# Is the linear span of the Neumann eigenfunctions dense in $C(\overline{D})$

Let $D\subset R^d$ be a bounded Lipschitz domain. We know that the Neumann eigenfunction lies in $C(\overline{D})$ (i.e. continuous up to the boundary). This can be seen from the fact that $\phi_k=e^{\lambda_kt}P_t\phi_k$ where $P_t$ is the semigroup of the reflected Brownian motion.

Is the linear span of the Neumann eigenfunctions dense in $C(\overline{D})$?

If yes, what's the reason? If not, what is the closure of the linear span with respect to the supremum norm?

Yes. First, note that $P_t$ is strongly continuous on $C(\bar D)$ so that, for every $f \in C(\bar D)$, $P_t f \to f$ in $C(\bar D)$ as $t \to 0$. For every $t > 0$, $P_t f$ belongs to every Sobolev space $H^{2k}$, viewed as the domain of the $k$th power of the Neumann Laplacian, so that it does belong to the closure of $span\{\phi_j\}$ in $H^{2k}$. For $k$ large enough, the $H^{2k}$ norm dominates the supremum norm by Sobolev embedding, so that $P_t f$ also belongs to the closure of $span\{\phi_j\}$ in $C(\bar D)$, which is all you need.
• Thanks Martin. I'm a bit confused about the notation. Is $H^{p}= \{f\in L^{p}(D): |\nabla f|\in L^p(D)\}$? You mentioned above (in the case $k=1$) that $H^{2}$ is "viewed as the domain of the Neumann Laplacian", but it is not the usual Sobolev space $W^{1,2}(D)$. Thanks! Commented Oct 4, 2013 at 8:41
• I used the notation $H^p$ for $\{f \in L^2\,:\, |\nabla^p f| \in L^2\}$. Commented Oct 5, 2013 at 9:23