I'm willing to believe (though I don't know a proof) that K has a well-defined surface area, by virtue of being convex. But surely L could be some weird fractal thing for which surface area is not defined. Does the theorem have a tacit assumption that L's boundary is integrable? On Mon, Jun 24, 2019 at 12:40 PM Dan Asimov <dasimov@earthlink.net> wrote:

I think Cauchy proved the 2-dimensional version of this by averaging orthogonal projections over all different directions. So there may be a proof like that in higher dimensions. Convex surfaces are almost everywhere C^2, which may help.

—Dan

Jim Propp wrote ----- I believe the two-dimensional version of this assertion was made by Archimedes, and was intended to serve as a way of characterizing arc-length.

On Sun, Jun 23, 2019 at 8:09 PM Adam P. Goucher <apgoucher@gmx.com> wrote: ----- Excellent proof!

By the way, for the 'obvious' lemma that the girth is an increasing function of the two semi-axes, it follows from:

Theorem: If L is a compact subset of R^n and K is a convex compact subset of L, then the surface area of K is bounded above by the surface area of L, with equality if and only if K = L. ... ----- -----

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