# Convergence of semi-concave functions on Alexandrov spaces

I will use the terminology of this paper by A. Petrunin: https://arxiv.org/pdf/1304.0292.pdf

Let a sequence of $n$-dimensional Alexandrov spaces $\{X_i\}$ of curvature at least $-1$ converges to an Alexandrov space $X$ in the Gromov-Hausdorff sense. Let $f_i\colon X_i\to \mathbb{R}$ be 1-Lipschitz $\lambda$-concave functions (see Def 1.1.1 and 1.1.2 on p.5 of the above paper) converging to a function $f\colon X\to \mathbb{R}$.

Is the function $f$ also $\lambda$-concave?

I think the question has the positive answer if $X$ has no boundary. My question is about the case when $X$ has a non-empty boundary.

• Perhaps a bit late to advise you on this, but the definition of semi-concavity for an Alexandrov space with boundary is that it would be semi-concave on the double, so the boundary makes no difference. Feb 1, 2018 at 12:28
• @JohnHarvey: It would be helpful to elaborate. First assume for simplicity that $\{X_i\}$ are smooth closed manifolds converging to $X$ which is a smooth manifold with boundary. How one could prove that the function $f\colon X\to \mathbb{R}$ is $\lambda$-concave? My problem is that I do not know how to relate the doubling of $X$ to the sequence $\{X_i\}$.
– makt
Feb 1, 2018 at 16:36

The answer is "yes".

Consider the gradient flow $$\Phi_n^t\colon X_i \to X_i$$ for $$f_n$$. Note that $$\Phi_n^t$$ converges to the gradient flow $$\Phi^t\colon X \to X$$ for $$f$$. Since $$\Phi^t_n$$ is onto for any $$t$$, so is $$\Phi^t$$. The latter implies the need condition at the boundary.

(Sorry --- I did not notice your question before.)