# Harmonic extension of $L^\infty$ function is in $L^\infty$?

Let $u \in H^{\frac 12}(\Omega)$ with $\int_\Omega u = 0$ and consider the solution $v \in H^1(C)$ where $C=\Omega \times (0,\infty)$ of $$-\Delta v(x,y) = 0$$ $$\partial_\nu v = 0$$ $$v(x,0) = u(x)$$ in the weak sense. We can write $v(x,y) = \sum_{k=1}^\infty e^{-\lambda_k^{\frac 12}y}(u,\varphi_k)_{L^2}\varphi_k$ where $\lambda_k$ and $\varphi_k$ are the eigenelements corresponding to the Neumann Laplacian.

If $u \in L^\infty(\Omega)$, does it follow that $v \in L^\infty(C)$?

• Wouldn't the last inequality applied to -v and -u also give $-v(x,y)\le -u(x)$? – Pietro Majer Feb 8 '15 at 22:36
• @Pietro unfortunately I made a mistake and cannot show that inequality at the moment. – TomBerry Feb 9 '15 at 10:30
• This is just an application of the maximum principle. The Hopf lemma takes care of the Neumann boundary. – Michael Renardy Feb 9 '15 at 16:37