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Let $S$ be a simply connected surface, possibly with boundary components, with a smooth complete metric of non-positive curvature. Let $X\subset S$ be a closed connected subset. I would like to know if the following procedure produces a convex subset of $S$ that contains $X$:

Construction. For any two points $x_1,x_2\in X$ let $[x_1,x_2]$ be the length minimizing path connecting $x_1$ and $x_2$. Now define $\Omega(X)$ as the union in $S$ of all such paths $[x_1,x_2]$ over all $x_1,x_2\in X$.

Question. Is $\Omega(X)$ convex in $S$?

PS. I have to add some information. A set $C$ is convex if for any two $x_1,x_2\in C$ and any minimizing path $[x_1x_2]$ $C$ contains it.

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    $\begingroup$ What is definition of convex set $C$ ? For any $x_i \in C$, if $[x_1x_2]$ is any minimizing path, then $C$ contains it ? $\endgroup$
    – Hee Kwon Lee
    Commented Apr 26, 2017 at 11:04
  • $\begingroup$ Thank you for your comment. You are right about convexity. I also added that $S$ is simply connected (it seems to me that otherwise the are counterexamples). $\endgroup$
    – aglearner
    Commented Apr 26, 2017 at 12:13
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    $\begingroup$ If $S$ is simply connected, then there is only one homotopy class of paths between two points, so the formulation can be simplified. $\endgroup$
    – Ivan Izmestiev
    Commented Apr 26, 2017 at 13:55
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    $\begingroup$ I would start by proving this for finite subsets of $S$: Prove that the construction yields a locally convex connected subset with polygonal boundary which, therefore, is convex. Then take the direct limit of convex hulls $H(F)$ for finite subsets of $S$. $\endgroup$
    – Misha
    Commented Apr 26, 2017 at 16:38

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Yes, it is true.

First note that $S$ is CAT[0] space. This can be proved along the same lines as in "The intrinsic geometry of a Jordan domain" by Richard Bishop (you may also check the proof of Theorem 4.4.1 in our paper).

Further note that any geodesic in $S$ can be extended to a maximal geodesic (infinite or with the ends on the boundary). This is a standard lemma — geodesics in a CAT[0] space might terminate only at the points which admit contactable punctured neighbourhood.

Assume $\Omega(X)$ is not convex, choose a point $w\notin \Omega(X)$ which lies between $u,v\in \Omega(X)$. Assume $u\in[xy]$ and $v\in [zt]$ for $x,y,z,t\in X$

Let $\gamma$ be a maximal extension of the geodesic $[uv]$. Since $X$ is connected, it has to intersect $\gamma$.

Since $w\notin \Omega(X)$, the set $X$ might intersect $\gamma$ only on one side from $w$. WLOG we can assume that it happens on the side of $u$.

Consider the maximal extension $\sigma$ of $[zw]$. Let $s$ be a point in $\sigma\cap X$ which lies in the closure of connected component of $t$ in the complement $X\backslash \sigma$. It remains to note that $w\in [zs]$

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