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strip containing convex set

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strip containing convex set


does anyone know if it is possible to contain given convex set in a strip. More precisely : given K\subset R^n convex, k<=n does there exist a change of coordinates (rotation), after which K is contained in strip (K\cap P1)\times P2, where P1=\R^k\times 0^{n-k} and P2=0^k\times \R^{n-k} are (linear) planes in \R^n. So the problem is finding the correct k-dimesional plane L(which is then rotated to \R^k\times 0).

The problem is easy for k=1, then you just take the line containing two points in K, which have the maximum possible distance from each other and rotate this plane to \R^times 0^{n-1}.

Maybe the correct choice of plane L would be the one, for which the diam(K\cap L) is maximal (diameter of set-maximal distance between pair of points in set), but how to prove that ?

Of course it can be done for a ball or a cube, a question is about general convex set.

Thank you for any suggestions, trom

The question is not just about being contained, it must be contained in correct way :

D x {0} < K < D x R^s

A picture :
Strip is the set D x R^s
K is made of x, D is made of D

S | S
S | S
S | S
S xxxxx S
S xxxxxxx S R^r
S xxxxxxx S
S xxxxx S
S | S
S | S
Even upper half of 2D plane is then contained in strip
{(x,0):x in R}x{(0,y): y>0}, even more : it is a strip. But I am only
interested in bounded sets. In my problem I am allowed to work with approximations - for example you can replace a convex set by a polyhedron which is very close to original set. So if the problem is solvable for this class of convex sets, if would be ok.

I can't say I completely understand you question, your notation is somewhat confusing. But consider an unbounded convex set, for example the upper half of a 2D plane. It is contained in no strip. Perhaps you want a bounded set, but then there is no need for it to be convex, just choose a plane that does not intersect it and another one parallel to it that is on the other side of K and doesn't intersect it either.

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