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Let $\Omega$ be a bounded domain of $\mathbb{R}^d$. We define the Sobolev space $H^1(\Omega)$ by \begin{align*} H^1(\Omega)=\{f \in L^2(\Omega,m) \mid |\nabla f| \in L^2(\Omega)\}. \end{align*} Here, $m$ denotes the Lebesgue measure on $\Omega$, and $\nabla f$ is the distributional gradient of $f$. The Neumann Laplacian $(L,\text{Dom}(L))$ on $\Omega$ is defined as \begin{align*} \text{Dom}(L)&=\left\{f \in L^2(\Omega,m) : H^1(\Omega) \ni g \mapsto \int_{\Omega}\nabla f\cdot \nabla g\,dm \text{ is continuous on $L^2(\Omega,m)$}\right\} \\ -\int_{\Omega}g Lf\,dm&=\int_{\Omega} \nabla f\cdot \nabla g\,dm,\quad f \in \text{Dom}(L),\,g \in H^1(\Omega). \end{align*}

Question. When $\partial \Omega$ is Lipschitz, we define \begin{align*} C=\left\{g \in C^2(\overline{\Omega}) :\frac{\partial g}{\partial n}(x)=0,\quad \sigma\text{-a.e.} \right\} \end{align*} Here, we denote by $n$ the inward unit normal vector on $\partial \Omega$, and $\sigma$ is the surface measure on $\partial D$. Can we show that $\{(g,Lg) \mid g \in C\}$ is a dense subspace of $\{(f,Lf) \mid f \in \text{Dom}(L)\}$ with respect to the $L^2$-norm? This means that $C$ is a core of $(L,\text{Dom}(L))$.

Although this fact should be valid, I do not know the proof.

Please let me know if you have any references.

sharpe
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  • Would this follow if you can show that $C$ is dense in $H^1(\Omega)?$ If so the following related question may help: https://math.stackexchange.com/a/3996523/149608. – ktoi Jan 23 '21 at 12:32

1 Answers1

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What helped me understand how things work with cores is the following consideration:

The set of finite linear combinations of eigenfunctions of a selfadjoint operator (say with compact resolvents, for simplicity) forms a core for the operator.

Now take your question for example. Let $$E=\{e\ : \ e \text{ is a finite linear combination of eigenfunctions of L}\}\subseteq \mathrm{D}(L).$$ Now, $$\{(e,L e) \ : \ e \in E\}$$ is a dense subset of $$\{(f,Lf) \ : \ f \in \mathrm{D}(L)\},$$ because of the density of the eigenfunctions in the domain. In particular, your question is now whether $$E \subseteq C =C^2_\mathrm{neu}(\Omega).$$ Now we have reduced this fairly abstract problem to a very concrete one. Pick any $\lambda\in(-\infty,0]$ (the Laplacian is a negative operator) and suppose that $e_\lambda \in L^2(\Omega), \|e_\lambda\|=1$ solves $$\Delta e_\lambda = \lambda e_\lambda.$$ What is the regularity of $e_\lambda$?

This kind of question is what Laplacians are build for, and maybe you already know that eigenfunctions are smooth. If not, to see this we can restate the problem as $$e_\lambda=(-\Delta +1)^{-1}\big( (1-\lambda) e_\lambda\big).$$ Now one can estimate $$\|e_\lambda\|_{H^2_{\mathrm{neu}}} = \bigg\|(-\Delta +1)^{-1} \big( (1-\lambda) e_\lambda\big)\bigg\|_{H^2_{\mathrm{neu}}} \leq C(1-\lambda) \|e_\lambda\|_{L^2} =C(1-\lambda) ,$$ where the constant $C>0$ depends on how you define the $H^2_{\mathrm{neu}}$ norm (choose your favourite version of Sobolev space). Iterating this inequality you find that $e_\lambda \in H^{2m}_{\mathrm{neu}}$ for any $m\in\mathbb{N}$. Then by Sobolev embedding you find that $e_\lambda \in C^{\infty}_{\mathrm{neu}}(\Omega)\subseteq C$.

A final note: this proof uses a lot of nontrivial tools, such as Sobolev embeddings, spectral decompositions etc.. With it I am not aiming for a simple direct proof (no doubt possible with some approximation via $C^2$ functions), but I hope it gives the idea of what's going on.

Kore-N
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  • Eigenfunctions of the Neumann Laplacian are in $C^2(\Omega)$ by the Sobolev embedding theorem, as you showed. But why are they in $C^2(\bar \Omega)$? I think that's a more delicate question, where one needs to take the boundary regularity into account. – MaoWao Jan 24 '21 at 16:01
  • The existence of derivatives on the boundary follows from the trace theorem: https://en.wikipedia.org/wiki/Trace_operator#Trace_theorem – Kore-N Jan 25 '21 at 08:38
  • The trace theorem does not imply $H^2(\Omega)\cap C^2(\Omega)\subset C^2(\bar \Omega)$, which you would need to answer my question. For example, let $\Omega={x\in \mathbb{R}^d : |x|<1,x_1,\dots,x_d>0}$. Then $u(x)=|x|^\gamma$ is in $H^2(\Omega)\cap C^2(\Omega)$ if $\gamma>2-\frac{d}{2}$, yet $u$ is not even in $C(\bar \Omega)$ if $\gamma<0$. – MaoWao Jan 25 '21 at 08:51
  • But $e_\lambda \in H^m_{\mathrm{neu}}(\Omega)$ for all $m\in\mathbb{N}$. See the "case $p=\infty$" in the wikipedia page. By the way, I understand your example, but your function does not lie in the \textbf{Banach} space $C^2(\Omega)$ in the sense of the Wikipedia page, and in the sense of my answer (the second derivative blows up at the boundary, so the norm is infinite). It would be more appropriate to replace $C^2(\Omega)$ with $C_{\mathrm{loc}}^2(\Omega)$ in your example. – Kore-N Jan 25 '21 at 13:12
  • @Kore-N Thank you for your comment (sorry for the rate reply). I found your proof using eigenfunctions interesting. – sharpe Jan 27 '21 at 17:50