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Symmetry and convexity · Question 5

Symmetry and convexity in geometric inequalities · Historical workshop question

Editorial mapping of AIMPL source category: Functional analysis. http://aimpl.org/symconvgeomineq/ Not an author-supplied MSC code.

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If we replace condition in the theorem with $\nabla^2 W(x) \geq F(x) \, Id$, can we obtain better results?

## Question context

\begin{theorem}[Caffarelli]
 	Let $T = \nabla \Phi: \mu \to \nu$ be a Brenier map, where $\mu$  and $\nu$ are probability measures with densities $e^{-V(x)}$ and $e^{-W(x)}$. Suppose that
 $$ 
 	\nabla^2 W(x) \geq K \, Id,
 $$	
for some constant $K$. Then for any $e \in S^{n-1}$
 	$$
 	\sup\limits_{ x \in \mathbb{R}^n } \Phi^2_{ e e } \leq \frac1K \sup\limits_{x \in \mathbb{R}^n}  V_{ee},
 	$$
 	where $\Phi_{e}$ denotes directional derivative.
 \end{theorem}

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Problem statement

If we replace condition in the theorem with ∇2W(x)≥F(x) Id\nabla^2 W(x) \geq F(x) \, Id, can we obtain better results?

Lean 4

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Background

Question context

\begin{theorem}[Caffarelli] Let T=∇Φ:μ→νT = \nabla \Phi: \mu \to \nu be a Brenier map, where μ\mu and ν\nu are probability measures with densities e−V(x)e^{-V(x)} and e−W(x)e^{-W(x)}. Suppose that

∇2W(x)≥K Id, \nabla^2 W(x) \geq K \, Id,

for some constant KK. Then for any e∈Sn−1e \in S^{n-1}

\sup\limits_{ x \in \mathbb{R}^n } \Phi^2_{ e e } \leq \frac1K \sup\limits_{x \in \mathbb{R}^n} V_{ee}, $$ where $\Phi_{e}$ denotes directional derivative. \end{theorem}

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References
  1. [1]Symmetry and convexity in geometric inequalities link

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