The Proximal Operator of the $ {L}_{1} $ Norm Function Write down explicitly the optimal solutions to the Moreau-Yosida regularization of the function $f(x)=\lambda\|x\|_1$, where $f:\mathbb{R}^n\to(-\infty,+\infty]$.
I have found that the answer is 
$x_i=sgn(x_i)\max\{|x_i|-1,0\}$
Here is my attempt to get the answer:

The proximal operator to $f(x)$ is $\min_{y\in\mathbb{R}^n}\lambda\|y\|_1+\frac{1}{2}\|y-x\|^2_1$.
  I need to minimize this over y.

I have no idea how to continue. I have read a lot of references but I cannot find an explicit step-by-step solution to this problem.
Any help would be appreciated!
 A: The Optimization Problem given by the Prox Operator:
$$ \operatorname{Prox}_{\lambda {\left\| \cdot \right\|}_{1}} \left( x \right) = \arg \min_{u} \left\{ \frac{1}{2} {\left\| u - x \right\|}^{2} + \lambda {\left\| u \right\|}_{1} \right\} $$
This problem is separable with respect to both $ u $ and $ x $ hence one could solve the following problem:
$$ \arg \min_{ {u}_{i} } \left\{ \frac{1}{2} {\left( {u}_{i} - {x}_{i} \right)}^{2} + \lambda \left| {u}_{i} \right| \right\} $$
Now, you can proceed using First Order Optimality Condition and the Sub Gradient of the $ \operatorname{abs} \left( \cdot \right) $ function or you can employ simple trick.
The trick is to understand that $ {u}_{i} $ can be either positive, zero or negative.
Assuming $ {u}_{i} > 0 $ the derivative is given by $ {u}_{i} - {x}_{i} + \lambda $ which vanishes for $ {u}_{i} = {x}_{i} - \lambda $ and holds for $ {x}_{i} > \lambda $.
The same procedure for the case $ {u}_{i} < 0 $ yields $ {u}_{i} = {x}_{i} + \lambda $ for $ {x}_{i} < -\lambda $.
For values of $ {x}_{i} $ in between, since ${u}_{i} = 0 $ and hence the derivative (Sub Gradient) of $ {u}_{i} $ can freely be chosen on the range $ \left[ -1, 1 \right] $ the value of $ {u}_{i} = 0 $ holds.  
In summary:
$$ \operatorname{Prox}_{\lambda {\left\| \cdot \right\|}_{1}} \left( x \right)_{i} = \operatorname{sign} \left( {x}_{i} \right) \max \left( \left| {x}_{i} \right| - \lambda, 0 \right) $$ 
As @NicNic8 noted, this operation is called Soft Threshold.
