# Version of the Poincaré Inequality

Let $\Omega\subset \mathbb{R}^n$ open and bounded. The Poincaré inequality $$\|u\|_p \le C \|\nabla u\|_p$$ ($\|\cdot\|_p$ denotes the usual $L^p(\Omega)$-norm; the Lebesgue measure shall be used here) is valid in the following cases:

• $u$ is zero on $\partial\Omega$ (in the $W^{1,p}$-sense)
• $u$ has an average value of zero (this implies an estimate of $\|u-\bar u\|_p$, but not of $\|u\|_p$ for general $u$)
• $\mu := |\{u=0\}| > 0$ (Lebesgue measure), with a constant that blows up as $\mu\to 0$.

Of course the inequality cannot hold in general (take $u$ to be a constant function), but the only obstruction seems to be that $u$ can be far away from zero. Therefore it should be true that the inequality holds whenever $u$ is zero somewhere in $\Omega$ (in a suitable sense, for example, zero is contained in the essential range of $u$).

So the question is: Is there a version of the inequality for this case?

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You need $u\in \mathcal{H}^1(\Omega)$. –  alext87 Sep 17 '10 at 20:36
On Lipschitz domains, this sounds reasonable (note that Nate's counterexample is on a domain with a cusp; even Sobolev embeddings fail on such domains, see Adams). How badly do you need this? –  Piero D'Ancona Sep 18 '10 at 0:08
@Piero D'Ancona: the shape of the domain is not the real problem, rather it's the smallness of the zero set. See my edit. –  Nate Eldredge Sep 18 '10 at 0:44
@Nate: you are right. One needs to anchor the function at zero (or at a constant) and one point is certainly not enough. On the other hand, if the function is zero at the boundary the zero set has measure zero, but Poincare' is ok, so requiring a zero set with positive measure seems overshooting. I should really think more carefully about this, but in simple cases it seems that a zero set with positive n-1 dimensional measure might suffice. Just thinking aloud. –  Piero D'Ancona Sep 18 '10 at 0:59

The Poincaré inequality need not hold in this case. The region where the function is near zero might be too small to force the integral of the gradient to be large enough to control the integral of the function.

For an explicit counterexample, let $$\Omega = \{(x,y) \in \mathbb{R}^2 : 0 < x < 1, 0 < y < x^2\}$$ be the region under the graph of a parabola, and take $p=2$. For $\epsilon > 0$, let $$u_\epsilon(x,y) = \begin{cases} x/\epsilon, & x < \epsilon \\\ 1, & x \ge \epsilon \end{cases}.$$ $0$ is in the essential range of each $u_\epsilon$, but one can easily verify $||\nabla u_\epsilon||_2^2 = \epsilon/3 \to 0$ as $\epsilon \to 0$, whereas $||u_\epsilon||_2^2 \to |\Omega| = 1/3$.

Edit: To address Piero's comment, the irregularity of the domain is not the main issue here. For another counterexample, let $\Omega$ be the unit ball in $\mathbb{R}^d$, $d \ge 3$, and take $$u_\epsilon(x) = \begin{cases} \frac{|x|^2}{\epsilon^2}, & |x| < \epsilon \\\ 1, & |x| \ge \epsilon \end{cases}.$$ Using polar coordinates, one easily computes $||\nabla u_\epsilon||_2^2 \sim \epsilon^{d-2}$ while again $||u_\epsilon||_2^2 \to |\Omega|$.

What's really the problem is that the set where $u$ vanishes has zero capacity. If you can control from below the capacity of the set where $u$ vanishes, then you can get a Poincaré inequality.

Indeed, the following theorem may be what the asker wants: for "reasonable" $\Omega$ (Lipschitz suffices), if $\mathrm{Cap}(\{u = 0\}) \ge \delta$, then $||u||_2^2 \le \frac{C}{\delta} ||\nabla u||_2^2$. The precise statement (for any value of $p$) can be found in section 4.5 of William P. Ziemer's Weakly Differentiable Functions (along with the definition of capacity).

Note in particular that $\mathrm{Cap}(E) \ge m(E)$, so it's enough to control the Lebesgue measure. This is your bullet point 3.

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Take $u$ and let $\overline{u}$ be the average value of $u$ in $\Omega$. Then $u-\overline{u}$ has an average value of zero in $\Omega$. Hence we have

$||u-\overline{u}||_p\leq C||\nabla u||_p$

Is this a 'version' of the inequality that you would accept?

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This is essentially bullet point 2 in my list. I'm looking for an estimate of $\|u\|$. I'm sorry, I should have made this explicit. –  Florian Sep 17 '10 at 21:31
Yes, but I thought I would state the obvious. –  alext87 Sep 17 '10 at 21:34