Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 62 / 5

Use the nonnegative-pairing dual cone $K^{\ast }=\{y:⟨y,s⟩\ge 0\text{ for all }s\in K\}$. The Lagrangian is $L(x,y)=c^{T}x-y^T(Ax-b)$ for $y\in K^{\ast }$. Its infimum over unrestricted $x$ is finite exactly when $A^{T}y=c$. Therefore the conic dual problem isThe final equality uses the self-dual cone assumption.

For the canonical self-concordant barrier, use its logarithmically homogeneous barrier normalizationAt parameter $\mu >0$, the primal barrier problem isWrite $F_{†}(y)=F^{\ast }(-y)$, the Legendre dual cone barrier. The matching dual barrier problem isDefining the dual barrier this way is valid generally; self-duality of the cone alone does not assert that an arbitrarily selected barrier equals its Legendre dual.

The joint central path characterization isThe primal stationarity equation is $c+\mu A^T\nabla F(s)=0$, exactly the dual feasibility equation after defining $y$. For the dual relation, logarithmic homogeneity gives $\nabla F(s/\mu )=-y$, so $\nabla F_{†}(y)=-s/\mu$. Thus dual stationarity is $-b+\mu \nabla F_{†}(y)+Ax=0$, giving the same primal equation. At these points,

Newton step. At a chosen target parameter $\hat{\mu }>0$, defineLinearizing the three equality conditions gives the central-path Newton systemEliminating the slack and dual directions yieldsThe barrier Hessian is positive definite. Full column rank of $A$ therefore makes $A^{T}HA$ positive definite and the reduced solve unique. Use a damped step with $s+\alpha \Delta s$ and $y+\alpha \Delta y$ remaining interior; a full Newton step need not do so.

At an exact point with parameter $\mu$, choosing $\hat{\mu }<\mu$ gives $r_p=r_d=0$ and $r_c=(\hat{\mu }-\mu )\nabla F(s)$, the usual predictor towards the next path point. Infinitesimally,

The path definition presupposes interior feasibility and attainment of the barrier problems; strict primal-dual feasibility is a standard sufficient setting. The printed full-rank condition by itself is insufficient. For example $K={\mathbb{R}}_{+}^2$, $A=(1,-1{)}^T$, $b=0$ and $c=0$ give the sole primal feasible point $x=0$, with no interior slack at all, despite full column rank. Rank guarantees the Newton solve at an interior point, not existence of the central path.

Put $d_i(u)=((G^{(1)}u{)}_i,(G^{(2)}u{)}_i)$. The image penalty is discrete isotropic total variation ${\sum }_i\Vert d_i(u){\Vert }_2$; the fidelity term is the L1 norm $\Vert u-f{\Vert }_1$. Introduce scalars $v_i,t_i$ and minimize the linear objectivesubject to the following affine residuals being cone-feasible:where ${\mathcal{L}}_3=\{(t,a,b):t\ge \sqrt{a^2+b^2}\}$ is the Lorentz cone. Every component displayed is affine in $(u,v,t)$, so stacking them gives exactly the requested $Ax-b\in K$ form, withThe first two inequalities imply $v_i\ge |u_i-f_i|$, so separate nonnegativity constraints for $v_i$ are unnecessary. The cone constraints imply $t_i\ge \Vert d_i(u){\Vert }_2$. Every lifted feasible objective bounds the original objective from above; choosing $v_i=|u_i-f_i|$ and $t_i=\Vert d_i(u){\Vert }_2$ gives equality. Since both objective weights are positive, an optimum uses these tight values. Thus the lift is an exact second-order cone program for box-constrained TV-L1 denoising.

The product cone is proper, closed, convex and self-dual. The stacked $A$ has full column rank: the box rows recover $u$, the fidelity rows then recover $v$, and the first coordinates of the Lorentz blocks recover $t$. Strict feasibility is also explicit: choose $u_i=1/2$, then take $v_i>|u_i-f_i|$ and $t_i>\Vert d_i(u){\Vert }_2$.

For generic cone slack $(r,q)$, a canonical barrier ison $r_j>0$ and $q_{i0}>\sqrt{q_{i1}^2+q_{i2}^2}$. The positive sheet condition matters: positivity of the quadratic expression alone would also include the wrong Lorentz nappe.

Composing the barrier with the affine residual map givesIts domain includes the strict inequalities above. Each orthant coordinate contributes one to the self-concordant barrier parameter and each Lorentz block contributes two. Therefore the product-cone logarithmically homogeneous barrier hasThe affine composition $\Phi$ is the barrier used in the primal variables; it need not itself be logarithmically homogeneous in $(u,v,t)$ because the image data and box offsets are affine constants.