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Methods of these type sometimes go under the name "Kaczmarz method". Kaczmarz method is a method for solving $Ax=b$ by iteratively (e.g. cyclic") projection onto the solutions of the equations given b …

First, you should restrict $x$ to be positive or use $|x|^\beta$ instead. Then I think that the answer is no: For $\beta\neq 1/2$ you can argue as follows: The special case of diagonal $\Omega = \ …

You could dualize the $h$ to get a saddle point problem. To be specific: Write $h(x) = H(Ax)$ with $H(y) = I_{\cdot\leq b}(y)$ and write $H(Ax) = \sup_y (Ax)^Ty - H^*(y)$. The resulting saddle point …

The claim in the paper is false. Since the problem is not convex, the claim does not follow from general results. However, there are some results in this direction in quite general cases: If $x^*$ i …

Your problem is also convex. Hence, a whole bunch of methods for convex optimization are available. Since projecting onto the constraints is not too difficult (project each row of $A$ onto the simplex …

The Wikipedia page has a counterexample: A continuous convex function for which coordinate descent fails to converge but getting stuck in a non-optimal point. Here are the level lines of this functi …

I would guess that the method is going to converge (weakly), even with constant stepsizes. Off the top of my head I don't know a precise reference. The method is close in spirit to the "hybrid project …

In general, small perturbations of the objective may change the set of local maxima drastically. Just think of a flat local maximum and adding a small wiggling (so also uniform approximation does not …

The function $\ell$ is concave (it's a minimum of linear functions). Its conjugate $\ell^*(\theta)$ is the supremum over a linear function minus $\ell$. If I see correctly, $\ell^*(\theta) = \infty$ f …

Here is an approach via Langange multipliers: The Lagrangian of the constrained problem is $$L(x,\lambda) = x_1\cdots x_n + x_2\cdots x_n + x_n - \lambda(\sum_{i=1}^n x_i - n-C).$$ The solution is a c …

$\newcommand{\Ag}{\mathcal{A}_\gamma}$ Not a full answer, put probably a fruitful pointer: In the discretization of infinite dimensional problems one faces (bi-)infinite matrices. There, the Jaffard …

Several splitting methods fit the bill: Often the non-convexity and the non-smoothness come from different parts of the objective and one can split the objective like $ f(x)=g(x) +h(x)$ with a convex …

Not a full answer, but some pointers on how to get a numerical method: If $\|\cdot\|_2$ denotes the spectral norm, then, by unitary invariance, your problem is equal to $$ \min_{T\in O(n)}\|AT-TB\|_2. …

The problem you are dealing with is of the form $$ \inf_{x\in H}\sup_{y\in H} F(x,y). $$ If $F$ is convex in $x$ and concave in $y$, this is a saddle point problem and you can find a lot of informatio …

No. The conjugate of the constant zero is the indicator of zero (and vice versa). More generally, the conjugate of the indicator of a closed convex set is positively one-homgeneous (and hence, not sup …