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In literature proximal operator $prox_{\lambda f} : R^n \rightarrow R^n$ of $f$ is defined as:

$prox_{\lambda f}(V) = argmin(X) (f(X) + (1/2 \lambda)||X-V||^2_2)$

Consider now $g(X) = \lambda||X||_1$, where $X$ is some $m$ by $n$ real matrix. Then using Moreau decomposition, the proximal operator of $g(X)$ is:

$prox_{\lambda f}(X) = $

\begin{cases} X_{i,j}-\lambda & X_{i,j}\ge \lambda, \\ 0 & |X_{i,j}|\le \lambda \\ X_{i,j} + \lambda & X_{i,j} \le -\lambda \end{cases}

Details of how exactly we use Moreau decomposition can be found in http://stanford.edu/~boyd/papers/pdf/prox_algs.pdf, sections 2.5 and then 6.5.2.

My question is: what will be proximal operator for the case when $g(X) = ||M-X||_1$ and $g(X) = ||M-XM||_1$, where $M$ is some constant real matrix?

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There are simple rules for the prox of translations (just do substitution in the minimization problem). For the last case I only expect a simple solution for invertable $M$ (again substitute). –  Dirk Jul 24 '14 at 18:30
For the first once, I suppose, you can take X := M-X, and then M - prox_g(M-X), since in the algorithm we ultimately are trying to find X satisfying X* = prox_g(X*). For the second one, it won't work, and my X is non-inverible. Can we use pseudo-inverse instead as an approximation? –  user3371282 Jul 25 '14 at 19:20

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