# L2 Trace of Harmonic Functions

I've been wondering about the following problem:

Let $\Omega \subset \mathbb{R}^3$ be an open, bounded, connected set with Lipschitz boundary $\partial \Omega$, let $u \in C^\infty(\overline{\Omega})$ be a function that is harmonic in $\Omega$, i.e., \begin{align*} \Delta u=0 \qquad \text{in } \Omega, \end{align*}

and let $\gamma u \colon \partial \Omega \rightarrow \mathbb{R}$ denote the usual boundary trace of the function $u$.

Question: Is there a sharp relationship between the $L^2(\Omega)$ norm of the function $u$ and the $L^2(\partial \Omega)$ of the trace function $\gamma u$? Specially, can we say that \begin{align*} \Vert u\Vert^2_{L^2(\Omega)} \leq \Vert \gamma u\Vert^2_{L^2(\partial \Omega)}\\[0.5em] \hspace{-5cm}\text{or alternatively,}\\[0.5em] \Vert \gamma u\Vert^2_{L^2(\partial \Omega)} \leq \Vert u\Vert^2_{L^2(\Omega)} \end{align*}

It is certainly possible to say something if the domain $\Omega$ is the unit ball (using Spherical harmonics) but I was wondering if a similar result holds for general domains.

Any help would be appreciated it! Thanks in advance!

Broader context: It's well known that Harmonic functions satisfy the $L^\infty$ maximum principle. I've been wondering if it's possible to obtain a similar maximum principle in terms of the $L^2$-norm.

If I understand the question correctly, perhaps you are looking for the trace theorem, that given your original assumptions and an arbitrary shaped open domain in $\mathbb{R}^n$, $$\Vert \gamma u\Vert^2_{L^2(\partial \Omega)} \leq \Vert u\Vert^2_{W^{1,2}(\Omega)}$$ and $W^{1,2}$ is compactly embedded in $L^2(\Omega),$ so $$\Vert \gamma u\Vert^2_{L^2(\partial \Omega)} \leq \Vert u\Vert^2_{L^{2}(\Omega)}$$
• Hi! Thanks a lot for the reply. Couple of quick questions: 1) In the above inequality, is there really no constant in the inequality? 2) Can you please provide a suitable reference for this version of the trace theorem? The version that I'm familiar with works for trace spaces $H^s(\partial \Omega)$ for $s > 0$... Many thanks for the help! – Hassan Aug 25 '17 at 13:02