# Does the Gaussian Poincare inequality hold for $p=1$ as well as $p=2$?

Let $$X$$ be a real-valued standard normal variable. Then, for any differentiable function $$f: \mathbb{R} \to \mathbb{R}$$ such that $$E[f(X)^2] < \infty$$ and $$E[\bigl( f'(X) \bigr)^2] < \infty$$, it is well-known that $$$$\text{Var}(f(X)) \leq E[\bigl(f'(X)\bigr)^2]$$$$ and is called the Gaussian Poincare inequality.

I can see that this is the Poincare inequality with $$p=2$$ according to the Wikipedia article https://en.wikipedia.org/wiki/Poincar%C3%A9_inequality

Now, I wonder if the Gaussian Poincare inequality holds for $$p=1$$. That is, do we also have $$$$E[\bigl \lvert f(X) - E[f(X)] \bigr \rvert ] \leq E[\bigl \lvert f'(X) \bigr \rvert ]$$$$ in general?

In the link above, the generalization of the Poincare inequality to general measure spaces is considered as well. I searched for papers myself but was not able to find anything specialized to Gaussian measures.

Could anyone please help me?

## 1 Answer

Yes, Gaussians also satisfy a Poincaré inequality with $$p = 1$$ (such an inequality is equivalent to what is called a "Cheeger inequality"). More generally, E. Milman has shown that for log-concave measures, all $$(p, q)$$-Poincaré inequalities are equivalent:

Milman, E. On the role of convexity in isoperimetry, spectral gap and concentration. Invent. math. 177, 1–43 (2009). https://doi.org/10.1007/s00222-009-0175-9

• Oh, thank you very much. I have a subsequent question where I clarify my further curiosities. I would deeply appreciate if you answer mathoverflow.net/questions/446903/… as well. Commented May 16, 2023 at 14:05
• I intended the underlying probability measure space of $X$ to be infinite dimensional. Sorry that I did not state it. Commented May 16, 2023 at 14:34