Starshaped surfaces

Given a positive differentiable function $f:{\mathbb S}^2\rightarrow{\mathbb R}$, define $$S(f)=\{f(p)p:p\in {\mathbb S}^2\}.$$ This surface is called a starshaped surface because the half-line starting from the origin of coordinates meets only at one point of $S(f)$. For example, if $f=2$, then $S(2)$ is the sphere cetered at the origin of radius $2$. In order to give more explicit examples, consider the parametrization of the sphere $X(t,\theta)=(\cos(t)\cos\theta,\cos(t)\sin\theta,\sin(t))$. Then define two functions $f$ by $f(X(t,\theta))=1+\cos(t)^2$ and $f(t)=1+\cos(s)^2$. The pictures are: 

We observe in the second surface that there appears 'strange point'. Exactly, points where the tangent plane is not well-defined. This occurs because $X$ is not a parametrization that cover the whole sphere, but only a part. Exactly, $s\in (0,2\pi)$, that is, except a meridian, exactly the points where appear the problem.

We give some properties of these surfaces.

1. The set $S(f)$ is, indeed, a surface. The map $$\phi:{\mathbb S}^2\rightarrow {\mathbb R}^3,\ \phi(p)=f(p)p$$ is differentiable and $d\phi_p$ is one-to-one. The proof is as follows. If $v\in T_p{\mathbb S}^2$, then if $0=d\phi_p(v)=(df_p(v))p+f(p)v$, we have a linear combination of $p$ and $v$. We know that $T_p{\mathbb S}^2=<p>^\bot$. If $v\not=0$, then $f(p)=0$, a contradiction. This proves that $\phi({\mathbb S}^2)=S(f)$ is a surface.

2. The surface $S(f)$ is compact and connected because  $S(f)=\phi({\mathbb S}^2)$.

3. The map $\phi$ is one-to-one. If $\phi(p)=\phi(q)$, then $f(p)p=f(q)q$. Taking the modulus, we have $f(p)=f(q)$, so $p=q$. 

4. The map $\phi:{\mathbb S}^2\rightarrow S(f)$ is a diffeomorphism because the inverse is   is $\phi^{-1}(p)=p/|p|$, so it is differentiable. 

A compact surface

Consider the set $$S^\{(x,y,z)\in{\mathbb R}^3: e^{x^2}+e^{y^2}+e^{z^2}=4\}.$$

We prove that $S$ is a compact surface diffeomorphic to the unit sphere ${\mathbb S}^2$.

Here it is the picture of $S$

First  $S$ is a surface because $S=f^{-1}(\{4\})$ and $4$ is a regular value of the function $f:{\mathbb R}^3\rightarrow{\mathbb R}$ given by $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$. To prove that $S$ is compact, the same fact $S=f^{-1}(\{0\}$ proves that $S$ is closed. On the other hand, $S$ is bounded because for any $(x,y,z)\in S$, we have

$e^{x^2}\leq 4$ so $x^2\leq\log(4)$. This shows that $|(x,y,z)|^2\leq 3\log(4)$.

Finally we prove that $\phi:S\rightarrow {\mathbb S}^2$, $\phi(p)=p/|p|$ is a diffeomorphism. We point out that $|p|\not=0$ for any $p\in S$. It is immediate that $\phi$ is differentiable.

We prove that $d\phi_p$ is an isomorphism. Since

$$d\phi_p(v)=\frac{v}{|p|}-\frac{\langle p,v\rangle}{|p|^3}p$$

if $d\phi_p(v)=0$, then $v$ is proportional to $p$, exactly $v=\langle p,v\rangle p/|p|^2$. If $\langle v,p\rangle=0$, then $v=0$. On the contrary, $p=\lambda v$ with $\lambda\not=0$. In particular, $p$ is a tangent vector. Because the tangent plane $T_pS$ is orthogonal to $\nabla f(p)=2(x e^{x^2},y e^{y^2},z e^{z^2})$,  $p=(x,y,z)$, then  $\langle p,\nabla f (p)\rangle=2(x^2 e^{x^2}+y^2 e^{y^2}+z^2 e^{z^2})=0$, which is not possible. Then the inverse function theorem proves that $\phi$ is a local diffeomorphism.

We prove that $\phi$ is tsurjective. Given $p\in {\mathbb S}^2$, we have to find $q\in S$ such that $q/|q|=p$. Then we take the half-straightline starting from the origin across $p$ until that we intersect with $S$ at one point. Then it is clear that $q$ is the desired point. Thus we have to find $\lambda\in {\mathbb R}$ such that if $(x,y,z)\in {\mathbb S}^2$, $$e^{\lambda^2 x^2}+e^{\lambda^2 y^2}+e^{\lambda^2 z^2}=4.$$ If we see the left hand-side as continuous function on $\lambda$, namely, $g(\lambda)$, we use the intermediate value theorem: if $\lambda\rightarrow 0$, $g(\lambda)\rightarrow 3$ and if $\lambda\rightarrow\infty$, then $g(\lambda)\rightarrow\infty$, obtaining the result.

Finally, and because $S$ is compact, it suffices to see that $\phi$ is injective. Suppose $\phi(p)=\phi(q)$. Then $p$ and $q$ are proportional, that is, $q=m p$, $m>0$. If $p=(x,y,z)$, then

$$ e^{x^2}+e^{y^2}+e^{z^2}= e^{m^2 x^2}+e^{ m^2y^2}+e^{m^2 z^2}.$$But it is clear that the function $g(m)=e^{m^2 x^2}+e^{ m^2y^2}+e^{m^2 z^2}$ is one-to-one because $g'(m)\not=0$ for $m>0$. This proves that in (*) $m$ is necessarily $1$.