Dual Problem of Projection onto the $ {L}_{1} $ Ball This is a very famous problem and there are many articles discussing it. 
For example:  https://stanford.edu/~jduchi/projects/DuchiShSiCh08.pdf  .
$$\min_{\|x\|_1 \leq \tau} \frac{1}{2}\|x-z\|^2$$  
My question is how to derive its dual problem. 

The following is my effort:  
he Lagrangian dual is the following:
\begin{align*}
L(x,u) = \frac{1}{2}\|x-z\|^2 + u(\|x\|_1 -\tau) 
\end{align*}So 
\begin{align*}
\nabla_x L = (x - z) + \bar{u}
\end{align*} where the $i$-th entry of $\bar{u}$ is
\begin{align*}
 \bar{u}_i=\begin{cases}
               u_i,  &x_i > 0\\
               [-u_i,u_i], &x_i=0 \\
      -u_i, &x_i< 0         
            \end{cases}
\end{align*}
So let $\nabla_x L = 0$, we have $x^* = z-\bar{u}$. So 
\begin{align*}
L(u) = \frac{1}{2}\|\bar{u}\|^2 + u(\|z-\bar{u}\|_1 - \tau)
\end{align*}

I am confused that there are three cases for $\bar{u}$, and they are imbedded in the norm functions. I have no idea how to write down a neat and clear dual problem.  
 A: Since the above it the Orthogonal Projection onto the Unit Simplex here is the solution using the Dual Function.
The Lagrangian in that case is given by:
$$ \begin{align}
L \left( x, \mu \right) & = \frac{1}{2} {\left\| x - y \right\|}^{2} + \mu \left( \boldsymbol{1}^{T} x - 1 \right) && \text{} \\
\end{align} $$
The trick is to leave non negativity constrain implicit.
Hence the Dual Function is given by:
$$ \begin{align}
g \left( \mu \right) & = \inf_{x \succeq 0} L \left( x, \mu \right) && \text{} \\
& = \inf_{x \succeq 0} \sum_{i = 1}^{n} \left( \frac{1}{2} { \left( {x}_{i} - {y}_{i} \right) }^{2} + \mu {x}_{i} \right) - \mu && \text{Component wise form}
\end{align} $$
Again, taking advantage of the Component Wise form the solution is given:
$$ \begin{align}
{x}_{i}^{\ast} = { \left( {y}_{i} - \mu \right) }_{+}
\end{align} $$
Where the solution includes the non negativity constrain by Projecting onto $ {\mathbb{R}}_{+} $
Again, the solution is given by finding the $ \mu $ which holds the constrain (Pay attention, since the above was equality constrain, $ \mu $ can have any value and it is not limited to non negativity as $ \lambda $ above).
The objective function (From the KKT) is given by:
$$ \begin{align}
h \left( \mu \right) = \sum_{i = 1}^{n} {x}_{i}^{\ast} - 1 & = \sum_{i = 1}^{n} { \left( {y}_{i} - \mu \right) }_{+} - 1
\end{align} $$
The above is a Piece Wise linear function of $ \mu $ and its Derivative given by:
$$ \begin{align}
\frac{\mathrm{d} }{\mathrm{d} \mu} h \left( \mu \right) & = \frac{\mathrm{d} }{\mathrm{d} \mu} \sum_{i = 1}^{n} { \left( {y}_{i} - \mu \right) }_{+} \\
& = \sum_{i = 1}^{n} -{ \mathbf{1} }_{\left\{ {y}_{i} - \mu > 0 \right\}}
\end{align} $$
Hence it can be solved using Newton Iteration.
I wrote MATLAB code which implements them both at Mathematics StackExchange Question 2338491 - GitHub.
There is a test which compares the result to a reference calculated by CVX.
