Is the Dirichlet boundary condition continuous? First, let's fix some notation:
Let $\mathbb D^n \subseteq \mathbb{R}^n$ be the closed $n$-dimensional unit ball. Given a real-valued function $f \in C^{\infty}(\mathbb D^n)$, we denote by $\omega_f$ the unique harmonic function satisfying $\omega_f|_{\partial \mathbb D^n}=f|_{\partial \mathbb D^n}$.

Now, suppose that $f_k \in C^{\infty}(\mathbb D^n)$ converge in $W^{1,2}(\mathbb D^n)$ to $f$, and let $\omega_{f_k}$ be the associated sequence of "harmonic projections". Is it true that $d\omega_{f_k} \to d\omega_f$ in $L^2$ or $L^1$?


The trace theorem implies that if $f_k \to f$ in $W^{1,2}(\mathbb D^n)$, then 
$f_k|_{\partial \mathbb D^n}\to f|_{\partial \mathbb D^n}$ in $L^2(\partial \mathbb D^n)$, so $\omega_k|_{\partial \mathbb D^n}\to \omega|_{\partial \mathbb D^n}$ in $L^2(\partial \mathbb D^n)$.
 A: If I'm not mistaken, the answer is yes: $d\omega_{f_k}$ converges to $d\omega_f$ in $L^2$.
Proof: W.l.o.g. we can assume that $f=0$ (everything here is linear). $f_k \to 0$ in $W^{1,2}(\Omega)$ implies that $f_k|_{\partial \Omega} \to 0$ in $W^{1/2,2}(\partial \Omega)$ (See here for example https://en.wikipedia.org/wiki/Sobolev_space#Traces, or the theorem I quote below).
Moreover, as a map $W^{1,2}(\Omega) \to W^{1/2,2}(\partial \Omega)$, the trace operator has a continuous inverse, that is, for every $v\in W^{1/2,2}(\partial \Omega)$ there exists $u\in W^{1,2}(\Omega)$ with $u|_{\partial \Omega} = v$ such that $\|u\|_{W^{1,2}(\Omega)} \lesssim \|v\|_{W^{1/2,2}(\partial \Omega)}$. See Leoni's "A first course on Sobolev spaces", second edition, Theorem 18.40.
Back to our case: the above theorem implies the existence of $g_k \in W^{1,2}$ with $g_k|_{\partial \Omega} = f_k$ such that $\|g_k\|_{W^{1,2}(\Omega)} \lesssim \|f_k\|_{W^{1/2,2}}(\partial\Omega)$. In particular, $\|dg_k\|_{L^2(\Omega)} \lesssim \|f_k\|_{W^{1/2,2}}(\partial\Omega)$. But $\|d\omega_{f_k}\|_{L^2(\Omega)}\le \|dg_k\|_{L^2(\Omega)}$ since the harmonic map is the minimizer of the Dirichlet energy with these boundary conditions. 
Therefore $f_k|_{\partial \Omega} \to 0$ in $W^{1/2,2}(\partial \Omega)$ implies $d\omega_{f_k} \to 0$ in $L^2$.
