lim
f
f(Tx)h(x) dx
T--+O
h(x) dx +
>
b
f
h(bi)(bi - a,)
'h(x) dx).
Hence,
sup(im
f E-S+
T
f
pm
O
f(Tx)h(x) dx) >
h(x)dx
and the proof is completedby observingthat fJ?h(x)dx <e. C[
Let 6RX denote the class of positive Borelmeasurablefunctionsf on [0, oo) such
that hinT,(1/ T)fIof(x)dx < oo. The following is the first main result of this
section.
THEOREM 4.5. Let h be a positive, continuousfunction on [0, oo). Assume that (i)
h(x) = sup,>xh(t) is integrableand C1 = Jfoh(x) dx, and (ii)f e %+. Then
Uin
f
T----
?0
f(Tx)h(x) dx <C1 iim
1
f(x) dx.
0
T-*oo
Moreover, C1 is the best estimation of the inequalityfor the class of functions f in
PROOF.For any functionp
him |
T--*oo
which is integrableon [0, p) and vanisheson [p, oo),
p(Tx)h(x)
dx = him
T--.oo
I
T .
< hm
p(t)h-t
\T/
dt
0
mo Pp(t) dt =
T
whereM = sup,>Oh(t).Letfp = fX[P0,),p > 1. It followsfrom(4.3) that
him J' f(Tx)h(x) dx = him 00fp(Tx)h(x) dx.
O
T-*oo
T-*oo
ApplyingProposition4.4 with (supT>P(l/ T)JoTf)-lfp E S +, we get
himf
00
I
T
fp(Tx)h(x)dx < C1sup - ff(x)
0
dx,
p > 0.
T>p
T-*oo
Hence by (4.4),
him
T-o*oo
|
f(Tx)h(x)
dx
< C1
T-oo
T
f(x) dx.
(4.3)
(4.4)
90
K.-S. LAU AND J. K. LEE
To prove the last assertion, it suffices to show that for any E > 0, there exists an ,
such that
lim
oo
T
I
dx < 1
f(x)
T
and
f f8(Tx)h(x)
hm
dx S (C1 -E).
But this inequality is obvious from Proposition 4.4. EO
Our second main result in this section is:
THEOREM4.6. Let h be a positive continuousfunction on [0, oo) such that h(x) =
supt>xh(t) is integrable. Suppose (i) there is an xo which satisfies xoh(xo) =
maxx>oxh(x) = C2 and h(x) > h(xo) for all x in [0, xo], (ii)f E 9)+. Then
C2lim
+fI
f(x)
f
| 0(Tx)h(x)
dx < fi
dx.
T-*oo
T-+oo
Moreover, C2 is the best estimation of the inequalityfor f
E '6)
.
PROOF.Letf be given as above; then
C2yT
I
Sf(x) dx =
<
-
Ty
f TO
xf(x)h(
f(x)h(x0) dx
x) dx
(since h(y) > h(xo) Vy E[0, x0])
fXof(Tx)h(x) dx
00
<
f(Tx)h(x)
dx.
By taking limit supremum on both sides, the first part of the theorem follows.
To prove the second part, we will construct, for a given 0 < e < 1, a positive f
such that
1~
T
oo
(X) dx =
and lim f f(Tx)h(x) dx < C2 +E.
j
and T ?-*oo
TWe will need the following statement, where the proof depends on the uniform
continuity of h: for any a > 0,
lim X0 + 6 fI(xo+S) h(x) dx = (f3x0)h(f3x0)
lim -T
uniformly for 0 S /3 < a.
(4.5)
For any 0 < e < 1, we choose a > x0 such that
f
p(Tx)h(x)
dx <8
Vp E S+ and T > 1
(4.6)
(Proposition 4.2(ii)). Since xoh(xo) = maxx>0xh(x), by (4.5), there exists 0 < 8 <
a - x0 which satisfies
x0 +
I~~~~x0~~
(xo+
?)
h(x) dx < xoh(xo) +
xxo~0j--
VO < P <
/ ~-
a
91
HARMONIC ANALYSIS
ON GENERALIZED
Let T1 > I and select Tn> T_1 so that
Tn-I
Tn
1
Tn
xo
a
I1
n
and
nI
e(7
+(4.7)
xo Ti/Tn6
h(x) dx <
|
Let
Sn
x0 +6
00
a
X[XOT,,,
(X0+8)T.,]Xf,=
68
fn
and
Note that the functions{fn}have disjointsupports.For any n,
|(xo + 8) Tnf(x)
f
(Xo + 86) Tn
1
d
n-I
+
+ 6)Tni
(x0(Xo+)Tn(XO
n-I
1
+
8)Tn
Y (XO
+
,8)Ti)
)7)
1
< 1+
=Ti
Since (1/ T)fOTf(x)dx has a local maximum at each (xo + 8) T, we have
sup 1
(X) dx < 2
and
I<T T
lim
T |f(x)
dx = 1.
Now for any T > 1, there exists an n such that xoTn< aT < xoTn+
and
o < Tn/T <a/xo
00
f(Tx)h(x)
(4.8)
T-*oo
dx < |
ae
fTx)h(x)dx
+ e
Hence
(by (4.6) and (4.8))
n
i(Tx)h(x) dx +
<-n
<x
x07/T
+6
< xoh(xo)
TX0 + 6 h(x) dx +
f(xo+8)T/T
+ e/4
+ e/2
h(x) dx +
(by (4.7))
(by (4.6))
< C2 + E.
The proof is completeby takingthe limit supremumon T. El
, x > 0, 1 <p < oo, satisfies
We remarkthat the function h(x) = 12sinex/7rxP
the hypothesesin Theorems4.5 and 4.6. We leave the simple verificationto the
reader.This functionwill be consideredthroughoutthe rest of the paper.
K.-S. LAU AND J. K. LEE
92
5. The Wiener transformation W.
A. The isomorphicnature of W on Dk(R).
Proposition2.1 impliesthat M2(R) C L2(R, dx/(l
+ x2));
hence forf E M2(R),
the integral
1+1? If(X)12dx
A
exists.This impliesthat
e- iuxd
oo1?
f(x)
I
dx
-oo
1
-
convergesin mean square.In [18],Wienerdefinedthe followingtransformationW:
forf E M2(R), let W(f) = g where
g(u)=
(f +
2
dx+
ff(x)e.
-1
f(x)
dx).
We call W the Wiener transformation.Now, for h > 0,
00
I
(Thg -
=
r-hg)(u)
2
eilm- e- ix
f(x)
f
T-g
e-eix dx
2sin(hx)e_dx.
= ff(x)
Thus rhg -
ix
is the Fouriertransformationof
f(x) sin(hx)
and the Planchereltheoremimplies
2h
f_
g(u + h)
-
g(u
-
h)2du =
If(x)I2sinhx dx
f7
Hence
IIW(f)I
lim
h-*O+
lim
= T-
-
h
(
.7X_
If(x)12sin2hx
|
-oo0
W7X2
dx.
If(Tx)I2sinx
Letting h(x) = (2sin2x)/ 7x2, x > 0, and f'(x) = I(If(x)12 + If(-X)l2), X > 0, f E
M2(R), Theorem 4.5 and (5.1) imply that W(f) E cV2(R)and W(f) = 0 for all
f E I2(R). Since O2(R)= M2(R)/12(R) (Proposition2.2), W inducesa map from
G)2(R) into S2(R).
THEOREM
5.1 (WIENER[18]). Letf E WVS2(R)
C )2(R). Then
Our main resultis:
||
W(f)l|V
=
|lf||62.
HARMONIC
ANALYSIS
ON GENERALIZED
93
THEOREM
5.2. The Wiener transformationW defines an isomorphismfrom 02(R)
onto V2(R)with
ii
=
00
(f
h(x)dx
~~1/2
and 11
W-
h()
Ma/
/
(maX xh(x))
=
where h(x) = (2sin2x)/I7X2,X > 0, and h(x) = supt>xh(t).
PROOF.
It is easy to show that h(x) = (2sin2x)/,Wx2,
x > 0, satisfiesthe hypotheses in Theorems4.5 and 4.6. By letting
f'(x)
=
x > 0 f E 9Th2(R)
+ If(_X)12),
2(If(x)2
the same theoremsyield
24f
C2 lim
-T
T-*oo
If(x)12dx < lim fIf(Tx)12h(x)
T--oo
dx
-00
< C1 iHm I f
T--oo
f
jf(x)12
e 933(R),
-T
where C1 = f1oh(x) dx and C2 = maxx>oxh(x). By (5.1), we have
C211f
112 < IIW(f)112 < Clllf 112, f e ,2(R)
Moreover,Theorems4.5 and 4.6 imply that C1 and C2 are the best constants to
estimatethe above inequalities.Hence we concludethat W is an isomorphismfrom
92(R)
into cV(R) with
IiWIi= C1/2 and IIW-1ll= C2-1/2
It remains to show that W is a surection. Let g E V2(R); by Theorem 3.3, we
may assumethatg E L2(R). Let g be the (inverse)Fouriertransformationof g, i.e.
g(x) = V and letf(x) =
g(u)eiux du
f
-iV5i x(x), x E R. We claim that (i)f E
GT2(R) and (ii) W() =
To prove(i), note that
g in cV2(R).
(,h(g)
T-h(g))
(x) = (e-ix
elhx)g(x)
-
= -2i(sin(hx))g(x).
As C2 = maxx>0(2sin2x)/-rx, Theorem 4.6 applied to f'(x)
If(-x)12),x > 0, yields
C2 lim
I
T
()1
2T|_If(x)I2dx
X<0
i2
-
lim f
If(Tx)2m
T-oo
.,-~2
00
h-1*
h-O+
-
h
= 11gI2vi
dx
7TX
00
I(00
= lim
2
(5.2)
= 2(If(x)12 +
_
-_
in2hx d
T
219g(x)2sin2hxdx
(by (5.2) and the Planchereltheorem)
K.-S. LAU AND J. K. LEE
94
Thusf E D12(R). To prove (ii), we observe that
W(f)(U)
+
=
J
1 (?
=
f(x)
e
e
dx + f_f(x)
fj^x(x)
)(x)-') i dx
^(
1 dx)
dx)
g(u) + C a.e.
where C = -(1 /27T)J1 g(x) dx. Since constant functionsin YV(R)are equivalent
to zero, we have W(f) = g in 512(R). O
B. The nonisometricnature of Won <KtS5(R)>.
Let h(x) = (2sin2x)/,fx2,x > 0. It follows from Theorem5.2 and elementary.
calculusthat
h(x)dx)
=
11WII
-
>
dx)
x2
=1
and
(
=1wli
max
X>O
2sin2x
7TX/
-1/2
>1
I
In view of the fact that W is an isometryon 'll9(R) (Theorem5.1), Masani [16]
the closed linear
asked whetherW is an isometrywhen it is restrictedon <K'OS(R)>,
subspace generated by 6l2(R).
Let GVj2(R) be the subset of Y (R) such that for g E 60(R),
lim
-h)l2du
+ h)-g(u
fIg(u
2
exists. Theorems5.1 and 5.2 imply that W is an isomorphismfrom <K'i2(R)>onto
<G12(R)>. In the following, we will give two examples 11 and 12 EK<62(R)> with
< 1. Hence neitherW nor W can
= 1 and 11
= 111211
111111
W(1j)II> 1 and 11W(12)II
respectively.These answerMasani'squesbe a contractionon <KQS2(R)>,
<K1G2(R)>
the
functions constructedin the proof of
tion negatively. Both examples refine
Lemma4.3 and Theorem4.6.
with ll11,II2= 1 and 1IW(l1)llV > 1.
5.3. Thereexists an 11in <KIS2(R)>
EXAMPLE
in
4.3 with h(x) = (2sin2x)/Tx2, k = 1,
Lemma
notation
as
will
use
the
same
We
and [a,, bl] = [a, bJ such that h(b) = h(a) > h(x), x E (a, b). We assume further
that the 8 we choose in Lemma 4.3 satisfies (b + 8 - a)/3 = (m + 1)2 for some
positiveintegerm. Let T",, andf be as in Lemma4.3. Define for n > 1, 1 < i <
2m + 1,
O < x < aT,,-,
O,
aTn- I
1(-)'+',
i
bTn A<x<(b
(-2
0,
<aTn,
aTn < X < bTn',
-(-l)i,
fn
(_)i+
x
1,
+8)Tn,
(b + 8)Tn < x
aTn x.
< aTn,
ON GENERALIZED
HARMONIC
ANALYSIS
95
Note that W Mj"(x) + 4)2 = f"(x). For each i, the functions {}f 1 have disjoint
supports. Letf' = : 1fn.Then Ifl
andf' is in W(R) for each i. Let
4
1
'1
11=V22m+1
f
=VI2E;fi+l
=
4.
Then 11EK1
<qe(R)> and
2m+1
2
II(x)12 =
=2
2
oo
2
+
fi(x)
(o
2
1 2
2m+1
E
wherex E supp(fk)
fk(x) +
i=1
=
= 2fk(x)
2f(x).
Hence
I%I2W= urm
T-*oo
fT| l(x)I2dx
-T
=
lim T ff(x)
T-*oo
?
dx =1
and by (5.1) and (4.1),
jjW(l1)1j = lim
J
Ili(Tx)I2
0
T-.oo
lim
X
dx
71TX
2f(T
dx
T-*oo
7TX2
2sin2xdx = 1.
>f00
EXAMPLE
5.4. Thereexists an 12 e <KW(R)>such that 1l1211%2= 1 and jjW(l,)IIv
<1.
We use the same notation as in the proof of Theorem 4.6 with h(x)=
(2sin2x)/rx2 and C2 = maxx;Oxh(x)< 1. We assumethat 0 < e < 1 - C2 and let
8 be as in Theorem4.6 and satisfythe condition:
n0+
6
-
(2m)2 for some integerm.
Let { Tn)be as in Theorem4.6 and for n > 1,1 < i < 2m, define
fn Ix
0O
x < XOTn- 1
(- 1l)i
XOTn-I
1,
< x < (xo +
XOTn
8)1Tn
+
(xo 8)TT < x.
0,
Then
(2m
)2
< x < XOT
=
XO+
68
K.-S. LAU AND J. K. LEE
96
Note that for each fixed i, the supportsof the functions {f, )%'. are disjoint.Let
21jfi. It follows that
n= Zn ..1f; then P1 1 andfi E 'lIS(R). Let 12=W
12E
and
<KVQ(R)>
_mm2
T-*oo
I
T
T
?T
dx = lim
112(X)12
T1T
T-_Too
x
dx=
Cf(x)
f.1
0
wheref is definedin Theorem4.6. By (5.1) and the constructionof f, we have
rim j
jW(I)l2)V2 =
2f(Tx)
-
1.
dx<C2+E<
2
fX
O
T -oo
C. Won MP(R).
It is well known that for 1 < p < 2, the Fouriertransformationis a contraction
from LP(R)into LP'(R),I/p + l/p' = 1. Letf E %P(R),1 <p < 2, and let g =
of
W(f). SinceThg- Thg is the Fouriertransformation
f(x) sin(hx)
we have
g(u + h)
(j
-
g(u
h)I- du
-
i
)i/
Pf(x)
dx
sin(hx)
)
This implies
g
im( 2h
=
Ig(u + h)
f
s
0+('7ThP'
<i-xf(x)
00
(|X
Tm cxATOO
g(u
-
f(
-
h)IP'du
x) | dx)
sinPx d \ i/P
'3IMP
dx)
(5.3)
5.5. For 1 <p < 2, the Wiener transformation W defines a bounded
THEOREM
linear operatorfrom 61?(R) into cV7"
(R) with
jj Wil <
where h(x)
=
I(2sinmx)/,rxPI, x
(f/ 0 h(x) dx
I/
'lp
> 0.
PROOF.The result follows from Theorem 4.5 and (5.3). E0
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ANALYSIS
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19. _
_,
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