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...However, what about if $\phi$ is not (necessarily) even?

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?

Let $\phi:[0,\infty) \rightarrow (-\infty,\infty]$ 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: $$ \operatorname{Prox}_{\Phi}(x):= \operatorname{argmin}_{z \in \mathbb{R}^n}\, \frac1{2}\|x-z\|^2 + \Phi(z). $$

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...However, what about if $\phi$ is not (necessarily) even?

Let $\phi:[0,\infty) \rightarrow (-\infty,\infty]$ be a some-where finite, lower semi-continuous, increasing, and convex function. It is easy to verify that for every $y \in \mathbb{R}^n$, the function $\Phi_y:\mathbb{R}^n\rightarrow (-\infty,\infty]$ defined by $\Phi_y(x):= \phi(\|x-y\|)$ is convex, lower semi-continuous and also somewhere finite.

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

Let $\phi:[0,\infty) \rightarrow (-\infty,\infty]$ 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: $$ \operatorname{Prox}_{\Phi}(x):= \operatorname{argmin}_{z \in \mathbb{R}^n}\, \frac1{2}\|x-z\|^2 + \Phi(z). $$