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Dear Convex Optimization Experts,

My question is related to this post: The Proximal Operator of the $ {L}_{\infty} $ (Infinity Norm), but not really same, I think, as I have a constraint. Apologies if it is obvious to extend the answer.

So, I am seeking a proximal operator $\textrm{prox}_{\lambda f}(x)$ of this function $$f(x) = I\left\{x \in C \right\} \ ,$$ $$C = \left\{x : \lVert x \rVert_{\infty} \leq \gamma \right\} \ ,$$ where $x \in \mathbb{R}^{n \times 1}$, $I\left\{\cdot \right\}$ is an indicator/characteristic function, $\gamma \in \mathbb{R}^{n \times 1}$, and the constraint $\lVert x \rVert_{\infty} \leq \gamma$ is element-wise.

EDIT: The desired constraint is: $$C = \left\{x : |x_i| \leq \gamma_i \right\} \ ,$$

Royi
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user550103
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    $| x |_{\infty}$ is a scalar, so $\gamma$ should be a scalar too. – Brian Borchers Aug 19 '18 at 16:29
  • The prox operator for the indicator function of a convex set is simply the projection onto the set $C$. Algorithms for the projection onto your particular set $C$ are well known. – Brian Borchers Aug 19 '18 at 16:32
  • It is an element-wise operation. Example, if I have a vector $x = [1, 2, 3, 2, 4] $ and $\gamma = [3, 3, 3, 3, 3] $ of course, for simplicity, I can consider a constant $\gamma$. Does that make sense? – user550103 Aug 19 '18 at 16:34
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    You're constraint would then be $| x_{i} | \leq \gamma_{i}$ for $i=1, 2, \ldots, n$. The projection onto this box is also very simple to compute. – Brian Borchers Aug 19 '18 at 16:40
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    @user550103 wrote "Does that make sense?" No, it doesn't. $|x|_\infty = 4$ for your sample $x$. – Mark L. Stone Aug 19 '18 at 16:41
  • @Brian: thanks for that. Yes, this is what I want to do actually. – user550103 Aug 19 '18 at 16:42
  • @BrianBorchers: May I know the projection onto such box please? – user550103 Aug 19 '18 at 16:48

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