I need to solve the following problem: $$\min._{0 \le \mu \le 1} \|\mu-z\|_2^2 + \alpha\|\mu\|_\infty$$ I indicate the box constraint on $\mu$ explicitly as a function and get $$\min._{\mu\in\mathbb{R^n}} \|\mu-z\|_2^2 + \alpha\|\mu\|_\infty + I_C(0 \le \mu \le 1)$$

Hence, my optimization problem is about computing the proximal operator of the function $\alpha\|\mu\|_\infty + I_C(0 \le \mu \le 1)$. Inspired by this thread https://mathoverflow.net/questions/221188/l-infinity-norm-regularized-proximity-problem, I thought I will use Moreau's decomposition: $I = \text{prox}_f(.) + \text{prox}_{f^*}(.)$

In order to derive the conjugate of the sum of two functions, I used the concept of infimal convolution given here: http://www.athenasc.com/convexdualitysol3.pdf

Subsequently, I am able to derive closed-form expressions for the proximal operator of the conjugate of the sum of functions and obtain the optimal $\mu$.

In the above notes, the conjugate of the sum of functions, is defined as the closure of the infimal convolution of the conjugates of the individual functions, provided the original sum of functions is proper. What is the practical consequence of dealing with the concept of closure?

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    $\begingroup$ I doubt there is any "practical" consequence of requiring closure here. For one thing, most functions one deals with in practice are already closed, and I believe the infimal convolutions of such functions are going to be closed, too. And the conjugate function is always closed anyway. It would be an interesting challenge to come up with a pair of functions whose infimal convolution is not already closed. I'm sure it's possible, but I suspect it would require a manufactured case. So again, in practice, I don't see it likely that the closure requirement changes anything. $\endgroup$ – Michael Grant May 29 '17 at 22:29
  • $\begingroup$ Thanks a lot for your inputs. $\endgroup$ – haripkannan May 30 '17 at 7:00

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