Let $A$ be an orthogonal matrix. Then the proximal mapping $prox_{f \circ A}(x)$ can be evaluated efficiently by $$ (I + \partial (f \circ A))^{-1}(x) = prox_{f \circ A}(x) = A^T prox_{f}(A x), $$ as described here:


I was wondering if this extends to a proximal mapping with a diagonal matrix $D$ as step size: $$ (I + D \partial (f \circ A))^{-1}(x) = ? $$ I'm asking if the above proximal operator is easily to evaluate, assuming that $(I + D \partial f)^{-1}$ is easy.

I have been trying quite a while to find some closed-form for the above, but it seems difficult. I do not want to use an iterative method like ADMM to evaluate the proximal mapping.

  • $\begingroup$ You can work directly with the induced-norm $v \mapsto \|D^{-1/2}v\|_2$, w.r.t the matrix $D$ disappears from the prox. Note that proximal algorithms are available for much more general Bregman distances than the euclidean-norm $\endgroup$ – dohmatob May 18 '16 at 21:29
  • $\begingroup$ Are you looking for a way to prove what you wrote? $\endgroup$ – Royi Jun 10 '16 at 7:01

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