The following problem arose for my collaborators and me when studying the computational complexity of the Maximum-Cut problem.


Let $f : \mathbb{R} \to \mathbb{R}$ be an odd function. Let $\rho \in [0,1]$. Let $X$ and $Y$ be standard Gaussians with covariance $\rho$. Prove that $\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[f(X)^2 \mathrm{sgn}(X) \mathrm{sgn}(Y)]$.


The quantity on the left-hand side arises naturally in many contexts; e.g., it is the integral of $f(x)f(y)$ against the Mehler kernel (with parameter $\rho$). 


I have some reason to believe this inequality is true. For one piece of evidence, suppose $f$'s range is $\pm 1$. Then the inequality reduces to 


$\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[\mathrm{sgn}(X) \mathrm{sgn}(Y)]$;


i.e., it's saying that $\mathrm{sgn}$ is the $\pm 1$-valued odd function maximizing the LHS. This is indeed true; it follows from a result of Christer Borell ("Geometric bounds on the Ornstein-Uhlenbeck velocity process"), proved by Ehrhard symmetrization. It was also given a different proof by Beckner, deducing it from a rearrangment inequality on the sphere. 


The second inequality generalizes to the case of functions $f : \mathbb{R}^n \to$ {$-1,1$}, but I believe the first inequality, which I would like to prove, is inherently $1$-dimensional.


Any ideas, or pointers to literature that might help?
Thanks!



Inequality in Gaussian space -- possibly provable by rearrangement?
Using the antisymmetry of $f$ and $\mathrm{sgn}$ to bring the expectations to integral expressions
over $[0, \infty) \times [0, \infty)$, the first expectation takes the form:


$const\times \int f(x) f(y) \exp\left(-\frac{x^2+y^2}{2(1-\rho^2)}\right)\sinh\left(\frac{2\rho xy}{2(1-\rho^2)}\right) dx dy$


while for the second case $f(y)$ is replaced by $f(x)$. 


The proof becomes an application of the Cauchy–Schwarz inequality.


