Let $\mathbb{S}^n$ be the unit sphere and choose $p_0$ as the north pole. Consider the function $d:\mathbb{S}^n \to [0, \infty)$ defined by $d(p) = d(p,p_0) = \cos^{-1}( \langle p, p_0 \rangle)$. It is the intrinsec distance to $p_0$ in the sphere. Is this function convex?

In order to answer this question, we have to show that $d \circ \gamma$ is a convex function of a real variable, for any geodesic $\gamma$ of the sphere. I showed it for geodesics issuing from $p_0$. Is it enough?

  • 1
    $\begingroup$ Sorry. Geodesics through $p_0$ tell you only how the function behaves as you move away from $p_0$. They do not tell you how it behaves in other directions. $\endgroup$ – Paul Sinclair Apr 3 at 0:53
  • $\begingroup$ Thank you for your comment. Do you believe that $d$ is convex? $\endgroup$ – Eduardo Longa Apr 3 at 1:08
  • $\begingroup$ Look at $\Bbb S^2$ first. Consider a great circle slanted at 45 degrees. Calculate your function as a function of $\theta$ around the circle. Is that function convex everywhere? $\endgroup$ – Paul Sinclair Apr 3 at 1:13

Consider $p,\ q$ in $\mathbb{S}^2$. When $c(t)$ is unit speed geodesic starting at $q$, then $f(t)=|p-c(t)|$ so that Cosine law is $$ \cos\ f=\cos\ t\cos\ l +\sin\ t\sin\ l\cos\ \alpha$$ where $l=|p-q|$ and $ \alpha =\angle\ ({\rm Dir}\ [qp],c'(0) )$ (Here ${\rm Dir}$ is a direction)

Hence $$ -\cos\ f=(\cos\ f)''=(-\sin\ ff')'=-\cos\ f(f')^2-\sin\ ff'' $$

Here $f'=-\cos\ \alpha$ so that $f'' =\cot\ f\sin^2 \alpha $

When $f <\frac{\pi}{2}$, then $f$ is convex.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.