2 and 3 Refinable Functions
David Malone
22 December 1999
1
Refinable Functions
A 2-refinable function satisfies:
X
f (x) =
ck f (2x − k).
k
The equation is a dilation equation.
Harr function
D4
People are usually interested in
compactly supported L1 solutions for
P
wavelets with
ck = 2. In this case
there is at most one solution.
2
L2 solutions are not unique. They are in
correspondence with L2 (±[1, 2)) or
L∞ (±[1, 2)). However, adding more
scales gave uniqueness in operator
results.
What functions are 2 and 3 refinable?
ie.
X
X
f (x) =
ck f (2x−k) =
dk f (3x−k).
k
k
Assumptions:
• Only finitely many non-zero ck and
dk .
• Functions compactly supported.
3
Left Hand End
Lemma 1 Suppose
P
g(x) = dk g(2x − k), and only finitely
many of the dk are non-zero. Then we
can find l so that when we translate g by
l to get f we find:
P
f (x) = ck f (2x − k), c0 6= 0, ck = 0
when k < 0 and ck = dk−l .
Lemma 2 If f is compactly supported
and satisfies a dilation equation
P
f (x) = ck f (2x − k), where c0 6= 0 and
ck = 0 when k < 0, then f is zero
almost everywhere in (−∞, 0).
4
f (2x − 2)
c2
f (2x − 1)
c1
c0
f (2x)
0 f (x) 1
f (x)
=
2
n
X
ck f (2x − k)
k=0
f (x)
=
n n+1
x∈
,
2
2
Pn
x
f 2 − k=1 ck f (x − k)
c0
x ∈ [n, n + 1).
5
If you do this for two scales, everything
lines up and we get:
f (x) = c0 f (2x) = d0 f (3x),
on [0, 1/2) ∩ [0, 1/3).
Theorem 3 Suppose f is 2 and 3
refinable, say:
X
X
f (x) =
ck f (2x−k) =
dk f (3x−k),
k
k
and c0 6= 0 and ck = 0 when k < 0.
Suppose further that f is integrable on
some interval [0, ], then f (x) = γxβ on
[0, 1) where β = − log2 c0 = − log3 d0 .
6
In Lp this will be an almost everywhere
relationship. So we integrate:
Z x
F (x) =
f (t) dt.
0
We can show this satisfies:
n m
2
3
n m
F (2 3 α) =
F (α)
c0
d0
Continuity forces log2 c0 = log3 d0 = −β.
So:
F (2n 3m α) = (2n 3m )β+1 F (α)
One continuous function does this:
β+1 F (α)
x
αβ+1
7
Giving f on [0, 1/3):
d
β F (α)
f (x) =
F (x) = (β + 1)x β+1 .
dx
α
From here it is easy so show f must
have the form:
f (x) =
n
X
al (x − l)β ,
l=0
on [n, n + 1). For this to be compactly
supported β ∈ N. This means that f
must be a B-spline.
8
The fact that f behaves like xβ on [0, 1)
is interesting, and actually holds in a
much looser sense if you have function
which just solves one dilation equation.
There is a measure of smoothness called
the Hölder exponent, where f ∈ C n+s if
f is n times differenciable and:
|f (x + h) − f (x)| < k|h|s .
Now, xβ ∈ C n+s for n + s < β, and it
can be shown that for f ∈ C n+s then
n + s < β.
I have examples which can produce f
which are almost this smooth, and in
some simple cases this gives the correct
smoothness.
9
10
|c0 |
0.683
0.470
0.325
0.2264
0.1577
0.1109
N
2
3
4
5
6
7
3.1831
2.6644
2.1429
1.6179
1.0878
0.550
− log2 |c0 |
Coefficents
1.682
1.432
1.177
0.913
0.636
0.339
p226
2.158
1.888
1.596
1.275
0.915
0.500
p232
TLoW
1.6179
1.0878
0.550
p239
2.4348
2.1637
1.9424
1.6066
1.0831
0.550
r20
1.6179
1.0878
0.550
r∞
SRCfSS
2.4604
2.1891
1.9689
1.6179
1.0878
0.550
UB