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Re: curvature and convexity
Posted:
Apr 13, 2011 9:36 AM


On Tue, 12 Apr 2011 06:50:58 0600, d <cliffordsmyth@gmail.com> wrote:
For the differential geometers out there: "Given B, a compact subset of R^n with nonempty interior is the condition that B is strictly convex equivalent to some other condition involving some sort of curvature on S the boundary of B?"
Depends on the meaning of "some sort" and "strictly convex".
I understand strictly convex to mean that a chord connecting any two points on the boundary has only its endpoints on the boundary and the rest in the interior of the set.
Consider the function: f(x) = \sum 2^{n}xr_n where {r_n : n=1,2,3,...} is an enumeration of the rationals. Then the supergraph {(x,y): y >= f(x)} is strictly convex, but f''(x)=0 almost everywhere.
However, f'(defined at every irrational) is strictly increasing, so "some sort" of curvature condition might be said to be satisfied, defined in terms of the turning of tangent lines.
Caveat: I am a geometer, but not a differential geometer.
Dan To reply by email, change LookInSig to luecking



