Wavelets with ompat
support
1. Introdution
In this hapter we study wavelets with ompat supp ort. It is quite easy
toseethatif thefatherwaveletorsaling funtion 'hasompatsupp ort,
then the lter has ompat supp ort as well, i.e., it is a nite sequene.
Atleast in theasewhere 'deayssuÆiently rapidly at1theonverse
also holds.
First we onsider some results that are somewhat more general than
whatwe atuallyneed for theanalysisof wavelets.
2. Dilation equations
InChapter 4we found thataruial prop ertyofthefatherwaveletorsal-
ingfuntion 'determining a multiresolution is thatit satisesthe dilation
equation
'=2 X
k 2Z
(k )'(2 k ):
(5.1)
Inthissetionwelo okatsomeprop ertiesofageneralizationofthisequation.
First we observe that every sequene ((k ))
k 2Z
an b e identied with
a Radon measure, i.e. alo al measure, bydening
(E) =
P
k 2E
(k ) for
every b ounded set E 2R. If we assume that 2l 1
(Z), then
is a nite
measure. Now (5.1)an b erewrittenas
'=2(
')(2)
41
wheredenotesonvolution. Thus thegeneralized equationthatwewillb e
lo oking athere is
f =(f)();
(5.2)
where is some real numb er > 1 and 2 M(R;C), i.e., is a omplex
measure onR.
Firstwe prove an auxiliary result on the onvergeneof pro duts. We
usethe notationjj
+
=maxf0;g.
Lemma 5.1. Let 2 M(R;C) be suh that (R) = 1 and R
R
(jlogjxjj
+ +
1)jj(dx) < 1. Then the produt Q
1
k =1
^ (
k
!) onverges uniformly on
ompat subsets of Rtowards a ontinuousfuntion.
Pro of. Sine (R)=1wehave ()^ 1= R
R (e
i2 x
1)(dx), andhene
j^() 1j Z
R
2jsin(x)jjj(dx); 2R:
Letmb eap ositiveintegerandlet!2R. Nowitislearfromthepreeding
inequality, Fubini's theorem, andthefat thatjsin (t)jmin f1;jtjgthat
1
X
k =m
^ (
k
!) 1
2 1
X
k =m Z
R jsin(
k
x!)jjj(dx)
2 Z
R dlog
( jx! j)e
X
k =m
1+
1
X
k =maxfdlog
( jx! j)e+1;mg
k
jx!j
jj(dx)
2 Z
R
dlog
(jx!j)e+1 m
+ +
1
1
jm dlog
( jx! j)e 1j
+
jj(dx):
From this inequality we get the uniform onvergene on ompat in-
tervals of the series and this implies the laim of the lemma by [1, Th.
15.4℄.
Wepro eed withan easyresult.
Prop osition 5.2. Assume >1andthat 2M(R;C) satisesj(R)j1
and R
R
jlog(jxjj
+ +1
jj(dx) < 1 if j(R)j = 1. If equation (5.2) has a
nontrivial solution f 2L 1
(R;C), then(R) =1 and this solutionis unique
in L 1
(R;C) up to a multipliative onstant.
Pro of. TakingFourier transformsof b oth sides of (5.2)we get
^
f(! )= (! )^
^
f(!):
(5.3)
If j(R)j < 1, then it is lear that lim
m!1 Q
m
j=1 j^(2
j
!)j = 0 for every
! 2R. Thus we seefrom (5.3)that
^
f(!)=0forall ! 2R,so we an have
no nontrivialsolution f.
Supp ose next that j(R)j=1. If
^
f(0)6=0,thenweonlude from(5.3)
that(R)= (0)^ =
^
f(0)=
^
f(0)=1. If
^
f(0)=0then we have
j
^
f(!)j= lim
m!1 m
Y
k =1 j(^
k
!)jj
^
f(
m
!)j=0; !2R:
b eause thepro dut Q
1
k =1 j^(
k
!)j onvergesbyLemma5.1. Thus we see
thatf isidentially 0.
Ifnow(R)=1,thenwehave bylemma 5.1and (5.3)that
^
f(!)=
^
f(0) 1
Y
k =1
^ (
k
!); !2R;
and we see thatf isunique uptothemultipliative onstant
^
f(0).
Nextwe onsider thease where in (5.2) hasompat supp ort. First
weprovean auxiliary resulton howthesupp ort of(f)()is related to
thesupp ortsof and f.
Lemma 5.3. Assume that >1, 2M(R;C) with supp() [M ;M
+
℄
and that f 2L 1
(R;C) with supp(f)[F ;F
+
℄. Then
supp
(f)()
h
F +M
; F
+ +M
+
i
: (5.4)
Pro of. Letx<(M +F )=. Thenx t<M +F tF iftM .
Similarly when x >(M
+ +F
+
)= we have x t>M
+ +F
+
t F
+ if
tM
+
. Thisgivesthe desiredonlusion.
Sine the previous result says that theop erator f ! (f)()fores
thesupp ortloser tothatof itisnaturaltoexp et thatif hasompat
supp ort and there is a solution of (5.2), then this solution has ompat
supp ort aswell. This turns outto b ethease,atleast if f isintegrable.
Prop osition 5.4. Assume that > 1, 2 M(R;C) has ompat support
ontained in the interval [M ;M
+
℄andthat (R)=1. Iff 2L 1
(R;C) sat-
sies (5.2), thenf has ompatsupportontainedin theinterval [ M
1
; M+
1
℄.
Pro of. Let f 2 L 1
(R;C) b e some nontrivial funtion that satises (5.2).
If we an show that f has ompat supp ort, then it follows from rep eated
appliations of Lemma 5.3 that thesupp ort is ontained in the desired in-
terval.
Let usfor simpliity assumethat M <0 and that M
+
>0Let m0
b e an integer and let f
m
= f f
[ m
M ; m
M+℄
. Moreover, we dene the
linearop eratorT :L 1
(R;C)!L 1
(R;C)byT(g)=(g)().Ifweapply
Lemma 5.3 m times we see that T m
(f f
m
) hassupp ort ontained in the
interval[ M
1
; M+
1
℄. OntheotherhandwehaveT m
(f
m
)=f T m
(f f
m ),
and this meansthat
f(x)=T m
(f
m
)(x); x62 h
M
1
; M
+
1
i
: (5.5)
Moreover,we easily seethat
\
T m
(f
m )(!)=
m
Y
k =1
^ (
k
!)
f
m (
m
! ):
(5.6)
Lethb esomeinnitely manytimesdierentiablefuntionwithsupp ort
ontained in [ 1;1℄ and let h
= h(), > 0. Now it follows from the
inversiontheorem forFourier transforms(Theorem2.3.(b)),(5.5)and(5.6)
that
Z
R h
(x t)T m
(f
m
)(t)dt= Z
R e
i2 x!
h
(!)
m
Y
k =1
^ (
k
!)
f
m (
m
!)d!:
(5.7)
NowweknowbyLemma5.1thatj Q
1
k =1
^ (
k
!)jisb oundedwhen j!j1.
Then itfollows forall m1and !2Rthat
m
Y
k =1
^ (
k
!)
( sup
! 2R j^(!j)
dlog
2 (j! j)e
sup
j j1 j
1
Y
k =1
^ (
k
)jC(j!j+1) C
whereC issomeonstant. Sine h isinnitely manytimes dierentiable, it
follows that
Z
R j
h
(!)j(j!j+1) C
d! <1;
andthereforeit followsfrom(5.7),thedominatedonvergenetheorem and
from the fat that f
m
! 0 in L 1
(R;C) and hene
f
m
! 0 in L 1
(R;C) as
m!1 that
lim
m!1 Z
R h
(x t)T m
(f
m
)(t)dt=0; x2R; k1:
But when we let ! 1 we see from (5.5) that f must have ompat
supp ort.
3. Constrution of wavelets with ompat
support
Whenp erformingalulationswiththelteronsomerealdataitislearly
advantageous tohave thesequene to b e real. This requirement we will
make throughout this setion where we want to nd suitable sequenes
thatgeneratemultiresolutions. FromTheorems 4.12and4.13we seethat
mustsatisfy (4.11),(4.46), and(4.48).
Weget thefollowing haraterization of theFourier transformof lters
withompat (i.e.,nite) supp ort.
Theorem5.5. Letf(k )g
k 2Z
beasequeneofrealnumberswithonlynitely
many nonzero terms. Then (4.11) and (4.46) hold if and only if
^ (! )=
1
2 1+e
i2 !
N
Q(e i 2 !
)e i2 L!
; (5.8)
whereN 1, L2Z and Q isa polynomial withreal oeÆientssuh that
(5.9)
Q(e i2 !
)
2
= N 1
X
k =0
N +k 1
k
sin(! ) 2k
+sin(!) 2N
R os(2! )
;
whereR isan odd realpolynomial.
Pro of. Supp ose rst that (4.11) and (4.46) hold. Sine we require that
^
(0)= 1, it follows from (4.11) that (^ 1
2
) =0. In order to see that ^ an
b e written in the form (5.8) we argue as follows: For some integer L the
funtion z L
P
k 2Z (k )z
k
is a p olynomial and this p olynomial vanishes in
thep ointz= 1. Thusitanb ewrittenin theform( 1
2
(1+z)) N
Q(z)where
N 1and Q isa realp olynomial. Substitutinge i 2 !
for zweget (5.8).
IfQ(z)= P
M
j=0 q
j z
j
,thenwe have
jQ(e i2 !
)j 2
= M
X
k = M
~ q
k e
i2 k !
=q~
0 +
M
X
k =1
~ q
k (e
i2 k !
+e i2 k !
)
=q~
0 +2
M
X
k =1
~ q
k
os(2k ! );
sine q~
k
= q~
k
for all k b eause q~
k
= P
minfM;M k g
j=maxf0; k g q
j q
j+k
and the o eÆ-
ientsq
j
in Qarereal. Sineeverytermos(2k ! )anb ewrittenasap oly-
nomialinos(2! )(useDeMoivre'sformulaandsin (2!) 2
=1 os(2! ) 2
)
orequivalently asa p olynomial in sin (! ) 2
we seethat there existsa p oly-
nomial P suh that
jQ(e i2 !
)j 2
=P(sin (! ) 2
):
(5.10)
Sinesin ((!+
1
2 ))
2
=os(! ) 2
=1 sin(! ) 2
andj 1
2 (1+e
i2 !
)j=os(! ) 2
itfollows from (4.11)that
(1 z) N
P(z)+z N
P(1 z)=1;
(5.11)
on the interval [0;1℄and therefore also on R. We an write P in the form
P(z) = P
N 1
j=0 p
j z
j
+ N
R
0
(z). Inserting this expression into (5.11) we get
thefollowing system ofequationsfor theo eÆients p
j ,
p
0
=1;
p
k
= k 1
X
j=0 ( 1)
k j 1
N
k j
p
j :
Nexthavetowehekthatthesolutionofthisreursivesystemofequations
is
p
k
=
N+k 1
k
; 0kN 1:
Fork=0thisisertainlytheaseandanindutionargumentworksb eause
wehave
k 1
X
j=0 ( 1)
k j+1
N
k j
N +j 1
j
= N
k ! ( 1)
k +1 k 1
X
j=0
k
j
( 1) j
(N+j 1)!
(N+j k )!
x N+j k
x=1
= N
k ! ( 1)
k +1 k 1
X
j=0
k
j
( 1) j d
k 1
dx k 1
x N+j 1
x=1
= N
k ! ( 1)
k +1
d k 1
dx k 1
x N 1
(1 x) k
( 1) k
x N+k 1
x=1
= N
k ! d
k 1
dx k 1
x N+k 1
x=1
=
N +k 1
k
:
Thuswe dene thep olynomial P
N by
P
N (z)=
N 1
X
k =0
N+k 1
k
x k
: (5.12)
Next we observe that this p olynomial is in fat a solution of (5.11),
b eause theonstrution oftheo eÆeientsp
k
guaranteesthat
(1 z) N
P
N
(z)+z N
P
N
(1 z) 1=z N
V(1 z);
(5.13)
whereV is p olynomial ofat mostdegree N 1. But thenit followsthat
z N
V(1 z)=(1 z) N
V(z):
(5.14)
Itfollowsfromaalulationsimilartotheoneusedforndingtheo eÆients
p
k
,thatV is identiallyzero sine itis of amostdegree N 1.
Theoriginalp olynomialP waswrittenintheformP =P
N (z)+z
N
R
0 (z).
Ifweinsert this expression in (5.11)weonlude that
(1 z) N
z N
R
0
(z)+(1 z) N
z N
R
0
(1 z)=0;
(5.15)
that isR
0
(z) = R
0
(1 z) and this implies that R
0
(z)=R(1 2z) where
Ris ano ddp olynomial. But this isexatly whatwe wantedtoprove.
The onverse go esin exatlythesameway.
If we want to onstrut a lter sequene , one p ossibility is to use
Theorem5.5. Butthenwemustb eabletondthetrigonometrip olynomial
Q(e i2 !
) if jQ(e i 2 !
)j 2
isknown. Thislassial result isgivenin thenext
lemma.
Lemma 5.6. Assume that A(!)= P
M
k = M a
k e
i2 k !
,wherea
k
=a
k 2R
for k=0;1;:::;M,isnonnegativeand A
M
6=0. Thenthe 2M zeros of the
polynomial P
M
k = M a
k
! k +M
areof the form w
j ,w
j ,w
1
j , w
j 1
2C nR, for
j=1;:::;J,and r
k , r
1
k
2R, for k=1;:::;K,and
B(!)= v
u
u
t
ja
M j
K
Y
k =1 jr
k j
1 J
Y
j=1 jw
j j
2
K
Y
k =1 e
i2 !
r
k
J
Y
j=1 e
i 4 !
2e i 2 !
Re(w
j )+jw
j j
2
;
isatrigonometripolynomialwithrealoeÆientssuhthatjB(!)j 2
=A(!).
Pro of. LetP
A (z)=
P
M
k = M a
k z
k +M
. Thisp olynomialhas2M zeros(ount-
ingmultipliities) andsine theo eÆients arerealwehaveP
A
(z)=P
A (z)
for all z, so that if z is a zero, then z is a zero as well. Moreover, sine
a
k
= a
k
for k = 1;:::;M, it follows that P
A
(z) = z 2M
P
A (
1
z
) and this
implies that if z is a zero of P
A
, then so is z 1
. (The assumption a
M 6= 0
guarantees that P
A
(0) 6= 0.) Moreover, every zero on the unit irle has
even multipliity b eause z M
P
A
(z) is by assumption nonnegative on the
unitirle. 1isazero,thenweseefromtherelationP
A
(z)=z 2M
P
A (
1
z )that
itisatuallyazeroofevenmultipliity. Thisgivestheonlusionab outthe
zerosof P
A .
Thuswe anwrite P
A
in theform
P
A
(z)=a
M
K
Y
k =1 (z r
k
)(z r 1
k )
J
Y
j=1 (z w
j
)(z w
j
)(z w 1
j
)(z w
j 1
)
:
Sine we forevery z2C n0 have
j(e i2 !
z)(e i2 !
z 1
)j=jzj 1
j(e i2 !
z)(z e i2 !
)j
=jzj 1
je i2 !
zj 2
;
itfollowsfromthenonnegativityofAandthefatthatjA(! )j=jP
A (e
i 2 !
)j
that
A(!)=jA(!)j=jP
A (e
i2 !
)j=ja
M j
K
Y
k =1 jr
k j
1 J
Y
j=1 jw
j j
2
K
Y
k =1 e
i2 !
r
k
J
Y
j=1 (e
i2 !
w
j )(e
i2 !
w
j )
2
=jB(!)j 2
:
This ompletesthepro of.
If we want to onstrut a wavelet with ompat supp ort, the simplest
approah aording to Theorem 5.5 isto ho ose a p ositive integerN,take
L = 0 for simpliity, sine another hoie only amounts to a translation,
ho osethep olynomial in (5.9)tob eidentially zero,and soon.
Weleaveitasan exerisetoshowthatinthiswaywe getalter thatin
addition to (4.11) and (4.46) also satises (4.48) and therefore generates a
father wavelet orsaling funtion thatturns outtob eontinuous if N >1.
Infatonean saymuhmoreab outthesmo othnessof thesefuntionsbut
this question will notb estudied here.
4. Propertiesof ompatly supported wavelets
FirstweonsiderbrieythequestionofhowoneaneÆiently alulatethe
values ofthefuntion '.
Prop osition 5.7. Assumethat ((k ))
k 2Z
issuh that(k )=0whenk<
a or k>a
+ ,
P
k =a a
+
(k )=1 and '2C
(R), with'60, is a solution
to the equation
'(x)=2 X
k 2Z
(k )'(2x k ):
Then the matrix A dened by A(i;j) =2(2i j), i;j =a =1;:::;a
+ 1
has the eigenvalue 1, ('(a +1);:::;'(a
+ 1))
T
is an eigenvetor for this
eigenvalue and the values of ' at the points 2 j
n, j 1 an be reursively
alulated fromthe equation
'(2 j
n)=2 a+
X
k =a
(k )'(2 j+1
n k ); n2Z; j1:
Observethatwedonotlaim thattheeigenvalue1frothematrixAhas
geometrimultipliity1soitismaynotb elearwhiheigenvtortoho ose,
but in mostasesthis turns outnottob ethe ase.
Our next result restrits the smo othness of the saling funtion ' in
termsof thesupp ort of thelter .
Theorem5.8. If m 0 and f 2 C m
(R;C), f 6 0, has ompat support
and satises
f =2 a+
X
k =a
(k )f(2 k );
(5.16)
for some numbers f(k )g, thenm<a
+
a 1.
Pro of. If we apply Lemma 5.3,we see thatthe supp ortof f mustb e on-
tained in the losed interval [a ;a
+
℄. Thus the supp ort of f (j)
must also
b e ontained in this interval for0 j m. Moreover,dierentiating b oth
sides of (5.16) weget
f (j)
=2 j+1
a
+
X
k =a
(k )f (j)
(2 k ):
(5.17)
Let A b e a matrix with elements A(i;j) = 2
2i j
for i, j = a +
1;:::;a
+
1 (the indexing is nonstandard but this is of no onsequene).
Nowweseefrom(5.17)thatifthevetor(f (j)
(a +1);f (j)
(a +2);:::;f (j)
(a
+
1)) T
is not the zero vetor, then it is an eigenvetor of the matrix A or-
resp onding to the eigenvalue 2 j
. We leave it as an exerise to show that
this vetor annot b e thezero vetor. Thus A has at least m+1 distint
eigenvalues so thatA mustb e atleast an m+1m+1 matrix. Thus we
seethat m+1a
+
a 1 andthis givesthedesired onlusion.
Next we show that exept for the Haar funtion, no father wavelet or
saling funtion foramultiresolutionan notb e symmetriwithresp et to
anyp oint.
Prop osition 5.9. Let (fV
m g
m2Z
;') be a multiresolution of L 2
(R;C) suh
that ' is real-valued and has ompat support. Then ' is not symmetri
(norantisymmetri) withrespetto anypoint unless'is theHaar funtion
[0;1℄
.
Pro of. Itislear thatweannothave'(+)= '( )forsome2R,
b eausethenwewouldhave R
R
'(x)dx=0whihisimp ossible byTheorem
4.9.
Supp ose onthe otherhand that'(+)='( ). Ifisan integer,
thenantakeanintegertranslationof',sowemaywithoutlossofgenerality
asumethat'isan evenfuntion. Itfollows thatthelter isevenaswell
andhas ompatsupp ort. Weshall showthatthisleads toaontradition.
Let us intro due the notation that if p is a trigonometri p olynomial
withp erio d 1,i.e.,p= P
k
^ p(k )e
i2 k
,then
N
+
(p)=maxfkjp(k )^ 6=0g;
N (p)=minfkjp(k )^ 6=0g:
It iseasytohekthat
N
+ (jpj
2
)= N (jpj 2
)=N
+
(p) N (p):
(5.18)
Let e
b e the sequene e
k
= 1
2
(1+( 1) k
)(k ) with nonzero even indies
and o
thesequene o
k
= 1
2
(1 ( 1) k
)(k )withnonzeroo ddindies. Sine
e
= 1
2
(()+^ (^ + 1
2 ))and
o
= 1
2
(()^ (^ + 1
2
)),itfollowsfrom(4.11)
that
j
e
j 2
+j
o
j 2
= 1
2 : (5.19)
Sineneither e
nor o
annot b eidentiallyzero(b eause
e
(0)=
o
(0)=
1
2
) itfollows from(5.18) that
N
+ (
e
N
+ (
e
=N
+ (
o
N
+ (
e
: (5.20)
From thedenition of e
and o
we get
N
+
()^ =maxfN
+ (
e
);N
+ (
o
)g;
N ()^ =min fN (
e
);N (
o
)g;
Ifweombinethis resultwith(5.20)we onlude that
N
+
()^ N ()^ =maxfN
+ (
e
) N (
o
);N
+ (
o
) N (
e
)g:
(5.21)
SineN
(
e
)areevennumb ersandN
(
o
)areo ddnumb ers,itfollowsthat
N
+
()^ N ()^ isano ddnumb er. But then annotb ean even sequene
and we have aontradition.
Assume next that '(+)= '( ) where is not an integer. We
mayagainshiftthefuntion'sothat2(0;1). TakingFouriertransforms
weget
^ '()=e
4 i
^ '( ):
(5.22)
Butthen itfollows from (4.10)thatwealso have
^
()=e 4 i
^ ( ):
(5.23)
Now ^ and (^ ) are b oth trigonometri p olynomials with p erio d 1 and
therefore wemusthave= 1
2
. Thus wehave '(+1)='( ). It follows
from (4.8)after somehanges of variables that
2k +1
=
2k
forall k2Z.
Sine 'isreal-valued, is real-valued aswell, andhene we have
e
=
o
; (5.24)
and ombining thisresult with(5.19) we get
j
e
j 2
= 1
4 : (5.25)
It followsthat thereexistsan index ksuhthat
j
= 1
2
when j=2k+1 or
j= 2k . If k=0,thenwe getthe Haarfuntion and otherwise we get
^ =e
i
os((4k+1)):
(5.26)
Butthen itfollowsfrom Prop osition 4.15that (4.48)annothold true,and
this ontraditsTheorem4.13.
