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In page 290 of this book, Evans prove the Poincare inequality (Theorem 1) arguing by contradiction. Is there a direct proof of this theorem (Theorem 1) without arguing by contradiction?

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up vote 5 down vote accepted

Any paper that estimates the value of the Poincaré constant must avoid this argument by contradiction. Here is such an estimate (sharp!) for the $L^1$ case and here is one for $L^p$ case. Both papers have references to older literature. As far as I can tell, all of them concern convex domains and need convexity in order to apply one-dimensional estimates. I have not seen such results for non-convex domains, but then again, I did not click on every link in Google search for poincare constant estimate.

[Added] For the sake of clarity, I'll draw the distinction between two versions of the Poincaré inequality: for $W^{1,p}_0$ and for $W^{1,p}/\{\mathrm{constants}\}$. The $W^{1,p}_0$ version is not very sensitive to the geometry of the domain, because one can extend a $W^{1,p}_0$ function to a larger domain (a ball, a cube or a suitable convex set) and apply the inequality there. If one uses a cube, then the estimate becomes an exercise with Fubini.

But the most useful form of the Poincaré inequality is for $W^{1,p}/\{\mathrm{constants}\}$. This inequality measures the connectivity of the domain in a subtle way. For example, joining two squares by a thin rectangle, we get a domain with very large Poincaré constant, because a function can be $-1$ in one square, $+1$ in the other, and all the gradient can be confined to the joining rectangle. This "poor connectivity" is not easy to express in purely geometric terms, which is why people are content with an implicit constant obtained from an argument by contradiction.

The above distinction can be put in a different language: Dirichlet eigenvalues are monotone with respect to the domain while Neumann eigenvalues are not.

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Pavel, I was looking for a demonstration like this one (take a look in Theorem 2.1, but here they do only case $N=1$):… – Tomás Dec 20 '12 at 19:12
@Tomás Yes, in the one-dimensional case explicit bounds are easy to get. That's why the papers I cited dissect convex domains by lines, apply one-dimensional estimates, and then put things back together. – user53153 Dec 20 '12 at 19:35
Now i understood, thank you Pavel – Tomás Dec 20 '12 at 19:38

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