Blaschke–Santaló inequality

Remarks

The product $\operatorname{vol}(K) \operatorname{vol}(K^\circ)$ is the Mahler volume of $K$; it does not change when $K$ is replaced by $AK$ for an invertible linear map $A$, because $(AK)^\circ = (A^T)^{-1} K^\circ$. A standard proof uses Steiner symmetrization: symmetrizing $K$ in a hyperplane does not decrease the Mahler volume, and suitable sequences of symmetrizations converge to a ball. The inequality holds without symmetry if the origin is replaced by the Santaló point of $K$, the unique interior point $z$ that minimizes $\operatorname{vol}((K - z)^\circ)$.

Example: $K = [-1, 1]^2$ has $K^\circ = \set{(y_1, y_2) : \abs{y_1} + \abs{y_2} \le 1}$, of area $2$. So the Mahler volume is $4 \cdot 2 = 8$, which is less than $\pi^2 \approx 9.87$.

Some centering is needed: for $K = [-1, 1]^2 - (1 - \varepsilon, 0)$, the origin is at distance $\varepsilon$ from the boundary, and $\operatorname{vol}(K^\circ) \to \infty$ as $\varepsilon \to 0$.

Going deeper

For a student

The inequality is a basic tool of asymptotic convex geometry and has functional forms for log-concave functions. The reverse question, how small $\operatorname{vol}(K) \operatorname{vol}(K^\circ)$ can be, is the Mahler conjecture. For symmetric $K$, the Bourgain–Milman inequality gives a lower bound $c^n \kappa_n^2$ with an absolute constant $c > 0$.