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If $X$ is a connected convex set in $S^{n-1}(1)$, then what is $\partial X$ ?

Recall the following definition.
Definition : $X$ is a convex subset of $S^{n-1}(1)$ if any two points in $X$ can be joined by distance minimizing geodesic which lies in $X$

Except finite points, is it a union of $C^1$-paths ? And $\partial X$ has a finite variation ? : Let $x_i^n \in \partial X$ with $i=1, \cdots, n$ such that $|x_i^n-x_{i+1}^n|=|x_1^n-x_n^n|$ for all $1\leq i \leq n-1$ where $|\cdot |$ is a distance function on $S^{n-1}(1)$ Then $lim_{n\rightarrow \infty} n|x_1-x_2| < \infty$.

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By the condition, $X$ is a singleton (or empty). If $a\ne b$ with $||a||=||b||=1$ then because $a+b\perp b-a$ we have $$\begin{align}f(s):=\Vert sa+(1-s)b\Vert^2&=\left\Vert\frac{a+b}2+\left(s-\frac12\right)(b-a) \right\Vert^2\\ &=\left\langle \frac{a+b}2+\left(s-\frac12\right)(b-a),\frac{a+b}2+\left(s-\frac12\right)(b-a) \right\rangle\\ &=\left\Vert \frac{a+b}2\right\Vert^2 +\left(s-\frac 12\right)^2\left\Vert b-a\right\Vert^2.\end{align} $$ This is a quadratic function in $s$ with $f(0)=f(1)=1$ and hence $f(s)<1$ for $0<s<1$. Thus $sa+(1-s)b\notin S^{n-1}=\{x\mid \Vert x\Vert=1\}$.

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You are right. So I editted my question. –  Hee Kwon Lee Nov 22 '12 at 14:25
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