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For a convex function $h$, its proximal operator is defined as: $$\operatorname{prox}_h(x)=\arg\min_u \Big(h(u)+\frac{1}{2}\|u-x\|^2\Big)$$ Can anyone provide an intuitive explanation/motivation of proximal mapping?

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    Not an answer, but... If you let $g(x) = \frac{1}{2} ||x||^2$, then the function $h \square g (x) = \inf_{u} h(u)+g(x-u)$ is called the infimal convolution and corresponds to the function you get from the set $\mathbb{epi}\, h + \mathbb{epi} \, g$. So there is a nice geometrical interpretation as a 'smoothing' of sorts. Rockafellar's classic Convex Analysis is a good read & reference. Predates the current popularization of convex analysis.2012-07-08
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    Oops, I should have been more explicit, $\mathbb{epi} \, h \square g = \mathbb{epi} \,h + \mathbb{epi} \, g$. The epigraph defines the function.2012-07-08
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    Thank you very much. I have ordered one copy.2012-07-08
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    I also don't have a definite answer, but one way to think of it is that you want a point that minimizes $h(u)$ but is also close to $x$ (the argument of prox). There is usually also a constant that dictates the tradeoff between these two components of the objective. The prox operator is very important in convex analysis and its properties are well studied. Take a look at [this monograph](http://www.stanford.edu/~boyd/papers/pdf/prox_algs.pdf) by Boyd for more details.2014-06-12

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