The minimizer of $2$-norm

Question

I am trying to compute $$ \operatorname{argmin}_{z\in\mathbb R^n}\frac12\|z-x\|_2^2+\|z\|_2 $$ where $\|\cdot\|_2$ is the usual 2-norm.

By doing sub-differential with respect to $z$ and let it equal to 0, I got $$ z(1+1/\|z\|_2)=x $$ I am wondering is this all I can do? or could I future simplify it?

Answer 1

You are computing the prox-operator of the $\ell_2$-norm.

The Moreau decomposition formula expresses the prox-operator of a proper closed convex function $f$ in terms of the prox-operator of $f^*$, the convex conjugate of $f$: $$ \text{prox}_{f}(x) = x - \text{prox}_{f^*}(x). $$ If $f(x) = \| x \|$, where $\| \cdot \|$ is any norm on $\mathbb R^n$, then $f^*$ is the indicator function for the dual norm unit ball. It follows that the prox-operator of $f^*$ simply projects onto the dual norm unit ball. Hence, $$ \text{prox}_{\| \cdot \|}(x) = x - P_B(x), $$ where $B$ is the dual norm unit ball and $P_B(x)$ is the projection of $x$ onto $B$.

In the special case where $\| \cdot \|$ is the $\ell_2$-norm, $B$ is the $2$-norm unit ball (because the $\ell_2$-norm is self-dual) . So, $$ \text{prox}_{\| \cdot \|_2}(x) = x - P_B(x), $$ where $B = \{u \mid \| u \|_2 \leq 1\}$.

The prox-operator of the $\ell_2$-norm simply shrinks $x$ towards by origin by a distance of one unit.

Answer 2

The formula implies that $z = tx$ for some $t$. Put that into your formula and cancels $x$, and help you solves for $t$. Then $z = tx$ is found explicitly.