TITLE: Differentiability of polytope shadow areas
QUESTION [5 upvotes]: Let $P$ be an opaque convex polyhedron containing the origin in $\mathbb{R}^3$,
and let $S$ be an origin-centered sphere strictly containing $P$: $S \supset P$.
For a point $x$ on $S$, let $\sigma(x)$ be the area of the shadow
of $P$ cast from a light at $x$ onto the plane tangent to $S$ at $-x$:

          


My question is:

Q. What is the differentiability class $C^k$ of $\sigma(x): S \to \mathbb{R}$?

I would be surprised if $\sigma$ is a smooth map, $C^\infty$, but it seems to be
at least $C^1$...
The question makes sense for a convex polytope in $\mathbb{R}^d$ for $d \ge 2$,
with $\sigma(x)$ the $(d{-}1)$-volume of the shadow cast on a $(d{-}1)$ hyperplane.

REPLY [4 votes]: Let me try to give a computation free (sketch of) proof that the shadow area is not $C^1$. The basic idea is the same as in the answer of Willie Wong: a problem happen when a corner shows up.
Consider a position $x_0$ where some corner $c$ of the polytope is projected right on a facet of the shadow. Then move $x_0$ along a smooth path $x_t$ such that for small negative $t$, $c$ is projected in the interior of the shadow, and for small positive $t$ it is projected "outside the shadow", i.e. it is projected to a vertex of the shadow. Then the area $\sigma(x_t)$ is given by the sum of a smooth function (corresponding to the shadow of the facets already contributing to the shadow at $x_0$) and a function which is zero for $t<0$ (when the facets of $c$ do not contribute to the shadow) and is linear for $t>0$ (when the facets of $c$ start contributing to the shadow). Thus $\sigma$ is not $C^1$ (but it is certainly Lipschitz).