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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 $C'$.

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    $\begingroup$ Which paper? (In case you are missing something there) $\endgroup$ – 40 votes Jul 16 '13 at 18:50
  • $\begingroup$ I am curious about the paper as well. $\endgroup$ – Shuhao Cao Jul 17 '13 at 3:04
  • $\begingroup$ The paper in question is "Rate of Convergence for Ergodic Continuous Markov processes: Lyapunov vs. Poincare", by Bakry et al. They only use this result in passing (Section 4.1) without the details. However, I have seen a similar result in several other paper, always stated without justification, which led me to believe it's obvious given the right assumptions. $\endgroup$ – Il-Bhima Jul 17 '13 at 9:02
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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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  • $\begingroup$ Hi, thanks for this answer. I managed to find a proof of the result for when $w$ is a smooth density. The approach is quite similar to that of Shuhao Cao below, though mine requires more assumptions. I also managed to obtain a bound for the weighted Poincare' constant $C'$. The counterexample you provided is extremely interesting as well as the Heinonen reference, I was not aware of this, nor of the relation to the Muckenhoupt condition. I came to this problem from a stochastic point of view, for me the constant $C'$ is a lower bound for the $L^2$-spectral gap of a diffusion process. $\endgroup$ – Il-Bhima Jul 17 '13 at 8:58
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Like 40 votes said, there is an industry of this. I happened to learn some of the results of the weighted Poincaré type inequality for some of my previous researches. There is a famous book by A. Kufner: Weighted Sobolev spaces. A more recent treatise on this is Nonlinear Potential Theory and Weighted Sobolev Spaces, please refer to a dedicated section 4.3 for Poincaré type inequality in weighted Sobolev space.

I myself learned the proof from this paper: An explicit right inverse of the divergence operator which is continuous in weighted norms. The section 3 of this paper proved the weighted Friedrichs' inequality for the weight $w = (|x-x_0|^2 + \theta^2)^{1/2}$.

Following is roughly the adaptation of the proof for any positive smooth weight $w>0$, $\int_{\Omega} w = 1$ (the weight can vanish on the boundary), under the same assumption "$\Omega$ is star-shaped with respect to a ball $B$".

Also what you wrote in your question is actually okay with a slight modification, which is the weighted Friedrichs' inequality (notice the following is much stronger version of 40 votes wrote):

For smooth function $f$ such that $\displaystyle \int_{\Omega} f w = 0$ with smooth weight $w>0$ in $\Omega$, and $\displaystyle \int_{\Omega} w = 1$: $$ \int_{\Omega}f^2 w \leq \int_{\Omega}f^2 \leq C\int_{\Omega} |\nabla f|^2w + (\text{Optional } L^2\text{-term}).\tag{$\star$} $$

A sketch of the proof: For simplicity we just assume $\Omega$ is a ball $B$ with radius 2 centered at the origin, and we can find some ball $\hat{B}\subset \Omega$ such that inside $B$, with positive radius $\hat{r} \geq c r_B$ for some positive $c$. For $x\in \hat{B}$, and $y\in B$: $$|y-x|\leq C w. \tag{1}$$ Why? Think a mollifier centered at origin, and this is a sufficient condition for the weight to satisfy for there to be a Poincaré type inequality. Let $\phi$ be a smooth weight vanishing at boundary of $\hat{B}$, and $\bar{f} = \int_{B} f\phi$ the weighted mean of $f$ vanishing outside this smaller ball then: for any $y\in B$ $$ f(y) - \bar{f} = \int_{B} \big(f(y) - f(z)\big)\phi(z)dx \\ = \int_{B} \int^1_0 (y-z)\cdot \nabla f(y + t(z-y)) \phi(z) dt \,dz, $$ which follows from the multi-dimensional Taylor formula. Now the trick is to rescale the variables: let $x = y + t(z-y)$ $$ f(y) - \bar{f} = \int_{B} \int^1_0 \frac{y-x}{t}\cdot \nabla v(x) \phi\left(y + \frac{x-y}{t}\right) \frac{1}{t^2} dx\,ds. $$ Now $\phi(y + \frac{x-y}{t})$ vanishes when $0< t < \gamma|x-y|$ for $\gamma$ relying on $\hat{r}$, hence for the kernel $$ K := \int^1_0 \frac{y-x}{t^3} \phi\left(y + \frac{x-y}{t}\right) dt, $$ we have $$ |K| \leq \sup|\phi| \int^1_{\gamma|x-y|} \frac{|y-x|}{t^3} dt \leq C |y-x|^{-1}. $$ Now use condition (1), $w^{1/2} \leq C |x-y|^{-1/2}$ inside the $\hat{B}$, and $K$ vanishes outside $\hat{B}$ we have $$ |f(y) - \bar{f}| \leq \int_B |K| |\nabla f| \\ = \int_B |K|w^{-1/2} w^{1/2}|\nabla f| \\ \leq C\int_B |y-x|^{-3/2} w^{1/2}|\nabla f|. $$ Now Young's inequality for convolution reads: $$ \|f - \bar{f}\|_{L^2(B)} \leq \Big\| |x|^{-3/2}\Big\|_{L^1(B)} \Big\|w^{1/2}|\nabla f|\Big\|_{L^2(B)}.\tag{2} $$ For $|x|^{-3/2}$ is $L^1$ when $B\subset \mathbb{R}^2$. Now if $\int_B fw = 0$, we have $$ \bar{f}_{B} := \frac{1}{|B|}\int_B f \phi = \frac{1}{|B|}\int_B f(\phi-w) \leq \frac{1}{|B|} \|f\|_{L^2(B)} \left(\int_B (\phi-w)^2\right)^{1/2}, $$ and the optional $L^2$-term in $(\star)$ is determined by how we can manipulate the weight so that $\|\bar{f}_{B} \|_{L^2(B)}$ can be absorbed by $ \|f\|_{L^2(B)}$, then we have: $$ \|f\|_{L^2(B)}^2\leq \|f-\bar{f}_{B}\|_{L^2(B)}^2 + \|\bar{f}_{B} \|^2_{L^2(B)}. $$ By (2) we have the result, for domain with radius not $O(1)$, a scaling argument will be in place.


I took these from some old notes I wrote to myself, there are maybe many loose ends. The $C$ in $(\star)$ depends on how this smooth weight $w$ "concentrates" itself with this domain, if it is very concentrated around some point, then $C$ in $(\star)$ has to be really large to control $\|f\|_{L^2(B)}$.

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  • $\begingroup$ I was just about to answer my own question with a proof of this result for when $w$ is smooth, but the sketch proof you provided is quite similar, using the Taylor to express the $L^2$ norm of $f$ in terms of its derivative. I need some time to process your proof, however, I noticed you managed to prove this without requiring a positive lower bound for $w$. In my approach this is an essential assumption in the final step of the proof to obtain the correct weighted Dirichlet form on the RHS of the inequality. Thanks again! I'll comment further after I read it carefully. $\endgroup$ – Il-Bhima Jul 17 '13 at 8:47

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