Search Results

No, you can't. Even in linear optimization with only two variables you may have several linear constraints that form a corner of your feasible domain. Removing some constraints may make the corner les …

Your problem is of the form $$ \min F(x)\quad \text{s.t.}\quad Ax=b $$ or $$ \min F(x) + G(Ax) $$ with $G$ being the indicator function of the single point $b$. Hence, you can try basically all method …

I do not think that this can be converted to a convex problem: There are multiple solutions in general and these do not form a convex set: If $c>0$ and $x^*$ is a solution, then $-x^*$ is also one. H …

A lot of things are known for the convergence of alternating projections for these convex feasibility problems. I suggest to start with H.H. Bauschke and J.M. Borwein: On projection algorithms for …

I think the claim is true under some additional assumption: Let $\bar x$ be another solution of $$\min_x f(x) ~~ s.t.\\ g_{t^*}(x)\leq0 ~ $$ for which some constraints are not fulfilled, i.e. $g_{\bar …

There is abundant literature about stuff like this but it can be hard to find the framework that is best suited for your case. For example there is quite general theory in "Incremental subgradients f …

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 = \ …

Not sure if this points in the right direction since I don't know enough context. It seems like you could in principle set $y= Ux$ and either work with the rotated variable all over the place or use " …

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 …

As remarked by Dirk Werner in a comment, it's not true that directionals derivatives need to vanish at a local minimum point. There is a whole zoo of conditions that can be used. If you want to see th …

Problems of this type may be solved with splitting methods. One very popular case if the proximal gradient method. If $h$ is convex and lower-semicontinuous and if you are able to caluculate the proxi …

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 …

Note that the conclusion is that the gradient of $g_r$ converges to zero, not $g_r$ itself! The first inequality and the fact that $r>0$ imply $$\ g_r(M(y^k)) - g_r(y^k) > r|\nabla g_r(y^k)|^2 \geq 0 …

You are looking for optimization in vector spaces (preferably normed spaces). There is a rich theory available. The choice of spaces and norms is not always clear and often the key to the solution. In …