[Math] Weighted Poincare Inequality

functional-analysisfunctional-inequalitiespartial differential equations

I'm trying to prove a result I found in a paper, and I think I'm being a bit silly.

The paper claims the following:
By the Poincare inequality on the unit square $\Omega \subset \mathbb{R}^2$ we have that
$$\int_{\Omega} f(x)^2 dx \leq C \int_{\Omega}|\nabla f|^2 dx + \left(\int_{\Omega}f(x)dx\right)^2.$$
Thus, if $w(x)$ is a weight satisfying $\int_\Omega w(x)\,dx =1$ and $0 < W_{-} \leq w(x)$ one can prove the Poincare inequality with respect to the measure $w(x)dx$.

I'm assuming that the author means the following result
$$\int_{\Omega} f(x)^2 w(x)dx \leq C' \int_{\Omega}|\nabla f|^2 w(x)dx + \left(\int_{\Omega}f(x)w(x)dx\right)^2.$$

Does anybody have any idea how to prove this? I'm particularly interested in the relationship between the Poincare constant $C$ and $C'$.

Best Answer

The natural form of Poincaré inequality is $$\int_\Omega |f-f_\Omega|^2 \le C\int_\Omega |\nabla f|^2\tag1$$ where $f_\Omega=\int_\Omega f$ is the mean of $f$. This is exactly your first inequality, but I think (1) captures the meaning better. The weighted Poincaré inequality would be $$\int_\Omega |f-f_{\Omega, w}|^2w \le C'\int_\Omega |\nabla f|^2w\tag2$$ where $f_{\Omega,w}=\int_\Omega fw$ is the weighted mean of $f$. Again, this is what you have but written in a more natural way.

The industry of weighted Poincaré inequalities is huge, but the most fundamental result is that the Muckenhoupt condition $w\in A_2$ is sufficient for (2). This is proved in detail, e.g., in Chapter 15 of Nonlinear Potential Theory of Degenerate Elliptic Equations by Heinonen, Kilpeläinen, and Martio (now published by Dover, $10). The constant $C'$ of course depends on the $A_2$ norm of $w$, but not in any explicit way. There has been some recent interest in estimates of the form $C'\le c\|w\|_{A_2}$ with weight-independent constant $c$, but I do not know if this particular one has been proved.

Anyway, the assumptions you stated are not enough for (2) to hold. Let $f(x)=\min(M, \log\log (e+|x|^{-1}))$ where $M$ is a large number to be chosen later. Spread one half of available weight uniformly on the square, and put the other half onto the set where $f=M$. Since on most of the square $f$ is much smaller than $M$, the weighted mean of $f$ is about $M/2$. Hence, the left hand side of (2) is of order $M$. But the right hand side of (2) is bounded independently of $M$.

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