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Let $D \subset \mathbb{R}^d$ be a bounded smooth domain. We consider the Neumann semigroup $\{T_t\}_{t>0}$ on $C(\overline{D})$. In other words, $\{T_t\}_{t>0}$ is the semigroup of the normally reflected Brownian motion on $\overline{D}$.

Let $(L,D(L))$ be the generator of $\{T_t\}_{t>0}$. That is, we define \begin{align*} D(L)&=\left\{f \in C(\overline{D}) : \lim_{t \to 0}\frac{T_tf-f}{t} \text{ exists in }C(\overline{D}) \right\},\\ Lf&=\lim_{t \to 0}\frac{T_tf-f}{t},\quad f \in D(L). \end{align*}

Define \begin{align*} C_{\text{Neu}}(\overline{D})=\{f \in C^2(\mathbb{R}^d)|_{\overline{D}} \mid \partial f /\partial \nu=0\text{ on }\partial D\}, \end{align*} where $\nu$ denotes the innward unit normal on $\partial D$.

I think $C_{\text{Neu}}(\overline{D})$ is a core for the Neumann Laplacian $(L,D(L))$. Is this correct? I do not know a full proof. I also suspect that there is a core smaller than $C_{\text{Neu}}(\overline{D})$. For example, define \begin{align*} A_{\text{Neu}}(\overline{D})=\{f \in C^2(\mathbb{R}^d)|_{\overline{D}} \mid \nabla f \in C_c(D)\}. \end{align*} Here, $C_c(D)$ is the continuous functions on $D$ with compact support. Isn't $A_{\text{Neu}}(\overline{D})$ is a core for $(L,D(L))$?

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  • $\begingroup$ To see that the first is a core you can use Schauder estimates. If $u_n-\Delta u_n=f_n$ with Neumann b.c. and $f_n \in C^1$, $f_n \to u-\Delta u$ in the sup norm, then $u_n \in C^2$ and $(u_n) \to u$ in the graph norm. The second is too small, I guess. I would substitute with functions for which only the normal derivative is zero near the boundary. $\endgroup$
    – Giorgio Metafune
    Commented Oct 16, 2021 at 11:21
  • $\begingroup$ @GiorgioMetafune Thank you for your kind answer. I was not aware that Schauder estimates are useful. Could you tell me the definition of "near the boundary"? $\endgroup$
    – sharpe
    Commented Oct 16, 2021 at 11:32
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    $\begingroup$ Yes, sure. I have been too vague. In the half plane $\{y>0\}$ I would say $u(x,y)=u(x,0)$ for $y \leq \delta$, for a certain $\delta>0$. In a regular domain one should reformulate using a tubolar neighborhood and requiring that $u$ does not depend on the distance from the boundary, for small distance. However I am not sure if it is true, $\endgroup$
    – Giorgio Metafune
    Commented Oct 16, 2021 at 11:42
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    $\begingroup$ Sorry, the second even modified is too small. For example in 1d consider $D^2$ with Neumann b.c at $x=0$. The domain consists of all $C^2$ functions $u$ with $u'(0)=0$ and if $u''(0) \neq 0$ such a $u$ cannot be approximated by functions which are locally constant near 0. However this works in Sobolev spaces. $\endgroup$
    – Giorgio Metafune
    Commented Oct 16, 2021 at 11:52
  • $\begingroup$ @GiorgioMetafune Thank you for your comment. I also think your idea is available for $L^2$-Neumann Laplacians. $\endgroup$
    – sharpe
    Commented Oct 16, 2021 at 11:56

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