Let $d\mu$ be a Gaussian measure on $\mathbb{R}$ with the center $a \in \mathbb{R}$ and variance $1$.

Let $B(a,r) \subset \mathbb{R}$ be the interval $[a-r,a+r]$.

Then, for any smooth mapping $f : \mathbb{R} \to \mathbb{R}$, is it true that \begin{equation} \frac{1}{\mu(B(a,r))}\int_{B(a,r)} f^2(x) d\mu(x) - \Bigl( \frac{1}{\mu(B(a,r))}\int_{B(a,r)} f(x) d\mu(x) \Bigr)^2 \leq \ \int_{B(a,r)} [f'(x)] ^2 d\mu(x)? \end{equation} for any $r>0$?

Due to smoothness of $f$, we observe that both sides of the above inequality tends to zero as $r \to 0^+$.

I guess that due to concentration of the Gaussian measure around the center, this inequality must hold for any $r>0$, but I cannot really find a relelvant reference to help justify my guess.

Could anyone please help me?


1 Answer 1


In "On how Poincaré inequalities imply weighted ones" which is interesting in itself in figuring out when a measure satisfies a Poincare-inequality, I came across the reference "An optimal Poincaré-Wirtinger inequality in Gauss space", they prove a Poincare inequality for the Gaussian measure over general Convex domains with a C2+accessibility boundary condition. Let

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  • $\begingroup$ Thank you. I read through your second link but now quite confused. The paper first assumes $N$ to be greater than $1$ but deals with one dimension at the last page. $\endgroup$
    – Isaac
    Commented Jun 22, 2023 at 3:02
  • $\begingroup$ Also, the "first Diricblet eigenvalue of the one dimensional Hermite operator" on the interval $(-a,a)$ coincides with the Gaussian measure of the interval?? I would like to divide the LHS of the Poincare inequality by the measure of the interval. $\endgroup$
    – Isaac
    Commented Jun 22, 2023 at 3:06
  • $\begingroup$ in the [3] reference i.e. (1.2) is proved for d=1 too and they get a comparison constant in terms of $1/diam(\Omega)^{2}$ that you ask. $\endgroup$ Commented Jun 22, 2023 at 4:20

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