Let $f \in L^{2}(\mathbb{R}^{d}, \gamma)$ be a square integrable function w.r.t to the Gaussian measure $\gamma$ such that $\int_{\mathbb{R}^{d}}{f(x) \,dx } = 0$.

I've bumped into a research paper in which author asserts that the inequality $$\int_{\mathbb{R}^{d}}{f^{2}\, d \gamma} \leq \frac{1}{2} \int_{\mathbb{R}^{d}}{\| \nabla f \|^{2} \,d \gamma}$$ holds for any such function $f$. On the first glance, it looks as an improved version of the classical Poincare inequality which states that $$\int_{\mathbb{R}^{d}}{f^{2}\, d \gamma } \leq \int_{\mathbb{R}^{d}}{\|\nabla f\|^{2}\, d \gamma}$$ for any $L^{2}$ function that is orthogonal to the constant function (equivalently, with mean $0$).

The questions are:

(1) Is it possible to prove these inequalities via the Fourier series techniques? (seems as if it is reasonable to consider the Fourier expansion of $f$ w.r.t to the orthogonal system consisting of Hermite polynomials and the try to obtain the inequality somehow, probably, with the use of the Parserval's identity). The same approach is known to be working quite well for the $1$-dimensional Wirtinger inequality.

(2) The general Poincare inequality states that for a given region $\Omega$ and $1 \leq p < \infty$ there exists a constant $C := C_{\Omega, p}$ such that for any $u \in W_{0}^{1, p}$ $$\|u\|_{L^{p}(\Omega)} \leq C \|\nabla u\|_{L^{p}(\Omega)}$$

So, under some mild assumptions (e.g. $\mathbb{E}(f) = 0$ ) is it possible to find the best lower bound for $C$? (as suggested, for $C=\frac{1}{2}$ the very first inequality holds)

  • $\begingroup$ In the first sentence, is the hypothesis $\int f(x)\,dx = 0$, or $\int f(x)\,\gamma(dx) = 0$? Is it mean zero with respect to Lebesgue or Gaussian measure? $\endgroup$ – Nate Eldredge Feb 10 '18 at 16:45

The first inequality in your post is false. Just consider the case $d=1$ and the function $f(x) = x^2+x-1 $, for which the LHS is 3 and the RHS is $5/2$. The claimed inequality is true (for every $d$) if $f$ is not only orthogonal to constants but also to any affine mapping on $\mathbb{R}^d$. A proof of this fact can be e.g. deduced by Theorem 6.2 in this recent paper (but other parts of the literature for sure imply this fact): https://arxiv.org/pdf/1411.6265.pdf

The proof in the above reference rests exactly on the 'spectral approach' hinted by your post.

By the way, the approach based on Hermite polynomials implies that one can further improve the constant in the Gaussian Poincaré inequality if one assumes that $f$ is orthogonal to polynomials of degree $\leq 2$, $\leq 3$, ... and so on.


In "The (B) conjecture for the Gaussian measure of dilates of symmetric convex sets and related problems" the authors prove exactly this fact using the $L^2$ (non spectral) method, except the assumption should be also that the gradient is mean zero, not just $f$ is mean zero. Without the assumption it is obviously false.

  • $\begingroup$ Welcome to MSE. This should be a comment, not a answer. $\endgroup$ – José Carlos Santos Feb 10 '18 at 15:39

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