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Let $S$ be a hypersurface enclosed inside the unit sphere in $R^n$. Assume that every ray $\{t x: t \geq 0 \}$ intersects $S$ at most once.

Is it always true that ${\rm Area}(S) \leq {\rm Area}(P(S))$?

Here $P$ is the radial projection map onto $S^{n-1}$, i.e. $P(x) = x/\|x\|$.

(I am mostly interested in the 2-dimensional case.)

Thanks.

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The answer is negative: the area of $P(S)$ is at most the area of the unit sphere, while the area of $S$ can be made arbitrarily high.

An $S$ contained in the unit sphere and star-shaped at $0$ can be parametrized by the radius in polar coordinates: $S=\{\phi(u)u : \lVert u\rVert=1\}$ where $\phi$ is any smooth function from the unit sphere to $(0,1)$. Now, the area of $S$ is something like $$\int \phi^{n-1}\sqrt{\lVert \nabla \phi\rVert^2+1}$$ (a bit late here, so I might have gotten the formula wrong but in any case the integrand goes to infinity with $\nabla\phi$). Taking $\phi$ with value in say $[\frac13,\frac23]$ and with a lot of variation (e.g. making fingers or wrinkles) we can easily make the area of $S$ arbitrarily high.

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  • $\begingroup$ Yea, definitely. I was hoping that the extra condition on $S$ can avoid those 'fingers'. $\endgroup$ – Thomas Feb 14 at 21:20
  • $\begingroup$ If the fingers are straight, slightly conical, with axis containing the origin, then $S$ is star-shaped at $0$ as you asked. $\endgroup$ – Benoît Kloeckner Feb 14 at 21:35
  • $\begingroup$ Actually, you can make things more explicit, I'll edit my answer. $\endgroup$ – Benoît Kloeckner Feb 14 at 21:35
  • $\begingroup$ I think you are right. I need a stronger condition on S in order to have a hope. $\endgroup$ – Thomas Feb 14 at 21:46
  • $\begingroup$ Adding curvature bounds would certainly do the trick. $\endgroup$ – Benoît Kloeckner Feb 14 at 21:48
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Here is another counterexample. This is in $\mathbb{R}^3$ for simplicity, but the same argument works in any dimension. Let $$ S_\epsilon=\{(x,y,z):\, x^2+y^2\geq 0.01,\ z=\epsilon,\ z^2+y^2+z^2\leq 1\}. $$ This is a disc parallel to the equator plane, $\epsilon$ above the equator, and with a small disc of radius $0.1$ removed. $P(S_\epsilon)$ is a small strip above the equator so $\operatorname{Area}(P(S_\epsilon))\to 0$ as $\epsilon\to 0^+$. Therefore $$ \lim_{\epsilon\to 0^+}\frac{\operatorname{Area(P(S_\epsilon))}}{\operatorname{Area}(S_\epsilon)}=0. $$

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