2
$\begingroup$

Let $ h \left( \cdot \right) $ be a convex function.
The proximal operator is defined by:

$$ \operatorname{Prox}_{ \lambda h( u ) } \left( x \right) = \arg \min_{u} \lambda h \left( u \right) + \frac{1}{2} {\left\| u - x \right\|}_{2}^{2} $$

Why is this operator well defined? We must see that exist that minimum and is unique. But I don't see how to do that and I don't find any information on the bibliography.

$\endgroup$
  • 2
    $\begingroup$ The sum of a convex function $h(u)$ and a strongly convex function $\frac12\|u-x\|^2$ is strongly convex. $\endgroup$ – Rahul Aug 11 '18 at 17:42
  • 1
    $\begingroup$ @Rahul is right, and with strong convexity you get a guarantee of a unique global minimum (except in the rather degenerate case of an empty domain) $\endgroup$ – Michael Grant Aug 11 '18 at 21:16
1
$\begingroup$

I will assume that the domain of $h$ is finite dimensional, non empty and open. As a result, $h$ is continuous and the epigraph of $h$ has a supporting hyperplane, that is, there is some $g, \alpha$ such that $h(x) \ge \langle g , x \rangle + \alpha$ for all $x$.

It is straightforward to show that the level sets of $u \mapsto \langle g , x \rangle + \alpha+ {1 \over 2} \|u-x\|^2$ are bounded, and hence the level sets of $u \mapsto h(u) + {1 \over 2} \|u-x\|^2$ are bounded. Since $h$ is continuous, the level sets are compact and it follows that the proximal function has a $\min$. Since the $\|u-x\|^2$ term is strictly convex, it follows that the minimiser is unique.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.