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What are the known sufficient conditions,analagous to the planar curvature condition, in terms of functions of theta and phi, on the support function h(theta,phi) of a surface in 3D which imply it is the boundary of a convex body?

[The origin of this question is an elementary approach to exploring 3D (convex) sets of constant width (and their volumes using the Divergence Theorem) in advance of the anniversary of the Blaschke-Lebesgue (1914) theorem.]

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up vote 3 down vote accepted

For me, the simplest way to figure out whether a function, which is often defined as a function of the unit sphere, is a support function or not is to extend it to all of $\mathbb{R}^n$ as a function homogeneous of degree $1$. Then the function is a support function if and only if it is convex.

You can figure out whether a smooth function of $\theta$ and $\phi$ is a support function or not by finding the formula for this function in terms of the extended function and differentiating twice.

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Thanks very much for your answer. As you suggest I was considering functions defined on the unit sphere such as h = 0.5 + (1/16)Cos(3Theta) + (1/8)Cos(Phi)Sin(Theta). I understand the extended function idea but I am still unclear that "differentiating twice" gives the simple solution I was, perhaps foolishly, seeking. I do know of the quadratic form tests for convexity of functions at a point in Eggleston's book "Convexity" –  Ian Calvert Mar 7 '13 at 16:06
    
To be honest, I haven't worked out the details myself, so some trial and error is probably needed. It seems to me easier to use the formulas that give $x$, $y$, and $z$ in terms of $r$, $\theta$, and $\phi$. These can be used to write the function of $\theta$ and $\phi$ in terms of the homogeneous function $x$, $y$, and $z$. If you calculate the Hessian of the function of $\theta$ and $\phi$, you will get a formula that involves the Hessian of the homogeneous function, which you know is positive definite. You use this to try to infer what the corresponding condition is for the first function. –  Deane Yang Mar 7 '13 at 16:10
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