On the $L_p$-Brunn–Minkowski and Dimensional Brunn–Minkowski Conjectures for Log-Concave Measures

Abstract

We study several of the recent conjectures in regards to the role of symmetry in the inequalities of Brunn–Minkowski type, such as the $L_p$-Brunn–Minkowski conjecture of Böröczky, Lutwak, Yang and Zhang, and the Dimensional Brunn–Minkowski conjecture of Gardner and Zvavitch, in a unified framework. We obtain several new results for these conjectures. We show that when $K\subset L,$ the multiplicative form of the $L_p$-Brunn–Minkowski conjecture holds for Lebesgue measure for $p\ge 1-Cn^{-0.75}$, which improves upon the estimate of Kolesnikov and Milman in the partial case when one body is contained in the other. We also show that the multiplicative version of the $L_p$-Brunn–Minkowski conjecture for the standard Gaussian measure holds in the case of sets containing sufficiently large ball (whose radius depends on p). In particular, the Gaussian Log-Brunn–Minkowski conjecture holds when K and L contain $\sqrt{0.5 (n+1)}B_2^n.$ We formulate an a-priori stronger conjecture for log-concave measures, extending both the $L_p$-Brunn–Minkowski conjecture and the Dimensional one, and verify it in the case when the sets are dilates and the measure is Gaussian. We also show that the Log-Brunn–Minkowski conjecture, if verified, would yield this more general family of inequalities. Our results build up on the methods developed by Kolesnikov and Milman as well as Colesanti, Livshyts, Marsiglietti. We furthermore verify that the local version of these conjectures implies the global version in the setting of general measures, and this step uses methods developed recently by Putterman.

Appendix

Lemma 11.1

Let K be an origin-symmetric convex body and w a continuous function on $S^{n-1}$. Then,

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\frac{\mu (W(h_K + \varepsilon w)) - \mu (K)}{\varepsilon }&= \int _{S^{n-1}} w(\theta ) \mathrm{{d}}\sigma _{\mu , K}(\theta ). \end{aligned}$$

Proof

Our proof follows the proof given in the appendix of [25]. Recall that for $H_{n-1}$-almost every $x \in \partial K$ there exists a unique normal vector $n_x$. Let us denote the subset of $\partial K$ where this occurs by $\widetilde{\partial K}$. Let $X:\widetilde{\partial K} \times [0,\infty ) \rightarrow {\mathbb {R}}^n\setminus K$ be defined by $X(x,t) = x + tn_x$, and let D(x, t) be the Jacobian of this map. Moreover, from properties of Wulff shapes, we have that $h_{A[h_K + \varepsilon w]}(n_x) \le h_K(n_x) + \varepsilon w(n_x)$ with equality for $H_{n-1}$-almost every $x \in \widetilde{\partial K}$. See Schneider [34, Sect. 7.5]. Let $\widetilde{\partial K}' \subset \widetilde{\partial K}$ be the subset where we have equality. Then,

$$\begin{aligned} \frac{1}{\varepsilon }(\mu (A[h_K + \varepsilon w]) - \mu (K))&= \frac{1}{\varepsilon }\int _{\widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}D(x,t)g(x+tn_x)\mathrm{{d}}t \mathrm{{d}}H_{n-1}(x). \end{aligned}$$

Observe that X(x, t) is an expanding map. Indeed, for $x_1, x_2 \in \widetilde{\partial K}$ and $t_1, t_2 \in [0,\infty )$ we have

$$\begin{aligned} \begin{aligned} |X(x_1, t_1) - X(x_2, t_2)|^2&= |x_1 + t_1 n_{x_1} - x_2 - t_2 n_{x_2}|^2 \\&= |x_1 - x_2|^2 + |t_1n_{x_1} - t_2 n_{x_2}|^2 \\&\quad + t_1 \langle x_1 - x_2, n_{x_1}\rangle + t_2 \langle x_2-x_1, n_{x_2} \rangle . \end{aligned} \end{aligned}$$

(52)

Since K is convex, we have $\langle x_1, n_{x_1}\rangle \ge \langle x_2, n_{x_1}\rangle$ and $\langle x_2, n_{x_2}\rangle \ge \langle x_1, n_{x_2}\rangle$. Therefore,

$$\begin{aligned} |X(x_1, t_1) - X(x_2, t_2)|&\ge |x_1-x_2|^2 + |t_1n_{x_1} - t_2 n_{x_2}|^2 \\&\ge |x_1-x_2|^2 + |t_1-t_2|^2 \end{aligned}$$

as desired. It follows that $D(x,t) \ge 1$, and so

$$\begin{aligned} \begin{aligned}&\liminf _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }(\mu (A[h_K + \varepsilon w]) - \mu (K)) \\&\quad \ge \liminf _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }\int _{\widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}g(x+tn_x) \mathrm{{d}}t \mathrm{{d}}H_{n-1}(x) \\&\quad =\int _{\widetilde{\partial K}'}w(n_x)g(x) \mathrm{{d}}H_{n-1}(x). \end{aligned} \end{aligned}$$

(53)

Since $\partial K\setminus \widetilde{\partial K}'$ has $H_{n-1}$-measure zero, we get that

$$\begin{aligned} \begin{aligned} \liminf _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }(\mu (A[h_K + \varepsilon w]) - \mu (K))&\ge \int _{\partial K} w(n_x) g(x) \mathrm{{d}}H_{n-1}(x) \\&= \int _{S^{n-1}}w(\theta ) \mathrm{{d}}\sigma _{\mu , K}(\theta ). \end{aligned} \end{aligned}$$

(54)

We now pursue the reverse inequality. For an arbitrary $\delta > 0$, define

$$\begin{aligned} (\partial K)_{\delta }&= \{x \in \partial K: \exists \ a\in {\mathbb {R}}^n \text { s.t. } x \in B(a,\delta ) \subset K\} \end{aligned}$$

where $B(a,\delta )$ is the Euclidean ball $\{y \in {\mathbb {R}}^n: |y-a|<\delta \}$. For a sufficiently small $\varepsilon >0$, take $0 \le t_1, t_2 \le \varepsilon$ and $x_1, x_2 \in (\partial K)_{\delta }$. From (52), we have

$$\begin{aligned}&|X(x_1, t_1) - X(x_2, t_2)| \le |x_1-x_2|^2 + |t_1 - t_2|^2 + \varepsilon ^2 |n_{x_1} - n_{x_2}|^2\\&\quad + \varepsilon \langle x_1-x_2, n_{x_1} - n_{x_2}\rangle . \end{aligned}$$

Now, it is a result of Hug [14] that the Gauss map is Lipschitz on $(\partial K)_{\delta }$. Let us denote the Lipschitz constant by $L(\delta )$. Then

$$\begin{aligned}&\frac{|x_1-x_2|^2+|t_1-t_2|^2+\varepsilon ^2|n_{x_1} - n_{x_2}|^2 + \varepsilon \langle x_1 - x_2, n_{x_1} - n_{x_2}\rangle }{|x_1-x_2|^2 + |t_1-t_2|^2}\\&\quad \le 1 + L(\delta )\varepsilon + L(\delta )^2 \varepsilon ^2. \end{aligned}$$

Hence,

$$\begin{aligned} D(x,t)&\le (1 + L(\delta )\varepsilon + L(\delta )^2 \varepsilon ^2)^{n-1} \le 1 + C(K,n,\delta ) \varepsilon . \end{aligned}$$

We have therefore

$$\begin{aligned}&\limsup _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }\int _{(\partial K)_{\delta } \cap \widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}D(x,t)g(x+tn_x)\mathrm{{d}}t \mathrm{{d}}H_{n-1}(x) \\&\quad \le \int _{(\partial K)_{\delta } \cap \widetilde{\partial K}'}w(n_x) g(x) \mathrm{{d}}H_{n-1}(x) \\&\quad = \int _{(\partial K)_{\delta }}w(n_x)g(x) \mathrm{{d}}H_{n-1}(x). \end{aligned}$$

Since $D(x,t) \ge 1$, we have as in (53) also that

$$\begin{aligned}&\liminf _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }\int _{(\partial K)_{\delta } \cap \widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}D(x,t)g(x+tn_x)\mathrm{{d}}t \mathrm{{d}}H_{n-1}(x) \\&\quad \ge \int _{(\partial K)_{\delta }}w(n_x) g(x) \mathrm{{d}}H_{n-1}(x). \end{aligned}$$

It follows that the limit in $\varepsilon$ exists and

$$\begin{aligned}&\lim _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }\int _{(\partial K)_{\delta } \cap \widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}D(x,t)g(x+tn_x)\mathrm{{d}}t \mathrm{{d}}H_{n-1}(x)\\&\quad = \int _{(\partial K)_{\delta }}w(n_x) g(x) \mathrm{{d}}H_{n-1}(x). \end{aligned}$$

By the dominated convergence theorem and lower semi-continuity,

$$\begin{aligned}&\limsup _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }\left( \mu (A[h_K + \varepsilon w]) - \mu (K)\right) \\&\quad = \limsup _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }\int _{\widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}D(x,t)g(x+tn_x)\mathrm{{d}}t \mathrm{{d}}H_{n-1}(x) \\&\quad = \limsup _{\varepsilon \rightarrow 0}\lim _{\delta \rightarrow 0}\frac{1}{\varepsilon }\int _{(\partial K)_{\delta } \cap \widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}D(x,t)g(x+tn_x)\mathrm{{d}}t \mathrm{{d}}H_{n-1}(x) \\&\quad = \lim _{\delta \rightarrow 0}\lim _{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }\int _{(\partial K)_{\delta } \cap \widetilde{\partial K}'}\int _{0}^{\varepsilon w(n_x)}D(x,t)g(x+tn_x)\mathrm{{d}}t \mathrm{{d}}H_{n-1}(x) \\&\quad = \lim _{\delta \rightarrow 0}\int _{(\partial K)_{\delta }}w(n_x)g(x) \mathrm{{d}}H_{n-1}(x) \\&\quad = \int _{\widetilde{\partial K}}w(n_x)g(x)\mathrm{{d}}H_{n-1}(x) \\&\quad = \int _{\partial K}w(n_x)g(x)\mathrm{{d}}H_{n-1}(x) \\&\quad = \int _{S^{n-1}}w(\theta ) \mathrm{{d}}\sigma _{\mu , K}(\theta ). \end{aligned}$$

Combining this with (54) gives us the desired conclusion. $\square$