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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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    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 $\Sigma_0$*?2012-06-04
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    @Leonid, nice observation!2012-06-04
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    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: http://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.2012-06-09
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    @Leonid: That looks like it wraps it up! Can you post it as an answer so I can accept it?2012-06-10
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    @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.2012-06-10
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    @yasmar, thanks for being careful about the question. I guess what I mean is that the surface can be approximated arbitrarily closely by a smooth surface of nonnegative Gaussian curvature. The angle defect at vertices is not quite enough to capture this. But I think it's sufficient to check that that the angle defects at all vertices *and* the dihedral angles at all edges have the right sign.2012-06-10
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    Ya, you could stipulate something like the absolute extrinsic curvature mentioned here: http://www.win.tue.nl/EWCG2005/Proceedings/18.pdf but it seems your observation can be expressed more cleanly and more generally along the lines you express at the beginning of your comment. From @Leonid's answer it seems like you may be able to conclude that the convex hull is some kind of Gromov-Hausdorff limit of smooth surfaces of positive Gaussian curvature.2012-06-10
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    @yasmar Just for clarity's sake: I did not try to apply Hadamard's theorem to the convex hull. My interpretation of the question was that we look for the infimum of areas of nonnegatively curved smooth closed surfaces enclosing $S$. For such surfaces the area of the convex hull provides a lower bound. To show that this lower bound is indeed the infimum, one needs a sequence of smooth surfaces converging to the convex hull. Unfortunately $1/n$ neighborhoods are not quite smooth enough ($C^{1,1}$ but not $C^2$).2012-06-10
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    Thanks @Leonid. I agree with what you said, but I didn't understand the last sentence of your comment: I'm not sure what you mean by $1/n$ neighbourhood, and the smoothness limitation.2012-06-10
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    @yasmar I meant the level sets of the distance function, $\Sigma_n=\{x\colon \mathrm{dist}\,(x,\Sigma_0)=1/n\}$, where $\Sigma_0$ is the convex hull of $S$.2012-06-10
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    @yasmar: Thanks for pointing out that there exist nonconvex polyhedra with positive angle defect, in particular ones whose area is less than that of the corresponding convex hull. I just realized that this is very relevant for my original motivation for this question.2012-06-11
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    @Leonid, I know what $C^1$ and $C^2$ mean, but what is $C^{1,1}$?2012-06-11
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    $C^{1,1}$: first order derivatives are [Lipschitz continuous](http://en.wikipedia.org/wiki/Lipschitz_continuity). More generally, $C^{k,\alpha}$ means that $k$th order derivatives satisfy the Holder condition with exponent $\alpha$.2012-06-11

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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 $S$.