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Operations research, linear programming, control theory, systems theory, optimal control, game theory
3
votes
Accepted
Maximize sum of largest eigenvalues
If I understand the question correctly, the answer is that no, optima need not occur at a unique point where some of the hyperplanes defined by tightness of the linear inequalities meet the boundary o …
3
votes
Accepted
Lagrangian duality
At the level of generality you asked about, the answer is no, the claim is not correct. Of course, your case of interest may rule out counterexamples like the one below.
It can happen that the prima …
4
votes
Accepted
Maximizing linear objective function with absolute values
In general such a problem is NP-hard, so not expressible by a polynomially-sized linear program. A special case of the problem you mention is computing the $\lVert A\rVert_{\infty,1}$ norm of a matri …
0
votes
Accepted
Best constant in a convex polynomial inequality.
This is just an answer to the new question added in the edit: no, $p\geq q$ need not hold everywhere. Let $p(x) = x^2 + (1+\alpha) x^6$ for some $\alpha\in (0,\frac{1}{4})$ and $q(x) = x^4 + x^6$. T …
1
vote
Convex optimization over vector space of varying dimension
You may be interested in some of the recent work of Bill Helton and his collaborators. The idea (very roughly) is to study convex problems which can be defined in some intrinsic way without reference …
1
vote
Finding an unfrustrated set of local linear constraints with given minimal value
As written it doesn't seem like this should happen very often. For example take $B$ to be a set with a smooth boundary, such as the $n$-disc and suppose that the sum $F$ is not the zero function. Th …
9
votes
Accepted
minimize the sum of absolute eigenvalues
The sum of the absolute value of the eigenvalues is the same (since the matrix is real and symmetric) as the sum of the singular values. This sum is called the nuclear norm of the matrix. So what yo …
11
votes
Accepted
Non-negative quadratic maximization
First, note that the condition that $A$ be positive semidefinite (PSD) doesn't buy you anything. Replacing $A$ by $A+kI$ changes the objective value of any feasible solution by $k$, so if we could so …
6
votes
Accepted
Analogue of PSD matrices for permanents?
The set $C$ is not convex, nor is its intersection with the symmetric matrices. To see this note that by linearly interpolating between each of the matrices below we maintain positive permanent and s …
7
votes
SDP Feasibility
Alex gave a good answer, but I would just like to highlight a subtle problem with your claim about polynomial time solvability of SDPs. This depends on having inner and outer bounding balls to the fe …
5
votes
Computational complexity of low rank SDP
Various NP-hard problems, such as MAX-CUT, can be formulated exactly as SDPs with rank constraints (just google e.g. "max cut sdp"). If you want to enforce such rank constraints anyway, a popular app …
4
votes
a different algebra/representation for convex sets
Of course it's hard to say without any information, but linear matrix inequalities may be what you're looking for. They are to semidefinite programs what linear inequalities are to linear programs. …
22
votes
Is all non-convex optimization heuristic?
I think you will be interested in the work of Parrilo, Lasserre, Putinar, Sturmfels, Nie, Helton, etc., on sums-of-squares and moment methods for polynomial optimization. They look at principled ways …
2
votes
What is the dual of an semidefinitely representable (SDR) cone?
Edited in response to Alex Monras's correction in the comments:
The cone $\mathcal{K}^*$ is always SDR: this is just the conic / homogeneous version of Theorem 5.57 in the new book "Semidefinite Opti …