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.



$H^1$-continuity of Laplace's equation with respect to boundary data
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.


