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.
Maybe somebody has a more elementary argument but I hope this will do...
