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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 …

Although this question sounds quite innocent, a systematic treatment of higher order derivatives of implicit functions is quite involved. On a second thought, this is no surprise if you think about ho …

Here is my two cents: In (unconstrained continuous) optimization you want to find minima of functions and in the differentiable case these have vanishing gradient. (In the convex case, vanishing gradi …

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 …

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 …

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 …

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 …

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 …

No, quasinorms are in general not quasiconvex. (Well, this is true if quasiconve means that the levelsets of the function are convex; other defintions of quasiconvexity may exist…) By positive homoge …

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. …

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 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 …

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 …

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 …

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 = \ …