Let $\Omega\subset \mathbb{R}^d$ be open and bounded with $C^\infty$ boundary $\partial\Omega$, $\phi\colon \partial\Omega \rightarrow \mathbb{R}$ continuous and $u^\phi$ the solution to Laplace's equation $\Delta u^\phi=0$ on $\Omega$ with Dirichlet boundary conditions $u^\phi|_{\partial\Omega}=\phi$. Is it true that there exists $C>0$ independent of $\phi$ such that $$ |u^\phi|_{H^1(\Omega)}:=\|\nabla u^\phi\|_{L^2(\Omega)}\leq C\|\phi\|_{L^\infty(\partial\Omega)}? $$ More generally, if $M\subset \mathbb{R}^n$ is a Riemannian submanifold of $\mathbb{R}^n$ and $$ u^\phi:=\mathop{argmin}_{\substack{v \in H^1(\Omega,M)\\v|_{\partial\Omega}=\phi} } |v|_{H^1(\Omega,\mathbb{R}^n)}, $$ is it true that $$ |u^{\phi_1}-u^{\phi_2}|_{H^1(\Omega,\mathbb{R}^n)}\leq C\|\phi_1-\phi_2\|_{L^\infty(\partial\Omega,\mathbb{R}^n)}? $$ I am trying to derive a discretization error estimate for a numerical scheme. Since the boundary condition can in general not be implemented exactly I would like to estimate the error due to errors in the boundary data.
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It is not true. An $H^1$ function has a boundary trace in $H^{1/2}$, and you cannot bound the $H^{1/2}$ norm by the $L^\infty$ norm.
answered Jun 20, 2016 at 15:33
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$\begingroup$ Here we have not just a H1 function but the harmonic map for which we can expect higher regularity. I do not see why it should not be possible. $\endgroup$ Commented Jun 20, 2016 at 17:15
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4$\begingroup$ Regardless of what equation is satisfied, the regularity of the boundary data does not improve over what is given! $\endgroup$ Commented Jun 20, 2016 at 17:28
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2$\begingroup$ Just to be clear: What I am saying it that if $\phi$ is in $L^\infty$ and not in $H^{1/2}$, then the solution to Laplace's equation (or to any other equation) is not in $H^1$. $\endgroup$ Commented Jun 20, 2016 at 17:39
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3$\begingroup$ Why was this downvoted? It may be a bit terse, but Michael is making a very valid point. You can also consider concrete examples such $u(z)=\textrm{Re}\exp ((z+1)/(z-1))$ on the unit disk (and approximate $\phi$ in $L^1$ by continuous functions) to see that what you were hoping for is false. $\endgroup$ Commented Jun 20, 2016 at 19:05
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$\begingroup$ Ok, now I understand your argument. Can we say that the inequality holds with $L^\infty$ replaced by $H^{1/2}$? $\endgroup$ Commented Jun 20, 2016 at 19:51