A Theory for Multiresolution Signal Decomposition: The Wavelet Representation

• p. 3 Theorem 1 statement

  There exists a unique function φ(x) ∈ L²(R), called a scaling function

• p. 3 text after Theorem 1

  there exists a unique scaling function φ(x) which satisfies (9)

• p. 10 Section IV, 2-D case

  The function Φ(x, y) is unique with respect to a particular multiresolution approximation

Only the authors can say what is right here.

Judge's note: Read on page 3. My own check: if φ satisfies (9), take φ'(x)=φ(x−1). Then φ'_{2^j}(x−2^{-j}n) = 2^jφ(2^jx−n−1) = φ_{2^j}(x−2^{-j}(n+1)). That is the same family with the index shifted, so it is again an orthonormal basis of V_{2^j}. −φ also works. In the paper's own Haar example both 1_[0,1] and 1_[1,2] give orthonormal bases of every V_{2^j}. So φ is unique only up to such transformations, e.g. a unimodular 2π-periodic factor on φ̂. The text gives only 'indications for the proof' (Appendix B). The existence part is unaffected.

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• p. 5 Sec. II-B, symmetric extension of A_1^d f

  if 0 < n < 0.

• p. 5 display defining α_n

  α_{2N−n} if 0 < n < 0.

• p. 5 border handling equation for α_n

  if 0 < n < 0

Judge's note: The page 5 image and the text layer ('if0 < II < 0') both show the upper bound printed as 0. The preceding sentence says A_1^d f is symmetric about n = 0 and n = N, which this branch is meant to supply. As printed, the condition can never hold.

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• p. 5 Sec. II-B, after symmetric extension

  with respect to n = 0 and n = 2^{-j}N

• p. 5 Section II-B, after symmetric extension

  symmetric with respect to n = 0 and n = 2^{−j}N

Judge's note: Page 5 (image; text layer '2-/N'). The same paragraph says A^d_{2^j}f (j<0) has 2^jN samples, so 2^{-j}N > N lies outside the signal. My own check with (16): A(n)=Σ_k h(2n−k)a_k. Then A(N−n)=Σ_k h(2N−2n−k)a_k. Substituting k=2N−k' and using h even and a_{2N−k'}=a_{k'} gives A(N−n)=A(n). So one step gives symmetry about N/2, and by induction about 2^jN. Example with N=256, j=−1: printed 512, correct 128.

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• p. 5 definition before Theorem 2

  A scaling function φ(x) must be continuously differentiable

• p. 5 Theorem 2, converse

  is the Fourier transform of a scaling function.

• p. 6 Haar example after Theorem 2

  the scaling function is the indicator function of the interval [0, 1]

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Judge's note: Page 5 shows the regularity condition ('We also impose a regularity condition... must be continuously differentiable') and the converse with (17a)–(17c) and (18). My own check with h(0)=h(1)=1/2, H(ω)=e^{-iω/2}cos(ω/2): |H(0)|=1; h has finite support; |H(ω)|²+|H(ω+π)|² = cos²+sin² = 1; cos(ω/2)≠0 on [0,π/2]. The product Π_{p≥1}H(2^{-p}ω) = e^{-iω/2}sin(ω/2)/(ω/2), the transform of 1_[0,1]. That function is not C¹, so the converse as printed does not hold under the paper's own definition. Page 6 also calls the indicator 'the scaling function' of the example.

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• p. 6 Sec. II-B, Haar example

  H(ω) = e^{−iω} cos(ω/2)

• p. 6 Sec. II-B, example filter

  H(w) = e^{-iw} cos(w/2)

• p. 6 Sec. II-B, Haar example

  H(w) = e-‘” cos

Judge's note: The crop of page 6 shows the exponent as −iω, with no /2. My own check: φ_{2^{-1}}(u) = (1/2)1_[0,2](u), so h(0)=h(1)=1/2 and H=(1+e^{-iω})/2=e^{-iω/2}cos(ω/2). The printed form is not 2π-periodic (H(ω+2π) = −H(ω)), so it cannot be a series of the form (17).

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• p. 7 Sec. III-A, right column top

  [−2^{−j+1}π, −2^{−j}π] ∪ [2^{−j}π, 2^{−j+1}π]

• p. 7 right column, before Eq. (23)

  describes f(x) in the frequency bands [−2^{−j+1}π, −2^{−j}π] ∪ [2^{−j}π, 2^{−j+1}π]

Judge's note: Page 7, right column top (image; text layer '2-‘+’x', '2-JT'). The FT of ψ_{2^j} is ψ̂(2^{-j}ω), concentrated where π ≤ 2^{-j}|ω| ≤ 2π, i.e. 2^jπ ≤ |ω| ≤ 2^{j+1}π. For j=−1 the printed band starts at 2π, the correct one at π/2. As printed, coarser resolutions would carry higher frequencies, which is inconsistent with sampling D_{2^j}f at rate 2^j.

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• p. 10 Sec. IV, right column

  has 2^jN pixels

• p. 12 Sec. IV, left column

  has 2^jN pixels ( j < 0)

• p. 12 Section IV, paragraph after the 3J+1 images

  has 2^jN pixels ( j < 0). The total number of pixels in this new representation is equal to the number of pixels of the original image

• p. 10 Section IV, right column

  proximation A^d_{2^j}f has 2^jN pixels

Judge's note: The page images on p10 and p12 show the exponent j, not 2j. In 2-D, sampling at interval 2^{-j} in both x and y gives 2^{2j}N pixels. With the printed 2^jN, the 3J+1 images add up to 2^{-J}N + 3N(1−2^{-J}), which for J=1 is 2N. With 2^{2j}N they add up to exactly N, as p12 states. N is defined as 'the number of pixels' of the image. On p12 the paper also says a pyramid grows by 4/3 in two dimensions, which fits 1/4 per level.

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• p. 14 Eq. (49)

  α = m_2Γ(1/β) / NΓ(3/β)

• p. 14 Equation (49)

  α = m2Γ(1/β) / NΓ(3/β)

• p. 14 Eq. (49)

  α = m₂Γ(1/β) / (NΓ(3/β))

Only the authors can say what is right here.

Judge's note: My own working from the p14 image: integrating h(u)=K exp(−(|u|/α)^β) gives N = 2KαΓ(1/β)/β, which matches (45). It also gives m2 = 2Kα³Γ(3/β)/β, which matches (47). Substituting K gives m2 = Nα²Γ(3/β)/Γ(1/β), so m2Γ(1/β)/(NΓ(3/β)) = α². I also verified (48). The image shows a plain fraction in (49) with no radical. Check with β=2, α=2: m2/N = 4·Γ(3/2)/Γ(1/2) = 2, and the printed (49) then gives 2·Γ(1/2)/Γ(3/2) = 4 = α².

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• p. 15 Sec. V-B

  F(x) and r^H F(rx) are statistically identical

• p. 15 Section V-B, definition of Brownian fractal noise

  ∀r > 0, F(x) and r^H F(rx) are statistically identical.

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Judge's note: The p15 image shows r^H with a positive exponent. My own working: by the stated definition, increments over a step Δ have scale |Δ|^H. For G(x)=r^H F(rx), the increment over Δ is r^H[F(rx+rΔ)−F(rx)], which has scale r^H·(r|Δ|)^H = r^{2H}|Δ|^H. That differs from |Δ|^H for every r≠1. With r^{−H} the scales match. Counterexample: Brownian motion with H=1/2 and r=2 gives Var[2^{1/2}B(2x)] = 4x ≠ x.

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• p. 16 Appendix A, first paragraph

  p times continuously differentiable and equal to a polynomial of order 2p + 1

• p. 16 Appendix A, first paragraph

  p times continuously differentiable and equal to a polynomial of order 2p + 1 on each interval [k, k + 1]

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Judge's note: From the p16–p17 images: (56) gives φ̂ ~ ω^{−(2p+2)}, and the text says p=1 corresponds to cubic splines. Cubic splines on integer knots are C² = C^{2p}, not C¹. A counterexample under the printed definition with p=1 is the cubic Hermite function (1−|x|)²(1+2|x|) on [−1,1]. It is C¹ and piecewise cubic, but its second derivative jumps at ±1, so it is not in the cubic-spline space spanned by integer translates of φ. Counting degrees of freedom, C^p piecewise polynomials of degree 2p+1 have p+1 free parameters per unit interval, so that space is not generated by translates of a single φ. The wording 'p times' should probably read '2p times'.

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• p. 17 Appendix A, formula for Σ_8(ω)

  105 (sin ω/2)^8

• p. 17 Appendix A, formula for Σ₈

  Σ₈(ω) = (N₁(ω) + N₂(ω)) / (105 (sin ω/2)^8)

Judge's note: From the p16–p17 images: (57) defines Σ_n(ω) = Σ_k 1/(ω+2kπ)^n, and the paper's own Σ2 = 1/(4 sin²(ω/2)) carries the factor 2². At ω=0, N1 = 35 and N2 = 70, so the printed formula behaves like 1/sin^8(ω/2) ≈ 256/ω^8, whereas Σ8 ~ 1/ω^8. At ω=π, the printed value is (17/3)/105 ≈ 0.05397, while the true value is 2·(255/256)·ζ(8)/π^8 = 510/(256·9450) ≈ 2.108e-4, a ratio of 256. H(ω) in (58) is a ratio, so the constant cancels and Table I is unaffected. φ̂ in (56) is affected.

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• p. 17 Table I, n=5 and n=6

  -0.030

• p. 17 Table I, n=6

  0.012

• p. 17 Table I

  5 -0.030 ... 6 0.012

Judge's note: On the p.17 image, Table I reads 0.542, 0.307, -0.035, -0.078, 0.023, -0.030 | 0.012, -0.013, 0.006, 0.006, -0.003, -0.002. The text says the filter is symmetric, so H(0) = h(0) + 2Σh(n). With the printed values H(0) = 0.928. One reviewer computed 0.908, which is an arithmetic slip, but the conclusion holds. The alternating sum gives H(π) = 0.168. (17a) requires |H(0)| = 1 (p.5), and (17b) gives H(π) = 0. Flipping only the signs of n = 5 and n = 6 gives 1.000 and 0.000 exactly. The tail beyond n = 11 is about 0.003 per term and cannot close a 0.07 gap. The identity of the fix also matches the usual ++−− pattern of the cubic Battle–Lemarié filter. That last point rests on outside knowledge, but the defect is shown by the paper's own (17a)/(17b).

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• p. 6 Sec. II-B, first line of p.679

  sponding scaling function with (16).

• p. 5 end of Sec. II-B, after Theorem 2

  we can then compute the Fourier transform of the corre-

• p. 5 Theorem 2 discussion, end of page / start of p.6

  Fourier transform of the corre-

• p. 6 first lines

  sponding scaling function with (16)

Judge's note: Pages 5–6: (16) is ⟨f,φ_{2^j}⟩ = Σ h̃(2n−k)⟨f,φ_{2^{j+1}}⟩, while (18) is φ̂(ω)=Π H(2^{-p}ω). Later on page 6 the paper itself says 'compute the corresponding scaling function φ(x) with equation (18)'.

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• p. 9 Sec. III-C, last paragraph

  Fig. 8(a) is a reconstruction of the signal A_1 f(x)

• p. 10 Fig. 8 caption

  (a) Original signal A_1 f(x) approximated at the resolution 1. (b) Reconstruction

• p. 9 Section III-C, last paragraph

  Fig. 8(a) is a reconstruction of the signal A1f(x) from the wavelet representation given in Fig. 6(b). By comparing this reconstruction with the original signal shown in Fig. 8(b)

• p. 10 Fig. 8 caption

  (a) Original signal A1f(x) approximated at the resolution 1. (b) Reconstruction of A1f(x)

Only the authors can say what is right here.

Judge's note: I read both places on the page images: the end of Sec. III-C on p9 and the Fig. 8 caption on p10. They give opposite panel labels. The two panels look the same, so the images cannot show which labelling is right; only the authors can say whether the text or the caption should change.

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• p. 13 Fig. 13 block diagram

  D^2_{2^j} f → [2↑1] → [H]

• p. 12 Fig. 12 block diagram

  D^2_{2^j} f

Judge's note: From the images: Fig. 12 (p12) has the rows-G̃ branch split into columns G̃→D^3 and H̃→D^2, and the rows-H̃ branch into G̃→D^1 and H̃→A. Fig. 13 (p13) has D^3→G and D^2→H on rows, summed and then sent to columns G, and D^1→G and A→H on rows, summed and then sent to columns H. So in Fig. 13, D^2 is rows H with columns G and D^1 is rows G with columns H, the opposite of Fig. 12. Fig. 12 agrees with (40), where D^1 uses φ(−x)ψ(−y). Fig. 13 as labelled does not invert Fig. 12.

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• p. 14 Fig. 15 caption

  characterized by (49)

• p. 14 Fig. 15 caption

  Graph of the function F^{-1}(x) characterized by (49).

• p. 14 Fig. 15 caption

  Graph of the function F-'(x) characterized by (49)

• p. 14 Eq. (48)

  β = F^{-1}(m1²/(m2 N)) where F(x)

Judge's note: On the p14 image, (48) contains F and (49) contains no F. The caption should point to (48).

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• p. 18 Appendix C, start of proof of (17b)

  We saw in Appendix A that a scaling function must satisfy

Judge's note: Appendix A (p.16–17) gives only the spline construction, eqs. (56)–(59), and Table I. It does not contain this condition. Appendix B (p.17) derives Σ|φ̂(ω+2kπ)|² = 1 as (60) from the Poisson formula.

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• p. 18 Appendix D

  By inserting (64), (66), and (74) in (77), we obtain

• p. 18 Appendix D

  By taking the inverse Fourier transform of (79), we prove

• p. 18 Appendix D

  of (79), we prove

Judge's note: The image shows the sentence citing (77) immediately before the display numbered (77), so the reference is circular. The orthogonality condition just stated is (76). The (79) reference that some reviewers also raised is handled under C037.

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• p. 18 Appendix D

  taking the inverse Fourier transform of (79), we prove

• p. 18 Appendix D

  taking the inverse Fourier transform of (79)

Judge's note: The p.18 image shows (78) as G(ω) = e^{−iω} conj H(ω+π) and (79) as g(n) = (−1)^{1−n}h(1−n). (79) is the result of the inverse transform, so citing (79) is self-referential.

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• p. 12 Sec. IV-A, left column

  We first convolve the rows of A^d_{2^{j+1}} f with a one-dimensional filter, retain every other row

• p. 12 Fig. 12 legend

  keep one column out of two

• p. 12 Section IV-A, first paragraph

  We first convolve the rows of A^d_{2^{j+1}}f with a one-dimensional filter, retain every other row, convolve the columns of the resulting signals with another one-dimensional filter and retain every other column.

• p. 12 Sec. IV-A

  filter, retain every other row, convolve the

Judge's note: I compared the left-column text on p12 with the Fig. 12 diagram and its legend on the same page. The text swaps row and column in the two subsampling steps. The reconstruction paragraph on the same page and Fig. 13 follow the figure's convention (put zeros between columns, then convolve the rows).

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• p. 18 Appendix D, first paragraph

  ψ_{2^{-1}}(x − 2^{-1}n))_{n∈Z} is an orthonormal basis of O_{2^{-1}}

Judge's note: Theorem 3 (p.6) gives (√(2^{−j}) ψ_{2^j}(x − 2^{−j}n))_{n∈Z}. Substituting j = −1 gives the factor √2 and translation 2n. The p.18 image shows 2^{−1} in both the normalisation factor and the translation, in the first paragraph, before (76) for ψ and φ, and after (77). Since ψ_{1/2} ∈ O_{1/2} ⊂ V_1 is used, j = −1 is intended.

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• p. 18 Eq. (69) and following sentence

  converges towards |φ(ω)|² almost everywhere

• p. 18 Appendix C, Eq. (69)

  g_q(ω) = Π_{p=1}^{q} H(2^{-p}ω) for |ω| < 2^qπ

• p. 18 Appendix C, Eq. (70) and preceding sentence

  converges towards |φ̂(ω)|² almost everywhere

Only the authors can say what is right here.

Judge's note: I reworked the step. By (67), Π_{p≤q} H(2^{−p}ω) tends to φ̂(ω), not |φ̂(ω)|². Counterexample to (70) as printed, with q = 1: ∫_{−2π}^{2π} H(ω/2)e^{ikω}dω = Σ_n h(n)∫e^{i(k−n/2)ω}dω. Terms with n odd vanish (half-integer frequency over a length 4π), as do terms with n ≠ 2k, so the integral is 4π h(2k). For k = 0 and Table I's h(0) = 0.542 this gives about 2.17π ≠ 2π. With |H|², the integral is 4π times the even autocorrelation coefficient, which (17b) sets to δ_k/2, giving 2πδ_k. So (69)/(70) as printed are false. With the modulus squared inserted, Theorem 2 stands. The hat is also missing in the printed "|φ(ω)|²".

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