# Is the radial projection map area increasing?

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.

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.
• Yea, definitely. I was hoping that the extra condition on $S$ can avoid those 'fingers'. – Thomas Feb 14 at 21:20
• If the fingers are straight, slightly conical, with axis containing the origin, then $S$ is star-shaped at $0$ as you asked. – Benoît Kloeckner Feb 14 at 21:35
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 $$\lim_{\epsilon\to 0^+}\frac{\operatorname{Area(P(S_\epsilon))}}{\operatorname{Area}(S_\epsilon)}=0.$$