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(!;)whe re A is the 1-periodic function (2t) + <1>(2t- l ). Prove that for some c,
From Parseval's identity for Fourier series. we have j 1 \A(i;) 1 2 dt; 0 Parseval's identity for Fourier transforms, it follows that
LnEC£:
\a, 1 2 . From
(6.4.4)
=
~ 1'~
=
~
11
=
1
IA(l;)l
1
2
2
IA
+
/)12 dl;
(.2: l(~
+ /)\ 2 )
\A(l;)\2\
1
2
0
i
d~.
To prove that (i) implies (ii) in Proposition 6.4.8, we simply note that the integrand in parenthese s in ( 6.4 .4) is bounded above by C and thus the integral is bounded above by 1 C .fr1 \A(.!;)\ 2 = C L,. Ia, I!. similarly for the lower bound, which proves (ii). To prove that (ii) implies (i). we use the abov~ transforma tions to rewrite the Riesz condition (6.4.2) in the form
(6.4.5)
cs
fol
\A(l;)\2
(L,Ez: l
\A(l;)\2dl;
S C.
This holds for every trigonomet ric polynomia l A(l;). Taking a sequence AN that converges boundedly to the indicator function of the interval [a, b] c (0, 1) (the partial sums of the Fourier series of 1 1" ;, 1 will suffice for this purpose), we obtain
This holds for every interval (a, b). Taking a sequence with (a, b) ---+ {x} and applying Lebesgue's differentiat ion theorem, we obtain (i).
•
Example 6.4.12. Returning to the tent function A (t), we have sin lT~ )4 (L \A(~+ /)12) = L ( 7r(~ + l) IE''l'.
fEZ
308
INTRODUCTION TO FOURIER ANALYSIS AND WAVELETS
This sum can be evaluated by repeated differentiation of the series from Example 6.4.10, namely rr 2 esc 2 rr~ = """' ~ (~ IEZ
1
+
l)2
to obtain the identity . sm rr 5t-
"""'
fa
(
JT(~
+
l)
)4
I
-
-(1
3
+
2cos 2 rr~).
which again reaffirms that c = 1/3, C = I for this Riesz system. Exercise 6.4.1 3. Check the details of this computation.
We can use Corollary 6.4.9 to estimate the support of the Fourier transform of a scaling function. Corollary 6.4.14. Suppose that E L 2 (JR) and that { (t - k) hEZ is an orthonormal set. Then Isupp I > 1, with equality if and only if = 1K for some measurable set K with iKi = 1. Proof. From Corollary 6.4.9 we have Lta lei>(~ From Parseval's identity
+ /)1 2
-
I a.e., hence lci>(~)l < I a.e.
which proves that 1supp ci> I ~ l. If equality holds, then the middle terms give
But the integrand is nonnegative a.e., hence I -lci>(~)l 2 means that ci> = IK a.e., where IKI = lsupp 1 = l.
=
0 a.e. on the support of ci>, which •
Proposition 6.4.8 allows us to obtain the following orthogonalizatio n procedure to generate scaling functions from a Riesz sequence. Proposition 6.4.15. Let EL 2 (JR) be such that {
+ I) 2 = 1
1 a.e.
INTRODUC TION TO WAVELETS
:= L lcPI (~ +
309
B (!;) cP (!;)
/)1 2 =LIB(~+ 01 2 lcP(~
feZ.
+ 01
2
IE:...
=
IB(l;)l 2
L
lcP(~
+
1)1
2
.
IEZ
Therefore we must choose the constants bn so that
1
IB(~)I2:=1I;:bne-hm
nEZ
lcP(l;+/) 12. /e2
Clearly there are many possible solutions . The simplest one is to take the positive square root, leading to (6.4.6)
This is clearly the Fourier transform of an L 2 function, since the denomina tor is bounded above and below by the Riesz condition . To prove the last statement , we need to study the equation (6.4.7)
L nEZ
an
=
L
Cn
neZ
and to show that, given (a,) E / 2 (';!,), we can solve for (c of Fourier transform s, this is written
11 )
E
l 2 (Z) and conversel y. In terms
Recalling the relation between <1> 1 and
an
= b,,
Cn
= 8,o. Making
(6.4.8) := B(l;)C(l;) .
But we have shown above that c-J ::::: I LnEZ bne-Zrrm<. I ::::: c- 1 from the Riesz property. Hence, given (a 11 ) E / 2 (Z), we may solve (6.4.8) uniquely by taking en as the Fourier coefficien ts of the !-periodi c function A(l;)/B(l;). Converse ly, given C , one simply refers 11 to (6.4.8) and chooses a 11 as the Fourier coefficien ts of the right side. •
The above proof shows that the set of function s describe d by the left side of (6.4.7) when (a,) E / 2 (2) is identical to the set of function s describe d by the right side when (c,) E / 2 (2). But the set on the right is a closed subspace of L 2 (lR), from the orthonor mality of {
310
INTRODUCTION TO I Ol'RIER At'.AI YSI'-; ,'\NO \V/\Vl-1 I'T<;
Example 6.4.16. In the case of the tent function <1> Fourier transform is obtained as ~
<1>,(.;)=
(sin rr.; jrr.;-> 2
.J(T +
2cos 2 rr~)/3
A, the orthogonalized
.
This corresponds to the MRA of Example 6.4.4, where the functions are continuous and piecewise linear on each dyadic interval. We write <1> 1 (t) = Ln bnAU- n), which is clearly a piecewise linear continuous function with <1> 1 (n) = b 11 • The coefficients are obtained from the Fourier expansion 1
(6.4.9) ./ (I
+
2 cns 2 rr E.) /3
-_ "" L..,; b ne 2:r ml; . nco:·.
Since the left side is a real analytic function, the Fourier coefficients have an exponential decay. Exercise 6.4.17. Show that there exist constants K > 0, Ke-fllnl and obtain an estimate for {3.
6.4.2
f3
>
0 so that Ibn I <
Scaling Equations and Structure Constants
The axioms describing an MRA system are not completely independent of one another, as we will show. First we note a simple consequence of properties (ii) and (iii) from Definition 6.4.2. Proposition 6.4.18. For each j basis ofV1•
E
Z. {21 12
Proof. From property (ii). ~ and \1( 1 are isomorphic by virtue of the map x --+ 2· 1x. The indicated functions are clearly orthonormal. Now we pull back to V 0 and use property (iii) .
•
sum
In order to proceed further, we discuss the consequences of the inclusion Vo c V 1 • Since V 1 is spanned by translates of { (2t - n)} 11 ,~::_:, we have the L 2 convergent
(6.4.10)
(f)=
L
a 11 <1>(2t- n)
nEG
where the structure constants satisfy LnE:L la 11 12 < CXJ. Relation (6.4.10) is called the scaling equation and will be instrumental in the sequel. Example 6.4.19. If
INTRODUCTION TO WAVELETS
Hint: Take the Fourier transform of both sides and iterate to solve for
311
.P.
The next exercise shows that a similar two-scale difference equation can have radically different solution behavior. L 1 (lR) satisfies the functional equation 3<1>(3t- 1). Prove that
Exercise 6.4.21. Suppose that
+
E
¥ 0 must agree with the Fourier transform of the Cantor measure.
The next example shows that the existence of a scaling equation does not follow from the orthogonalit y properties.
Example 6.4.22. Let
E / 2 (Z)
1r-~.~J(t).
Wesuppose that(6.4.JO )holdsforso me
and obtain a contradictio n.
To see this, we have from the orthogonalit y of { (2t - n) }nE:Z
a, =
2l
But this integral is nonzero unless n = 0, ± 1, in which case we obtain a 0 = 1, a± 1 = But this leads to a contradictio n on the interval ~ < t _::::: ~, where the left side of (6.4.1 0) is zero but the right side is nonzero. We record some properties of the structure constants.
!.
Proposition 6.4.23. The structure constants obey the following properties: (6.4.11)
ak =
2l
L
(6.4.12)
lakl
2
kEZ
= 2
kE:Z
L
(6.4.13)
ak'a2k+k' = 28k0
(Kronecker delta).
k'E:Z
/falso
E
L 1 (lR), JTJ?. # Oand(6.4.1 0) converges inL 1 (lR), then
(6.4.14) Proof. Since a, I h are the Fourier coefficients of <1> E V 1 with respect to the orthonormal basis vtz
l
=
00k·
312
INTRODUC TION TO I"OURIER ANALYSIS AND WAVELETS
Substitut e (6.4.10) and use Parseval' s identity and orthogon ality to write 8ok
=
L 1
2
f
ak'ak"
k'.k''
z=
JTR
ak·ak"·
2k+k'=k"
which is the same as (6.4.13). In particular , taking k = 0 we retrieve a new proof of (6.4.12). If, in addition, we have
r <~>
JTR
which we divide by
kEZ
•
JrR
It is often useful to work with (6.4.1 0) in the Fourier domain. The Fourier transform of
l
l
n)e- 2 rra~ dt = & =
-2rri~ ( n; u) Jdu
&e-inrr~~(~)
so that ( 6.4.1 0) is written (6.4.15) where the scaling filter is defined by (6.4.16)
mo(O :=
~ L ane-2rrin~. nEZ
The existenc e of a scaling equation can be formulat ed in the Fourier domain as follows where L 2 (1R/Z) denotes 1-period ic function s that are square-i ntegrabl e on any period. Proposi tion 6.4.24.
INTRODUCTION TO WAVELETS
313
Example 6.4.25. The Shannon scaling function is I
sin(.rrt) =
!2
-~
e 2rrtti'; dl:: ., .
The scaling equation can be obtained at the level of Fourier transforms by solving (6.4.15) as follows: 1r-t.1J(/;) = mo
(~) 1 r-t.~J (~) ·
Therefore we need to choose the coefficients so that mo(l;) = 1 for II; I < ~and mo(/;) = 0 for ~ < II; I < The structure constants are obtained from ( 6.4.1 0) as the Fourier coefficients
!.
Thusao = 1 andan = (-2/n.rr)sin( n.rr/2). Exercise 6.4.26. Consider the spline function A(t) = (1 - lt1)1r- 1• 11 (t) with cP(/;) = (sin .rrl; /.rr/;) 2 . Show that A satisfies a scaling equation (6.4.10) and exhibit the scaling filter m 0 (1;). Use this to infer that its orthogonaliz ation, defined
by cP(/;) = A(/;)/ )Lt;EZ IA(/;)1 2 • also satisfies a scaling equation (6.4.10).
6.4.3
From Scaling Function to MRA
We now prove an important theorem, showing the existence of MRA systems under useful hypotheses. Theorem 6.4.27. Suppose that
(i) The translates {
CXl.
Define\')= span {
PJ =
=
2JI 2 ct>(21t- k)
L (/' ct>jk) ct>jk. kEZ
These projection operators satisfy the bounds IIP1 /II ::=:: 11/11. To prove the MRA property, it suffices to show that limj~oo P1 / = f and lim1 ~-oc P1 f = 0 for allf E L 2 (~). This is done in two separate lemmas.
314
INTRODUCTION TO FOURIER ANALYSIS AND\,, ' :' : ·.,.,
Proof. Since IIP1 11 = 1, it suffices to prove the result on a dense set, e.g., L 2 functions with compact support. Iff has support in [ -R, R], then
fu 2 (f: = 11/112 fu f_~::J2~R
=
11111:
1
14>(2
1
s- k)l ds) 2
j
~, then these integrals are over disjoint intervals whose union is written U 1 UkEJ::(-k- 2 1 R, -k + 2 1 R), with n1 U1 = Z, which has Lebesgue measure zero. Therefore
If 2 1 R <
by Lebesgue's dominated convergence theorem.
•
To proceed further, we now tum to the Fourier domain and prove a useful identity. L 2 (JR) with a Fourier transform j that is bounded and supported in [-R, R]for some R > 0. Thenfor 2j-i > R we have Lemma 6.4.29. Let f
E
(6.4.17) Proof. We use Parseval's identity to write
L I(PJ,
IIPJII 2 =
KE.Z
2
kEZ
where we have used the fact that the Fourier transform of the function
1:--.:TRODUCTION TO WAVELETS
Fourier series, we have for 2'-
1
315
> R 1 '
2
IIPJII =
I
~~~ lf
JR lf
=
1
2
dt;.
- R
•
which completes the proof.
Corollary 6.4.30. Suppose that the scaling function satisfies the additional condition that (l;) is continuous at l; = 0 with l(O)I = 1. Then for any f E L 2 (JR), IIP1 f - !II ---+ 0 when} ---+ CXJ. Proof. Since P, is a contraction, it suffice~ to prove this on the dense set off whose Fourier transforms have compact support and are bounded. Furthermore, from the projection property, we have IIP,/11 2 = 11/11 2 - llf- PJII 2 • so we must show that IJP1 /II---->- 11/11. Using the hypothesis, we see that 1<1>(2-'/;)l converges uniformly to l on compact sets, so that (6.4.17) give;, for 21 1 > R
IIP,/11 2 = ---->-
1:
J:
llcO(r1 )/;l 2 dt; 2
1lct;)l dt;
= llfll".
•
which completes the proof.
Combining the above lemmas and corollaries completes the proof of the theorem. The following exercise provides a one-parameter generalization of the Shannon wavelet. Exercise 6.4.31. Let K = [a - 1, a] where 0 < a < 1 and set that is the scaling function of an MRA.
1K. Prove
Hint: First check that has orthonormal translates. To find the scaling relation, it suffices to find m 0 E L 2 (1R/Z), by solving <1>(2/;) = m 0 (t;)(t;).
Example 6.4.32. One should not infer from the continuity at I; = 0 that is continuous elsewhere, much less that E L 1 (lR). Consider the Shannon scaling function, where