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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?

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.

Thanks in advance.

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  • $\begingroup$ In general, the norm of the function on the boundary can be estimated by the norm of the function in $H^1$, so this may be what you are looking for. The answer below shows how just $L^2$ norms may fail. $\endgroup$
    – Daniel Spector
    Commented Jan 24, 2013 at 10:02
  • $\begingroup$ well, estimating the $L^2$-norm of a trace of $u$ by the $H^1$-norm of $u$ simply means that the trace operator is bounded from $H^1(\Omega)$ to $L^2(\partial \Omega)$ (which, by the way, only holds under certain smoothness assumption on $\partial \Omega$, as far as i know). My question was motivated by my hope that, perhaps, in this regard (weakly) harmonic function may perform better than mere H1-functions. $\endgroup$
    – Delio Mugnolo
    Commented Jan 27, 2013 at 8:48

1 Answer 1

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

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