Proximal Operator / Mapping of $ g \left( x \right) = {\left\| x \right\|}_{1} $ ($ {L}_{1} $ Norm) in the Complex Domain The $ \operatorname{Prox}_{g \left( \cdot \right)} $ operator is given by:
$$ \operatorname{Prox}_{\lambda g \left( \cdot \right)} \left( x \right) = \arg \min_{u} \left\{ g \left( u \right) + \frac{1}{2 \lambda} {\left\| u - x \right\|}^{2} \right\} $$
For $ g \left( x \right) = {\left\| x \right\|}_{1} $ in the $ {\mathbb{R}}^{n} $ the is given by:
$$ \operatorname{Prox}_{\lambda {\left\| \cdot \right\|}_{1}} = \begin{cases}
{x}_{i} - \lambda & \text{ if } {x}_{i} \geq \lambda \\ 
0 & \text{ if } \left| {x}_{i} \right| \leq \lambda \\ 
{x}_{i} + \lambda & \text{ if } {x}_{i} \leq -\lambda \\ 
\end{cases} $$
My question is, what would be the result if the domain is $ {\mathbb{C}}^{n} $?
Namely:
$$ \operatorname{Prox}_{\lambda {\left\| \cdot \right\|}_{1}} \left( x \right) = \arg \min_{u} \left\{ {\left\| x \right\|}_{1} + \frac{1}{2 \lambda} {\left\| u - x \right\|}^{2} \right\}, \; u ,x \in {\mathbb{C}}^{n} $$
Thank You.
 A: Let $\phi(u,x) = \|u\|_1 + {1 \over 2\lambda} \|x-u\|_2^2$. With a slight abuse
of notation, note that 
$\phi(u,x) = \sum_k \phi(u_k,x_k)$, so we may as well assume that $u,x \in \mathbb{C}$.
If $x=0$, we see that $\operatorname{prox}_{\lambda {|\cdot|}} (x) = 0$, so assume $x \neq 0$.
Note that if $|\theta|=1$, then $\phi(u,x) = \phi(\theta u, \theta x)$ and
also $\phi(u,x) \ge \phi(\operatorname{re} u, \operatorname{re} x)$.
In particular, 
$\phi(u,x) = \phi({\bar{x} \over |x|} u, |x|) \ge \phi(\operatorname{re}({\bar{x} \over |x|} u), |x|)$, from which we see that (i) we need only
optimise over real $u$ and (ii)
$\operatorname{prox}_{\lambda {|\cdot|}} (x) = {|x| \over \bar{x}} \operatorname{prox}_{\lambda {|\cdot|}} (|x|)$, where the latter is
computed over a real domain (we have
$\operatorname{prox}_{\lambda {|\cdot|}} (|x|) = \max(0,|x|-\lambda)$).
Notes: 
(i) ${|x| \over \bar{x}} = { x \over |x| }$.
(ii) The real (one or higher dimension) problem
$\min_u \|u\|_1 + {1 \over 2 \lambda } \|x-u\|_2^2$ has a nice solution
via a minor extension of the von Neumann mimimax theorem.
We have $\|u\|_1 =\max_{\|h\|_\infty \le 1} \langle h, u \rangle$ and
the problem can be written as
$\min_u \max_{\|h\|_\infty \le 1} \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|_2^2$. Applying the aforementioned theorem (with a minor extension to
deal with the non compact domain for $u$) we have
$\min_u \max_{\|h\|_\infty \le 1} \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|^2 = \max_{\|h\|_\infty \le 1} \min_u  \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|_2^2$ and solving the inner quadratic problem
gives
$\min_u \max_{\|h\|_\infty \le 1} \langle h, u \rangle + {1 \over 2 \lambda } \|x-u\|_2^2 = \max_{\|h\|_\infty \le 1} \langle h, x \rangle - {\lambda \over 2} \|h\|_2^2$ (with $\lambda h - (x-u) = 0$).
This is separable, and reduces to solving
$\max_{|h_k| \le 1} h_k x_k  - {\lambda \over 2} h_k^2$, which
has solution $h_k = \operatorname{sgn} x_k \min(1,{|x_k| \over \lambda})$, and
substituting into $u_k= x_k-\lambda h_k$ 
gives the solution $u_k=\operatorname{sgn} x_k \max(0,|x_k|-\lambda)$
