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.

  • 2
    $\begingroup$ 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. $\endgroup$ Feb 1, 2018 at 12:28
  • $\begingroup$ @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 $\{X_i\}$. $\endgroup$
    – makt
    Feb 1, 2018 at 16:36

1 Answer 1


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.)


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