# Proximal mapping for composition of functions

Suppose I have a convex function $f(x)$ for which I can easily compute the proximal mapping

prox$_f(z) = \arg\min_{x} f(x) + \frac{1}{2}||x-z||^2_2$

is there a simple expression for the proximal mapping of $f(\mathcal{A}(x))$ where $\mathcal{A}(x)$ is linear? That is, can I express

$\arg\min_{x} f(\mathcal{A}(x)) + \frac{1}{2}||x-z||^2_2$

in terms of prox$_f(z)$?

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You might want to check the paper 'Proximal Splitting Methods in Signal Processing' by Combettes and Pesquet (*) which reviews the prox and the prox of typical operations on functions (translation, scaling, …).

In that paper, you will find that if $\mathbf A$ is a semi-orthogonal linear transform (i.e.: $\mathbf A\mathbf A^* = \nu \mathbf{Id}$ with $\nu>0$) then there is a closed form of the prox of $f\circ \mathbf A$. I don't think there is one for the general case (I'd be tempted to say that it would be in that paper if it existed!)

The expression is the following:

$$\mathrm{prox}_{f\circ \mathbf A}(x) = x + \nu^{-1} \mathbf{A}^*\left(\mathrm{prox}_{\nu f}(\mathbf A x) - \mathbf A x\right) \tag{1}$$

Hope this helps a little!

EDIT if you're interested in proving the above expression, here are the main steps of a possible proof:

• Express the definition of the problem in a dual form (Fenchel-Rockafellar) and write the first order optimality conditions
• Use Moreau's decomposition you should get something like $$\nu^{-1}\mathbf A \mathbf d - y^* = \mathrm{prox}_{(\nu^{-1}f^\star)^\star}(\nu^{-1}\mathbf A\mathbf d)$$ where $y^*$ is the dual variable, and $^\star$ denotes the convex conjugate.
• express the convex conjugate of $(\nu^{-1}f^\star)$
• Shake the whole a little and using the optimality conditions, it should boil down to (1).

Another proof is using the fact that $\partial f\circ \mathbf A = \mathbf A^*\circ\partial f\circ \mathbf A$ (equality holds since $\mathbf A$ is surjective and hence the strong relint qualification condition holds, cf. Rockafellar). With Moreau decomposition $$x - \mathrm{prox}_{f\circ \mathbf A}(x) \in \mathbf A^*\partial f(\mathbf Ax)\in Im(\mathbf A^*)$$ multiply everything by $\mathbf A$ and it boils down to $$\mathbf A \mathrm{prox}_{f\circ\mathbf A}(x) = \mathrm{prox}_{\nu f}(\mathbf A x)$$ you can then use the decomposition of the prox in its orthogonal projection on $Im(\mathbf A^*)$ and on $ker(\mathbf A)$ and using the last equation you should end up with (1).

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