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Nonlinear objectives, nonlinear constraints, non-convex objective, non-convex feasible region.

2 votes
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Argmax of a function of $n$ variables under linear constraint

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 …
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2 votes
Accepted

Calculating derivatives of arbitrary-order at an operator's root

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 …
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4 votes

Reference request: importance of Lipschitz continuity

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 …
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0 votes
Accepted

Does coercivity/supercoercivity conjugates?

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 …
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1 vote
Accepted

Gradient-descent "type" Methods for non-convex and non-smooth functions

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 …
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1 vote

Is there a general guideline for minimizing $\sup_{y\in H}F(\;\cdot\;,y)$?

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 …
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2 votes

Optimization with weaker oracle than projection

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 …
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3 votes
Accepted

Adding constraints as penalty with $\| \cdot \|_0$ norm

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 …
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1 vote
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Quasiconvexity property of quasinorms

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 …
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6 votes

Nearest matrix orthogonally similar to a given matrix

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. …
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0 votes
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Optimization of non-smooth convex function in a polytope

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 …
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0 votes
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Calculate $k:=\sup\left\{\left\|\theta\right\|_{*} \: |\: \ell^{*}(\theta)<\infty\right\}$ f...

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 …
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1 vote

Block coordinate descent convergence rate

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 …
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1 vote
Accepted

Question about optimizing a given function by optimizing an approximation

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 …
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5 votes

Can this optimization problem be transformed into or approximated by a SOCP?

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 = \ …
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