= Solution
{c}
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> solves
$$
\min_{u\in\mathbb R^k}\left\{\tau\|u\|+\frac12\|u-z_\ell\|^2\right\}.
$$
At a nonzero $u$, the <gradient> of the <Euclidean norm> is $u/\|u\|$. The <subgradient optimality condition> therefore becomes
$$
z_\ell=u+\tau\frac{u}{\|u\|}.
$$
Thus $u$ points in the direction of $z_\ell$ and $\|z_\ell\|=\|u\|+\tau$. This yields $u=(1-\tau/\|z_\ell\|)z_\ell$ when $\|z_\ell\|>\tau$.
At zero, the <subdifferential> of the <Euclidean norm> is the closed <Euclidean unit ball>: the <subgradient inequality> is $q^Tw\leq\|w\|$ for every $w$, which is equivalent to $\|q\|\leq1$ by the <Cauchy-Schwarz inequality> and the choice $w=q$. Hence zero solves the <subgradient optimality condition> exactly when $\|z_\ell\|\leq\tau$. This also covers $z_\ell=0$ and avoids division by zero.
The resulting <radial soft thresholding> is
$$
T_\tau(z_\ell)=
\begin{cases}
0,&\|z_\ell\|\leq\tau,\\
\left(1-\dfrac{\tau}{\|z_\ell\|}\right)z_\ell,&\|z_\ell\|>\tau.
\end{cases}
$$
If $y_\ell$ is the corresponding block of $x$, the required <proximal gradient mapping> is
$$
\boxed{[G_\alpha(x)]_\ell=\frac{y_\ell-T_{\alpha\lambda}(z_\ell)}{\alpha},\qquad z_\ell=y_\ell-\alpha[\nabla f(x)]_\ell.}
$$
\b[Whole 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).
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