Truncation error on Wavelet Sampling Expansions
N.Atreas and C.Karanikas
Abstract
We estimate the truncation error of sampling expansions on translation invariant
spaces, generated by integer translations of a single function and on wavelet subspaces of
L2 (R). As a byproduct of the main result, we get the classical D.Jagerman’s bound for
Shannon’s sampling expansions. We also examine this error on certain wavelet sampling
expansions.
Key Words: Translation invariant spaces, sampling, Truncation error, wavelet subspaces, multiresolution analysis.
1.
Introduction
Let V be a closed subspace of L2 (R) generated by a continuous real valued function φ(x), such
that {φ(. − n), n ∈ Z} is an orthonormal basis on V ; it is clear that V is a shift invariant space
1
and its reproducing kernel K(x, t) is given by:
K(x, t) =
X
φ(x − n) φ(t − n),
(x, t) ∈ R2 ,
n∈Z
(see [9] and [10]).
Throughout this work, we assume that the function φ satisfies the following conditions:
(i) |φ(x)| ≤ (cons.)
(ii)
X
|B(x)|
, where ε ≥ 0 and |B(x)| is bounded and 1-periodic function on R.
|x|1+ε
φ(n) e−inγ converges absolutely to a function which has no zeros on [−π, π].
n∈Z
It is known that conditions (i) and (ii) imply that {K(x, n), n ∈ Z} is a Riesz basis on
V , with a unique biorthogonal Riesz basis {S(x − n), n ∈ Z}, (see [10] Proposition (9.1) and
[11]); moreover, the function S is determined by its Fourier transform
Ŝ(γ) = X
φ̂(γ)
, γ ∈ R.
φ(n) e−inγ
n∈Z
It is also well known that for each f ∈ V we may write
f (x) =
X
< f, K(. , n) > S(x − n)
n∈Z
and since f (n) = < f, K(. , n) >, n ∈ Z we have
f (x) =
X
f (n) S(x − n), x ∈ R,
(1.1)
n∈Z
where the convergence is in the L2 (R) sense. It is clear that S has the following interpolation
property
S(n) = δ(0 , n), n ∈ Z,
2
where δ(0 , n) is the Kronecker’s delta. Thus S(x) is a sampling function of V ; other properties
of sampling functions can be found in [3] and [10].
The idea to derive sampling series from reproducing kernels on Hilbert spaces is not new;
see [6], [9] and [13]. Moreover from the theory of frames (a notion more general than the notion
of a basis in a Hilbert space), one can obtain general sampling series involving regular as well
irregular sample points; see [1] and [2]. Notice that in case where the space V is generated by
a Riesz basis (which is a special case of frames), one can apply the orthonormalization trick to
obtain an orthonormal basis (see [5] p. 139-140).
We may obtain a more general form of (1.1). Consider the spaces VW , (W > 0) determined
by the orthonormal basis {W 1/2 φ(W. − n), n ∈ Z}, i.e. VW is a translation invariant space
under the action of n/W, n ∈ Z. As above, any f ∈ VW has a unique sampling expansion:
X
f (x) =
n∈Z
f(
n
) S(W x − n), x ∈ R.
W
(1.2)
Some particular translation invariant spaces satisfying (1.2) for W = 2m , m ∈ Z, are obtained by a multiresolution analysis of L2 (R). Recall that a multiresolution analysis is a nested
sequence {Vm ⊂ Vm+1 , m ∈ Z} of closed subspaces in L2 (R) and a scaling function φ such
that:
(iii) for any m ∈ Z, {2m/2 φ(2m x − n), n ∈ Z} is an orthonormal basis of VW ,
(iv) f (x) ∈ V0 ⇔ f (2m x) ∈ Vm
(v) closure
[
Vm = L2 (R) and
T
m
Vm = {0}.
m
3
For more details about multiresolution analysis see [5] and [10]; in particular see [10] for an
overview on wavelet sampling expansions.
Since under some weak conditions on S(x) the L2 -convergence of the series (1.2) can be
strengthened, it is natural to examine the truncation error of these series. For example, if
X
n∈Z
|S(x − n)|2 and
X
|φ(x − n)|2 are bounded, the series (1.2) converges absolutely and
n∈Z
uniformly (see [10] p. 137-138).
For f ∈ VW and N ∈ N, the truncation error of the sampling expansion (1.2) is defined to
be
RN f (x) := f (x) −
X
|n|≤N
f(
n
) S(W x − n), |x| < N W −1 .
W
Truncation errors, which occur naturally in applications, have been studied rather extensively in engineering literature. For the case of band-limited functions f (i.e. fˆ(γ) = 0 outside
a bounded interval), this error has been estimated by D. Jagerman in [8]. By imposing some
extra conditions on f besides being band-limited, the truncation error can be decreased; see
[4], [7], [8], [14] and [15] for an overview of the results concerning this type of error.
In this work we get Jagerman-type estimates for |RN f (x)|, in case where f is in a translation
invariant space VW as above. In order to present the main Theorem, we need to express the
sampling function in terms of the sequence {bn }n∈Z described below:
Lemma 1. Let {φ(. − n), n ∈ Z} be an orthonormal basis of V satisfying conditions (i) and
(ii) and let S be the corresponding sampling function, then there holds
4
X
(a) S(x) =
bn φ(x + n), where
n∈Z
1 Zπ
1
X
e−inγ dγ, n ∈ Z.
−ikγ
2π −π
φ(k) e
bn =
(1.3)
k∈Z
(b) The sequence {bn } is in l1 .
(c)
X
φ(n) e−inγ = C ⇔ C bn = δ(0, n) ⇔ φ(x) = C S(x), where C is a non zero constant.
n∈Z
Theorem 1. We suppose that φ satisfies conditions (i) and (ii) . Let the constants C1 , C1+ , C1− ,
be C1 =
X
|bµ |, C1+ =
X
|bµ |, C1− =
"
N , C = supy∈[0,1)
X
|bµ | respectively. For any integer N > 0, |h−1 x| <
µ<0
µ>0
µ∈Z
X
#1/2
|φ(y − n)|2
"
, KN =
h
n
X
|f (nh)|2
1/2

#1/2
X
, LN =  h
n>N
|f (nh)|2 
,
n<−N
{bn }n∈Z as in (1.3), h = W −1 , and f ∈ VW we have:
|RN f (x)| ≤ (cons.) |B(h−1 x)| √
³
hε
1 + 2ε
´
ε
ε
KN FN,h
(x, C1 − C1+ , C1+ ) + LN FN,h
(−x, C1 − C1− , C1− )
+ C h−1/2
X
{|bν | (KN δ(ν, |ν|) + LN δ(ν, −|ν|))} ,
[h−1 x]
N +δ(1,λ)
|ν+ 2 |>
2
where for x, y, ω ∈ R

Ã
1
2 (1 + 2 ε)
ε
 y + 21/2+ε 1 +
FN,h
(x, y, ω) =
1/2+ε
(N h − x)
N − h−1 x
!1/2

ω ;
finally λ = 0 or 1 according to whether N + [h−1 x] is even or odd.
Next in Section 1, we prove Theorem 1. In Corollary 1, we give a simple version of the truncation error, in case where the sampling function coincides with φ. We also give in Proposition
5
1 an estimate of this error for the case of regular sampling functions. In Remark 2, we see that
Corollary 1 is a natural generalization of D. Jagerman’s result.
Finally, in Section 2 we apply the results of Section 1 on certain wavelet sampling expansions.
2.
The Main Theorem
Proof of Lemma 1.
(a) Since S ∈ V using Parseval’s formula and the fact that {φ(· − n), n ∈ Z} is an orthonormal
basis in V , i.e.
X
|φ̂(γ + 2 n π)|2 = 1, we get:
n
S(x) =
X
n∈Z
|φ̂(γ)|2
1 X Z∞
X
< S, φ(. − n) > φ(x − n) =
einγ dγ φ(x − n)
−ikγ
2π n∈Z −∞
φ(k) e
k∈Z
=
1 X X Z π |φ̂(γ + 2λπ)|2 inγ
1 X Zπ
X
X
e
dγ
φ(x
−
n)
=
−ikγ
2π
2π
φ(k) e
−π
n∈Z λ∈Z
n∈Z
k∈Z
e
−inγ
−π
1
φ(k)e−ikγ
k∈Z
dγ φ(x + n).
(b) It is a direct application of Wiener’s Lemma (on the absolutely convergent Fourier series
of a non-vanishing function).
(c) Clearly the condition
X
φ(n) e−inγ = C, implies that C bn = δ(0, n) and we get φ(x) =
n
C S(x). The converse follows easily.
Proof of Theorem 1. Let x be such that |h−1 x| < N and S be as in Lemma 1, then we have
RN f (x) := f (x) −
X
f (nh)
X
µ
|n|≤N
6
bµ φ(h−1 x + µ − n)
X
=
f (nh)
X
µ
n>N
X
+
bµ φ(h−1 x + µ − n)
f (nh)
X
bµ φ(h−1 x + µ − n)
(2.1)
µ
n<−N
1
2
= RN
f (x) + RN
f (x),
1
2
1
where RN
f (x), (or RN
f (x)) is the first (or the second) term of (2.1). First we deal with RN
f (x)
and we observe that


1
RN
f (x) =
X
f (nh) 
X
bµ φ(h−1 x + µ − n) +
µ>0
n>N
X
bµ φ(h−1 x + µ − n)
(2.2)
µ≤0
11
12
= RN
f (x) + RN
f (x),
11
12
12
where RN
f (x), (or RN
f (x)) is the first (or the second) term of (2.2). For the term RN
f (x) we
have
12
|RN
f (x)| ≤ (cons.)
X
|f (nh)|
X
|bµ |
|B(h−1 x + µ − n)|
|h−1 x + µ − n|1+ε
|bµ |
|B(h−1 x)|
.
|h−1 x + µ − n|1+ε
µ≤0
n>N
≤ (cons.)
X
|f (nh)|
X
µ≤0
n>N
(2.3)
Since h−1 x < N < n and µ ≤ 0, we have that |h−1 x + µ − n| = −h−1 x − µ + n, so (2.3) implies
12
|RN
f (x)| ≤ (cons.)|B(h−1 x)|
X
|f (nh)|
X
|bµ |
µ≤0
n>N
≤ (cons.)|B(h−1 x)|
X
|bµ |
µ≤0
X
|f (nh)|
n>N
1
(−h−1 x − µ + n)1+ε
1
,
(n − h−1 x)1+ε
where the inequality (≤) holds whenever bµ = δ(0, µ), µ ∈ Z, or at the zeros of B(h−1 x),
otherwise we have a strict inequality. Next we apply the Cauchy-Schwarz inequality and we
use the monotonicity of (y − h−1 x)−2−2ε for y > h−1 x to get
12
f (x)|
|RN
−1
≤ (cons.)|B(h x)|
X
µ≤0
Ã
|bµ |
X
n>N
7
!1/2 Ã Z
|f (nh)|
2
∞
N
1
(y − h−1 x)2+2ε
!1/2
≤ (cons.)|B(h−1 x)| √
X
hε
KN
|bµ |
.
(N h − x)1/2+ε
1 + 2ε µ≤0
(2.4)
11
f (x) be as in (2.2), then for ν = µ + [h−1 x] and y = h−1 x − [h−1 x] ∈ [0, 1) we have
Now let RN
X
11
RN
f (x) =
n>N
bν−[h−1 x] φ(y + ν − n)
ν>[h−1 x]
X
=
X
f (nh)
X
bν−[h−1 x]
f (nh) φ(y + ν − n).
(2.5)
n>N
ν>[h−1 x]
We split the first sum of (2.5), while for the second sum we take k = n − ν, thus if we define
λ = 0 or 1 according to whether N + [h−1 x] is even or odd, (2.5) implies
N +[h−1 x]+δ(1,λ)
2
X
11
f (x) =
RN
f ((k + ν)h) φ(y − k)
k>N −ν
ν=[h−1 x]+1
X
=
X
bν−[h−1 x]
X
bν−[h−1 x]
f ((k + ν)h) φ(y − k)
(2.6)
k>N −ν
N +[h−1 x]+δ(1,λ)
ν>
2
111
112
= RN
f (x) + RN
f (x)
111
112
where RN
f (x), (or RN
f (x)) is the first (or the second) term of (2.6). Now we have:
111
|RN
f (x)|
≤ (cons.)|B(h−1 x)|
N +[h−1 x]+δ(1,λ)
2
X

X
|bν−[h−1 x] | 
1/2 
|f ((k + ν)h)|2 
k>N −ν
ν=[h−1 x]+1

1/2
X
k>N −ν
1

|y − k|2+2ε
≤ (cons.)|B(h−1 x)|
N +[h−1 x]+δ(1,λ)
2
X
Ã
|bν−[h−1 x] |
≤ (cons.)|B(h x)|
X
Ã
X
|bν |
ν>0

|f (nh)|2
n>N
ν=[h−1 x]+1
−1
X
!1/2
2
X



k>
N −[h−1 x]−δ(1,λ)
2

1


2+2ε
|y − k|
!1/2
|f (nh)|
2
n>N



 N −[h−1 x]
1/2

1
+1−
1/2
2+2ε
δ(1,λ)
2

−y
X
+
k>
8
N −[h−1 x]−δ(1,λ)
+1
2

1


2+2ε
|y − k|
Ã
X
−1
≤ (cons.)|B(h x)|
|bν |
ν>0
!1/2
|f (nh)|
2
n>N


 N −[h−1 x]
2
X
2+2ε
1
δ(1,λ)
2
+1−

−y
+
1/2
Z ∞
N −[h−1 x]−δ(1,λ)
+1
2
1
dω 

2+2ε
|ω − y|
and since
N − [h−1 x]
δ(1, λ)
N − [h−1 x]
δ(1, λ) y
N − [h−1 x]
+1−
−y =
+1−
+ −y ≥
2
2
2
2
2
2
we get
111
|RN
f (x)| ≤ (cons.)|B(h−1 x)|
2
1/2
(2h)ε X
KN
√
|bν |
(N h − x)1/2+ε
1 + 2ε ν>0
Ã
2(1 + 2ε)
1+
N − h−1 x
!1/2
.
(2.7)
112
For the term RN
f (x) in (2.6) we have
Ã
X
112
|RN
f (x)| ≤
|bν−[h−1 x] |
N +[h−1 x]+δ(1,λ)
ν>
2
< C h−1/2 KN
X
X
1/2
!1/2 
X

|f (nh)|2
|φ(y − k)2 |
n>N
k>n−ν
|bν |.
(2.8)
N +[h−1 x]+δ(1,λ)
ν>
2
2
Similarly working for RN
f (x) we get
2
RN
f (x) ≤ (cons.)|B(h−1 x)| √
hε
1 + 2ε

X
LN

|bµ | + 21/2+ε
1/2+ε
(N h + x)
µ≥0
Ã
2(1 + 2ε)
1+
N + h−1 x
X
+C h−1/2 LN
!1/2
|bν |.

X
|bµ | 
µ<0
(2.9)
N +[h−1 x]+δ(1,λ)
ν<−
+1
2
ε
(x, y, ω) and combining together (2.4), (2.7), (2.8) and
Defining appropriately the function FN,h
(2.9) we have the result.
9
Remark 1. It is easy to see that the inequality (≤) in Theorem 1 occurs whenever bµ =
δ(0, µ) µ ∈ Z, or at the zeros of B(h−1 x).
Let us now assume that
X
φ(n) e−inγ = 1. Then we have:
n∈Z
Corollary 1. Let φ(x) = S(x), then for any integer N > 0 and ε ≥ 0, we obtain:
hε
|RN f (x)| ≤ (cons.)|B(h−1 x)| √
1 + 2ε
Ã
!
KN
LN
+
,
1/2+ε
(N h − x)
(N h + x)1/2+ε
(2.10)
where h, KN , LN are as in Theorem 1.
Proof. We apply Lemma 1 (c) and we modify the proof of Theorem 1 as follows: clearly, the
first term in the right hand side of (2.2) vanishes, while the second term is less or equal to the
right hand side of (2.4) (for bµ = δ(0, µ)). Similarly we obtain a simpler form of the right hand
side of (2.9) and so (2.10) follows.
Remark 2. In Shannon’s case (where the spaces VW consist of πW − bandlimited functions),
we have that φ(x) = sin(π x)/(π x) = S(x) satisfies conditions (i) and (ii), thus for ε = 0 we
apply Corollary 1, where (cons.) = 1/π, B(x) = sin(π x), KN and LN are as in Theorem 1, to
obtain the exact Jagerman’s bound (see also [8])
| sin(π h−1 x)|
|RN f (x)| ≤
π
Ã
!
KN
LN
+
.
1/2
(N h − x)
(N h + x)1/2
In case we know the decay of the sampling function directly (rather than of φ(x)), we have
Proposition 1. If |S(x)| ≤ (cons.)|Γ(x)||x|−1−ε , where |Γ(x)| is bounded and 1-periodic function satisfying |Γ(x)| = |Γ(x − n)|, n ∈ Z, x ∈ R, ε ≥ 0 and h = W −1 , KN and LN are as in
10
Theorem 1, then for N = 1, 2, . . ., we have
hε
√
|RN f (x)| ≤ (cons.)|Γ(h x)|
1 + 2ε
Ã
−1
!
KN
LN
+
.
(N h − x)1/2+ε (N h + x)1/2+ε
Proof. It is similar to previous arguments.
3.
Applications
It is well known that most multiresolution analuses induce sampling expansions for continuous
functions f , in the space L2 (R) of all square integrable functions. In this section, we apply the
results of Section 1 on certain wavelet sampling expansions, as for example is the wavelet of
Meyer. Recall that the scaling function φ(x) of Meyer’s wavelet (see [5]), is given by its Fourier
transform
φ̂(γ) =










1
|γ| ≤
3|γ|
cos[ π2 ν( 2π − 1)],









0
2π
3
2π
3
< |γ| <
|γ| ≥
4π
,
3
4π
3
where ν(γ) is a real valued function satisfying ν(γ) + ν(1 − γ) = 1.
Example 1. Let ν(γ) = γ, then
φ(x) =
9 sin(2 π x/3) + 12 x cos(4 π x/3)
.
16 π x (9/16 − x2 )
Without loss of generality we consider x ≥ 0. We observe that |φ(x)| ≤
bn =
1 Zπ
1
sin(2 π n/3)
−inγ
X
e
dγ
=
2π −π
πn
φ(k) e−ikγ
k∈Z
11
3
2.216 x−2 and
4π
1 Zπ
1
+
cos(nγ) dγ, n ∈ Z;
π 2π/3 sin(3 γ/4) − cos(3 γ/4)
using twice integration by parts we get
Ã
!
3
3 Zπ
3 + sin(3 γ/2)
bn =
cos(2 π n/3) −
cos(nγ) dγ ,
4πn2
4 2π/3 (sin(3 γ/4) − cos(3 γ/4))3
so for n 6= 0, |bn | <
X
3
and b0 ≈ 0.9311. Now we apply Theorem 1 for ε = 1, C1 =
n−2
2
4πn
ν∈Z
and we take into account that C1+ = C1− = π 2 /6 to obtain
|RN f (x)| <
h
3
2.216 √
4π
3
Ã
!
1
KN FN,h
(x, 0.9311
+
3
C h−1/2
4π
π2 π2
π2 π2
1
+ , ) + LN FN,h
(−x, 0.9311 + , )
6 6
6 6
n
X
|ν+
o
n−2 (KN δ(ν, |ν|) + LN δ(ν, −|ν|)) .
N +δ(1,λ)
[h−1 x]
|>
2
2
Example 2. Let δ ∈ (1/2, 1] and



γ−δ

 π2 arccot[ 1−γ−δ
] 1/2 ≤ γ ≤ δ
ν(γ) =




1
γ>δ
Using ν(γ) + ν(1 − γ) = 1, we extend ν(γ) on [0, 1]. It is not difficult to see that
S(x) = 3
It is clear that |S(x)| ≤
sin(π x) sin(2 π (δ − 0.5) x/3)
.
2 π 2 (δ − 0.5) x2
3
| sin(π x)||x|−2 , so we apply Proposition 1 to get
2 π (δ − 0.5)
3
h
|RN f (x)| ≤ 2
| sin(h−1 πx)| √
2π (δ − 0.5)
3
12
Ã
!
LN
KN
+
.
3/2
(N h − x)
(N h + x)3/2
Example 3. Let
ν(γ) =
Then S(x) = 36





2
arccot[ 2γ12
π
− 1]
0 ≤ γ ≤ 1/2

2


 2 arccot[ 2(γ−1) 2 ]
π
(1−2(γ−1) )
1/2 < γ ≤ 1
sin(πx) sin2 (π x/6)
36
. Since |S(x)| ≤ 3 | sin(πx)||x|−3 , we apply Proposition
3
3
π x
π
1 and we have
36
h2
|RN f (x)| ≤ 3 | sin(h−1 π x)| √
π
5
Ã
!
LN
KN
+
.
(N h − x)5/2 (N h + x)5/2
Example 4. Let ψ(x) = [sin(2πx) − sin(πx)]/(πx) be the Shannon wavelet, such that for fixed
j ∈ Z, {2j/2 ψ(2j . − n), n ∈ Z} is an orthonormal basis of the orthogonal complements Wj of
Vj in Vj+1 . It is easy to see that ψ(x) = S(x), thus we apply Corollary 1 to get
| sin(2j+1 π x) − sin(2j π x)|
|RN f (x)| ≤
π
Ã
!
KN
LN
+
.
1/2
(N h − x)
(N h + x)1/2
Example 5. In this example, we estimate the truncation error of a class of band-limited
wavelets such that φ(x) = S(x); this class looks similar to Meyer’s original function, but φ̂(γ)
has the interpolating property which Meyer’s function does not share; this is made possible by
allowing φ̂(γ) to take complex values. An extensive analysis on this class is originally given in
[12]. Let θ(γ) be a real valued function such that, for some k ∈ Z and 0 < ² ≤ π/3 we have
(a1) θ(−γ) = −θ(γ) + 2 k π, k ∈ Z
(b1) |θ(γ)| = (2 k + 1) π, |γ| ≥ π + ²
(c1) θ(γ) + θ(2 π − γ) = (2 k + 1) π, 0 ≤ γ ≤ 2π.
13
It is easy to see that if φ̂(γ) = (1 + ei θ(γ)/2 )/2, |γ| < 2π, then φ(x) is an orthonormal scaling
and sampling function, i.e. (φ(x) = S(x)). Let π − ² < α ≤ π; we define the following odd
function
θα,² (γ) =















π
γ
2(α−π+²)
−
π(π−²)
2(α−π+²)
π/2
π−²≤γ <α
α ≤ γ ≤ 2π − α



π

γ + π(2α−3π+²)


2(α−π+²)
2(α−π+²)








π
2π − α < γ ≤ π + ²
γ ≥π+²
Obviously θα,² (γ) satisfies (a1), (b1) and (c1) for k = 0; the corresponding sampling function
is the following
Sα,² (x) =
sin(π x) [cos(² x) + sin((α − π) x))]
.
x [π + 2(α − π + ²) x]
For α = π, we get Volkmer’s example (see [12]). By straight calculation using Corollary 1, we
obtain the truncation error
Ã
!
| sin(h−1 π x)|
cos(² x) + sin((α − π) x)
KN
LN
|RN f (x)| ≤
sup |
|
+
.
3/2
π
π + 2 x (α − π + ²)
(N h − x)
(N h + x)3/2
x
Acknowledgments. We would like to thank Professor Gilbert Walter for his invaluable suggestions; in particular for designating us example 5 and that the main result can be extended
on translation invariant spaces.
This paper is Research supported by GSRT (Greece) Pened program No 1914 (1996-98).
14
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15
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