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.

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    $\begingroup$ 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
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    $\begingroup$ @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

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.


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