 Solution

Write $A_1={\mathrm{Div}}_1$ and $A_2={\mathrm{Div}}_2$, with Euclidean adjoints $A_1^{\ast },A_2^{\ast }$ determined by the chosen discretization and boundary conditions. Take the usual positive total generalized variation weights ${\alpha }_1,{\alpha }_2$. Introduce a primal variable $w\in {\mathbb{R}}^{n\times 2}$ throughThe support function of the row-ball product is a sum of row norms. Convex duality gives the equivalent augmented saddle problemThe equality follows by dualizing the row-ball constraint. With positive radii, $v=0$ strictly satisfies both row constraints, supplying the finite-dimensional qualification for this splitting. The original feasible dual set is compact and nonempty, and the quadratic primal term is coercive; saddle points exist. The TGV divergence splitting avoids the difficult projection onto $\{v:v\in C,A_{1}v\in D\}$.

Use the Chambolle–Pock algorithm. Choose $\tau ,\sigma >0$ with $\tau \sigma \Vert K{\Vert }^2<1$; the sufficient bound $\tau \sigma (\Vert A_2{\Vert }^2+\Vert A_1{\Vert }^2)<1$ is convenient. Initialize $u^0,w^0,v^0$ and ${\bar{u}}^0=u^0$, ${\bar{w}}^0=w^0$. For $k=0,1,\dots$, computeThe dual update is a Euclidean projection onto a convex set onto $C$; the two primal updates are the quadratic proximal operator and radial soft thresholding. Their signs follow from $K^{\ast }v=(A_{2}v,-A_{1}v)$. All substeps are closed form, and the standard finite-dimensional primal-dual convergence result applies to this saddle problem with the stated step-size condition. The iterates $v^k$ satisfy $C$; the additional $D$ constraint is enforced through the splitting at convergence, not claimed for every intermediate iterate. If a weight is zero, the corresponding row projection or support-function proximal step is interpreted directly rather than by division by zero.