7
$\begingroup$

Suppose $f: \mathbb{R}^d \to \mathbb{R}$ is a smooth convex function.

Consider the level sets of the function, namely $M_s = \{x: f(x) = s\}$. Is it true/known that the surface areas of $M_s$ are log-concave as a function of $s$?

(This feels awfully like a Brunn-Minkowski style inequality, but I'm unsure if it follows from known results. If so, references are highly appreciated!)

$\endgroup$
2
  • $\begingroup$ It is true for smooth convex radial functions. $\endgroup$ Oct 30, 2018 at 1:48
  • $\begingroup$ Indeed, that's where my intuition comes from that this may be true. $\endgroup$
    – Andy Mack
    Oct 30, 2018 at 8:08

1 Answer 1

9
$\begingroup$

Yes, this is true, and you are right, this follows from a generalization of the Brunn-Minkowski inequality.

Let $K_s = \{x \mid f(x) \le s\}$, so that $M_s = \partial K_s$. We have $K_s \supseteq (1-s)K_0 + sK_1$, thus the surface area of the former is $\ge$ the surface area of the latter.

The surface area of a convex body can be written as a mixed volume: $$ \mathrm{vol}_{n-1}(\partial K) = n \mathrm{vol}(\underbrace{K, \ldots, K}_{n-1}, B). $$ A general version of the Brunn-Minkowski inequality says that the function $$ \mathrm{vol}(\underbrace{(1-t)K_0 + tK_1, \ldots, (1-t)K_0 + tK_1}_{n-i}, L_1, \ldots, L_i)^{\frac1{n-i}} $$ is concave for any convex bodies $L_1, \ldots, L_i$.

It follows that the $(n-1)$-st root of the surface area of $K_t$ is concave. Concavity of $f^{\frac{1}{n-1}}$ implies concavity of $\log f$, and we are done.

References:

Gardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008, Section 17.

Burago, Yu. D.; Zalgaller, V. A., Geometric inequalities. Transl. from the Russian by A. B. Sossinsky, Grundlehren der Mathematischen Wissenschaften, 285. Berlin etc.: Springer-Verlag. XIV, 331 p.; DM 184,- (1988). ZBL0633.53002, p. 146.

Schneider, Rolf, Convex bodies: the Brunn-Minkowski theory, Encyclopedia of Mathematics and Its Applications. 44. Cambridge: Cambridge University Press. xiii, 490 p. (1993). ZBL0798.52001, Theorem 6.4.3.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.