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Given a smooth bounded open subset $\Omega$ of $\mathbb{R}^n$, does there exist $A >0$ such that if $f\in BV(\Omega)$ with zero trace on $\partial \Omega$, and $\int_\Omega |Df| = 1$, then $\|f\|_{L^1(\Omega)} \leq A$? This is not a homework problem.

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You mean $A$ should be independent of the choice of $f \in BV(\Omega)$ ? – begeistzwerst Apr 11 '12 at 17:13
I found a possible reference: Integral Inequalities of Poincaré and Wirtinger Type for BV Functions Norman G. Meyers and William P. Ziemer American Journal of Mathematics Vol. 99, No. 6 (Dec., 1977), pp. 1345-1360 – Stefan Smith Apr 11 '12 at 17:15
Yes, A is independent of f – Stefan Smith Apr 11 '12 at 17:15

Let $\epsilon > 0$ and consider the set $$ \Omega_\epsilon = \{x\in \Omega : \text{dist}(x,\partial \Omega) > \epsilon\}. $$ with characteristic function $\chi_{\Omega_\epsilon}$. By the coarea formula the function $f_\epsilon = \frac{1}{\mathcal{H}^{n-1}(\partial \Omega_\epsilon)} \cdot \chi_{\Omega_\epsilon}$ has zero trace and fulfils $\int_\Omega|Df_\epsilon| = 1$ and $|| f_\epsilon ||_{L^1(\Omega)} = \frac{\lambda(\Omega_\epsilon)}{\mathcal{H}^{n-1}(\partial \Omega_\epsilon)}$, where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}^n$ and $\mathcal{H}^{n-1}$ the $(n-1)$-dimensional Hausdorff measure. By letting $\epsilon \rightarrow 0$ you get the bound $A = \frac{\lambda(\Omega)}{\mathcal{H}^{n-1}(\partial \Omega)}$.

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but this function has trace $C$, not $0$, on $\partial \Omega$. – Stefan Smith Apr 11 '12 at 18:14
I appreciate the time you put into your post. This proves $A \geq \frac{\lambda(\Omega)}{\mathcal{H}^{n-1}(\partial \Omega)}$ but the opposite inequality is not clear to me, although it very well may be true. I can use properties of BV functions to convert this into a question about $C^\infty$ functions and post a new question. – Stefan Smith Apr 11 '12 at 21:58
it always seemed clear to me (from intuition in 1 and 2 dimensions) that the sequence thus constructed tends to a function of maximal $L^1$ norm among all admissible functions. but you are right. this should be proved. at the moment, i don't see how to do so. i am not sure either, if a "conversion" to $C^\infty$ makes it easier in any way. – begeistzwerst Apr 12 '12 at 9:39
@begiestzwirst: For $f \in BV(\Omega)$ with trace $0$ on the boundary, there exists $g \in C^\infty(\Omega)$ with $g \equiv 0$ on $\partial \Omega$, $\|g\|_{L^1(\Omega)} \geq \frac{1}{2}\|f\|_{L^1(\Omega)}$ and $\int_\Omega |\nabla g| \leq 2\int_\Omega |Df|$, by a theorem of Giusti. Then you use Holder's inequality and a Poincare inequality to get $$\|f\|_{L^1(\Omega)} \leq 2\|g\|_{L^1(\Omega)} \leq A\|g\|_{L^\frac{n}{n-1}(\Omega)} \leq B\|\nabla g\|_{L^1(\Omega)} \leq 2B\int_\Omega |Df|$$ – Stefan Smith Apr 12 '12 at 19:58

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