Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 340 (Topics in Mathematics of Information)

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$. An orthonormal multiresolution analysis is a family of closed vector subspaces $V_j\subset L^2(\mathbb{R})$, indexed by integers, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, and ${\bigcap }_j{V}_j=\{0\}$. Its dilation condition is $f\in V_j$ if and only if $f(2\cdot )\in V_{j+1}$. The space $V_0$ is invariant under integer function translations and admits a scaling function $\phi$ whose integer function translations form an orthonormal basis. Consequently ${\phi }_{j,k}=2^{j/2}\phi (2^j\cdot -k)$ form an orthonormal basis of $V_j$.

Since $\phi \in V_1$, expansion in that orthonormal basis gives the scaling refinement equationThe series converges in the L2 norm. The associated low-pass filter of a multiresolution analysis is the periodic Fourier series symbolInitially the symbol is defined almost everywhere. Orthonormal integer function translations give the quadrature mirror filter identity $|m(\xi ){|}^2+|m(\xi +\pi ){|}^2=1$ almost everywhere. For a Lebesgue integrable scaling function, its Fourier transform is continuous and $|\hat{\phi }(0)|=1$; choosing a constant phase makes $\hat{\phi }(0)=1$ and the continuous representative near zero has $m(0)=1$. Different conventions absorb $\sqrt{2}$ into the refinement coefficients; the displayed convention fixes that ambiguity.

Let $W_j=V_{j+1}\ominus V_j$ denote the orthogonal complement of $V_j$ inside $V_{j+1}$. The Meyer-Mallat theorem constructs a wavelet whose integer function translations form an orthonormal basis of $W_0$. Dilation then gives the orthogonal decomposition into orthogonal complements $L^2(\mathbb{R})={⨁}_{j\in \mathbb{Z}}{W}_j$, so its translates and dilates form an orthonormal wavelet basis.

Using the MRA low-pass filter, choose the high-pass symbol $n(\xi )=e^{-i\xi }\overline{m(\xi +\pi )}$. The quadrature mirror filter identity makes the two analysis channels orthonormal. The resulting Fourier transform formula isEquivalently, with the refinement coefficients from the scaling refinement equation, put $g_k=(-1{)}^{k-1}\overline{h_{1-k}}$ and $\psi (x)=\sqrt{2}{\sum }_k{g}_k\phi (2x-k)$. This index choice gives precisely the displayed high-pass symbol. Changing all $g_k$ by one constant unit phase gives the same orthonormal wavelet construction. In particular the often-used $(-1{)}^k\overline{h_{1-k}}$ convention differs only by a global minus sign. The basis is ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ for $j,k\in \mathbb{Z}$.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

Under the smoothness hypotheses in the smooth-mask vanishing-moment criterion, write any nonzero integer $j$ as $j=2^r\ell$, where $r\ge 0$ and $\ell$ is odd, possibly negative. Iterating the scaling refinement equation exactly $r+1$ times givesThe last MRA low-pass filter factor is $m(\pi \ell +t/2^{r+1})$. Its derivatives of all orders less than $p$ vanish at $t=0$, because $\ell$ is odd and the MRA low-pass filter is $2\pi$-periodic. If the remaining factors are $C^{p-1}$ at the displayed arguments, the product rule forces every derivative of order less than $p$ of the product to vanish. In particular compact support and a finite MRA low-pass filter supply all these hypotheses. Therefore the integer-frequency zeros of a scaling function are

Under only the printed assumptions, the exact same finite refinement argument gives the Peano zero $\hat{\phi }(2\pi j+t)=o(|t{|}^{p-1})$. Indeed, the last MRA low-pass filter factor has that estimate, the other MRA low-pass filter factors are bounded by one almost everywhere by the quadrature mirror filter identity, and the last Fourier transform factor is bounded because $\phi \in L^1$. The estimate initially holds almost everywhere and extends to every $t$ by continuity of $\hat{\phi }$. It does not imply arbitrary ordinary higher derivatives. In the lacunary scaling-phase regularity counterexample, $\hat{\phi }=a{\hat{\phi }}_0$ is nondifferentiable at dense dyadic points next to the isolated integer-frequency zeros of the smooth base transform. Thus the same regularity qualification is necessary here.

Fix the ordering of the orthonormal basis and write $c_m=⟨f,g_m⟩$, taking indices from $1$. The linear N-term approximation retains the first $N$ coefficients:Its index set is independent of $f$, so the map is an orthogonal projection and is linear.

For the nonlinear N-term approximation, let ${\Lambda }_N(f)$ be any set of $N$ indices with greatest $|c_m|$, resolving ties arbitrarily and padding with zero coefficients if necessary. Since $(c_m)\in {\ell }^2$, such a selection exists. ThenThe Parseval identity shows that for any fixed index set the displayed coefficients minimize the Hilbert space error. Choosing the largest squared magnitudes minimizes the omitted sum over every set of at most $N$ indices. This is the best N-term approximation; dependence of the selected indices on $f$ makes the operation nonlinear.

For $0<\alpha \le 1$, uniform Lipschitz-$\alpha$ means that one constant $M$ satisfies $|f(x)-f(y)|\le M|x-y{|}^{\alpha }$ for every $x,y\in [0,1]$. The resulting Hölder space has normFor the subsequent estimates at general $\alpha >0$, use the higher-order Hölder class convention: write $\alpha =r+\beta$ with $r\ge 0$ an integer and $0<\beta \le 1$, and require $f\in C^r([0,1])$ with $f^{(r)}$ uniformly Lipschitz-$\beta$. ThusAt integer $\alpha =k$, this uses $C^{k-1,1}$; the alternative classical $C^k$ convention is stronger on this compact closed interval and also suffices for the estimates.

For exponents above one, a first-difference inequality alone forces the function to be constant. Partition $[x,y]$ into $n$ equal pieces and use the triangle inequality: $|f(y)-f(x)|\le M|y-x{|}^{\alpha }{n}^{1-\alpha }\to 0$. Consequently a nontrivial higher-order definition is needed for the full range of exponents in the later wavelet estimates. Equivalently, higher-order Hölder class gives a local Taylor polynomial with remainder bounded by $C\Vert f{\Vert }_{C^{\alpha }}|x-x_0{|}^{\alpha }$.

Use localization and cancellation of the assumed interval-adapted wavelet basis: at scale $j$, every detail wavelet has support diameter at most $B{2}^{-j}$, L1 norm at most $B^{\prime }{2}^{-j/2}$, and annihilates polynomials of degree less than $q$. These bounds also apply to the modified boundary wavelets; a restriction of an arbitrary whole-line wavelet without the cancellation-preserving boundary construction would not suffice. The finitely many coarse scaling functions can be treated separately.

Write $\alpha =r+\beta$ with $0<\beta \le 1$ as above. Because $\alpha <q$, we have $r<q$. Choose $x_0$ in the support of a function of ${\psi }_{j,n}$, and let $T_{x_0}$ be the degree-$r$ Taylor polynomial of $f$. The Hölder-Taylor remainder bound givesFor $r=0$ this is the Hölder seminorm bound. For $r\ge 1$, the integral Taylor remainder is bounded using $f^{(r)}(t)-f^{(r)}(x_0)$, whose magnitude is at most $[f^{(r)}{]}_{C^{0,\beta }}|t-x_0{|}^{\beta }$.

The vanishing moments remove $T_{x_0}$ from the inner product. Applying the remainder estimate and the L1 norm bound on the localized support yields the wavelet coefficient decay for Hölder functionsThusHere $C$ depends on the fixed wavelet family, the exponent and boundary construction, but not on $f,j,n$. The proof uses the assumed regularity and moments; it does not identify the minimal Daubechies wavelet order with its differentiability order, which need not coincide.

Let $M$ bound the piecewise Hölder norms, and let $B=\Vert f{\Vert }_{\infty }$. At scale $j$, classify a wavelet as bad when its localized support of a function meets a discontinuity, and as good otherwise. Compact support and bounded overlap imply at most $C_{0}K$ bad wavelets per scale and at most $C_1{2}^j$ total wavelets. A good support lies in a single smooth piece, so the preceding wavelet coefficient decay for Hölder functions gives $|c_{j,n}|\le CM{2}^{-j(\alpha +1/2)}$. The permitted bounded-function estimate gives $|c_{j,n}|\le CB{2}^{-j/2}$ for bad coefficients.

Consider a specific N-term approximation: retain all coarse scaling functions and all wavelets through scale $J$, and also retain the bad coefficients at scales $J<j\le L$, where $L=\lceil 2\alpha J\rceil$. Since $\alpha >1/2$, $L\ge J$ for large $J$. Its term count is bounded byBy the Parseval identity, its squared L2 norm error is the sum of omitted squared coefficients. The good tail satisfiesand the remaining bad tail satisfiesChoose $J=\lfloor {\log }_2(N/(2C_2^{\prime }))\rfloor$ with a fixed sufficiently large $C_2^{\prime }$; then the $O(KJ)$ extra terms fit the remaining budget for all sufficiently large $N$, while $2^J$ is comparable to $N$. Padding to $N$ terms does not increase the error. The best N-term approximation is at least as accurate as this constructed selection, soThe constant may depend on the finite number of jumps, the piecewise Hölder norms, $B$ and the fixed wavelet family. Values exactly at the jumps are immaterial in $L^2$. The improvement comes from retaining only a bounded number of jump-crossing coefficients at each finer scale, as in best N-term wavelet approximation of piecewise Hölder functions.

The strict null space property of order $s$ isHere $v_S$ agrees with $v$ on $S$ and is zero elsewhere. Suppose a nonzero null vector had at most $2s$ nonzero entries. Split its support of a vector into disjoint sets $S,T$ of size at most $s$. Applying the null space property to both sets would give $\Vert v_S{\Vert }_1<\Vert v_T{\Vert }_1$ and $\Vert v_T{\Vert }_1<\Vert v_S{\Vert }_1$, a contradiction. Empty parts cause the same contradiction. Thus no such nonzero null vector exists.

For an $s$-sparse $x$, the feasible vector $x$ has $\Vert x{\Vert }_0\le s$, where the zero-subscript quantity counts nonzero entries and is not a norm. Any feasible competitor $z$ with $\Vert z{\Vert }_0\le \Vert x{\Vert }_0$ also has at most $s$ nonzero entries. The difference $z-x\in \ker A$ has at most $2s$ nonzero entries, so $z=x$. Competitors with larger support have strictly larger objective. Therefore every s-sparse vector is the unique sparsest feasible vector. This is sparse injectivity; for $s=0$ the only sparse vector is zero and the conclusion is immediate.

It is equivalent to minimize $\Vert z{\Vert }_q$ or its $q$th power $F_q(z)={\sum }_i|z_i{|}^q$, because taking the $q$th root is strictly increasing. For $q<1$, this is an Lq quasi-norm, while $q=1$ gives the L1 norm.

Assume the Lq null space property of order $s$. Let $x$ have support of a vector $S$ with $|S|\le s$, and let $z=x+v$ be any distinct feasible vector. Then $0\ne v\in \ker A$. The triangle inequality for complex magnitudes and the subadditivity $(a+b{)}^q\le a^q+b^q$ giveFor $q=1$ the same inequality follows directly from the triangle inequality. Since $x=0$ off $S$,Thus $x$ is the unique minimizer.

Conversely, suppose every $s$-sparse vector is the unique minimizer for its measurements. Fix $0\ne v\in \ker A$ and any $S$ with $|S|\le s$. The vectors $x=-v_S$ and $z=v_{S^c}$ are distinct and satisfy $Az=Ax$, since $Av=0$. Uniqueness for this $x$ therefore implies $F_q(x)<F_q(z)$, exactly the required strict inequality. HenceThe quantifier is uniform over the entire sparse class. Recovery of one particular signed vector alone would not imply this null space property.

Use the Lq null space property established above. Fix $0\ne v\in \ker A$ and order its magnitudes $a_1\ge \cdots \ge a_N\ge 0$. For a fixed exponent, the sum over the largest $s$ entries is the greatest sum over any support of a vector of size at most $s$, so it suffices to verify the property for this ordered support.

If $s\ge 1$, put $t=a_s$. We have $t>0$: otherwise $v$ would have fewer than $s$ nonzero entries, and applying the Lq null space property to its support would assert a positive number is less than zero. For $0<p<q$, the exponent $p-q$ is negative. ThusThe first inequality uses $a_i\ge t$ on the top part; the last uses $0<a_i\le t$ in the tail. Zero tail entries contribute zero without invoking a negative power of zero. Every other set of size at most $s$ has no larger top sum and no smaller complementary sum, so it too satisfies the strict $p$ inequality. The preceding equivalence provesThis is monotonicity of uniform sparse recovery in the exponent. If $\ker A=\{0\}$ the feasible vector is unique regardless of the objective; if $s=0$, only the zero vector is relevant. These cases need no threshold argument.

For real $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe vector fields have compact support inside $\Omega$, so no boundary term is charged. Taking both $\xi$ and $-\xi$ makes this equivalent to a supremum of absolute pairings. When finite, the distributional derivative $Du$ is a finite vector-valued Radon measure, and the displayed supremum is its total variation norm of a measure. For a smooth function it equals ${\int }_{\Omega }|\nabla u{|}_{2}dx$.

The function of bounded variation on a domain belongs to the Banach spaceThe L1 norm is needed because variation alone vanishes on constant functions and is only a seminorm. No smoothness of $u$ is part of this definition.

Take the indicator function $u(x_1,x_2)={\chi }_{\{x_1>1/2\}}$ inside $(0,1{)}^2$. For a compactly supported test vector field, integration in $x_1$ and then $x_2$ givesThe vertical derivative term integrates to zero. The absolute pairing is at most one, and test vector fields with ${\xi }_1=-1$ along all but arbitrarily short endpoint pieces approach one. Thus $|Du|(\Omega )=1$. More explicitly, the distributional derivative is $Du=e_1{\mathcal{H}}^1\upharpoonright \{(1/2,x_2):0<x_2<1\}$, where ${\mathcal{H}}^1$ is Hausdorff measure along the segment. Also $\Vert u{\Vert }_1=1/2$.

A Sobolev space function in $W^{1,1}(\Omega )$ has each weak derivative represented by an $L^1$ density with respect to planar Lebesgue measure. The nonzero segment measure here is singular with respect to that measure, so this requirement fails. Equivalently, the Sobolev fundamental theorem of calculus on lines would require an absolutely continuous representative on almost every horizontal slice, whereas each slice has a jump. ThereforeThis is the bounded-variation step outside W11 example.

Work in the real Hilbert space $L^2({\mathbb{R}}^2)$, with $g\in L^2$ and $\alpha >0$. The penalty $J$ is a proper convex function: it is finite at zero, and the bounded-variation space domain is convex. Its subdifferential at a finite-penalty $u$ consists of $q\in L^2$ satisfying $J(v)\ge J(u)+⟨q,v-u⟩$ for every $v\in L^2$.

If $q=(g-u)/\alpha \in \partial J(u)$, expand the quadratic term and use the subgradient inequality:Thus $u$ is the unique minimizer.

Conversely, let $u$ be a minimizer. Its penalty is finite since comparison with zero gives a finite objective. For any $v$ with $J(v)<\infty$, set $w_t=u+t(v-u)$, where $0<t\le 1$. Convexity gives $J(w_t)-J(u)\le t(J(v)-J(u))$. Minimality and quadratic expansion then implyLetting $t\downarrow 0$ yields $J(v)\ge J(u)+⟨(g-u)/\alpha ,v-u⟩$. The inequality is automatic if $J(v)=\infty$. HenceThis also follows from the subdifferential sum rule, since the quadratic term is everywhere continuous and differentiable, and the subgradient optimality condition. The direct proof above needs no unproved existence theorem.

On the entire plane, the printed global BV domain is not closed in the L2 geometry. The usual $BV({\mathbb{R}}^2)$ definition includes an $L^1$ condition. For $1<a\le 2$, $f(x)=(1+|x|{)}^{-a}$ belongs to $L^2$, has finite distributional variation $2\pi a{\int }_0^{\infty }r(1+r{)}^{-a-1}dr$, and fails to belong to $L^1$. The truncated $f_T=f{\chi }_{B(0,T)}$ lies in $BV\cap L^2$ and tends to $f$ in $L^2$. Its variation is the interior variation plus $2\pi T(1+T{)}^{-a}$, and tends to a finite limit. Thus $J(f_T)$ stays bounded but the literal $J(f)=\infty$, establishing failure of L2 closure of the global BV domain. One must not infer universal minimizer existence from an inapplicable closed-penalty proximal operator theorem. The standard closed extension uses the homogeneous bounded-variation space, allowing finite distributional variation without global $L^1$. The optimality equivalence just proved is valid for the literal penalty whenever a minimizer exists; the next datum has an explicit certified minimizer in its domain.

Let $B=B(0,R)$, with $R>0$. Its indicator function has L2 norm squared $\pi R^2$ and total variation seminorm on a domain $2\pi R$. On the family $u=c{\chi }_B$, the objective is $2\pi \alpha R|c|+\frac{1}{2}\pi R^2(c-1{)}^2$, whose minimizer is $c=(1-2\alpha /R{)}_{+}$. To prove global optimality, a total variation calibration is needed.

Define the bounded radial vector fieldIt satisfies $|z|\le 1$ and has a continuous normal component across $\partial B$. Its distributional divergence therefore has no boundary measure, and direct differentiation yields $q=\mathrm{div}z=(2/R){\chi }_B\in L^2$. Although $z$ is not compactly supported, multiply it by a smooth radial cutoff equal to one through radius $T$ and zero beyond $2T$. The extra divergence has magnitude $O(R/T^2)$ on an annulus of area $O(T^2)$, hence L2 norm $O(R/T)$. Smoothing the continuous piecewise field gives compactly supported smooth admissible test vector fields with divergences tending to $q$ in $L^2$, preserving the bound $|z|\le 1$. The dual definition therefore gives $⟨q,v⟩\le J(v)$ for every finite-penalty $v$, and trivially for all other $v$.

Moreover $⟨q,{\chi }_B⟩=(2/R)\pi R^2=2\pi R=J({\chi }_B)$. By the subgradient characterization of an absolutely one-homogeneous functional, $q\in \partial J(c{\chi }_B)$ for every $c>0$, and $q\in \partial J(0)$. If $0<\alpha <R/2$, choose $c=1-2\alpha /R$; then $(g-c{\chi }_B)/\alpha =q$. If $\alpha \ge R/2$, choose $u=0$; then $g/\alpha =(R/(2\alpha ))q$ belongs to $\partial J(0)$, since multiplying the dual bound by a number in $[0,1]$ preserves it. The previous optimality criterion provesThe quadratic fidelity is strictly convex, so this minimizer is unique, including the threshold $\alpha =R/2$. The disk retains its radius on the positive branch and disappears on the zero branch; this is total variation denoising of a disk.

Write $D=\nabla$ and use the real pixelwise inner products. Let $D^{\ast }$ be the adjoint operator, defined by $⟨Du,p⟩=⟨u,D^{\ast }p⟩$. For each nontrivial grid direction its one-dimensional formula iswith the analogous formula in $j$ for $D_y^{+\ast }{p}^2$. Their sum is $D^{\ast }p$; unused components $p_{N,j}^1$ and $p_{i,N}^2$ do not contribute. For $N=1$, $D=D^{\ast }=0$. If discrete divergence is defined, its sign convention is $\mathrm{div}=-D^{\ast }$, as in the adjoint of a discrete forward gradient.

Define the compact convex set $P_{\lambda }=\{p\in X^2:|p_{i,j}{|}_2\le \lambda \}$ and its image $C=D^{\ast }{P}_{\lambda }$. The image is compact and convex, hence closed, and contains zero. The finite-dimensional Euclidean duality formula $\lambda |a{|}_2={\max }_{|p{|}_2\le \lambda }a\cdot p$, applied independently at every pixel, givesThus the penalty is the support function ${\sigma }_C$. Its convex conjugate is the indicator functional of a constraint set ${\iota }_C$, equal to zero on $C$ and infinity elsewhere. For $w\in C$, the conjugate supremum is zero; for $w\notin C$, the separation theorem supplies a direction $u$ with $⟨u,w⟩>{\sigma }_C(u)$, and scaling that direction makes the supremum infinite.

The proximal operator is ${\mathrm{prox}}_{{\sigma }_C}(g)=\arg {\min }_u\{{\sigma }_C(u)+\frac{1}{2}\Vert u-g{\Vert }_2^2\}$. It exists uniquely because the objective is continuous, coercive and strictly convex. The Moreau decomposition for a proper closed convex function gives ${\mathrm{prox}}_{{\sigma }_C}(g)+{\mathrm{prox}}_{{\iota }_C}(g)=g$. The second term is the Euclidean projection onto a convex set, so the projection residual for discrete total variation isAlternatively, the variational characterization of convex projection says $w=P_{C}g$ exactly when $⟨g-w,z-w⟩\le 0$ for every $z\in C$. Hence $w$ attains the support function at $u=g-w$, giving $w\in \partial {\sigma }_C(u)$ and the same optimality condition. This supplies a direct verification of the Moreau decomposition step in this instance.

The printed wording needs “the residual after a projection”, rather than “the projection itself”. In general this denoising map is not a projection onto any fixed closed convex set. A metric projection onto a closed convex set is idempotent, while denoising a sufficiently large jump twice shrinks it twice. Concretely, for $N=2$ and data with rows $(0,0)$ and $(6\lambda ,6\lambda )$, the minimizer has rows $(\lambda ,\lambda )$ and $(5\lambda ,5\lambda )$. Reapplying the map gives rows $(2\lambda ,2\lambda )$ and $(4\lambda ,4\lambda )$, so it is not idempotent. These formulas have a global certificate: take $p_{1,j}^1=\lambda$, all other dual entries zero, giving $D^{\ast }p$ with rows $(-\lambda ,-\lambda )$ and $(\lambda ,\lambda )$; it saturates every positive row difference and yields the primal optimality condition. The qualification is therefore not merely a consequence of a restricted two-level ansatz.

Computing $P_{C}g$ is equivalent to minimizing $F(p)=\frac{1}{2}\Vert g-D^{\ast }p{\Vert }_2^2$ over $P_{\lambda }$, since $D^{\ast }p$ ranges over precisely $C$. Differentiating with the adjoint operator gives $\nabla F(p)=D(D^{\ast }p-g)$, whose Lipschitz constant is $L=\Vert D{D}^{\ast }\Vert =\Vert D{\Vert }^2$. Starting with any $p^0\in P_{\lambda }$, for example zero, the projected-gradient dual total variation algorithm isThe block formula is the Euclidean projection onto a convex set $P_{\lambda }$, since it clips each two-dimensional block to a Euclidean ball of radius $\lambda$. Both components must be clipped together for the isotropic block mixed norm.

For the unscaled forward differences, $\Vert D_x^{+}u{\Vert }_2^2\le 4\Vert u{\Vert }_2^2$ and likewise in the other direction, so $L\le 8$. More exactly, for $N>1$ the one-dimensional path difference has eigenvalues $4{\sin }^2(k\pi /(2N))$, $0\le k<N$, and the square-grid operator sums two such spectra. Thus $L=8{\cos }^2(\pi /(2N))$. Finite-dimensional projected gradient descent for a convex objective with an $L$-Lipschitz gradient converges from every feasible starting point to a dual minimizer for fixed $0<\tau <2/L$. The theorem assumes a nonempty closed convex set, a convex differentiable objective with globally Lipschitz gradient, and a nonempty minimizer set. All hold here: $P_{\lambda }$ is compact and convex, and $F$ is a continuous convex quadratic. Finite-dimensional convergence is of the full sequence, rather than only its objective values.

Consequently any fixed step with $0<\tau <1/4$ is safe on this grid; the more conservative $0<\tau \le 1/8$ is also safe. The dual minimizer need not be unique because $D^{\ast }$ has a null space, but $D^{\ast }{p}^k\to P_{C}g$ and $u^k\to u$ uniquely. For $N=1$, the gradient penalty vanishes and $u=g$, so no positive-$L$ step restriction is needed.

Absolute values are missing in the printed formula. For signed real blocks, the intended block mixed norm must beWithout them, at $q=1$ a block $(-1,1)$ has value zero although it is nonzero, and a negative scalar has negative value. For nonintegral $q$, a negative component may not even have a real power. The literal expression therefore is not a norm and does not define the claimed general convex problem. Even integer exponents do not fix all other $q$ in the stated range.

With the corrected norm, let $q^{\prime }$ be the Holder conjugate exponent: $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Blockwise Holder inequality, with equality from a norming block, givesThe duality of block mixed norms identifies the closed compact convex set and the projection residual from the Moreau decompositionFor $q=2$, the dual blocks are Euclidean balls; for $q=1$, they are cubes given by componentwise bounds $|p_{i,j}^r|\le \lambda$. No injectivity or surjectivity of $A$ is required: compactness of the block product makes its linear image compact, and the strictly convex fidelity still gives a unique primal minimizer. For general $q^{\prime }$, the Euclidean projection onto a dual block ball is not generally obtained by radial scaling, so the special clipping formula from the isotropic case should not be reused without further analysis.