# Proximity operator of lower semi-continuous and convex functions pre-composed with norm

Let $$\phi:(-\infty,\infty) \rightarrow (-\infty,\infty]$$ be a some-where finite, lower semi-continuous, increasing, and convex function. It is easy to verify that the function $$\Phi:\mathbb{R}^n\rightarrow (-\infty,\infty]$$ defined by $$\Phi(x):= \phi(\|x\|)$$ is convex, lower semi-continuous and also somewhere finite.

Can we express the proximity operator of $$\Phi$$ in terms of the proximity operator of $$\phi$$? Where: $$\operatorname{Prox}_{\Phi}(x):= \operatorname{argmin}_{z \in \mathbb{R}^n}\, \frac1{2}\|x-z\|^2 + \Phi(z).$$

An expression can be found in the case where $$\phi$$ is even in Combette's book (as pointed out by Christian in the comments: the result is Example 24.51 (page 433, second edition))...However, what about if $$\phi$$ is not (necessarily) even?

• Can you give an exact reference to the expression in Combette's book so the reader doesn't have to hunt through all 620 pages of it? Also, since norms are nonnegative, you can always consider $\phi$ to be even without loss of generality, can't you? Feb 18, 2021 at 11:01
• Ah, found it, it's Example 24.51 (page 433, second edition). I guess the main question is whether the differentiability requirement can be relaxed if you're only interested in the expression for the proximity operator and not whether it is a proximal thresholder? Feb 18, 2021 at 11:05
• @ChristianClason Can it? I haven't managed so far. Feb 18, 2021 at 14:31
• I don't know; I'd have to go through the proof carefully to see where it is used and whether those arguments can be adapted. I'll leave that to someone with more time, as these proofs are quite dense. Feb 18, 2021 at 17:28