Exercise 3.4.2

We follow the example 1, where the set $K$ is $x^{T}v=1$, which is a closed convex set. Let’s rewrite this as a minimizationof $g$. We assume a new function $g(z)$, which is an indicator function of the set $x^{T}v=1$, $g(z)=0$ or $g(z)=+\infty$ for $z$ in or out of $x^{T}v=1$.

So the problem changes to

So our scaled augmented Lagrangian is

If we start with $x_0=0,z_0=0,u_0=0$, then we have

• $x_1=\text{argmin}[|x{\text{|}}^2+\frac{1}{2}\rho \parallel x-z_0+u_0{\text{∥}}^2]=\text{argmin}[\parallel x{\text{∥}}^2+\frac{1}{2}\rho \parallel x{\text{∥}}^2]=\text{argmin}[(1+\frac{1}{2}\rho )\parallel x{\text{∥}}^2]$

  • Solve for the minimal, we find $x_1=0$
• $z_1=\text{projection of}{x}_1+u_0\text{onto }K$, since $x_1+u_0=0$, we find $z_1=0$
• $u_1=u_0+x_1-z_1=0$

It seems that if we start with $x=z=u=0$, the ADMM won’t find the solution. Is this right?