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Let $\Omega$ be a smooth bounded open subset of $\mathbb{R}^n$. Does there exist $A = A(\Omega)$ with the property that for any $f \in C^\infty(\bar{\Omega})$ with $f = 0$ on $\partial \Omega$, $\int_\Omega |f(x)|\,dx \leq A\int_\Omega |\nabla f(x)|\,dx$? I think that this is known as some version of ``Poincare's inequality''.

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Did you forget a square on those absolute values? $$\int_{\Omega} \lvert f\rvert^2\, dx \le A\int_{\Omega}\lvert \nabla f \rvert^2\, dx.$$ –  Giuseppe Negro Apr 11 '12 at 23:33
    
No, I don't want a square on the absolute values. –  Stefan Smith Apr 11 '12 at 23:39
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This is Friedrichs' inequality. –  timur Apr 22 '12 at 12:44
    
@timur: thank you –  Stefan Smith Apr 23 '12 at 22:10

2 Answers 2

Technically no, but only because the gradient operator has a nontrivial kernel - for constant functions $f(x)=c$, $\nabla f =0$. If you somehow "tie down" the function at a point - for example by specifying a point $p$ such that $f(p)=0$, or by requiring that the average value of the function is zero, or by quotienting your space out by the constant functions, then yes it is true (for connected domains).

The intuition for this can be found by chopping the function along it's level sets - then the integral of the function is the sum of the areas of the level sets, whereas the integral of the gradient is the sum of the perimeters of the level sets. Since the circle (sphere/hypersphere..) has the largest area for a given perimeter, the area sum is bounded by the perimeter sum, up to a geometric constant.

I have a short graphical explanation about this on my website, which you might be interested in, http://maze5.net/?page_id=790

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I think there is a "tie down" condition in OP's question: it is requested $f \equiv 0 $ on $\partial \Omega$. –  Giuseppe Negro Apr 11 '12 at 23:31
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I found the answer at http://anhngq.wordpress.com/2010/02/22/the-poincare-inequality/. By Holder's Inequality and Poincare's Inequality for $W_0^{1,p}$ spaces, there exist $A(\Omega), B(\Omega) > 0$ such that $$\|f\|_{L^1(\Omega)} \leq B\|f\|_{L^{\frac{n}{n-1}}(\Omega)} \leq C\|\nabla f\|_{L^1(\Omega)}.$$

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True, but be careful about the chain of logic. Sometimes the statement in your original question is used to prove the $W_0^{1,p}$ Poincare inequality (though in the linked article they used a compactness argument so it's OK there). –  Nick Alger Apr 12 '12 at 0:52

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