Formulate the dual problem for primal problem with absolute value constraint Let $y \in R$, the goal is to find the dual problem to:
$$\min y\\
s.t. |y| \leq 0$$
The lagrangian of the problem is: 
$$L(y, \lambda) = y + \lambda|y|$$
The dual function is:
$$g(\lambda) = \inf_y y + \lambda |y|, \lambda \geq 0$$
But the constraint $|y| \leq 0$ is only satisfies at $y = 0$
Then $g(\lambda) = 0$
So the dual problem is:
$$\max 0 \\ s.t. \lambda \geq 0$$
Weird, probably incorrect, can anyone see where I went wrong?
 A: Your mistake is that you're still trying to enforce the $|y|\leq 0$ constraint after you have built the Lagrangian. Don't do that. Once the Lagrangian is built, the constraint on $y$ goes away. Furthermore, the dual function doesn't include $\lambda\geq 0$, just the dual problem.
So the dual function is
$$g(\lambda) = \inf_y y + \lambda |y|$$
nothing more. But we can simplify. Note that when $y\leq 0$, $y+\lambda|y|$ is $(1-\lambda) y$. When $\lambda<1$, this quantity is unbounded below. So we have
$$g(\lambda) = \inf_y y + \lambda |y| = \begin{cases} 0 & \lambda \geq 1 \\ -\infty & \lambda < 1 \end{cases}$$
In other words, $g(\lambda)=-I_{[1,+\infty)}(\lambda)$, the negative of indicator function for the interval $[1,+\infty)$, so the dual problem is
\begin{array}{ll}
\text{maximize} & - I_{[1,+\infty)}(\lambda) \\
\text{subject to} & \lambda \geq 0
\end{array}
This is the correct Lagrangian dual. However, it is commonplace to remove any constraints implied by the domain of the dual function and make them explicit. In this case, we have $\lambda \geq 1$, so
\begin{equation*}
\begin{array}{ll}
\text{maximize} & 0 \\
\text{subject to} & \lambda \geq 0 \\
& \lambda \geq 1
\end{array} \quad\Longrightarrow\quad
\begin{array}{ll}
\text{maximize} & 0 \\
\text{subject to} & \lambda \geq 1
\end{array}
\end{equation*}
