most recent 30 from http://mathoverflow.net 2013-05-23T05:48:15Z http://mathoverflow.net/feeds/question/119686 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119686/on-an-inequality-of-brezis-lieb Delio Mugnolo 2013-01-23T18:55:04Z 2013-01-23T19:46:51Z

<p>In their 1983 JFA-paper Brezis and Lieb have shown, among many other things, a Poincaré-type inequality: in the case of a harmonic function $f$ on a bounded domain $\Omega$, their inequality ((3.14) in the paper) states that the $L^2(\Omega)$-norm of $f$ can be estimated by the $L^2(\partial \Omega)$-norm of its trace on $\partial \Omega$ (times a constant only depending on $\Omega$). My question: Is it possible to reverse this inequality, viz estimating the $L^2(\partial \Omega)$-norm of the trace of $f$ by the $L^2(\Omega)$-norm.</p> <p>This is indeed possible if $\Omega\subset R$ ($\Omega$ an interval), but this clearly follows simply from the fact that on an interval both the space of harmonic functions and the space of their boundary values are $2$-dimensional (then using equivalence of any two norms on a finite dimensional space). I have no clue whether this may extend to higher dimensions - in fact, I am pessimistic.</p> <p>Thanks in advance.</p>

http://mathoverflow.net/questions/119686/on-an-inequality-of-brezis-lieb/119688#119688 jjcale 2013-01-23T19:46:51Z 2013-01-23T19:46:51Z

<p>No, choose $\Omega={z \in \mathbb{C} : |z| \leq 1 } ,\ f(z)=Re\ z^{n}$ and let $n\rightarrow \infty$ .</p>