Exercise 3.4.4

We assume a new function $g(z)$ along with constraint $x-z=0$, where $g(z)=0$ when $z$ is in the set $z^{T}w=1$ and $g(z)=\infty$ when $z$ is out of $z^{T}w=1$.

Following example 2, we split the problem into two functions: * $f_1(x_1)=\lambda \left|x_1\right|+\frac{1}{2}\rho {(x_1-v_1)}^2$ with $v_1=z_1-u_1$ * $f_2(x_2)=\lambda \left|x_2\right|+\frac{1}{2}\rho {(x_2-v_2)}^2$ with $v_2=z_2-u_2$

From example 2, we have the solution $x_i^{\text{∗}}=v_i{(1-\frac{\lambda }{\rho \left|v_i\right|})}_{\text{+}}$

If we start with $x_0=(1,1)$, also with $z_0=(1,1)$ and $u_0=(0,0)$ then we have $v_0=(1,1)$, and the next $x$:

$x_i^{\text{∗}}={(1-\frac{\lambda }{\rho })}_{\text{+}}$