Correction of the exercises from the book A Wavelet Tour of Signal Processing Gabriel Peyré Ceremade Université Paris-Dauphine [email protected]Nov. 2009 Abstract These corrections refer to the 3 rd edition of the book A Wavelet Tour of Signal Processing – The Sparse Way by Stéphane Mallat, published in December 2008 by Elsevier. If you find mistakes or imprecisions in these corrections, please send an email to Gabriel Peyré ([email protected]). More information about the book, including how to order it, numerical simulations, and much more, can be find online at wavelet-tour.com. 1 Chapter 2 Exercise 2.1. For all t, the function ω → e -iωt f (t) is continuous. If f ∈ L 1 (R), then for all ω, |e -iωt f (t)| 6 |f (t)| which is integrable. One can thus apply the theorem of continuity under the integral sign which proves that ˆ f is continuous. If ˆ f ∈ L 1 (R), using the inverse Fourier formula (2.8) and a similar argument, one proves that f is continuous. Exercise 2.2. If |h| =+∞, for all A> 0 there exists B> 0 such that B -B |h| >A. Taking f (x)=1 [-A,A] sign(h(-x)) which is integrable and bounded by 1 shows that fh(0) = B -B sign(h(t))h(t)dt > A. This shows that the operator f → fh is not bounded on L ∞ , and thus the filter h is unstable. Exercise 2.3. Let f u (t)= f (t - u), by change of variable t - u → t, one gets ˆ f u (ω)= f (t - u)e -iωt dt = f (t)e -iω(t+u) dt = e -iωu ˆ f (ω). Let f s (t)= f (t/s), with s> 0, by change of variable t/s → t, one get ˆ f s (ω)= f (t/s)e -iωt dt = f (t)e -iωst |s|dt = |s| ˆ f (sω). 1
Transcript
Correction of the exercises from the bookA Wavelet Tour of Signal Processing
AbstractThese corrections refer to the 3rd edition of the book A Wavelet Tour of Signal Processing
– The Sparse Way by Stéphane Mallat, published in December 2008 by Elsevier. If youfind mistakes or imprecisions in these corrections, please send an email to Gabriel Peyré([email protected]). More information about the book, including howto order it, numerical simulations, and much more, can be find online at wavelet-tour.com.
1 Chapter 2
Exercise 2.1. For all t, the function ω 7→ e−iωtf(t) is continuous. If f ∈ L1(R), then for all ω,|e−iωtf(t)| 6 |f(t)| which is integrable. One can thus apply the theorem of continuity under theintegral sign
∫which proves that f is continuous.
If f ∈ L1(R), using the inverse Fourier formula (2.8) and a similar argument, one proves thatf is continuous.
Exercise 2.2. If∫|h| = +∞, for all A > 0 there exists B > 0 such that
∫ B−B |h| > A. Taking
f(x) = 1[−A,A] sign(h(−x)) which is integrable and bounded by 1 shows that
f ? h(0) =∫ B
−Bsign(h(t))h(t)dt > A.
This shows that the operator f 7→ f ? h is not bounded on L∞, and thus the filter h is unstable.
Exercise 2.3. Let fu(t) = f(t− u), by change of variable t− u→ t, one gets
fu(ω) =∫f(t− u)e−iωtdt =
∫f(t)e−iω(t+u)dt = e−iωuf(ω).
Let fs(t) = f(t/s), with s > 0, by change of variable t/s 7→ t, one get
fs(ω) =∫f(t/s)e−iωtdt =
∫f(t)e−iωst|s|dt = |s|f(sω).
1
Let f by C1 and g = f ′, the by integration by parts, since f(t)→ 0 where |t| → +∞,
where Conj(a) = a∗ is the complex conjugate. The same computation leads to
fi(ω) = [f(ω)− f∗(−ω)]/2.
Exercise 2.5. One hasf(0) =
∫f(t)dt = 0.
If f ∈ L1(R), one can apply the theorem of derivation under the integral sign∫and get
ddω f(ω) =
∫−itf(t)e−iωtdt =⇒ f ′(0) = −i
∫tf(t)dt = 0.
Exercise 2.6. If f = 1[−π,π] then one can verify that
f(ω) = 2 sin(πω)ω
.
It result that ∫ sin(πω)πω
= 12π
∫f(ω)dω = f(0) = 1.
If g = 1[−1,1] then g(ω)/2 = sin(ω)/ω. The inverse Fourier transform of g(ω)3 is g ? g ? g(t) so∫ sin3(ω)ω3 dω = 1
8
∫g(ω)3dω = 2π
8 g ? g ? g(0) = 3π4 ,
where we used the fact thatg ? g ? g(0) =
∫ 1
−1h(t)dt = 3
where h is a piecewise linear hat function with h(0) = 2.
Exercise 2.7. Writing u = a− ib, and differentiating under the integral sign∫, one has
f ′(ω) =∫−ite−ut
2e−iωtdt.
By integration by parts, one gets an ordinary differential equation
f ′(ω) = −ω2u f(ω)
2
whose solution isf(ω) = Ke−
ω24u
for some constant K = f(0). Using a switch from Euclidean coordinates to polar coordinates(x, y)→ (r, θ) which satisfies dxdy = rdrdθ, one gets
K2 =∫e−ux
2dx∫e−uy
2dy =
∫∫e−u(x2+y2)dxdy
=∫ 2π
0
∫ +∞
0e−ur
2rdrdθ = 2π
∫ +∞
0re−ur
2dr = π
u,
which gives the result.
Exercise 2.8. If f is C1 with a compact support, with an integration by parts we get
f(ω) = 1iω
∫f ′(t)e−iωtdt
so that|f(ω)| 6 C
ωwith C =
∫|f ′(t)|dt < +∞,
which proves that f(ω)→ 0 when |ω| → +∞.Let f ∈ L1(R) and ε > 0. Since C1 functions are dense in L1(R), one can find g such that∫|f − g| 6 ε/2. Since g(ω) → 0 when |ω| → +∞, there exists A such that |g(ω)| 6 ε/2 when|ω| > A. Moreover, the Fourier integral definition implies that
|f(ω)− g(ω)| 6∫|f(t)− g(t)| dt
so for all |ω| > A we have |f(ω)| 6 ε which proves that f(ω)→ 0 when |ω| → +∞.
Exercise 2.9. a) For f0(t) = 1[0,+∞)(t)ept, we get
f0(ω) =∫ +∞
0e(p−iω)tdt = 1
iω − p.
For fn(t) = tn1[0,+∞)(t)ept, an integration by parts gives
fn(ω) =∫ +∞
0tne(p−iω)tdt = n
iω − pfn−1(ω),
so thatfn(ω) = n!
(iω − p)n .
b) Computing the Fourier transform on both sides of the differential equation gives
g = f ? h where h(ω) =∑Kk=0 ak(iω)k∑Mk=0 bk(iω)k
.
We denote by pkLk=0 the poles of the polynomial∑Mk=0 bkz
k, with multiplicity nk. If K < M ,one can decompose the rational fraction into
h(ω) =L∑k=0
Qk(iω)(iω − pk)nk
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where each Qk is a polynomial of degree strictly smaller than nk. It results that h(t) is a sum ofderivatives up to a degree strictly smaller than nk of the inverse Fourier transform of
fpk,nk(ω) = 1(iω − pk)nk
which isfpk,nk(t) = 1
nk! tnk1[0,+∞)(t)epkt.
Each filter fpk,nk is causal, stable and nk times differentiable. It results that that h is causal andstable.
If, there exists l with Re(pl) = 0 then for the frequency ω = −ipl we have |h(ω)| = +∞ so hcan not be stable.
If, there exists l with Re(pl) > 0 then by observing that fpl,nl(−ω) = (−1)nl(iω+ pl)−nl andby applying the result in a) we get
fpl,nkl (t) = 1nl!
tnl1(−∞,0](t)e−plt
which is anticausal. We thus derive that h is not causal.c) Denoting α = eiπ/3, one can write
Exercise 2.17. If |t| > 1, the ray ∆t,θ does not intersect the unit disc, and thus pθ(t) = 0. For|t| < 1, the Radon transform is computed as the length of a cross section of a disc
pθ(t) = 2√
1− t2.
Exercise 2.18. We prove that the Gibbs oscillation amplitude is independent of the angle θ and
is equal to a one-dimensional Gibbs oscillation. Let us decompose f(x) into a continuous partf0(x) and a discontinuity of constant amplitude A:
f(x) = f0(x) +Au(cos(θ)x1 + sin(θ)x2)
where u(t) = 1[0,+∞)(t) is the one-dimensional Heaviside function. The filter satisfies hξ(x1, x2) =gξ(x1) gξ(x2) with gξ(t) = sin(ξt)/(πt). The Gibbs phenomena is produced by the discontinuitycorresponding to the Heaviside function so we can consider that f0 = 0. Let us suppose that|θ| 6 π/4, with no loss of generality. We first prove that
f ? hξ(x) = f ? gξ(x) (1)
where gξ(ω1, ω2) = 1[−ξ,ξ](ω2). Indeed f(x) is constant along any line of angle θ, one can thusverify that its Fourier transform has a support located on the line in the Fourier plane, of angleθ+π/2 which goes through 0. It results that f(ω)hξ(ω) = f(ω)gξ(ω) because the filtering limitsthe support of f to |ω2| 6 ξ. But gξ(x1, x2) = δ(x1) sin(ξx2)/(πx2). The convolution (1) is thusa one-dimensional convolution along the x2 variable, which is computed in the Gibbs Theorem2.8. The resulting one-dimensional Gibbs oscillations are of the order of A× 0.045.
2 Chapter 3
Exercise 3.1. One has φs,n(t) = s−1/21[ns,(n+1)s), which satisfies ||φs,n|| = 1 and 〈φs,n, φs,n′〉 = 0for n 6= n′ because [ns, (n + 1)s) and [n′s, (n′ + 1)s) are disjoint. If f(x) = an on each interval[ns, (n+ 1)s), then
f(x) =∑n
an1[ns,(n+1)s) =∑n
〈f, φs,n〉φs,n
So φs,nn is an orthonormal basis of functions that are piecewise constant on each interval[ns, (n+ 1)s).
Exercise 3.2. If Supp(f) ⊂ [−π/s, π/s], then
f(ω) = f(ω)1[−π/s,π/s](ω) = 1sf(ω)φs(ω)
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and hence using the Fourier convolution theorem and the fact that φs is symmetric,
f(u) = 1sf ? φs(u) = 1
s〈f(t), φs(t− u)〉.
Exercise 3.3. a) The function f is an interpolation function if and only if∑n
f(n)δ(t− n) = δ(t)
in the distribution sense. Using the sampling Theorem 3.1 with s = 1, one gets equivalently theequality of the Fourier transform ∑
k
f(ω + 2kπ) = 1.
b) One hasf(ω) =
∑n
h[n]θ(ω)e−inω = h(ω)θ(ω).
and thus ∑k
f(ω + 2kπ) = h(ω)A(ω) where A(ω) =∑k
θ(ω + 2kπ)
If for all ω, A(ω) 6= 0, one can set h(ω) = 1/A(ω) and f(ω) = θ(ω)/A(ω). Thus f ∈ L2(R) if itexists B > 0 such that |A(ω)| > B, in which case ||f || 6 ||θ||/B.
Exercise 3.4. Let A(t) =∑n f(t− n), then in the sense of distribution
A(ω) = f(ω)∑n
e−inω = f(ω)∑n
δ(ω − 2nπ) =∑n
f(2nπ)δ(ω − 2nπ).
Taking the inverse Fourier transform leads to∑n
f(t− n) =∑n
f(2nπ)e2inπω.
Exercise 3.5. The orthogonal projection of f on Us is defined by
∀ g ∈ Us, 〈f − f, f − g〉 = 0.
It thus satisfies∀ g ∈ Us, ||f − g||2 = ||f − f ||2 + ||f − g||2 > ||f − f ||2
and hence f minimizes ||f − f || subject to f ∈ Us.
Exercise 3.6. The sufficient condition comes from
||Lf ||∞ 6 ||f ||∞||h||1.
If ||h||∞ = +∞, then for any A > 0 it exists B such that∑|k|<B |h[k]| > A. Taking f [k] =
sign(h[−k])1[−B,B][−k] that satisfies ||f ||∞ 6 1 shows that
Lf [0] =∑|k|6B
|h[k]| > A.
7
The operator f 7→ f ? h is not bounded on `∞, so that the filter is unbounded.
Exercise 3.7. One has
g ? h(ω) =∫ π
−π
(∑n
h[n]e−inξ)(∑
p
g[p]e−ip(ω−ξ))
dξ
=∫ π
−π
∑n,p
h[n]g[p]e−iξ(n−p)e−iωpdξ.
Exchanging signs∫and
∑, and using the fact that∫ π
−πe−iξ(n−p)dξ = δ[n− p]
shows thatg ? h(ω) =
∑n
h[n]g[n]e−inω = f(ω).
Exercise 3.8. Let ek[n] = e2iπN kn = ωkn where ω = e
2iπN . If k 6= k′, one has a geometrical sum
〈ek, ek′〉 =∑n
ωknω−k′n =
∑n
(ωk−k′)n = 1− ωN(k−k′)
1− ωk−k′ = 0
because ωN = 1. Since ||ek|| =√N , the family ek/
√Nk is an orthonormal basis of CN .
Exercise 3.9. Denotingfd(t) =
∑k
f(ks)δ(t− ks),
one has using Theorem 3.1fd(ω) = 1
s
∑k
f(ω − 2kπ/s).
Since f is supported in In = [−(n+1)π/s,−nπ/s]∪ [nπ/s, (n+1)π/s], and the intervals In+2kπare disjoint, one has
f(ω) = fd(ω)φs(ω) where φs(ω) = s1In(ω),
which corresponds to the reconstruction formula
f(t) =∑n
f(ns)φs(t− ns)
with the kernel obtained by inverse Fourier transform formula
φs(t) = s
∫ −nπ/s−(n+1)π/s
eiωtdω + s
∫ (n+1)π/s
nπ/s
eiωtdω
= 1πω/s
[sin((n+ 1)π/s)− sin(nπ/s)]
Exercise 3.10. a) One has f(ns) = f ? φs where φs = 1[−s/2,s/2].b) Since ˆf(ω) = f(ω)φs(ω), supp( ˆf) ⊂ [−π/s, π/s], and hence f is recovered using Shannoninterpolation formula.
8
c) One hasˆf(ω) = f(ω) sin(ωs/2)
ωs/2and thus
f = f ? ψs with ψs(ω) = ωs/2sin(ωs/2) .
d) For ω ∈ [−π/s, π/s], one hasφs(ω) > 2/π
and hence ||f || 6 π/2||f || which shows that the reconstruction is stable.
Exercise 3.11. a) φ is supported in [−1, 1], on t ∈ [−1, 0], φ(t) = t+ 1, on t ∈ [0, 1], φ(t) = 1− t.b) If f(t) is linear on each interval [n, n+ 1], then
f(t) =∑n
f(n)φ(t− n).
One has, using (7.21),
∑k
|φ(ω − 2kπ)|2 =∑k
sin4(ω/2)(ω/2 + kπ)2 = 1
3(1 + 2 cos2(ω/2)) > 13 ,
so using Theorem 3.4, φ(t − n)n is a Riesz basis of the space of piecewise linear functions oneach interval [n, n+ 1].c) The dual basis satisfies
ˆφ(ω) = 3 sin2(ω/2)(ω/2)2(1 + 2 cos2(ω/2)) = h(ω)
−ω2
where h(om) = −12 sin2(ω/2)/(1+2 cos2(ω/2)) is the Fourier series of a discrete filter. It resultsthat φ(t) is obtained by integrating twice h(t) =
∑n h[n]δ(t − n). The Fourier series h(ω) is a
rational fraction of e−iω which is not reductible to a polynomial so h[n] has an infinite support,which proves that φ(t) also has an infinite support.
Exercise 3.12. One has
|f [k]| = |∑n
f [n]e− 2iπN nk| 6
∑n
|f [n]||e− 2iπN nk| =
∑n
|f [n]|.
Exercise 3.13. a) Let h[0] = 1, h[−1] = −1 and h[n] = 0 for n /∈ −1, 0. Then
||f ||V =∑n
|f ? h[n]|
andh[k] = 1− e 2iπ
N k =⇒ |h[k]| = 2| sin(kπ/N)|.
b) By applying the result of the exercise 3.12 we have that the Fourier transform f [k] h[k] off ? h[n satisfies
|f [k] h[k]| 6 ||f ||V
9
so for |k| 6 N/2 we verify that
|f [k]| 6 ||f ||V2| sin(kπ/N)| 6
||f ||V2| sin(kπ/N)| 6
N ||f ||V2 k
since sin(x) > 2x/π for |x| 6 π/2.
Exercise 3.14. Since (−1)n = e−iπn, one has
g(ω) =∑n
h[n]eiπneinω = h(ω + π).
If h is a low-pass filter, then g is a high-pass filter.
Exercise 3.15. If g = f ? h then ||g||1 6 ||f ||1||h||1 and we can exchange the summations order asfollow
g(ω) =∑n
e−iωn∑p
f [p]h[n− p] =∑p
f [p]e−iωp∑n
e−iω(n−p)h[n− p]
=∑p
f [p]e−iωp∑n
e−iωnh[n] = f(ω)h(ω)
where the second line is obtained by change of variable n− p→ p in the summation.
Exercise 3.16. a) Since h(ω)h−1(ω) = δ(ω) = 1, one has h−1 = 1/h.b) Up to translation we suppose that h is causal. The filter h−1 and h have finite support if andonly if h(ω) = P (e−iω) where P (z) and zk/P (z) are polynomial for some k ∈ N. This can onlyhappens if P (z) is a monomial P (z) = azp, so that h[n] = aδ[n− p].
Exercise 3.17. a) One has
|h(ω)|2 =K∏k=1
|a∗k − e−iω|2
|1 + e−iω|2
and one verifies that |a∗k − e−iω|2 = 1 + |ak|2 + 2Re(ake−iω) = |1 + e−iω|2.b) h[n−m]m is an orthogonal basis if and only if for all m,m′
〈h[n−m], h[n−m′]〉 = h ? h[m−m′] = δ[n−m′].
Taking the Fourier transform of this relation leads to ||h(ω)||2 = 1 for all ω.
Exercise 3.18. a) For h0[n] = an1[0,+∞)[n], one has the following geometrical sum
h0(ω) =∑n
(ae−iω)n = 11− ae−iω .
Let hp(ω) = (1− ae−iω)−p. Observe that
h′p(ω) = −paie−iω
(1− ae−iω)p+1 =∑n
hp[n](−in)e−inω .
The inverse Fourier transform of hp+1(ω) = (1− ae−iω)−p−1 thus satisfies
hp+1[n] = (n+ 1)hp[n+ 1]a p
.
10
Iterating on this relation with h1[n] = an1[0,+∞)[n] gives
hp[n] =h1[n+ p− 1] ,
∏p−1k=1(n+ k)
ap−1 (p− 1)! .
b) Taking the Fourier transform of the recursion formula shows that
g(ω) = h(ω)f(ω) where h(ω) =∑Kk=0 ake
−ikω∑Mk=0 bke
−ikω.
c) Let ak be the roots of the polynomial∑Mk=0 bkz
−k with multiplicity pk. Then
h(ω) = P (e−iω)∏k
(1− ake−iω)−pk ,
where P is a polynomial. If one has |ak| < 1 for all k, then using question a), each (1−ake−iω)−pkis the Fourier transform of a stable causal filter, and so is h.
If there exists l such that |al| = 1, then it can be written al = eiα so |h(α)| = +∞ and thefilter can therefore not be stable.
If, there exists l with |al| > 1 then let hl(ω) = (1− ale−iω)−pl . Observe that
Its inverse Fourier transform is hl[−n] and with question (a) we verify that it is not a causalfilter. To prove that h is then not a causal filter, one can decompose
h(ω) =L∑k=0
Qk(e−iω)(1− ake−iω)nk
where each Qk is a polynomial of degree strictly smaller than nk and observe that the componentfor k = l is not causal so h can not be causal.
Exercise 3.19. One has, using the inverse Fourier transform,
f [2n] = 12N
2N−1∑k=0
ˆf [k]e 2iπ2N k2n
= 1N
N/2−1∑k=0
f [k]e 2iπN kn + 1
N
2N−1∑k=3N/2+1
f [k]e 2iπN kn + 1
Nf [N/2]e 2iπ
N N/2n,
and one thus has
f [2n] = 1N
N−1∑k=0
f [k]e 2iπN kn = f [n].
Exercise 3.20. a) One has
x(ω) =∑n
y[Mn]e−iωn =∑n
(y · h)[n]e−iωn/M = 12π y ? h(ω/M)
where we have use the convolution result of Exercise 3.7 and where
h[n] =∑k
δ[n− kM ]
11
which satisfies, in the sense of distributions
h(ω) = 2πM
∑k
δ(ω − 2πk/M).
This shows that
x(ω) = 1M
∫ π
−πy(ξ)
∑k
δ(ω/M − ξ − 2πk/M)dξ = 1M
M−1∑k=0
y((ω − 2kπ)/M).
b) If for ω ∈ [−π, π], y(ω) = 0 for ω /∈ [−π/M, π/M ], then
y(ω) = x(Mω)φM (ω) where φM (ω) = M1[−π/M,π/M ](ω).
Let (x ↑ M)[Mn] = x[n] and (x ↑ M)[k] = 0 if k mod M 6= 0 be the up-sampled signal. TheFourier transform of x ↑M is x(Mω), so that
y = (x ↑M) ? φm where φM [n] = sin(nπ/M)nπ/M
.
Exercise 3.21. a) One has, using the change of variable r = n− pN ,
fp[k] =∑p∈Z
N−1∑n=0
fd[n− pN ]e− 2iπN kn =
∑r∈Z
fd[r]e−2ikπN r = fd
(2kπN
).
Using the sampling Theorem 3.1, one gets
fp[k] = fd
(2kπN
)= 1s
∑`
f
(2kπNs− 2`π
s
).
b) In order to have
sfp[k] ≈ f(
2kπNs
),
one needs that ω0 π/s. To be able to interpolate without too much aliasing f from the valuesf( 2kπNs
), one needs that t0 Ns. Since a function cannot be compactly supported in both time
and space, no exact interpolation formula is possible.c) The Fourier transform f(ω) is proportional to the convolution of the indicator of [−π, π]convolved with itself 4 times. Its support is thus [−4π, 4π].
Exercise 3.22. a) One hasf [`] = N
2 (δ[`− k] + δ[`+ k]) .
Sofa[n] = e
2iπN kn.
b) One has g[n] = (f [n] + f∗[n])/2, and using a change of variable n → −n in the summationgives
g[k] = f [k]/2 +∑n
f∗[n]e− 2iπN kn = f [k]/2 + f [−k]∗/2.
12
c) The definition of fa shows that g = Re(fa) satisfies
∀ 0 6 k1, k2 < L, ∀ 0 6 n1, n2 < M, fk[n] = f [n+ kM ].
One hasf [n] =
∑k
fk[n− kM ],
so thatf ? h[n] =
∑k
fk[n− kM ] ? h[n] =∑k
(fk ? h)[n− kM ]
where we have denoted by fk the image of size (2M − 1)× (2M − 1) obtained by zero-paddingfrom fk. This allows one to compute f ? h using L2 circular FFT of size (2M − 1)2, followed by4N additions to reconstruct the full image. The overall complexity of this overlap add methodis thus approximately 8KN log(M).
The complexity of a direct evaluation of the convolution is K ′NM2, where K ′ ∼ 2 (1 additionand 1 multiplication).
For K = 6 and K ′ = 2, using the overlap-add algorithm is better in the range of M such that
48N log2(M) 6 2NM2 ⇔ M2/ log2(M) > 24
which we found numerically to be M > 9.
Exercise 3.25. The computation of f is performed by applying a 1D FFT to each direction ofthe 3D array. This requires
Using Parseval conservation of energy and the fact that ||g|| = 1,∫|Sf(u, ξ)|2dξ = 2π
∫|h(t)|2dt = 2π
∫|g(t)|2 = 2π.
b) One has, using the fact that h′(ξ) = iξh(ξ) and Parseval conservation of inner product,∫ξ|Sf(u, ξ)|2dξ =
∫ξh(ξ)h∗(ξ)dξ = 2iπ
∫h′(t)h∗(t)dt,
and thus, expanding h′(t)h∗(t),∫ξ|Sf(u, ξ)|2dξ = 2π
∫φ′(t)|g(t− u)|2dt+ 2π
∫g′(t− u)g∗(t− u)dt,
and the second integral vanishes because g′ is an odd function.If one interpret Pu(ξ) = |Sf(u, ξ)|2/(2π) as a probability density then this result proves that
the average frequency of this density for a fixed t is equal to the averaged instantaneous frequencyφ′(u) over a neighborhood of t defined by the density |g(t− u)|2.
Exercise 4.2. The formula (2.32) for the Fourier transform of the Gaussian implies
Ag(τ, γ) = 1√πσ2
∫exp
(1
2σ2
[(v + τ/2)2 + (v − τ/2)2]) e−iγvdv
= 1√πσ2
e−τ2
4σ2
∫e−v
2e−iγvdv = exp
(− τ2
4σ2 −γ2σ2
4
)
Exercise 4.3. The reconstruction (4.66) is exact if and only if for all signal f ,
f = log(a)Cψ
J∑j=1
1ajf ? ψ∗j ? ψj + 1
aJCψf ? φ∗J ? φJ .
Taking the finite discrete Fourier transform of this relation leads to
∀ k, f [k] = f [k] ·
log(a)Cψ
J∑j=1
1aj|ψj [k]|2 + 1
aJCψ|φJ [k]|2,
and hence the result.
Exercise 4.4. One can write the discrete Fourier transform in (4.27)
Sf [m, l] = e−i2πlm/N f ? gl[m] with gl[n] = g[−n] e−i2πln/N .
It is thus computed with N convolutions. Since gl has a support of size L, the fast overlapp-addconvolution algorithm of Section 3.3.4 computes each convolution with O(N logL) operations.The windowed Fourier transform is thus computed with O(N2 logL) operations.
Exercise 4.5. a) Applying (4.19) to f = gu0,ξ0 proves
gu0,ξ0 = 12π
∫∫〈gu0,ξ0 , gu,ξ〉gu,ξdξdu.
14
Taking the inner product with gv,ν proves
K(u0, v, ξ0, ν) = 12π
∫∫K(u0, u, ξ0, ξ)K(u, v, ξ, ν)dudξ,
which implies PPΦ = PΦ so that P is a projector on Im(P ), which is exactly, by Theorem 4.2the functions Φ(u, ξ) that are windowed Fourier transform of some f ∈ L2.
The projector P is symmetric because
K(u0, u, ξ0, ξ) = K(u, u0, ξ, ξ0)∗.
A projector P is orthogonal (which means that Ker(P )⊥Im(P )) if and only it is a symmetricoperator P = P ∗ where P ∗ is the dual (conjugated transposed) operator of P for the innerproduct
〈Φ1, Φ2〉 = 12π
∫∫Φ1(u, ξ)Φ2(u, ξ)∗dudξ.
Indeed for all Φ1 ∈ Ker(P ) and Φ2 = PΦ2 ∈ Im(P ),
〈Φ1, Φ2〉 = 〈Φ1, PΦ2〉 = 〈PΦ1, Φ2〉 = 0.
b) Since an orthogonal projector P is 1-Lipschitz, computing PSf reduces the quantization errorbecause
||Sf − PSf || = ||P (Sf − Sf)|| 6 ||Sf − Sf ||.
Exercise 4.6. One has
b(t) = 1Cψ
∫ s0
0f ? ψ∗s ? ψs(t)
dss2 + 1
Cψs0f ? φ∗s0
? φs0 .
Taking the Fourier transform leads to
b(ω) = f(ω)Cψ
[∫ s0
0|ψs(ω)|2 ds
s2 + 1s0|φs0 |2
].
Using the fact that |ψs(ω)|2 = s|ψ(sω)|2 and
|φs0 |2 =∫ +∞
1|ψ(s0sω)|2 ds
s=∫ +∞
s0
|ψ(sω)|2 dss
leads to the result.
Exercise 4.7. Using Plancherel formula and then Fubini,
||φ||2 = 12π
∫|φ(ω)|2dω = 1
2π
∫ ∫ +∞
1|ψ(sω)|2 ds
sdω
= 12π
∫ +∞
1
(∫|ψ(sω)|2dω
)dss.
The change of variable sω → s in the inner integral, and the fact that ||ψ|| = 1 leads to
||φ||2 =∫ +∞
1
(1
2π
∫|ψ(ω)|2dω
)dss2 =
∫ +∞
1
dss2 = 1.
15
Exercise 4.8. Letb(t) = 1
C
∫ +∞
0f ? ψs(t)
dss3/2 .
Its Fourier transform reads, after a change of variable ξ = sω,
b(ω) = f(ω)C
∫ +∞
0
√sψ(sω) ds
s3/2 = f(ω)C
∫ +∞
0ψ(ξ)ds
s= f(ω).
Exercise 4.9. a) If f is Cp, then f (p)(ω) = (iω)pf(ω) and thus the inverse Fourier transformformula gives the result.b) On the set Ωρ = z \ Im(z) > ρ, one has
|(iω)pf(ω)eizω| 6 |ω|p|f(ω)|e−ρω,
which is integrable. Using classical result of holomorphy under the sign∫, one sees that f (p)(z)
is holomorphic on Ωρ for all ρ > 0.c) One has
f (p)(x+ iy) = 12π
∫ +∞
0(iω)pf(ω)eixωe−yωdω.
A wavelet transform of f is written over the Fourier domain
Wf(u, s) = 12π
∫f(ω)
√sψ(sω)eiωudω
so one sees that f (p) is indeed a Fourier transform if one sets
ψ(ω) = 1[0,+∞[(ω)(iω)pe−ω.
Using the inverse Fourier transform, and the fact that h(p)(ω) = (iω)ph(ω), one has
ψ(t) = h(p)(t) where h(t) = 12π
∫ +∞
0e−ωeiωt = 1
2π1
1− it ,
soψ(t) = ip!
2π(t+ i)p .
Exercise 4.10. For f(t) = cos(θ(t)) with θ(t) = a cos(bt), the width s of the window is smallenough if
s2|θ′′(t)| = s2ab2| cos(bt)| 1and thus s2ab2 1.
For θ(t) = cos(θ1(t)) + cos(θ2(t)) with θ1(t) = a cos(bt) and θ2(t) = a cos(bt) + ct, there isenough frequency resolution if
|θ′1(t)− θ′2(t)| = |c| > ∆ω
s
and enough spacial resolution if
s2|θ′1(t)| = s2|θ′′2 (t)| = s2ab2 1.
This shows that one needs √c
ab2 ∆ω ' 1.
16
Exercise 4.15. One has,∫Pf(u, ξ)du = 1
||f ||2
∫|f(u)|2du|f(ω)|2 = |f(ω)|2
Using Plancherel conservation of energy formula, one has
12π
∫Pf(u, ξ)dξ = |f(u)|2 1
||f ||2
∫|f(ω)|2dω = |f(u)|2.
Since Pf is not quadratic in f (it is of degree 4), one cannot apply Theorem 4.11.
Exercise 4.16. Example 4.20 shows that Pθf is a spetrogram with a Gaussian window gµ if
θ(u, ξ) = gσ(u)gβ(ξ) = PV gµ(u, ξ) = gµ/√
2(u)g√2µ(ξ),
where we have used (4.125) for the last equality. If σβ = 1/2, then one can take µ =√
2σ.Otherwise, if σβ > 1/2, one decomposes θ = θ0 ? θ1 where
θ0(u, ξ) = gµ/√
2(u)g√2µ(ξ) and θ1(u, ξ) = gσ−µ/√
2(u)gβ−√2µ(ξ)
where µ is chosen so that σ − µ/√
2 > 0 and β −√
2µ > 0. This shows that Pθ is a smoothingwith θ1 of Pθ0 which is a spectrogram and hence positive.
Exercise 4.17. Since gn(t)n∈N is an orthonormal basis,
+∞∑n=0
g∗n(t) gn(t′) = δ(t− t′) .
Indeed, for all f ∈ L2
∫f(t)
+∞∑n=0
g∗n(t) gn(t′) dt =+∞∑n=0〈f, gn〉gn(t′)
= f(t′) =∫f(t) δ(t− t′)dt .
It results that+∞∑n=0
g∗n(u− τ/2) gn(u+ τ/2) = δ(τ) .
Since PV gn(u, ξ) is the Fourier transform of g∗n(u− τ/2) gn(u+ τ/2) with respect to τ , it resultsthat
In the sense of distribution, one has the following equality∫(ξ − φ′(t))2e−iξτdξ = 2π[−δ′′(τ) + 2iφ′(t)δ′(τ) + φ′(t)2δ(τ)]
where δ, δ′, δ′′ are the Dirac distribution and its first and second derivatives. This shows that
A(t) = 2π[−h′′(0) + 2iφ′(t)h′(0) + φ′(t)2h(0)].
After computing the derivatives of h and simplification, one finds
A(t) = −π(a(t)a′′(t)− a′(t)2),
which is the result.
4 Chapter 5
Exercise 5.1. One has a union of K orthogonal bases
D = φp06p<KN =K−1⋃i=0φKp+i06p<N ,
so that D is tight frame of frame bound KN sinceKN−1∑p=0
|〈f, φp〉|2 =K−1∑i=0
N
(N−1∑p=0|〈f, φpK+i/
√N〉|2
)=K−1∑i=0
N ||f ||2 = KN ||f ||2.
Exercise 5.2. A change of variable t→ Kt gives
〈f, φp〉 =∫ 1
0f(t)e2iπpt/Kdt = K
∫ 1/K
0f(Kt)e2iπpt =
∫ 1
0fK(t)ep(t)dt,
where ep(t) = e−2iπtp is an orthonormal basis of L2[0, 1]. Let us define fK(t) = f(Kt) fort ∈ [0, 1/K] and fK(t) = 0 otherwise. For K > 1, we get∑
p
|〈f, φp〉|2 =∑p
|〈fK , ep〉|2 = ||fK ||2 = K||f ||2
So φpp is a tight frame of L2[0, 1] of frame bound K.
Exercise 5.3. The operator ΦΦ∗ is bounded on H = Im(Φ) which is of finite dimension, soB < +∞. If A = 0, then it exists f 6= 0 in H such that
∑n |〈f, φn〉|2 = 0 so that 〈f, φn〉 = 0
for all n. Since Φnn is generator of H, one has f = 0 which is a contradiction.
Exercise 5.4. One has
tr(U1U2) =∑i,j
U1[i, j]U2[j, i] = tr(U2U1).
Exercise 5.5. One has Φf [p] = f ? h where h = δ[·] − δ[· − 1]. As the discrete Fourier basisdiagonalizes Φ, the frame bounds are
A = minω 6=0|h[ω]|2 = min
ω 6=04 sin2
( πN
ω)
= 4 sin2(π/N)
18
and B = 4 (if N is even). One has A/B → 0 when N → +∞, so the frame is unstable when Nis large.
Exercise 5.6. If one does not constrain ||φp||, one can choose δ0/N,Nδ1, δ2 . . . , δN−1. If oneconstrains ||φp|| = 1, one can choose φp[n] = φ[n + p mod N ] so that Φf = f ∗ φ and thus oneshould impose ∑
ω
|φ[ω]|2 = N and |φ[ω]| > 0.
This can be achieved by setting φ[ω] = 1/N for ω 6= 0 and φ[0] =√N − (N − 1)/N2. This
impliesA = min
ω|φ[ω]|2 = 1/N2 → 0
andB = max
ω|φ[ω]|2 = N − (N − 1)/N2 → +∞.
Exercise 5.7. If x ∈ Null(U∗) and y = Uz ∈ Im(U), then
〈x, y〉 = 〈x, Uz〉 = 〈U∗x, z〉 = 0
so that Null(U∗) = Im(U)⊥.
Exercise 5.8. One has
Φmf [p] = 〈f, φm+p〉 = f ? φm[p] =⇒ Φ∗Φ =∑m
Φ∗mΦm
where Φm is a convolution operator, whose eigenvalues are φm[ω], and eigenvectors are thediscrete Fourier vectors. The eigenvalues of Φ∗Φ are thus
∑m |φm[ω]|2.
Exercise 5.9. Let
φk,p(t) = g(t− 2pπ/ω0)eikω0t so φk,p(ω) = g(ω − kω0)eiω/ω02pπ.
Since eiω/ω02pπp is an orthonormal basis of each interval [−ω0/2, ω0/2] + kω0, 1/√
2πφk,pk,pis an orthogonal basis of L2(R). Since f 7→ 1/
√2πf is an isometry, this prove that φk,pk,p is
also an orthogonal basis.
Exercise 5.10. a) Since 1/√Ke
2iπK nk06k<K is an orthogonal basis of the signal supported in
I = [−K/2,K/2− 1] +mM , and g[n−mM ]f [n] is supported in I, one has
∑n
|g[n−mM ]|2|f [n]|2 =K−1∑k=0|〈f, 1√
Kgmnk〉|2.
b) Summing over m leads to
K∑n
|f [n]|2N/M−1∑m=0
|g[n−mM ]|2 =∑k,m
|〈f, gm,k〉|2,
and hence the result of Theorem 5.18.
Exercise 5.11. For m = 2, the filters are written, with the convention z = e−iω
Applying Bezout algorithm to the polynomials (1 + 3z + 3z2 + z3)/8 and (z2 − z)/2 leads to
(8z − 7z2)× P (z) + (174 z + 5z2 + 7
4z3)Q(z) = 1
and the reconstruction filters are given by
ˆh(ω)/√
2 = 8z−1 − 7z−2 and ˆg(ω) = 174 z−1 + 5z−2 + 7
4z−3.
Exercise 5.12. For m = 3, one has
h(ω)/√
2 = cos4(ω/2)4 = P (z) = (z/2 + z−1/2)4
= (z2 + 4z + 6 + 3z−1 + z−2)/24
and one can choose
g(ω)/√
2 = sin2(ω/2) = Q(z) = (z − 2 + z−1)/4.
If h = h, then
ˆg(ω) = 2− |h(ω)|2
g∗(ω) = 21− cos4(ω/2)√2 sin2(ω/2)
=√
2[1 + cos2(ω/2)] =√
2Q(z)
where Q(z) = 1 + (z + 2 + z−1)/4, so g/√
2 = [1/4, 3/2, 1/4].
Exercise 5.13. Using the same proof as Theorem 5.18 and exercise 5.10, with M = 1, K = N ,one sees that
gm,`[n] = g[n−m] exp(
2iπN`n
)m,`
is a tight frame of frame bound
N∑n
|g[n−m]|2 = N ||g||2 = N.
Theorem 5.5 implies that the dual frame is gm,`/Nm,`.The reproducing kernel reads
K(m,m′, `, `′) = 〈gm,`, gm′,`′〉 =∑n
g[n−m]g[n−m′] exp(
2iπNn(`− `′)
).
Exercise 5.14. As βη → 2π, the frame becomes less orthogonal (A/B → +∞), so the dualvectors become more and more different from the primal ones, and hence the windows g and gdiffer more and more.
Since W is a white noise of variance σ2 independant of x
J(h) = E(||x ? (h− δ)||2) + E(||W ? h||2).
Since W is a white noise of variance σ2 its power spectrum is σ2. Let Rx(ω) be the powerspectrum of the stationary random vector x[n],
J(h) = 12π
∫ π
−π(Rx(ω)|1− h(ω)|2 + σ2|h(ω)|2) dω .
To minimize J(h) for each ω we minimize the value under the integral, which is obtained with
h(ω) = Rx(ω)σ2 + Rx(ω)
and J(h) = 12π
∫ π
−π
σ2Rx(ω)σ2 + Rx(ω)
dω 6σ2
K.
c) In this case x = x ? hp +W so
J(h) = E(||x ? (h ? hp − δ)||2) + E(||W ? h||2)
= 12π
∫ π
−π(Rx(ω)|1− h(ω) hp(ω)|2 + σ2|h(ω)|2) dω .
The minimum is obtained with
h(ω) = Rx(ω) hp(ω)σ2 + Rx(ω) |hp(ω)|2
and J(h) = 12π
∫ π
−π
σ2Rx(ω)σ2 + Rx(ω)|hp(ω)|2
dω.
Since hp(ω) = (1− e−iω)−p,
J(h) 6 σ2
2π
∫ π/K
−π/K|2 sin(ω/2)|2pdω 6
σ2
K
π2p
K2p (2p+ 1) .
For a fixed K > π, increasing p decreases the error.
Exercise 5.16. a) Each φs(t − ns − ks/K)n is an orthogonal basis of Us, the set of signalswith Fourier support included in [−π/s, π/s]. This proves that φs−n/Ksn is a tight frame withframe bound K.b) One has
Pa[n] =∑n′
a[n′]H[n, n′] where H[n, n′] = 〈φs(t− n/Ks),1Kφs(t− n′/Ks)〉,
Since φ(0) = 1 and φ(ω) tends to zero as ω increases, letting α and β go to −∞ and +∞ provesthat
∑j |ψ(2jω)|2 6 B. A similar proof applies to the left hand side.
22
Exercise 5.19. a) One has
Zgn,k(u, ξ) =∑`
e2iπξ`1[0,1](u− `− n)e2iπk(u−`).
Since 1[0,1](u− `− n) is zero unless ` = −n, one obtains
Zgn,k(u, ξ) = e−2iπξne2iπuk.
Since e−2iπξne2iπukn,k is an orthogonal basis of L2[0, 1]2, the Zak transform transforms anorthogonal basis into an orthogonal basis, and is thus an unitary transforms.b) One has ∫ 1
0Zf(u, ξ)dξ =
∑`
∫ 1
0e2iπξ`dξf(u− `) = f(u).
Note that this formula is valid to recover f(u) for u ∈ [0, 1], but it can be extended to arbitraryu ∈ R using the fact that
Zf(u+ 1, ξ) = e2iπξZ(u, ξ).
c) Let gn,k(t) = e2iπkg(t− n). Similarly to question a), one proves that
Zgn,k(u, ξ) = e−2iπξne2iπuk(Zg)(u, ξ).
One has, using the fact that the Zak transform is unitary, and then Plancherel formula for Fourierseries of functions defined on [0, 1]2,
hn,k(t) = e2iπkh(t− n) where (Zh)(u, ξ) = (Zg)∗(u, ξ)−1.
Exercise 5.21. We use the change of variables
x = ρ cos(α)− u sin(α) and y = ρ sin(α) + u cos(α).
For k + ` < p, one can expand the monomial P (x, y) = xky` as a polynomial in u of degree lessthan p
P (x, y) =∑t<p
A(ρ)ut.
This shows that
〈P, ψ〉 =∑t<p
∫A(ρ)
( ∫utψ(ρ cos(α)− u sin(α), ρ sin(α) + u cos(α))du
)dρ = 0.
Exercise 5.22. a) One has ∑k
nkn∗k =
(A CC B
),
where, denoting αk = 2kπ/K,
A =∑k
cos2(αk), B =∑k
sin2(αk) and C =∑k
cos(αk) sin(αk).
One has C = 0 andA = B = K
2 +∑k
cos(2αk) = K
2 ,
so∑k nkn
∗k = K
2 Id2, which proves that nkk is a tight frame with frame bound K/2.b) One has
ψk = 〈[ ∂θ∂x,∂θ
∂y], nk〉 =⇒ ψk(ω) = θ(ω)〈i[ωx, ωy], nk〉.
This shows thatK∑k=0
∑j
|ψk(2jω)|2 =∑j
|θ(ω)|222j∑k
|〈[ωx, ωy], nk〉|2
and hence the result since ∑k
|〈[ωx, ωy], nk〉|2 = K
2 ||ω||2.
24
5 Chapter 6
Exercise 6.1. a) It is an immediate consequence of Definition 6.1 of Lipschitz and uniformLipschitz regularity.b) For t 6= 0 f is C1 so it is 1-Lipschitz, and |f(t)| 6 |t| so f is also 1-Lipschitz at 0.
If f is uniformly Lipschitz α then there exists K > 0 such that for all (u, v) ∈ [−1, 1]2,|f(u)− f(v)| 6 K |u− v|α. For tn = (n+ 1/2)−1π−1 we have f(tn) = (−1)n tn. So
|f(tn)− f(tn−1)| = tn + tn−1 = π−1(
(n+ 1/2)−1 + (n− 1/2)−1)∼ 2π−1 n−1 .
Since tn − tn−1 = π−1 (n + 1/2)−1 (n − 1/2)−1 ∼ π−1n−2 it results that |f(tn) − f(tn−1)| ∼|tn − tn−1|1/2 and hence that α > 1/2.
We now prove that f is indeed Lipschitz 1/2. If u and v have same sign and |1/u− 1/v| > 1then there exists C > 0 with
|f(u)− f(v)| 6 |f(tn)− f(tn−1)| 6 C |tn − tn−1|1/2 6 C |u− v|1/2 .
If u and v of same sign with |1/u− 1/v| 6 1 and |u| > |v|. It result that |u− v| 6 u2, and
|f(u)− f(v)| 6 |u− v|+ |v|(| sin u−1| − | sin v−1|) 6 |u− v|+ |v|(v−1 − u−1|.
Since |u− v| 6 1
|f(u)− f(v)| 6 |u− v|1/2 + |u− v||u|
6 2|u− v|1/2 .
If u and v have different signs, and |f(u)| > |f(v)|, since |u− v| 6 2 it results that
|f(u)− f(v)| 6 2|f(u)| 6 2|u| 6 2|u− v| 6 2√
2|u− v|1/2.
It results that for any (u, v) ∈ [−1, 1]2 there exists C ′ > 0 with
|f(u)− f(v)| 6 C ′ |u− v|1/2
and hence that f is uniformly Lipschitz 1/2 on [−1, 1].
Exercise 6.2. a) The definition of f being α Lipschitz can be written
f(x) = f(u) + a(u)(x− u) +K(x, u)(x− u)α (2)
with |K(x, u)| uniformly bounded in both x and u, so
[f(x)− f(u)]/(x− u) = a(u) +K(x, u)(x− u)α−1
which implies that f has derivatives at u with a(u) = f ′(u), so (2) is re-written
f(x) = f(u) + f ′(u)(x− u) +K(x, u)(x− u)α (3)
Then summing (2) for x = a+ d, x = a and x = a− d shows that
(f(a+ d)− 2f(a) + f(a− d))/d = O(dα−1)
then applying (3) for x = a and both u = a− d and u = a+ d shows that
|f ′(a− d)− f ′(a+ d)|/d = O(dα−1)
25
this shows that f ′ is Lipschitz α− 1 and that this is uniform with the same bound.b) If f(t) = t2 cos(1/t) then f ′(t) = 2t cos(1/t)− sin(1/t). One has |f(t)− f(0)| 6 t2 so that fis 2-Lipschitz at 0, but sin(1/t) is discontinuous at 0 of f ′ is not Lipschitz 1 at 0.
Exercise 6.3. One can take a hat function that is zero outsize [−1, 1], that is linear over [−1, 0]and [0, 1], with f(0) = 1. It is 1-Lipschitz. We verify that |f(ω)| = | sin(ω/2)|2 |ω|−2 so (6.4) isnot satisfied.
Exercise 6.4. a) One has, for f(t) = cos(ω0t), and using Plancherel formula
Wf(u, s) = 1√s
∫ 1√sψ
(t− us
)f(t) dt
=√s
∫ψ(sω)e−iuω 1
2(δω0(ω) + δ−ω0(ω))dω
=√sψ(sω0) cos(uω0),
because ψ(−ω) = ψ(ω) because ψ(t) is symmetric.b) One has,
∂Wf(u, s)∂u
= 0 ⇐⇒ u = uk = kπ
ω0.
If ψ has p vanishing moments then |ψ(ω)| = O(|ω|p) so
|Wf(uk, s)| 6√s |ψ(sω0)| = O(sp+1/2).
Exercise 6.5. For f(t) = |t|α, one has, after the change of variable t→ t/s,
Wf(u, s) = 1√s
∫|t|αψ(t/s− u/s)dt = sα+1/2
∫|t|αψ(t− u/s)dt = Wf(u/s, 1).
If ψ(t) is an antisymmetric wavelet, ψ(−t) = ψ(t), since f(t) is even the wavelet transfrom isantisymmetric and henceWf(0, s) = 0 for all s. In this case we thus can not derive the Lipschitzregularity at t = 0 from Wf(0, s).
Exercise 6.6. One has |f(t)| 6 |t|α so f is α-Lipschitz. One has for t > 0
f ′(t) = αtα−1 sin(1/tβ) + βtα−β−1 cos(1/tβ),
so f ′ is α− β − 1-Lipschitz.The signal can be written f(t) = a(t) cos θ(t) with a(t) = |t|α and θ(t) = |t|−β . The instanta-
neous frequency is θ′(t) = β |t|−β−1. For such amplitudes and instantaneous frequencies one canapply the wavelet ridge calculations. Equation (4.109) shows that the larges wavelet coefficientsare along ridge curves (u, s(u)) defined by
s(u) = ηθ′(u)−1 = η β−1 |u|β+1
where η is the center wavelet frequency.Since f(t) is Lipschitz α at 0, we derive from (6.20) in Theorem 6.4 that equation (6.21) is
satisfied for α′ = α. One can not decrease α′ because at the ridge locations, (4.108) proves that
|Wf(u, s)| = O(s1/2a(u)) = O(s1/2|u|α).
26
Exercise 6.7. a) We proved in exercice 6.5 that if f(t) = |t|α thenWf(u, s) = sα+1/2Wf(u/s, 1).It results that the complex phase Φ(u, s) of Wf(u, s) satisfies Φ(u, s) = Φ(u/s, 1). The lines ofconstant phase are thus along curves (u, s) which satisfy u/s = cst, and which converge to 0when s goes to 0.b) A modulus maxima point u0 is an extrema of Wf(u, s) = f ? ψs(u) with ψs(t) = s−1/2ψ(−st)and thus satisfies ∂Wf(u0, s)/∂u = 0. But
∂Wf(u, s)∂u
= −sW 1f(u, s) = −sf ? ψ1s(u)
with ψ1(t) = ψ′(t). The modulus maxima computed with ψ are thus zeros of a wavelet transformcomputed with ψ′. Let us now consider the analytic wavelet transform W af(u, s) = f ? ψas (u)computed with the analytic complex wavelet
ψa(t) = ψ′(t) + iH(ψ′)(t)
where H(ψ′) is the Hilbert transform of ψ′. The modulus maxima of Wf(u, s) correspond topoints (u, s) where W af(u, s) has a real part equal to zero and hence a phase equal to π. Themodulus maxima points thus correspond to points of constant phase (equal to π) on this analyticcomplex wavelet tranform.
Exercise 6.8. One has
Wf(u, s) = 1√s
∫ +∞
0ψ
(u− ts
)dt =
√s
∫−u/s
ψ(t)dt
so that∂Wf
∂u(u, s) = 0 ⇐⇒ ψ(−u/s) = 0.
Since ψ is continuous and orthogonal to polynomials of degree less than p − 1, it has at least pzeros. Indeed, if this were not the case, there would exist less than p − 1 zero crossings tkkwhere ψ changes of sign at tk. Denoting P (t) =
∏k(t− tk) which is a degree less than p, Pψ is
non zero and of constant sign, so 〈f, P 〉 6= 0, which is a contradiction.
Exercise 6.10. If f(t) =∫ t
0 dµ∞(t) is a Cantor devil’s staircase then its singularities are locatedon the support of the Cantor positive measure dµ∞. Let t0 be on the support of dµ∞. Byconstruction, explained page 245, for any p > 0 there exists an integer q such that t0 = q3−p,and either (q + 1)3−p or (q − 1)3−p is also on the support of dµ∞. Suppose that it is the firstcase (the second one is treated similarly). The constrution shows that∫ q3−p
(q−1)3−pdµ∞(t) = 0 ,
∫ (q+1/9)3−p
q3−pdµ∞(t) > 0 ,
∫ (q+2)3−p
(q+1)3−pdµ∞(t) = 0 . (4)
If ψ = −θ′ then according to (6.32)
Wf(u, s) = s(f ′ ? θs)(u) = s
∫s−1/2θ(s−1(t− u)) dµ∞(t) . (5)
LetK be such that θ has a support included in [−K,K]. To simplify explanations we suppose thatθ > 0 for |t| < K. If s−1
p = K3p+1 then θ(s−1p (t−u)) has a support included in [−3−p−1, 3−p−1].
So necessarily, |Wf(u, sp)| has at least one local maxima at position up in the interval [(q −1/2)3−p, (q + 3/2)3−p] and since t0 = q3−p
limp→+∞
up = t0 .
Exercise 6.14. a) One has ψ(ω) = ||ω||2θ(ω) so that
F (ω) =+∞∑j=−∞
|ψ(2jω)|2 =+∞∑j=−∞
24je−22j+1||ω||2 .
One has ∑j60|ψ(2jω)|2 6
∑j60
24j = C < +∞.
Using the fact that 2j+1 6 1 + (2j + 1) log(2), one has∑j>0|ψ(2jω)|2 6
∑j>0
e4j log(2)−(1+(2j+1) log(2))||ω||2 6 C ′∑j<0
ej(4 log(2)−2||ω||2 log(2)).
This shows that for 0 6 ||ω||2 6 α < 2, F (ω) 6 B and also
F (ω) >∑j
24je−22j+1α = A > 0.
This proves that for 0 6 ||ω||2 6 α, A 6 F (ω) 6 B. Using the fact that F (2ω) = F (ω), thisbound is extended to all ω ∈ R2.b) For ψ = ∆θ, where ∆ is the Laplacian, one has
Wf(u, 2j) = 0 ⇐⇒ ∆(f ? θj) = 0,
so this is a multiscale crossing of the Laplacian edge detector.Suppose that the image can locally be approximated by a straight-edge profile
f(x1, x2) = ρ(x1 cosα− x2 sinα) (6)
where ρ(t) is a monotonous function with an inflection point at t = 0. A direct computationshows that if ψ is the Laplacian of a Gaussian θ then Wf(u, s) = 0 when u is an inflection pointof
f ? θs(x) = ρs(x1 cosα− x2 sinα)
which is located along a straight edge curve of angle α, x1 cosα− x2 sinα = t0 with ρ′′s (t0) = 0.It thus corresponds to the position of modulus maxima point computed with the two partialderivative wavelets ψ1 and ψ2.
Suppose that f has a curved edge, which can be modeled with an angle α which has a slowvariation as a function of (x1, x2). Then the location of the zero-crossings of Wf(u, s) are notidentical to the modulus maxima of the first derivative wavelets because of second order termsthat are not identical in both wavelet transforms. The zero-crossings are not exactly located atthe inflection points of f ? θs(x) but close to these points if the curvature is small.
28
Exercise 6.15. One has, denoting ψs(t) = ψ(t/s)/√s,
which gives the result after the change of variable t′/s→ t′ in the convolution.
6 Chapter 7
Exercise 7.1. a) One hasφ(ω) =
∏j>0
h( ω
2j+1
).
For each j > 0 and ` ∈ Z, h(ω/2j+1) has a zero of order p at location 2π(2j(2`+ 1)). Remarkingthat any k ∈ Z∗ can be written k = 2j(2` + 1) with j > 0 and ` ∈ Z, one sees that φ(ω) has azero of order p at each ω = 2kπ, k 6= 0.b) One has, using Theorem 3.1
A(ω) =∑n
nqφ(n)einω = 1iq
dq
dωq
(∑n
φ(n)einω)
= 1iq
dq
dωq
(∑n
φ(ω + 2nπ))
= 1iq
∑n
φ(q)(ω + 2nπ)
One thus have, for ω = 0,∑n
nqφ(n) = A(0) = 1iq
∑n
φ(q)(2nπ) = 1iqφ(q)(0) =
∫tqφ(t)dt,
since φ(q)(2nπ) = 0 for n 6= 0.
Exercise 7.2. Let A(t) =∑n φ(t − n). Its Fourier transform satisfies, using Theorem 2.4, in
the sense of distributions
A(ω) = φ(ω)∑n
e−inω = φ(ω)2π∑n
δ(ω − 2kπ) = 2π∑n
φ(2kπ)δ(ω − 2kπ).
Using Exercise 7.1, one has that φ(2kπ) = 0 for k 6= 0, and φ(0) = 1 so that A(ω) = 2πδ(ω) andhence A(t) = 1.
29
Exercise 7.3. We choose m to be even so that φm is a symmetric function
φm(ω)−2 = 1 +∑k 6=0
1(1 + 2kπ/ω)2m+2 .
For 0 < ω 6 a < π, one has |2nπ/ω| > |n|(1 + ε) with 0 < ε < 1 so that
|φm(ω)−2| 6 1 + 2∑n>1
1(n(2 + ε)− 1)(2m+2) .
We split the sum into 1 6 n 6 1/ε and 1/ε < n. For the first part, the sum is finite and one has
1/ε∑n=1
1(n(2 + ε)− 1)2m+2 −→ 0
when m −→ +∞. For the second part, one has
∑n>1/ε
1(n(2 + ε)− 1)2m+2 6
∑n>1/ε
1(2n)2m+2 6
∫ +∞
1/ε
1(2x)2m+2 dx −→ 0
when m −→ +∞. This shows that φm(ω) −→ 1 for m ∈ (−π, π). A similar derivation showsφm(ω) −→ 0 for |ω| > π, so that φm → φ almost everywhere. Using dominated convergence andPlancherel formula, this shows that ||φm − φ|| → 0.
Exercise 7.4. a) If K = 2, then Supp(φ) ⊂ [0, 1] and
If ε0 = 1, then φ(ε0.ε1 · · · εi) = 0 and if ε0 = 0 then φ((ε0 − 1).ε1 · · · εi) = 0 so that one has
φ(0.ε0ε1 · · · εi) = m[ε0]φ(0.ε1 · · · εi)
and by recursion one obtains
φ(0.ε0ε1 · · · εi) = m[ε0] · · ·m[εi]φ(0).
b) If t = 0.ε0ε1 . . . is a dyadic number, then there exists some K such that εk = 0 for k > K.For m[0] > 1, if φ(0) 6= 0, this implies that φ(t) is infinite.c)
Exercise 7.5. a) The recursion formula is written over the Fourier domain as
φk+1(ω) = 1√2h(ω/2)φk(ω/2)
so that, using Plancherel formula
ak+1[n] = 14π
∫|h(ξ/2)|2|φk(ξ/2)|2einξdξ.
30
Using the Poisson summation formula, Theorem 2.4, one has the following equality, in the senseof distributions
ak+1(ω) = 14π
∫|h(ξ/2)|2|φk(ξ/2)|2
(∑n
e−in(ω−ξ)
)dξ
= 12∑n
|h(ω/2− nπ)|2|φk(ω/2− nπ)|2
b) If φk → φ, then ak → a where a[n] = 〈φ(t), φ(t − n)〉. If |h(ω)| is bounded, then P is abounded linear operator on L2(R). Hence Pak → Pa and thus Pa = a, which means that a isan eigenvector of P for the eigenvalue 1.
One has
φk(ω) =k∏p=1
2−1/2h(2−pω)φ0(ω/2k)
Since φ0(ω/2k)→ 1 when k → +∞ and φk converges to φ, one has
φ(ω) =∏p>1
2−1/2h(2−pω).
Exercise 7.6. a) Since f ∈ VL, one has
f(x) =∑k
〈f, φL,k〉φL,k
b[n] =∑k
aL[k]2−L/2φ(k − n)
so that b = 2−L/2aL ? φd where φd[n] = φ(n).b) Using Theorem 3.1, one has
φd(ω) =∑k
φ(ω + 2kπ).
The filter φ−1d whose Fourier transform is 1/φd(ω) is stable if φd(ω) > A > 0.
d) If φL,nn is an interpolation basis of VL for the points 2Ln, then
f =∑n
b[n]φL,n =∑n
aL[n]φL,n.
The change from b to aL is stable if and only if the basis φL,nn is stable.
Exercise 7.7. a) A computation similar to the proof of Theorem 7.11 shows that the recon-struction property is equivalent to having
h(ω + π)ˆh(ω) + g(ω + π)ˆg(ω + π) = 0
andh(ω)ˆh(ω) + g(ω)ˆg(ω) = 2e−ilω.
31
One verifies that for a given h, the proposed filters satisfy the first condition, and that the secondcondition is equivalent to
h2(ω)− h2(ω + π) = 2e−ilω.
b) Denoting z = e−iω, this last condition is re-written as
(∑k
h[k]zk)2 − (∑k
(−1)kh[k]zk)2 = 2zl.
One verifies that in the left hand side of this equation, all terms of even degrees cancel out, sothat ` is necessarily odd.c) For the Haar filter,
√2h(ω) = 1 + e−iω, so that
h2(ω)− h2(ω + π) = (1 + z)2/2− (1− z)2/2 = 2z
which corresponds to the quadrature mirror filter condition with ` = 1.
Exercise 7.8. A Daubechies wavelet ψj,n with p vanishing moments has a support of size2j(2p− 1). One thus concentrates on the scales 2j such that 2j(2p− 1) < mink |τk − τk−1| suchthat each wavelet intersects only one discontinuity. Since the wavelets are translated by 2j withsupport of size 2j(2p − 1), each singularity generates exactly 2p − 1 non zero coefficients. Oneshould thus choose p = q+ 1. If p < q+ 1, then there are 2−j coefficients at each scale, that canbe all non zero.
Exercise 7.9. a) Denoting u = 1[0,+∞) and v = 1(1,+∞), one has θ1 = u− v. Denoting f (k) theconvolution k times of f with itself, one has
θm = (u− v)(m+1) =m+1∑k=0
(−1)k(m+ 1k
)u(k)v(m−k).
One has u(2) = [t]+, u(3) = ([t]+)2/2, and more generally, by induction, one proves that
u(k)(t) = ([t]+)k−1
(k − 1)! and v(k)(t) = ([t− k]+)k−1
(k − 1)! .
One then verifies thatu(i) ? v(j)(t) = 1
(i+ j + 1)! ([t− j]+)i+j+1,
so that
θm(t) = 1m!
m+1∑k=0
(−1)k(m+ 1k
)([t− j]+)m.
b) One has, using the Fourier convolution theorem
|θm(ω)| = |θ0(ω)|m+1 = |2 sin(ω/2)/ω|m+1
One has using (7.9)
Bm = maxω
∑k
|θm(ω + 2kπ)|2
= maxω
2m| sin(ω/2)|2m+2∑k
1|ω + 2kπ|2m+2 .
32
By comparison with an integral, one obtains
Bm > | sin(ω/2)|∑k>0
1|(ω/(2π) + k)π|2m+2 > | sin(ω/2)|
∫ +∞
ω/(2π)
dx(πx)2m+2
= | sin(ω/2)| (2m+ 1)/π(ω/2)2m+1 −→ +∞.
when m→ +∞ as long as 2 < ω < 2π.
Exercise 7.10. This proof is quite technical, and we detail a proof inspired by “UnimodularWavelets for L2 and the Hardy Space of H2” by Young-Hwa Ha, Hyeonbae Kang, Jungseob Lee,and Jin Keun Seo.
If ψj,nj,n is an orthogonal basis of L2(R), then for all f ∈ L2(R), using Parseval formula
f = 12π∑j,n
〈f , ψj,n〉ψj,n.
This corresponds to, for all ξ,
f(ξ) = 2j
2π∑j,n
∫f(ω)ψ∗(2jω)ψ(2jξ)e−ik2j(ξ−ω)dω.
One has, using Poisson formula, Theorem 2.4, in the sense of distribution with respect to thevariable ω, the equality ∑
n
e−ik2j(ξ−ω) = 2π2j∑k
δ(ω − ξ + 2−j2kπ),
and thusf(ξ) =
∑j,n
f(ξ − 2−j2kπ)ψ∗(2jξ − 2kπ)ψ(2jξ)
and hencef(ξ)(1− θ(ξ)) =
∑j,k 6=0
f(ξ − 2−j2kπ)ψ∗(2jξ − 2kπ)ψ(2jξ)
whereθ(ξ) =
∑j
|ψ(2jξ)|2.
For ξ0 > 0 (the same derivation applies for ξ0 < 0) and for some ε > 0 small enough, we usef = 1I with I = [ξ0 − επ, ξ0 + επ], and integrate with respect to ξ to obtain∫
I
|θ − 1| 6∑
|k2−j |<ε
∫I
|ψ(2jξ)| |ψ(2jξ − 2kπ)|dξ,
where we have used the fact that f(ξ − 2−j2kπ) 6= 0 on I only if |k2−j | < ε. Using Cauchy-Schwartz, and a change of variable 2jξ → ξ one obtains
∫I
|θ − 1| 6 2−j∑
2−j<ε
Aj(ξ)
∑|k|<ε2j
Bj,k(ξ)
,
33
whereAj(ξ)2 =
∫2jI|ψ(ξ)|2dξ and Bj,k(ξ)2 =
∫2jI|ψ(ξ + 2kπ)|2dξ.
We note that, if |k2−j | < ε, then all the intervals 2jI+ 2kπ are included in 2j [ξ0−3επ, ξ0 + 3επ],so that ∑
|k|<ε2jBj,k(ξ) 6 2ε2j
∫ 2j(ξ0+3επ)
2j(ξ0−3επ)|ψ(ξ)|2dξ.
If fallows that ∫I
|θ − 1| 6 2ε∑
2−j<ε
∫ 2j(ξ0+3επ)
2j(ξ0−3επ)|ψ(ξ)|2dξ.
If ε is small enough with respect to ξ0, ε < ξ0/(12π), then all the intervals 2j [ξ0 − 3επ, ξ0 + 3επ]are disjoint and thus
1|I|
∫I
|θ − 1| 6 1π
∫ +∞
(ξ0−3επ)/ε|ψ(ξ)|2dξ.
If ψ is continuous, then the left part of this inequality tends to θ(ξ0)− 1 whereas the right parttends to 0 when ε→ 0. This shows that θ(ξ0) = 0 for all ξ0 6= 0. When ψ ∈ L2(R), this resultsholds only for almost all points ξ0 6= 0 that are Lebesgue points of θ.
To show that the condition ∑j
|ψ(2jω)|2 = 1 (7)
is not sufficient, let us note that the requirement that ψ(t − n)n is an orthogonal system iswritten ∑
k
|ψ(ω + 2kπ)|2 = 1. (8)
We note that if ψ(ω) satisfies (7) and (8), then ψ(ω/2) still satisfies (7) but not (8) anymore,take for instance
ψ(ω) = 1[−2π,−π]∪[π,2π].
Exercise 7.11. If ψ = 1I , is an indicator function, ψj,k is an orthogonal basis if it satisfies∑j
|ψ(2jω)|2 = 1 and∑k
|ψ(ξ + 2kπ)|2 = 1.
This means that one has the disjoint unions
R =⋃j
2jI =⋃k
(I + 2kπ).
One verifies that this is the case for the set
I/ω = [−32,−28] ∪ [−7,−4] ∪ [4, 7] ∪ [28, 32].
where in the following we denote ω = π/7. The base approximation space is
Since ψ is a real wavelet, ψ(−ω)∗ = ψ(ω) and since h is also real, |h(−ω)| = |h(ω)|, which leadsto
2ψra(ω) = ψ(ω)(|h(ω/4− π/2)|2 + |h(ω/4 + π/2)|2
)= 2ψ(ω)
where we have used the fact that |h(ω)|2 + |h(ω+π)|2 = 2 because h is a quadrature filter. Thisproves that Re(ψa) = ψ.c) Denoting g1 as the finite filter such that g1(ω) = h(ω)|h(ω − π/2)|2, one has to switch fromthe ordinary computation of the detail coefficients
dj+1 = (aj ? g) ↓ 2
where a ↓ 2 is the sub-sampling by 2 operator, to the following computation
dj+1 =(
[(aj−1 ? g1) ↓ 2] ? g)↓ 2.
d) One can compute the coefficients aj [n] = 〈f, φj,n1(x1)φj,n2(x2)〉 from aj−1 by applying fil-tering by h[n1]h[n2] and subsampling along each direction. Then the detail coefficients d1
j [n] =〈f, ψ1
j,n〉 are computed by filtering/sub-sampling aj−2 two times by h along the x2 direction and
36
then by filtering/sub-sampling by g1 and then by g. Similar computations allow one to obtainthe other details coefficients dkj [n] = 〈f, ψkj,n〉 for k = 2, 3, 4.e) We denote by
ψ[1](x) = ψ(x1)φ(x1), ψ[2](x) = φ(x1)ψ(x2) and ψ[3](x) = ψ(x1)ψ(x2)
the classical 2D wavelets. One has for a real signal f ,
〈f, ψ[1]j,n〉 = Re(〈f, ψ1
j,n〉), 〈f, ψ[2]j,n〉 = Re(〈f, ψ2
j,n〉),
and〈f, ψ[3]
j,n〉 = Re(〈f, ψ3j,n〉) + Re(〈f, ψ4
j,n〉).
This implies that
|〈f, ψ1j,n〉|2 > |〈f, ψ[1]
j,n〉|2, |〈f, ψ2
j,n〉|2 > |〈f, ψ[2]j,n〉|
2,
and|〈f, ψ3
j,n〉|2 + |〈f, ψ4j,n〉|2 >
12(〈f, ψ3
j,n〉+ 〈f, ψ4j,n〉)2 >
12 |〈f, ψ
[3]j,n〉|
2.
This shows that ∑j,n,k
|〈f, ψkj,n〉|2 >12∑j,n,k
|〈f, ψ[k]j,n〉| = ||f ||
2/2,
and hence the frame is redundant.The reverse inequality is more technical, see [108], Appendix A, for a proof. It uses the fact
that if a function ψ satisfies |ψ(ω)| = O((1 + ||ω|||s) with s > 1/2 and |ψ(ω)| = O(||ω||α) withα > 0, then ψj,nj,n is a stable frame of its span. It is easy to show that the almost analyticwavelets ψk(ω) satisfy these decay conditions.
Exercise 7.16. a) One notes that φ1 = 1[0,1) satisfies the scaling equation
One has 〈φ1, φ2〉 = 0 so that φ1(t− n), φ2(t− n)n is an orthogonal basis of the function thatare affine on each interval [n, n+ 1).b) One needs to apply the Gram-Schmidt orthogonalization process to
φ1, φ2, t21[0,1)(t), t31[0,1)(t)
to obtainφ1, φ2, ψ1, ψ2.
They correspond to the Legendre polynomials on [0, 1), and
φ1(t− n), φ2(t− n), ψ1(t− n), ψ2(t− n)n∈Z
is an orthogonal basis of the functions that are polynomials of degree 3 on each interval [n, n+1).Since they are orthogonal to φ1, φ2, they are orthogonal to polynomial of degree 1 on theirsupport, and hence they have two vanishing moments (and ψ2 has 3 vanishing moments).
37
Exercise 7.17. a) One has∫ 1
0αrepl(t)βrepl(t)dt =
∑k
∫ 1
0α(t− 2k)βrepl(t)dt+
∑k
∫ 1
0α(2k − t)βrepl(t)dt
=∑k
∫ −2k+1
−2kα(t)βrep(t)dt+
∑k
∫ 2k
2k−1α(t)βrep(t)dt
=∫ +∞
−∞α(t)βrepl(t)dt
=∑k
〈α(t), β(t− 2k)〉+∑k
〈α(t), β(2k − t)〉 = 0.
b) We treat the case the case where φ, φ are symmetric about 1/2 and ψ, ψ are ansi-symmetricabout 1/2, which implies ψ(1−t) = −ψ(t). Since the folded signals are 2-periodic, one only needsto consider, at a scale 2j with j < 0, indexes 0 6 n < 2−j+1. One thus has, if (j, n) 6= (j′, n′),
and similarly with the other inner products with scaling functions.For a given function f ∈ L2([0, 1]), we extend it by zero outside [0, 1] and decompose it in
the wavelet basis of L2(R)
f =∑j6J,n
〈f, ψj,n〉ψj,n +∑n
〈f, φJ,n〉ψreplJ,n
so that for t ∈ [0, 1],
f(t) = f repl(t) =∑j6J,n
〈f repl, ψreplj,n 〉ψ
replj,n +
∑n
〈f repl, φreplJ,n 〉φ
replJ,n .
Exercise 7.18. a) In the following, we denote z = e−iω, and denote
p(ω) = P (z) = R(z)R(z−1)Q(z)Q(z−1) .
The perfect reconstruction property reads P (z) + P (−z) = 2, so that P (z) =∑k∈Z pkz
k neces-sarily satisfies p2k = 0 for k 6= 0. Hence one can decompose
P (z) = 1 + zC(z2)D(z2) = C(z2) + zD(z2)
D(z2) = R(z)R(z−1)Q(z)Q(z−1)
where C and D are polynomials with no root in common. Thus by identification 2D(z2) =R(z)R(z−1) +R(−z)R(−z−1), which corresponds to
p(ω) = 2|r(ω)|2
|r(ω)|2 + |r(ω + π)|2
38
where r(ω) = R(z).b) The constraint on r is rewritten R(z−1) = zk−1R(z), so that
where P and P are polynomials. This shows that hnew has p + 1 vanishing moments, whereashnew has p− 1 vanishing moments.b) One has the following matrix expression for the balancing operations(
hnew(ω)ˆhnew(ω)
)=(a(ω) 0
0 1/a(ω)
)(h(ω)ˆh(ω)
)where a(ω) = (1 + eiω)/2.
Denoting A = a(ω), and using the following decomposition of a scaling matrix(A 00 1/A
)=(
1 A−A2
0 1
)(1 0−1/A 1
)(1 A− 10 1
)(1 01 1
),
one sees that the balancing operation can be implemented with 4 lifting steps.c) We start by the original filters
h = [1] and h = [−1, 0, 9, 16, 9, 0,−1]/16,
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after one step of balancing, one obtains
h1new = [1, 1]/2 and h1
new = [−1, 1, 8, 8, 1,−1]/8,
after one other step of balancing, one obtains
h2new = [−1, 2, 6, 2,−1]/4 and h2
new = [1, 2, 1]/4,
which corresponds to the 5/3 biorthogonal wavelets.
Exercise 7.21. After a simple factorization, applying an odd-length filter of size K such thath[n] = h[K − n− 1] requires K − 1 additions and (K − 1)/2 multiplications.
The direct application 5/3 filters requires 6 additions and 3 multiplications (times N for asignal of length N), whereas the lifting implementation requires 4 additions and 4 multiplications.
The direct application 9/7 filters requires 14 additions and 7 multiplications (times N for asignal of length N), whereas the lifting implementation requires 8 additions and 4 multiplication.
Exercise 7.22. For an interpolation wavelet, one has ψ(x) = φ(2x−1) and hence h(ω) = e−iω/2.One verifies that φ = δ and
ˆg(ω) = 2g(ω + π)e−iω
satisfy the biorthogonality relations, so that
ψ(t) =∑k∈Z
(−1)kh[k]δ(t− k + 1
2
).
For interpolation of order 4, one has
h = [−1, 0, 9, 16, 9, 0,−1]/16.
One verifies that∀q < 4,
∑k
(−1)kh[k]kq = 0
so that the dual wavelet has 4 vanishing moments.
Exercise 7.23. Denoting φ0 the Daubechies orthogonal wavelet with p vanishing moments, onehas φ(ω) = |φ0(ω)|2. Following Y. Meyer in “Wavelets with Compact Support”, one has thefollowing formula
|φ0(ω)|2 = 1− (2p− 1)![(p− 1)!]222p−1
∫ ω
0sin2p−1(t)dt
which is an even trigonometric polynomial of order 2p− 1. This can be explicitly computed as
|φ0(ω)|2 = 12 + 1
2
((2p− 1)!
(p− 1)!4p−1
)2 p∑k=1
(−1)k−1 cos((2k − 1)ω)(p− k)!(p+ k − 1)!(2k − 1)
by expanding sin2p−1(t). When p→ +∞, one has for |ω| 6= π/2,
|φ0(ω)|2 −→ 12 + 2
π
∑k>1
(−1)k−1
2k − 1 cos((2k − 1)ω),
which is the Fourier series expansion of 1[−π/2,π/2](ω) for ω ∈ [−π, π]. This thus shows that
|φ0(ω)|2 −→ 1[−π/2,π/2](ω),
40
and hence φ(t) converges in L2(R) toward sin(πt)/(πt).
Exercise 7.24. We follow the proof of Theorem 7.22. For each t = 2jn+ h where |h| 6 2j , wewrite
Exercise 7.25. Let φ(x) be a 1D interpolation function. We define the 1D transforms τ0(x) = 0and τ1(x) = 2x− 1. For each set εkpk=1 of p binary values εk ∈ 0, 1, we define
ψε(x1, . . . , xp) =p∏k=1
φ(τεk(xk)).
where we denote 0 6 ε < 2p the number whose binary expansion is εkpk=1. Then ψ0 is aninterpolating scaling function, and ψε0<ε<2p defines 2p − 1 interpolating wavelets that can beused to analyze continuous functions defined on Rp.
Exercise 7.28. We do the derivation here in 1D. The foveated signal Tf(x) is obtained fromf(x) by a spatially varying convolution with a filter scaled by t around position t, g(·/t)/t, whereg is a symmetric smooth low pass function
Tf(x) =∫ +∞
−∞f(t)g (x/t− 1) dt
t.
The operator is written over the wavelet domain as
Tf =∑
j,j′,n,n′
Kj,j′,n,n′〈f, ψj,n〉ψj′,n′
whereKj,j′,n,n′ = 〈Tψj,n, ψj′,n′〉 =
∫∫ψj,n(t)ψj′,n′(x)g (x/t− 1) dt
tdx.
41
It is shown in “Wavelet Foveation” by Chang, Mallat and Yap that if g and ψ have compactsupport, with ψ regular with vanishing moments, then Kj,j′,n,n′ decay fast when |j−j′| or |n−n′|increases. Together with the fact that Kj,j′,n,n′ depends only on j − j′, this shows that one canapproximate the operator as a diagonal one
Tf =∑j,n
λn〈f, ψj,n〉ψj,n
One can show that if g is of class Cα, then λn = O(|n|−(α+1)).
7 Chapter 8
Exercise 8.1. For A = 2j−L, one has that θj [m− nA]n is orthonormal if and only if
〈θj [m], θj [m− nA]〉 = θj ? θj [An] = δ[n].
Taking the Fourier transform of this equality, and using exercise 3.20, one has that this isequivalent to
A−1A−1∑k=0|θj(A−1(ω − 2kπ))|2 = 1. (9)
With a similar derivation, for B = 2j+1−L, the families θ0j+1[m−Bn]n and θ1
j+1[m−Bn]nare orthogonal sets of vectors orthogonal to each other if and only if
B−1∑B−1k=0 θ
0j (B−1(ω − 2kπ))θ1
j (B−1(ω − 2kπ))∗ = 0,B−1∑B−1
k=0 |θsj (B−1(ω − 2kπ))|2 = 1,∀ s = 0, 1.
Similarly to the proof of Theorem 8.1, one verifies these relationships using (9) and
θ0j+1(ω) = θj(ω)h(Aω) and θ1
j+1(ω) = θj(ω)g(Aω).
Exercise 8.3. Each node in the tree has 2p children. Fixing two quadrature mirror filters(h0, h1), for each 0 6 s < 2p, one defines the filters
hs[n] =p−1∏i=0
hsi [ni]
where (si)i is the binary expansion of s. The computation of the wavelet packet coefficients isperformed by the following filtering process from the top to the bottom of the tree
d2pk+sj+1 [n] = dkj ? hs[2n].
The number of scale is log2(N)/d, the number of nodes per scale is 2pj , the number of coefficientsper node is N/2jp, so that the number of coefficients per scale is N . The complexity per scale isO(KN) where K is the length of the filter, so that the overall complexity is O(KN log(N)).
Exercise 8.4. There are√N log(N) horizontal and vertical atoms, so N log(N)2 atoms in total.
One applies the wavelet packet decomposition algorithm along each of the√N rows in
O(K√N log(
√N)) operations per row. Then one applies the same process to the column. The
overall complexity is thus O(KN log(N)) operations.
j,kj,k is a lapped sine basis.The Meyer wavelet coefficients can be computed by applying an O(N log(N)) FFT followed
by log(N) lapped transform, and thus the complexity of this algorithm is O(N log(N)2). Thecomputation of the filter bank wavelet algorithm with FFT would take O(N log(N)) operation,but would not provide as accurate results.
Exercise 8.12. a) An admissible tree is a an admissible binary tree (called root tree) with acollection of admissible binary trees indexed by the leafs of the tree (called leaf trees).b) We denote as Cj,k the number of admissible double trees with a root tree of depth at most jand with leaf trees of depth at most k. One has
Cj,j = C2j−1,j + 1 > C2
j−1,j
and C0,j = Bj where Bk is the number of admissible binary tree, so that
Using the frame property, and denoting (A,B) the frame bounds, one has
A∑m>N
|〈f, gm〉|2 6 ||f − fN ||2 6 B∑m>N
|〈f, gm〉|2.
44
Hence
AC 6+∞∑N=0
N2s−1εl(N, f) 6 BC where C =∑m>0|〈f, gm〉|2
∑N>0
N2s−1.
The following is similar to the proof of Theorem 9.1.b) We denote
fM =∑k<M
|〈f, gmk〉|gmk ,
and fM the best M term approximation. One has, using the frame property
||f − fM ||2 6 ||f − fM ||2 6= ||∑k>M
|〈f, gmk〉|gmk ||2
6 B∑k>M
|〈f, gmk〉|2 = O(∑k>M
k−2s) = O(M1−2s).
Exercise 9.4. An M -term approximation of the multi-channel signal is defined as
fΛ = (fk,Λ)k where fk,Λ =∑m∈Λ
〈fk, gm〉gm,
where Λ is a set of M = |Λ| indexes. One has
||f − fΛ||2 =∑m/∈Λ
Am where Am =∑k
|〈fk, gm〉|2.
The best support Λ that minimize the approximation error is thus the one that selects the Mlargest values of Am.
Exercise 9.5. One has, for N = 2−j
fN (t) = PVj (f)(t) = f ? Φ(Nt)
where Φ is a smooth low frequency function.We assume that Φ is smooth and compactly supported in [−K/2,K/2], and that
∫Φ = 1,
||Φ||∞ 6 C. One then checks that the approximation error is localized in [−K/2,K/2]/N∫ 1
0|f(t)− f ? Φ(Nt)|2dt 6 2CKN−1.
Exercise 9.6. a) For a smooth function, one has
|f(x)| = |∫ x
0f ′(x)dx+ f(0)| 6
∫ x
0|f ′(x)|dx+ |f(0)| 6 TV (f) + |f(0)| < +∞.
This result caries over to arbitrary function by approximating by a smooth function.b) We consider f(x) = ||x||−αφ(x) where φ is some smooth localizing function, that is 1 insidethe disc of radius 1. One has, for ||x|| 6 1
∇f(x) = −α x
||x||α+2 .
45
Using a polar change of variables,∫||∇f(x)||dx = C + α
∫ dx||x||α+1 = C + 2πα
∫ 1
0
drrα
where C account for the total variation outside the unit disk. One can see that for 0 < α < 1, fis of bounded variation, but f is not bounded.
Exercise 9.7. Formally, we define
g(t) =∫RH(ω, t)dω where H(ω, t) = (iω)pf(ω)eiωt.
One has|H(ω, t)| 6 |ω|p|f(ω)|
Using Cauchy-Schartz inequality, one has
∫|ω|>η
|ω|p|f(ω)|dω 6
(∫|ω|>η
|ω|2(p−s)dω)1/2
||f ||Sob(s)
which is bounded if s > p+ 1/2. So using classical Theorem of derivation under the sign∫, this
shows that f is Cα and f (α) = g.
Exercise 9.8. One can verify numerically that Fourier and orthogonal polynomial approxima-tions behave similarely. An orthogonal polynomial of degree p has p oscillations and is similar to asinusoid function. The linearN -term approximation with orthogonal polynomial of a Cα functionis better that the error produced by a Taylor approximation, and is thus ||f − fN ||2 = O(N−2α)
Exercise 9.9. One has
||f − fM ||2 6M∑m=0
(tk+1 − tk)∆ = ∆.
We use the fact that any bounded variation function can be decomposed as f = f1 − f2 wheref1, f2 are increasing functions. Without loss of generality, we assume f1(0) = f2(0) = 0 andf1(1) = f2(1) = 1.
For each function fi, we build a set Ti = tik = f−1i (i/M)M−1
k=0 of points which guaranteesthat fi varies of at most ∆ = 1/M over each [tik, tik+1]. Defining tk2M−1
k=0 = T1 ∪ T2 guaranteesthat f varies of less than ∆ = 1/M over each [tk, tk+1]. One thus has ||f − fM ||2 = O(M−1).Exercise 9.11. We consider a function f such that |〈f, ψj,n〉| = 2js. For dyadic N = 2J , and
ordering wavelet coefficients from coarse to fine scales, εn[N ] = εl[N ].For α < s− 1/2, one has∑
j60
∑n
|〈f, ψj,n〉|22−2jα =∑j
2−j2−2αj22sj =∑j60
22j(s−α−1/2) < +∞
and Theorem 9.4 shows that f is in Wα.
Exercise 9.12. b) Denoting | supp(ψ)| = C = 8, one has for the CK wavelets whose supportintersect a singularity at scale 2j
|〈f, ψj,n〉| 6 ||f ||∞||ψ||12j/2,
46
the other coefficients being zero. One has
εl[M ] 6 2CK||f ||2∞||ψ||21M−1.
Since there is only KC non-zero coefficients per scale, the non-linear approximation selects allnon-zero coefficients corresponding to wavelets whose support intersects a singularity at scale 2jup to scale −J = M/(KC)
b) We call Vp,` the set of signals supported on [0, . . . , `], that assumes less than p different values.Any signal fk ∈ Vp,k can be decomposed as
fk = f` + a1[`,k]
where f` ∈ Vp−1,`−1 and a ∈ R. One thus has
minfk∈Vp,k
||f − fk||2[0,k] = min`∈[0,k−1]
minf∈Vp,`−1
||f − f`||2[0,`−1] + mina∈R||f − a||2[`,k],
which gives the result.An algorithm can computes εp,k for increasing p and k, and Σp,k which are the locations of
discontinuities in the optimal signal in Vp,k.One initializes:
• For all k, ε1,k = c1,k and Σ1,k = ∅.
• For all p, εp,1 = 0 and Σp,0 = 0, . . . , 0 (p times).
For p = 2, . . . ,K, for k = 1, . . . , N − 1, compute
εp,k = min`∈[0,k−1]
εp−1,` + c`,k
and denoting `∗ such that εp,k = εp−1,`? + c`?,k, update
Σp,k = Σp−1,`? ∪ `?.
After running this algorithm, fK,N has discontinuities in ΣK,N . The main numerical cost is thecomputation of the cp,k, which takes O(N2) for each k, so the overall complexity is O(KN2).
Exercise 9.14. a) Performing a first order approximation of the phase near point 2jn, andassuming that the amplitude is nearly constant over the support of size 2j , one gets
and thus performing the change of variable t/2j−n 7→ t/2j−n in the inner product, one obtains
|〈f, ψj,n〉| ≈ a(2jn)2j/2|ψ(2jφ′(2jn))|.
b) For f(t) = sin(1/t), one has φ(t) = 1/t, a = 1, φ′(t) = −1/t2, and thus
|〈f, ψj,n〉| ≈ 2j/2|ψ(2−jn−2)|.
The `p norm of the coefficients reads∑j60,06n62−j
|〈f, ψj,n〉|p ≈∑
j60,06n62−j|2jp/2|ψ(2−jn−2)|p. (10)
The wavelet function is close to a band pass filter (Shannon wavelet), so |ψ(ω)| is nearly constantequal to A > 0 in some interval [C1, C2], and small outside. As an approximation, we can thusconsiders that for each j the sum in (10) is well approximated by restricting it to indexes n thatsatisfies
C1 6 2−jn−2 6 C2 =⇒ C−1/22 2−j/2 6 n 6 C
−1/21 2−j/2.
For each j, the number of elements in the sum (10) is thus approximately C2−j/2 for someconstant C, so that ∑
j60,06n62−j|〈f, ψj,n〉|p ≈ CA
∑j60
2j/2(p−1). (11)
This `p norm is thus finite if and only if p > 1.c) Theorem 9.10 tells us that for all p < 1, εn[M ] = O(M1−2/p), so that εn[M ] = O(M−α) forany α < 1.
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