Revista Colombiana de Matemáticas Volumen 32 (1998), páginas 93–99

Wavelet transforms and singularities of L2-functions in ℝ^{n}

Jaime Navarro

Universidad Autónoma Metropolitana, México D. F.

Abstract. For a function f in L^{2}(ℝ), a wavelet transform with respect to an admissible function is defined such that its singularities are precisely the points where f fails to be smooth. Key words and phrases. Admissible functions, wavelet transform. 1991 Mathematics Subject Classification. 42A38, 44A05.

Introduction

In this paper a group structure on {(a, b) : a ∈ ℝ^{+}, b ∈ ℝ^{n}} is used to define a wavelet transform of a function f ∈ L^{2}(ℝ^{n}) with respect to an admissible function h ∈ C_{0}^{∞}(ℝ^{n}). For (a, b) in the group and letting (U(a, b)h)(x) = \frac{1}{a^{n/2}} h(\frac{x−b}{a}), a representation U of the group acting on the Hilbert space L^{2}(ℝ^{n}) is defined.

By means of this representation, I. Daubechies [4] established the following resolution of the identity: for f, h in L^{2}(ℝ^{n}), where h is radially symmetric (i.e., h(x) = η(|x|), so that h(x) depends only on |x|), we have

f = \frac{1}{C_{h}} ∫_{ℝ^{+}} ∫_{ℝ^{n}}⟨f, U(a, b)h⟩ U(a, b)h \frac{1}{a^{n+1}} db da,  (1)

where ⟨ , ⟩ is the inner product in L^{2}(ℝ^{n}) and C_{h} = ∫_{ℝ^{+}} |η̂(k)|^{2} \frac{1}{k} dk < ∞, η̂ being the Fourier transform of η.

With the help of this resolution of the identity, and for (a, b) in the group, a wavelet transform (L_{h}f)(a, b) of a function f in L^{2}(ℝ^{n}) is defined with respect to an admissible function h in L^{2}(ℝ^{n}) satisfying ∫_{ℝ^{+}} |η̂(k)|^{2} \frac{1}{k} dk < ∞, such that the singularities of (L_{h}f)(a, b) are precisely the singularities of f.

Notations and definitions. With G we denote the set {(a, b) : a ∈ ℝ^{+}, b ∈ ℝ^{n}}. In G we define (a_{1}, b_{1}) · (a_{2}, b_{2}) = (a_{1}a_{2}, a_{1}b_{2} + b_{1}). With this operation G becomes a group in which (1, 0) is the identity and (a, b)^{−1} = (a^{−1}, −a^{−1}b). Moreover, G turns out to be a locally compact topological group with d(a, b) = \frac{1}{a^{n+1}} da db and d_{1}(a, b) = \frac{1}{a}da db as the left and right Haar measures, respectively.

Definition 1. For h in L^{2}(ℝ^{n}) and b in ℝ^{n}, the traslation operator T_{b} is (T_{b}h)(x) = h(x − b), where x ∈ ℝ^{n}.

Definition 2. For h in L^{2}(ℝ^{n}) and a in ℝ^{+}, the dilation operator J_{a} is (J_{a}h)(x) = \frac{1}{a^{n/2}} h(\frac{x}{a}), where x ∈ ℝ^{n}.

Definition 3. For h in L^{2}(ℝ^{n}) and c in ℝ^{n}, the rotation operator E_{c} is (E_{c}h)(x) = e^{2πix·c}h(x), where x ∈ ℝ^{n}.

Definition 4. For (a, b) in G, define U(a, b) = J_{a}T_{b}. This family of operators is a representation of G acting on the Hilbert space L^{2}(ℝ^{n}) by

(U(a, b)h)(x) = (J_{a}T_{b}h)(x) = \frac{1}{a^{\frac{n}{2}}} h(\frac{x − b}{a}).  (2)

Definition 5. A function h in L^{2}(ℝ^{n}) is said to be admissible if

∫_{G}|⟨h, U(a, b)h⟩|^{2}d(a, b) < ∞.  (3)

Lemma 1. A radially symmetric function h in L^{2}(ℝ^{n}) is admissible if and only if

C_{h} ≡ ∫_{ℝ^{+}} |η̂(k)|^{2} \frac{1}{k} dk < ∞,  (4)

where ĥ(y) = η̂(|y|).

See the Appendix for the proof.

Definition 6. For a function f in L^{2}(ℝ^{n}) and (a, b) in G, the wavelet transform of f with respect to the admissible function h in L^{2}(ℝ^{n}) is defined as

(L_{h}f)(a, b) = ⟨f, U(a, b)h⟩.  (5)

We now state and prove the main result of this paper.

WAVELET TRANSFORMS AND SINGULARITIES OF L_{2}–FUNCTIONS IN ℝ^{n} 95

Theorem. Suppose that h in C_{0}^{∞}(ℝ^{n}) is radially symetric, non-identically vanishing and such that ∫_{ℝ^{n}} h(x) dx = 0. For f in L^{2}(ℝ^{n}) and (a, b) in G, let L_{α}(a, b) = a^{−1}a^{− \frac{n}{2}} D_{b}^{α}(L_{h}f)(a, b). Then, for each multi-index α, L_{α} is continuous at any point (a_{1}, b_{1}) in G. Furthermore, f is C^{∞} in a neighborhood of x = b_{0} if and only if for each multi-index α, lim_{(a,b)→(0,b_{1})} L_{α}(a, b) exists for each b_{1} in a neighborhood of b_{0}.

Proof. First we show that L_{α} is continuous at (a_{1}, b_{1}) for a_{1} > 0. Note that

(L_{h}f)(a, b) = ∫_{ℝ^{n}} \frac{1}{a^{n2}} f(x)\overline{h (\frac{x − b}{a})}dx = (f ∗ (J_{a}\overline{h})^{∼})(b)

where ψ^{∼}(x) = ψ(−x) and ∗ means convolution. Now, since f ∈ L^{2}(ℝ^{n}) and h ∈ C_{0}^{∞}(ℝ^{n}), it follows that f ∗ (J_{a}\overline{h})^{∼} ∈ C^{∞}(ℝ^{n}) and D_{b}^{α}(f ∗ (J_{a}\overline{h})^{∼})(b) = (f ∗ D_{b}^{α}(J_{a}\overline{h})^{∼})(b). Thus, L_{α}(a, b) = a^{−1}a^{− \frac{n}{2}} \frac{(−1)^{|α|}}{a^{|α|}} (f ∗ (J_{a}\overline{D^{α}h})^{∼})(b) is continuous at (a_{1}, b_{1}) for a_{1} > 0.

Next we show that the smoothness of f implies the existence of the limit of L_{α}(a, b) as (a, b) → (0, b_{1}). Suppose that f is C^{∞} in a neighborhood of x = b_{0} containing the closed ball \overline{B_{Δ}(b_{0})}, where Δ > 0. Take b, b_{1} in the open ball B_{\frac{Δ}{2}} (b_{0}). Note that if L > 0 is such that supp h ⊂ B_{L}(0), then

(L_{h}f)(a, b) = ∫_{B_{L}(0)} a^{\frac{n}{2}} f(b + ay)\overline{h(y)} dy.

Thus, for a such that 0 < a < \frac{Δ}{2L},

D_{b}^{α}(L_{h}f)(a, b) = ∫_{B_{L}(0)} a^{\frac{n}{2}} D_{b}^{α}f(b + ay)\overline{h(y)} dy.

Now, since f is C^{∞} at the points in the region of integration, it follows from Taylor’s formula that

D_{b}^{α}f(b + ay) = D^{α}f(b) + ∫_{0}^{1} ∑_{|β|=1} \frac{1}{β!}D_{b}^{β+α}f(b + tay)a^{|β|}y^{β}dt,

for y in B_{L}(0), so that

L_{α}(a, b) =a^{−1}a^{− \frac{n}{2}} D_{b}^{α}(L_{h}f)(a, b)
=a^{−1} ∫_{B_{L}(0)} D^{α}f(b)\overline{h(y)}dy

+ ∫_{B_{L}(0)} ∫_{0}^{1} ∑_{|β|=1} D_{b}^{β+α}f(b + tay)y^{β}\overline{h(y)}dtdy.

Then, since ∫_{B_{L}(0)} h(y)dy = 0 and D^{β+α}f is continuous near b_{1}, it follows that

\begin{aligned}lim_{(a,b)→(0,b_{1})} L_{α}(a, b) & = ∫_{B_{L}(0)} ( ∑_{|β|=1} D_{b}^{β+α}f(b_{1})) y^{β}\overline{h(y)}dy \\ & = ( ∑_{|β|=1} D_{b}^{β+α}f(b_{1})) ∫_{B_{L}(0)} \overline{h(y)}y^{β}dy. \end{aligned}  (6)

Therefore, lim_{(a,b)→(0,b_{1})} L_{α}(a, b) exists for each b_{1} in B_{\frac{Δ}{2}} (b_{0}).

Now we show that the existence of the limit implies the smootness of f. Suppose that L_{α}(b_{1}) := lim_{(a,b)→(0,b_{1})} L_{α}(a, b) exists for each b_{1} in an open neighborhood containing the closed ball \overline{B_{R}(b_{0})}, where R > 0.

For fixed x in the open ball B_{R}(b_{0}), let

I_{α}(a, x, y) = \{ \begin{matrix}h(−y)L_{α}(a, x + ay) & if a > 0 \\ h(−y)L_{α}(x) & if a = 0, \end{matrix}  (7)

where supp h ⊂ B_{L}(0), L > 0. Note that for such x, I_{α} is well-defined for all a and y. Furthermore, for fixed y and a ≠ 0, I_{α}(a, x, y) is infinitely differentiable in the variable x, and we have the following three claims.

Claim 1. I_{α} is continuous at (a_{1}, x_{1}, y_{1}) for all a_{1} in ℝ^{+}, x_{1} in \overline{B_{R}(b_{0})} and y_{1} in ℝ^{n}.

In fact, if a_{1} ≠ 0, I_{α}(a, x, y) is continuous at (a_{1}, x_{1}, y_{1}). Thus, we only need to consider the limit as (a, x, y) → (0, x_{1}, y_{1}). But

lim_{(a,x,y)→(0,x_{1},y_{1})} I_{α}(a, x, y) = lim_{(a,b)→(0,x_{1})} h(−y)L_{α}(a, b)
=h(−y)L_{α}(0, x_{1}) = I_{α}(0, x_{1}, y_{1}).

Then, I_{α} is continuous at all (a_{1}, x_{1}, y_{1}) in ℝ^{+} × \overline{B_{R}(b_{0})} × ℝ^{n}. This proves Claim 1.

Claim 2. I_{α} is in L^{1}(ℝ^{+} × ℝ^{n}) for fixed x in B_{R}(b_{0}).

In fact, for a ≠ 0,

I_{α}(a, x, y) = h(−y)a^{−1}a^{− \frac{n}{2}} D_{x}^{α}(L_{h}f)(a, x + ay).

Then

\begin{aligned}|I_{α}(a, x, y)| & = |h(−y)| a^{− \frac{2+n}{2}} |\frac{(−1)^{|α|}}{a^{|α|}} ⟨f, T_{x+ay}D^{α}J_{a}h⟩| \\ & ≤ |h(−y)|a^{− \frac{2+n}{2}} a^{−|α|}∥f∥_{2}∥D^{α}h∥_{2}. \end{aligned}

Now let

G_{α}(a, y) = \{ \begin{matrix}|I_{α}(a, x, y)| & if 0 < a ≤ 1 \\ |h(−y)| a^{− \frac{2+n}{2} −|α|} ∥f∥_{2} ∥D^{α}h∥_{2} & if a > 1. \end{matrix}  (8)

Then |I_{α}(a, x, y)| ≤ G_{α}(a, y) for all (a, y) in ℝ^{+} × ℝ^{n}, and we can see that G_{α} is in L^{1}(ℝ^{+} × ℝ^{n}) as follows:

∫ℝ^{+} ∫ℝ^{n}|G_{α}(a, y)|dyda
= ∫_{0}^{1} ∫_{B_{L}(0)} |I_{α}(a, x, y)|dyda
+ ∫_{1}^{∞} ∫_{B_{L}(0)} |h(−y)| a^{− \frac{2+n}{2} −|α|} ∥f∥_{2} ∥D^{α}h∥_{2}dyda

= ∫_{0}^{1} ∫_{B_{L}(0)} |I_{α}(a, x, y)|dyda

+ ∥f∥_{2} ∥D^{α}h∥_{2} (∫_{B_{L}(0)} |h(−y)|dy) (∫_{1}^{∞} a^{− \frac{2+n}{2} −|α|}da) .

Since I_{α}(·, x, ·) is continuous on [0, 1] × \overline{B_{L}(0)} and ∫_{1}^{∞} a^{− \frac{2+n}{2} −|α|}da < ∞, it follows that G_{α} ∈ L^{1}(ℝ^{+} × ℝ^{n}). Hence, I_{α}(·, x, ·) ∈ L^{1}(ℝ^{+} × ℝ^{n}). This proves Claim 2.

Claim 3. For x in the open ball B_{R}(b_{0}), let w(x) = ∫_{ℝ^{+}} ∫_{ℝ^{n}} I_{0}(a, x, y) dyda and I_{α}(x) = ∫_{ℝ^{+}} ∫_{ℝ^{n}} I_{α}(a, x, y) dyda. Then D^{α}w(x) = I_{α}(x) for any multiindex α.

In fact, let x be in the open ball B_{R}(b_{0}). By Claim 1, I_{α} is continuous on ℝ^{+} × \overline{B_{R}(b_{0})} × ℝ^{n}, and by Claim 2, |I_{α}(a, x, y)| ≤ S a^{− \frac{2+n}{2} −|α|} ∥f∥_{2} ∥D^{α}h∥_{2}, for a ≠ 0, where S = Sup{|h(−y)| : y ∈ B_{L}(0)}. Thus,

Sup{|I_{α}(a, x, y)| : a ∈ ℝ^{+}, x ∈ B_{R}(b_{0}), y ∈ B_{L}(0)}

exists.

Note that, by Claim 2, for x in B_{R}(b_{0}), I_{α}(a, x, y) is integrable and D_{x}I_{α} (a, x, y) exists and is uniformly bounded for (a, y) in ℝ^{+}×ℝ^{n}. It follows that for each x in B_{R}(b_{0}), D_{x}I_{α}(a, x, y) is integrable and D_{x} ∫_{ℝ^{+}} ∫_{ℝ^{n}} I_{α}(a, x, y)dyda = ∫_{ℝ^{+}} ∫_{ℝ^{n}} D_{x}I_{α}(a, x, y)dyda. Thus, D^{α}w(x) = I_{α}(x) for any multi-index α. This proves Claim 3.

Now, for l > 0 and any x, define

U_{l}(x) = ∫ l_{\frac{1}{l}} ∫_{ℝ^{n}} h(−y)a^{−1}a^{− \frac{n}{2}} (L_{h}f)(a, x + ay) dyda.  (9)

Then, by Claim 3, for every x in B_{R}(b_{0}), lim_{l→∞} U_{l}(x) = w(x). That is, U_{l} → w pointwise on B_{R}(b_{0}) as l → ∞. On the other hand, by (1), U_{l} → C_{h}f weakly in L^{2}(ℝ^{+} × ℝ^{n}) as l → ∞. Then f = C_{h}^{−1}w almost everywhere on B_{R}(b_{0}), and because of Claim 3, f is C^{∞} on B_{R}(b_{0}). This completes the proof of our main theorem. □ ✓

98 JAIME NAVARRO

Appendix

Proof of Lemma 1. Suppose that h in L^{2}(ℝ^{n}) is admissible. Then

∫_{G} |⟨h, U(a, b)h⟩|^{2}d(a, b) < ∞,

and we have

∫G|⟨h, U(a, b)h⟩|^{2}d(a, b)

= ∫_{ℝ^{n}} ∫_{ℝ^{+}} |⟨ĥ, { \atop J_{a}T_{b}h}⟩|^{2} \frac{1}{a^{n+1}} da db = ∫_{ℝ^{n}} ∫_{ℝ^{+}} |⟨ĥ, E_{−b}J_{\frac{1}{a}} ĥ⟩|^{2} \frac{1}{a^{n+1}} da db

= ∫_{ℝ^{n}} ∫_{ℝ^{+}} |∫_{ℝ^{n}} ĥ(ξ)\overline{E_{−b}J_{\frac{1}{a}} ĥ(ξ)} dξ|^{2} \frac{1}{a^{n+1}} da db

= ∫_{ℝ^{n}} ∫_{ℝ^{+}} |∫_{ℝ^{n}} ĥ(ξ)\overline{e^{−2πib·ξ}}\overline{J_{\frac{1}{a}} ĥ(ξ)} dξ|^{2} \frac{1}{a^{n+1}} da db

= ∫_{ℝ^{n}} ∫_{ℝ^{+}} |\overline{∫_{ℝ^{n}} e^{−2πib·ξ}\overline{ĥ(ξ)}J_{\frac{1}{a}} ĥ(ξ) dξ}|^{2} \frac{1}{a^{n+1}} da db

= ∫_{ℝ^{n}} ∫_{ℝ^{+}} |∫_{ℝ^{n}} e^{−2πib·ξ} (\overline{ĥ}J_{\frac{1}{a}} ĥ) (ξ) dξ|^{2} \frac{1}{a^{n+1}} da db

= ∫_{ℝ^{n}} ∫_{ℝ^{+}} | ; (\overline{ĥ}J_{\frac{1}{a}} ĥ)(b)|^{2} \frac{1}{a^{n+1}} da db = ∫_{ℝ^{+}} (∫_{ℝ^{n}} |{ \atop (\overline{ĥ}J_{\frac{1}{a}} ĥ)}(b)|^{2} db) \frac{1}{a^{n+1}} da

= ∫_{ℝ^{+}} (∫_{ℝ^{n}} |(\overline{ĥ}J_{\frac{1}{a}} ĥ)(y)|^{2} dy) \frac{1}{a^{n+1}} da

= ∫_{ℝ^{+}} (∫_{ℝ^{n}} |\overline{ĥ(y)}|^{2} |J_{\frac{1}{a}} ĥ(y)|^{2}dy) \frac{1}{a^{n+1}} da

= ∫_{ℝ^{+}} (∫_{ℝ^{n}} |ĥ(y)|^{2} |a^{\frac{n}{2}} ĥ(ay)|^{2}dy) \frac{1}{a^{n+1}} da

= ∫_{ℝ^{+}} (∫_{ℝ^{n}} |ĥ(y)|^{2} |ĥ(ay)|^{2}dy) \frac{1}{a}da = ∫_{ℝ^{n}} |ĥ(y)|^{2} (∫_{ℝ^{+}} |ĥ(ay)|^{2} \frac{1}{a}da) dy

Since h is radiallly symmetric, so is ĥ. Then

∫_{G} |⟨h, U(a, b)h⟩|^{2}d(a, b) = ∫_{ℝ^{n}} |ĥ(y)|^{2} (∫_{ℝ^{+}} |η̂(a|y|)|^{2} \frac{1}{a}da) dy

= ∫_{ℝ^{n}} |ĥ(y)|^{2} (∫_{ℝ^{+}} |η̂(k)|^{2} \frac{1}{k} dk) dy

= (∫_{ℝ^{n}} |ĥ(y)|^{2}dy) C_{h}, where C_{h} = ∫_{ℝ^{+}} |η̂(k)|^{2} \frac{1}{k} dk < ∞.

By working backwards, it is proved that if C_{h} = ∫_{ℝ^{+}} |η̂(k)|^{2} \frac{1}{k} dk < ∞ then h is admissible. This completes the proof of Lemma 1. □ ✓

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