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 $L^2(\partial \mathbb D^n)$.

  • $\begingroup$ what does $d\omega_{f_k}$ mean? (and $L^2$ of what?) $\endgroup$ – mathworker21 Sep 12 '19 at 13:43
  • $\begingroup$ This is the differential (or the gradient if you prefer) of the function $\omega_{f_k}$. Perhaps I should write $d(\omega_{f_k})$ to make it clear. If you think in terms of gradients, then $d(\omega_{f_k}) \in L^2(\mathbb D^n,\mathbb R^n)$. Equivalently, you can think of each component of the gradient as an element in $L^2(\mathbb D^n)$. $\endgroup$ – Asaf Shachar Sep 12 '19 at 13:45

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 $L^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.