Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 340 / 5

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.