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The dimensional Brunn–Minkowski inequality in Gauss space
Journal of Functional Analysis ( IF 1.7 ) Pub Date : 2021-03-01 , DOI: 10.1016/j.jfa.2020.108914
Alexandros Eskenazis , Georgios Moschidis

Let $\gamma_n$ be the standard Gaussian measure on $\mathbb{R}^n$. We prove that for every symmetric convex sets $K,L$ in $\mathbb{R}^n$ and every $\lambda\in(0,1)$, $$\gamma_n(\lambda K+(1-\lambda)L)^{\frac{1}{n}} \geq \lambda \gamma_n(K)^{\frac{1}{n}}+(1-\lambda)\gamma_n(L)^{\frac{1}{n}},$$ thus settling a problem raised by Gardner and Zvavitch (2010). This is the Gaussian analogue of the classical Brunn-Minkowski inequality for the Lebesgue measure. We also show that, for a fixed $\lambda\in(0,1)$, equality is attained if and only if $K=L$.

中文翻译:

高斯空间中的维数 Brunn-Minkowski 不等式

令 $\gamma_n$ 是 $\mathbb{R}^n$ 上的标准高斯测度。我们证明对于 $\mathbb{R}^n$ 中的每个对称凸集 $K,L$ 和每个 $\lambda\in(0,1)$, $$\gamma_n(\lambda K+(1-\lambda )L)^{\frac{1}{n}} \geq \lambda \gamma_n(K)^{\frac{1}{n}}+(1-\lambda)\gamma_n(L)^{\frac {1}{n}},$$ 从而解决了 Gardner 和 Zvavitch (2010) 提出的问题。这是 Lebesgue 测度的经典 Brunn-Minkowski 不等式的高斯类比。我们还表明,对于固定的 $\lambda\in(0,1)$,当且仅当 $K=L$ 时才达到相等。
更新日期:2021-03-01
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