What is the proximal operator of $ {\left\| x \right\|}_{2}^{3} $ where $ {\left\| x \right\|}_{2} $ is the $ {L}_{2} $ norm?

Using Moreau Decomposition (Someone needs to create a Wikipedia page for it) one could solve it as following:

$$ \operatorname{Prox}_{\lambda f \left( \cdot \right)} (v) = v - \Pi_B \left( v \right) $$

Where $ \Pi_B \left( \cdot \right) $ is the projection of onto the unit ball of the dual norm.
Yet I'm not sure how to derive for the case mentioned above.

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    $\begingroup$ It's not clear to me why this was closed without any sort of comment. It's certainly not a trivial question to answer. I do not think there is a nice simple answer that you can offer for a general norms, however. $\endgroup$ – Michael Grant Dec 26 '16 at 19:51
  • $\begingroup$ @MichaelGrant, I voted it for reopen. Certainly can be amswered. $\endgroup$ – Royi Aug 1 '19 at 11:03
  • $\begingroup$ @Royi alas, I think a question of this age is not likely to be reopened regardless of merit. $\endgroup$ – Michael Grant Aug 1 '19 at 15:17
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    $\begingroup$ Well, You can believe as it happened :-). $\endgroup$ – Royi Aug 1 '19 at 16:21
  • $\begingroup$ Pay attention that your function isn't a vanilla norm. So I'm not sure its support is a norm ball which means its dual function is the dual norm. $\endgroup$ – Royi Aug 6 '19 at 7:40

The solution isn't trivial.
The problem is given by:

$$ \operatorname{Prox}_{\lambda f \left( \cdot \right)} = \arg \min_{x} \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \lambda f \left( x \right) = \arg \min_{x} \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \lambda {\left\| x \right\|}_{2}^{3} $$

I succeeded using identity of Euclidean Norm Composition (Calculus of the Prox Operator, From Amir Beck - First Order Methods in Optimization):

Norm Composition: Let $ f : E \to R $ be given by $ f \left( x \right) = g \left( \left\| x \right\| \right) $, where $ g : R \to \left(−\infty, \infty \right] $ is a proper closed and convex function satisfying $ \operatorname{dom} \left( g \right) \subseteq \left[0, \infty \right) $. Then:

$$ \operatorname{Prox}_{\lambda f \left( \cdot \right)} \left( y \right) = \begin{cases} \operatorname{Prox}_{ \lambda g \left( \cdot \right) } \left( \left\| y \right\| \right) \frac{y}{\left\| y \right\|} & \text{ if } y \neq 0 \\ \left\{ x \in \mathbb{E} : \left\| x \right\| = \operatorname{Prox}_{ \lambda g \left( \cdot \right)} \left( 0 \right) \right\} & \text{ if } y = 0 \end{cases} $$

So in our case:

$$ g \left( t \right) = \begin{cases} {t}^{3} & \text{ if } t \geq 0 \\ \infty & \text{ if } t < 0 \end{cases} $$

It is easy to derive that

$$ \operatorname{Prox}_{\lambda g \left( \cdot \right)} \left( t \right) = \frac{-1 + \sqrt{ 1 + 12 \lambda {\left[ t \right]}_{+} }}{6 \lambda} $$

So for $ \operatorname{Prox}_{\lambda g \left( \cdot \right)} \left( 0 \right) = 0 $.
Hence the composition becomes:

$$ \operatorname{Prox}_{\lambda f \left( \cdot \right)} \left( y \right) = \begin{cases} \frac{-1 + \sqrt{ 1 + 12 \lambda \left\| y \right\| }}{6 \lambda} \frac{y}{\left\| y \right\|} & \text{ if } y \neq 0 \\ 0 & \text{ if } y = 0 \end{cases} = \frac{2}{1 + \sqrt{1 + 12 \lambda \left\| y \right\|}} y $$

I validated the result comparing it to CVX solution. The MATLAB code is available at my Mathematics StackExchange Question 2071774 Repository.


Your idea about using the Moreau Decomposition is valid (Though I don't think it will yield solution in this case).
Yet you must pay attention that your function $ {\left\| \cdot \right\|}_{2}^{3} $ isn't a norm (Hence it is not the support of a norm etc...). You can't use the projection in this case but you need to use the Dual Function which as I wrote, I don't think will create a simpler solution.


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