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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?

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    $\begingroup$ 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
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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.

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