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. 



Convex hull of a connected subset on a complete surface of non-positive curvature
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"](https://arxiv.org/abs/math/0512622) by Richard Bishop (you may also check the proof of Theorem 4.4.1 in [our paper](https://arxiv.org/abs/1701.03483)).


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 $w\in [zs]$


