# Convex spherical neighborhood in Alexandrov spaces

Let $$X$$ be an Alexandrov space with curvature $$\ge k$$. For any point $$p \in X$$, can we find a constant $$r>0$$ such that $$B(p,r)$$ is geodesically convex?

Here, geodesically convex means for every two points $$x,y\in B(p,r)$$, there exists a minimizing geodesic between them which lies in $$B(p,r)$$.

• Doubled disk is a counterexample. – Anton Petrunin Jan 6 at 3:44
• (But, in the finite dimensional case, every point has a convex neighborhood.) – Anton Petrunin Jan 6 at 3:48

There is a variational criterion to determine whether geodesic balls are convex in finite dimensional open Alexandrov space $$(X,d)$$. Otherwise, if $$X$$ is compact, then my answer says nothing.
If $$X$$ is connected, let $$M^*=M^*(x_0)$$ be the set of noncompact maximal minimizing geodesic rays. For $$\lambda \in M^*$$, let $$h_\lambda(x)$$ be unique horofunction centred at $$\lambda$$ and satisfying $$h_\lambda(x_0)=0$$. Then $$b(x,y):=\sup_{\lambda \in M^*} |h_\lambda(x)-h_\lambda(y)|$$ defines a possibly degenerate metric on $$b:X\times X\to [0,+\infty)$$ satisfying $$b(x,y)\leq d(x,y)$$ for every $$x,y \in X$$. From the $$d$$-geodesic convexity of $$h_\lambda$$, we find the $$b$$-metric balls of arbitrary radius are $$d$$-geodesically convex subsets of $$X$$. The $$d$$ geodesic convexity of the horofunction $$h_\lambda$$ is well-known property of Alexandrov metrics $$d$$.
If $$(X,d)$$ has the property that equality is obtained everywhere $$b(x,y)=d(x,y)$$ we conclude all $$d$$-metric balls in $$X$$ are $$d$$-geodesically convex.
So in your case, we need further information on the set of infinite minimizing rays $$M^*=M^*(x_0)$$ at a basepoint $$x_0$$. In plain terms, for every pair of points $$x,y$$, you need determine whether there exists a horosphere $$H_\lambda$$ which separates $$x,y$$. In such case $$b$$ is nondegenerate distance function with $$b(x,y)=0$$ iff $$x=y$$.
If equality is attained everywhere and $$b=d$$, then $$b$$ is Alexandrov. But if the inequality is strict $$b , then I don't know whether the metric is again Alexandrov, and this appears interesting question.