Projection onto $\ell^{2,1}$ ball Does anyone have an idea how to project onto the $\ell^{2,1}$ ball efficiently, i.e. how to solve
$$ u = \arg \min_u  \|u-f \|^2    \text{such that } \left(\sum_i \big(\sum_j |u_{i,j}|\big)^2 \right)^{1/2} \leq 1$$
for a matrix $u \in \mathbb{R}^{n \times m}$ efficiently?
 A: Since the projection on to the $\ell_{1,2}$ ball doesn't have a simple solution, I would be surprised if there is one for the $\ell_{2,1}$ ball. The difficulty is that this case involves the squared $\ell_1$ norm rather than the simple $\ell_2$ norm. Maybe the following can be helpful if you would like to avoid splitting the problem.
First, note that your constraint can be equivalently expressed as 
$$ \|u\|_{2,1}^2 = \sum_i \|u_{i}\|_1^2 \leq 1,$$
where $u_i$ is the $i$th row of $u$.
Introducing a Lagrange multiplier $\lambda\geq 0$, the metric projection problem can be formulated as
$$\min_{u}\sup_\lambda \frac12\|u-f\|_{2,2}^2 + \lambda (\|u\|_{2,1}^2 -1).$$
Assume now that $\|f\|_{2,1}>1$ (otherwise we would have $u=f$), hence we can assume that for the minimizers $\|u\|_{2,1} = 1$ (the necessary optimality with respect to $\lambda$) and $\lambda >0$ holds (needs to be shown using a proof by contradiction, see below).
For optimality with respect to $u$, first note that the reformulated minimization problem completely decouples for the $u_i$. Hence, for every $i$, we have the optimality condition 
$$0\in \partial\left(\frac12\|u_i-f_i\|_2^2 + \lambda \|u_i\|_1^2\right).$$
Applying a sum rule ($0$ is a regular point), this is equivalent with
$$\frac1\lambda (f_i -u_i) \in \partial(\|\cdot\|_1^2)(u_i) = 2 \|u_i\|_1 \partial(\|\cdot\|_1)(u_i)$$
(using the fact that the subdifferential of one-half of the squared norm is the duality mapping, see Schirotzek, Nonsmooth Analysis, Remark 4.6.3).
If $f_i\neq 0$, then $\|u_i\|_1\neq 0$ (otherwise we obtain a contradiction).
Now we use the equivalence for every $\gamma>0$
$$-p\in\partial F(u) \Leftrightarrow u = \mathrm{prox}_{\gamma F}(u-\gamma p),$$
where the proximal mapping for the $1$-norm is the soft-thresholding operator
$$\mathrm{prox}_{\gamma\|\cdot\|_1}(w) = S_\gamma(w) = (|w|-\gamma)^+\mathrm{sign}(w)$$
with $(w)^+ = \max\{0,w\}$ (all of which is to be understood componentwise if $w$ is a vector).
Setting $\gamma = 2\lambda \|u_i\|_1$, we thus obtain the relations
$$ 
\begin{aligned}
&u_i = (|f_i|-2\lambda \|u_i\|_1)^+\mathrm{sign}(f_i) \qquad \text{for all }i,\\
&\sum_i \|u_i\|_1^2 = 1.
\end{aligned}
$$
(Here you get the desired contradiction: if $\lambda =0$, then $u_i = f_i$, which violates the second condition.)
Inserting the first relations into the second, you can solve for given $u_i$ for $\lambda$ via root finding, and for given $\lambda$, you can apply a fixed point iteration to (hopefully) compute $u_i$.
