 Solution

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).