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This is inspired by this previous question on physical processes that might give rise to convex hulls.

Consider the problem of gift-wrapping a three-dimensional object using an inextensible material, like paper. We can make the material conform to any surface with nonnegative Gaussian curvature by cutting and folding it. (At least, we can make an arbitrarily good polyhedral approximation.) But if we want to have negative Gaussian curvature, we have to make a cut and glue some extra material in there, which is awkward and cumbersome, so we forbid it. Now we want to perform this gift wrapping as tightly as possible, which suggests using the least amount of material.

Formally, given a set $S$ of points in $\mathbb R^3$, we want to find the surface with minimum area that encloses all the points in $S$, subject to an additional condition that the Gaussian curvature of the surface is nonnegative everywhere. In a comment on the previous question, I conjectured that this would be precisely the convex hull of $S$. But I have no idea if that's actually true, and if so, how to begin proving it.

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    $\begingroup$ Let $\Sigma_0$ be the convex hull of $S$. If $\Sigma$ encloses $\Sigma_0$, then consider the nearest-point projection of $\Sigma$ onto $\Sigma_0$: the projection is a contraction, hence $|\Sigma|\ge |\Sigma_0|$... True or False: every surface of nonnegative curvature that encloses $S$ must also enclose $\Sigma_0$? $\endgroup$
    – user31373
    Commented Jun 4, 2012 at 22:29
  • $\begingroup$ @Leonid, nice observation! $\endgroup$
    – user856
    Commented Jun 4, 2012 at 23:10
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    $\begingroup$ The answer is yes, by one of the forms of Hadamard's ovaloid theorem. If a surface has nonnegative curvature which is not identically zero, then it is the boundary of a convex set. This is stated in many places, e.g. here: people.math.gatech.edu/~ghomi/Talks/LocConvexSlides.pdf It follows that the surface encloses the convex hull of $S$, and the argument in the previous comment applies. $\endgroup$
    – user31373
    Commented Jun 9, 2012 at 23:34
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    $\begingroup$ @RahulNarain The convex hull of $S$ is not a smooth surface in general, so it would be prudent to clarify what you mean by Gaussian curvature. A theorem from differential geometry may not apply. For example, if $S$ is a finite set, then the convex hull is a polyhedral surface. In this case all the curvature is concentrated at the vertices. If you define the curvature at a vertex $v$ to be the angle defect (i.e., $2\pi - \sum \alpha_i$, where the $\alpha_i$ are the face angles at $v$), then I believe I can construct a counter-intuitive counter example. $\endgroup$
    – yasmar
    Commented Jun 10, 2012 at 15:56
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    $\begingroup$ $C^{1,1}$: first order derivatives are Lipschitz continuous. More generally, $C^{k,\alpha}$ means that $k$th order derivatives satisfy the Holder condition with exponent $\alpha$. $\endgroup$
    – user31373
    Commented Jun 11, 2012 at 14:46

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By Hadamard's ovaloid theorem, any positively curved surface (without boundary) in $\mathbb R^3$ is the boundary of a convex set. Therefore, such a surface encloses the convex hull of $S$. The projection onto the convex hull does not increase the area, being a 1-Lipschitz map. It follows that the area of any positively curved surface enclosing $S$ cannot be smaller than the area of the surface of the convex hull of $S$.

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