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Optimization with convex constraints and convex objectives; notions related to convex optimization such as sub-gradients, normal cones, separating hyperplanes

1 vote
0 answers
68 views

Fundamental regions in convex programming

In linear programming, the fundamental regions are polyhedra, since those are the intersection of half-spaces defined by linear inequalities. In semidefinite programming, the fundamental regions are s …
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1 vote
0 answers
133 views

Matrix completion in $2\times2$ case by nuclear norm minimization to guarantee rank $1$?

Does fixing diagonal entries and minimizing nuclear norm under weighted sum of entries conditions produce a rank $1$ matrix? I think the answer for this is no. At least could it be true in $2\times2$ …
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0 votes
1 answer
276 views

Quasiconvexity property of quasinorms

Schatten $p$ norm is convex when $p\geq1$ holds and if $p\in(0,1)$ it is quasinorm. If $p\in(0,1)$ then is Schatten $p$ norm quasi convex? I am interested in definition of quasi convexity here http …
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1 vote
1 answer
121 views

Convex integer programming on totally unimodular polytope?

If $$\min x'Qx + Rx$$ $$Ax\leq b$$ $$x\in\mathbb Z^n$$ is a quadratic program with $x'Qx$ is convex is there a polynomial time algorithm for this if $A$ is totally unimodular? In particular if we as …
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2 votes
1 answer
90 views

Linear program with one quadratic condition convex in domain of interest polynomial time sol...

$c\leq xy$ is not a convex condition. However we know $c\leq xy$ is convex in domain $x,y>0$ in $\mathbb R^2$ for any fixed $c\in\mathbb R$. Is $c\leq x_1y_1+\dots+x_ny_n$ with $0\leq x_1,\dots,x …
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1 vote
0 answers
88 views

On convex quadratic programming clarification

We know convex quadratic programming is in $P$. Is it also in $P$ if the function of interest is only convex in the domain of interest?
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7 votes
0 answers
216 views

Characterizing matrices with rank constraint

Given matrix $M\in\Bbb\{0,1\}^{n\times n}$, is there a nice method to characterize $$\mathscr{D}[M]=\{Q\in\Bbb\{0,1\}^{n\times n}:\mathsf{rank}(M-Q)= \mathsf{rank}(Q),\quad M-Q\in\Bbb\{0,1\}^{n\times …
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