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Let $D \subset \mathbb{R}^d$ be a bounded $C^1$ domain. We consider a reflected Brownian motion $X=(\{X_t\}_{t \ge 0},\{P_x\}_{x \in \overline{D}})$ on $\overline{D}$. Let $\{p_t\}_{t>0}$ denote the semigroup of $X$ (in other words, $\{p_t\}_{t>0}$ is the Neumann semigroup). It is known that $\{p_t\}_{t>0}$ is a feller semigroup. That is, we have $p_t f \in C(\overline{D})$ for any $t>0$ and $f \in C(\overline{D})$. Here, $C(\overline{D})$ is the space of continuous functions on $\overline{D}.$ We always equip $C(\overline{D})$ with the sup-norm so that it becomes a real Banach space.

Let $(L,D(L))$ be the Neumann Laplacian on $C(\overline{D})$. That is, \begin{align*} D(L)&=\left\{ f \in C(\overline{D})\mid \lim_{t \to 0}\frac{p_t f-f}{t} \text{ exists in }C(\overline{D})\right\},\\ Lf&=\lim_{t \to 0}\frac{p_t f-f}{t}. \end{align*}

Let $\nu$ denote the inward unit normal vector on $\partial D$, and define \begin{align*} \mathcal{C}=\{f \in C^2(\mathbb{R}^d)|_{\overline{D}} \mid \langle \nabla f(x),\nu(x) \rangle=0\text{ for every $x \in \partial D$}\}. \end{align*} Can we prove that $\mathcal{C}$ is a core for $(L,D(L))$?

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    $\begingroup$ I guess that you mean $f \in C^2(\mathbb R^n)$ and $\nabla f\cdot \nu=0$. Do you want $D \in C^1$, no more? $\endgroup$ Oct 31, 2022 at 10:23
  • $\begingroup$ @GiorgioMetafune Thank you for your comment (I already fixed the typo). Yes, $D$ is just a bounded $C^1$ domain (perhaps, is it necessary to assume $D \in C^{1,\alpha}$?). $\endgroup$
    – sharpe
    Oct 31, 2022 at 11:49
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    $\begingroup$ If $D \in C^{2,\alpha}$ I know it is true....with less regularity I do not know. $\endgroup$ Oct 31, 2022 at 12:07
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    $\begingroup$ In such a case, by the Schauder theory, for every $f \in C^\alpha$ the solution of $u-\Delta u=f$ with Neumann b.c. belongs to $C^{2,\alpha}$ up to the boundary (and can be extended to the whole space). Since $C^\alpha$ is dense in $C$, $(I-\Delta)^{-1})(C^\alpha)$ is dense in the domain, with respect to the graph norm. $\endgroup$ Oct 31, 2022 at 13:07
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    $\begingroup$ Yes, true. However I think that the result could be true with $D \in C^{1,\alpha}$. In such a case the solution $u$ should be $C^2$ in the interior and $C^{1,\alpha}$ up to the boundary and one should modify accoringly the core $\mathcal C$. I checked M. Giaquinta "Introduction to regularity theory for nonlinear elliptic systems", Chapter 3 (which is a bit vague) for similar results at least in the Dirichlet case. $\endgroup$ Oct 31, 2022 at 14:19

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I think $\mathcal C$ is not even dense in the space of continuous functions!

To be specific: consider a 2-D domain $D$ lying above the graph of a $C^1$ function $\phi : \mathbb R \to \mathbb R$, and assume that $\phi'$ is continuous, but nowhere differentiable — say, a generic sample path of the Wiener process.

Let $f$ be a $C^2$ function on $\mathbb R^2$ such that $\nabla f$ is orthogonal to the normal vector at each boundary point. This means that $$\partial_y f(x, \phi(x)) = \phi'(x) \partial_x f(x, \phi(x)).$$ On every interval where $\partial_x f(x, \phi(x)) \ne 0$, we find that $$\phi'(x) = \frac{\partial_y f(x, \phi(x))}{\partial_x f(x, \phi(x))}$$ is a $C^1$ function, a contradiction. Thus, $\partial_x f(x, \phi(x)) = 0$ for all $x$, and consequently also $\partial_y f(x, \phi(x)) = 0$ for all $x$. In particular, $f(x, \phi(x))$ is constant.

But that means that $\mathcal C$ only contains functions $f$ which are constant on the boundary, so $\mathcal C$ is not even dense in the space $C(\overline D)$ of continuous functions.

The same argument works for $C^{1,\alpha}$ domains, at least when $\alpha < 1$.

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  • $\begingroup$ Thank you for your very helpful comment. As you say, $\mathcal{C}$ is too small... If $D \in C^2$, then $\mathcal{C}$ may be core a for the Neumann Laplacian. $\endgroup$
    – sharpe
    Nov 3, 2022 at 11:46

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