Toponogov's theorem in the Alexandrov space with respect to a compact set

Question

Let $X$ be an Alexandrov space with curvature $\ge 0$ and $A \subset X$ a compact set. Suppose $p,q\in X$ satisfying $|Ap|=|Aq|=r$ and $p_1,q_1$ are points on the geodesics $Ap,Aq$ respectively such that $|Ap_1|=|Aq_1|=s<r$. Can we prove $$ |p_1q_1| \ge \frac{s}{r}|pq|? $$

Answer 1

Yes it is true. It can be proved by applying distance estimates to the gradient flow of the function $f=\mathrm{dist}_A^2/2$. See our draft.

Answer 2

Yes. It is also true (with different distortion constant) for Alexandrov space with curvature $\ge k$.

We define $$ \rho_k(x)= \begin{cases} \frac{1}{k}(1-\cos(x\sqrt{k}))\quad &\text{if}\quad k>0,\\ \frac{x^2}{2},\quad &\text{if}\quad k=0,\\ \frac{1}{k}(1-\cosh(x\sqrt{-k}))\quad &\text{if}\quad k<0.\\ \end{cases} $$

Then it follows from the Toponogov's theorem that for any $p \in X$, $$ \rho_k(\text{dist}_p)''\le 1-k\rho_k(\text{dist}_p). $$

By taking the minimum for all $p \in A$, we also have $$ f''\le 1-kf, $$ where $f:=\rho_k(\text{dist}_A)$.

Let $\gamma_1(t)$ and $\gamma_2(t)$ be the geodesics of $Ap$ and $Aq$ respectively. It is not hard to see that $\tilde \gamma_1(z):=\gamma_1(t(z))$ and $\tilde \gamma_2(z):=\gamma_2(t(z))$ are the gradient flows of $f$ starting from $p_1$ and $q_1$ respectively. Here the new parameter $z$ is determined by $$ dz=\frac{dt}{\rho_k'(t)} \quad \text{and} \quad t(0)=s. $$ If we set $l(t):=|\gamma_1(t)\gamma_2(t)|$, then it follows from the distance estimate of the gradient flow (see Lemma 1.3.3 in this paper) that $$ (\log l(t))' \le \frac{1-k\rho_k(t)}{\rho_k'(t)}. $$

From integration, we obtain $$ |pq|=l(r) \le \sigma_k(r,s) l(s)=\sigma_k(r,s)|p_1q_1|, $$ where $$ \sigma_k(r,s)= \begin{cases} \frac{\sin(r\sqrt k)}{\sin(s\sqrt k)}\quad &\text{if}\quad k>0,\\ \frac{r}{s},\quad &\text{if}\quad k=0,\\ \frac{\sinh(r\sqrt{-k})}{\sinh(s\sqrt{-k})}\quad &\text{if}\quad k<0.\\ \end{cases} $$