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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.

Again write $z=x-\alpha \nabla f(x)$ and $\tau =\alpha \lambda$, and let $z_{\ell }$ denote its $\ell$th block of $k$ coordinates. The block mixed norm in this part makes the proximal operator separate over blocks. For a single block, its minimiser solvesAt a nonzero $u$, the gradient of the Euclidean norm is $u/\Vert u\Vert$. The subgradient optimality condition therefore becomesThus $u$ points in the direction of $z_{\ell }$ and $\Vert z_{\ell }\Vert =\Vert u\Vert +\tau$. This yields $u=(1-\tau /\Vert z_{\ell }\Vert )z_{\ell }$ when $\Vert z_{\ell }\Vert >\tau$.

At zero, the subdifferential of the Euclidean norm is the closed Euclidean unit ball: the subgradient inequality is $q^{T}w\le \Vert w\Vert$ for every $w$, which is equivalent to $\Vert q\Vert \le 1$ by the Cauchy-Schwarz inequality and the choice $w=q$. Hence zero solves the subgradient optimality condition exactly when $\Vert z_{\ell }\Vert \le \tau$. This also covers $z_{\ell }=0$ and avoids division by zero.

The resulting radial soft thresholding isIf $y_{\ell }$ is the corresponding block of $x$, the required proximal gradient mapping isWhole blocks can become zero, while nonzero blocks retain their direction after the gradient step. For a block of size one, radial soft thresholding agrees with the soft-thresholding operator in part (a).