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.
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.