# Gaussian Poincare inequality in $1$ dimensions together with localization issue

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 $$$$\frac{1}{\mu(B(a,r))}\int_{B(a,r)} f^2(x) d\mu(x) - \Bigl( \frac{1}{\mu(B(a,r))}\int_{B(a,r)} f(x) d\mu(x) \Bigr)^2 \leq \ \int_{B(a,r)} [f'(x)] ^2 d\mu(x)?$$$$ for any $$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.

• 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. Commented Jun 22, 2023 at 3:02
• 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. Commented Jun 22, 2023 at 3:06
• 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. Commented Jun 22, 2023 at 4:20