Proximal Algorithms 1 Introduction

Proximal Algorithms

definition

Order \ (f: \ mathrm {R } ^ n \ rightarrow \ mathrm {R} \ cup \ {+ \ infty \} \) is closed convex function, i.e. on the mirror of FIG:
\ [\ mathbf {EPI} F = \ {(x, t) \
in \ mathrm {R} ^ n \ times \ mathrm {R} | f (x) \ le t \} \] non-empty closed convex subset of the domain:
\ [\ mathbf {dom} f = \ {x \ in \ mathrm {R} ^ n | f (x) <+ \ infty \} \]

Proximal operator (is so translated?) Proximal operator \ (\ mathbf {} _f Prox: \ mathrm {R} ^ the n-\ rightarrow \ mathrm {R} ^ the n-\) is defined as:
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We often have to add a proportion coefficient \ (\ the lambda \) , and the care \ (\ lambda f \) of the proximal end of the operator:
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Note: the right hand side by a constant \ (\ the lambda \) is \ (\ lambda f \) form, so it is equivalent.

Explanation

Graphic interpretation

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NOTE: the figure is a thin black line function \ (F \) contour, whereas bold lines represent boundaries of the domain. Which is estimated at a blue point \ (\ mathbf {prox} _f \) to give a red dot.

可以发现,\(\mathbf{prox}_f(v)\)实际上是对点\(v\)附近的一个估计。

梯度解释

假设\(\lambda\)很小,且\(f\)可微,那么,容易知道\(f(x) + \frac{1}{2\lambda}\|x-v\|_2^2\)取得极值(实际上也是最值)的条件是:
\[ \nabla f(x) +\frac{x-v}{\lambda}=0 \Rightarrow x=v-\lambda \nabla f(x) \approx v-\lambda \nabla f(v) \]
可以看到,\(\mathbf{prox}_f(v)\)近似为在\(v\)点的梯度下降,而\(\lambda\)为步长。

一个简单的例子

有一个问题,就是,如果我们的目的是最小化\(f(x)\),那么利用\(\mathbf{prox}_f\)会不会太愚蠢了,既然我们能求解\(\mathbf{prox}_f\),那么直接最小化\(f(x)\)应该也不是难事吧。这个问题留到以后再讨论吧,我也不知道能否找到一个恰当的例子来反驳。

When \ (F \) is an indicator function:
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where \ (\ mathcal {C} \ ) nonempty convex set, we look at this time \ (\ mathbf} {Prox _F (V) \) :
\ [\ mathbf {prox} _ { \ lambda f} (v) = \ mathrm {argmin} _x \: I _ {\ mathcal {C}} (x) + \ frac {1} {2 \ lambda} \ | xv \ | _2 ^ 2 \]
first of all, we can determine \ (the X-\ in \ mathcal {C} \) , otherwise the result is infinite, so the problem can be transformed into a Euclid norm projection problem:
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it is a problem is that if \ (\ mathbf {prox} _f \ ) tail terms do not \ (\ ell_2 \) norm, use other norm will become what?

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Origin www.cnblogs.com/MTandHJ/p/10969361.html