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.