The article under review gives an estimate for the upper bound of volume of a closed positively curved Alexandrov space with strictly convex boundary.
In particular the following result is shown:
Theorem 0.2. Let \(X \in \text{Alex}^n(0)\) be an Alexandrov space with 1-convex boundary \(\partial X\). For \(0 < r \leq 1\) we have \[ \mu_n(B(\partial X, r)) \leq \mu_{n-1}(\partial X) \int_0^r (1-t)^{n-1}\, dt, \] with equality if and only if \(B(\partial X,r)\) is isometric to the warped product space \(\partial X \times_{1-t} [0,r]\). Let \(Y \in \text{Alex}^n(1)\) be an Alexandrov space. For \(0 < r \leq \pi / 2\), we have \[ \mu_n(B(\partial Y, r)) \leq \mu_{n-1}(\partial Y) \int_0^r \cos^{n-1}(t)\, dt, \] with equality if and only if \(B(\partial Y, r)\) is isometric to the warped product space \(\partial Y \times_{\cos(t)} [0,r]\).
Let \(\rho_X = \text{d}_{\partial X}(\cdot) = \text{d}(\partial X, \cdot)\). The convexity of the level set of \(\rho_X\) is estimated as follows:
Theorem 0.7 (Hessian comparison). Let \(X\) and \(Y\) be as in Theorem 0.2. Let \(G_X(t) = \rho_X^{-1}(t)\) be the level set of \(\rho_X\) and \(\Omega_X' := \rho_X^{-1}([t,a])\) be the super-level set of \(\rho_X\) in \(X\). The sets \(G_Y(t)\) and \(\Omega_Y'\) are similarly defined for \(Y\). Then we have the base-angle lower bounds

1.

\(\text{BA}(\Omega_X', G_X(t)) \geq \frac{1}{1-t}\) in \(X\),

2.

\(\text{BA}(\Omega_Y', G_Y(t)) \geq \tan(t)\) in \(Y\),

where we view \(\Omega_X'\) and \(\Omega_Y'\) as stand-alone Alexandrov spaces with boundary.