  Several domination results have been obtained for maximal outerplanar graphs
(mops). The classical domination problem is to minimize the size of a set $S$
of vertices of an $n$-vertex graph $G$ such that $G - N[S]$, the graph obtained
by deleting the closed neighborhood of $S$, contains no vertices. In the proof
of the Art Gallery Theorem, Chv\'{a}tal showed that the minimum size, called
the domination number of $G$ and denoted by $\gamma(G)$, is at most $n/3$ if
$G$ is a mop. Here we consider a modification by allowing $G - N[S]$ to have a
maximum degree of at most $k$. Let $\iota_k(G)$ denote the size of a smallest
set $S$ for which this is achieved. If $n \le 2k+3$, then trivially $\iota_k(G)
\leq 1$. Let $G$ be a mop on $n \ge \max\{5,2k+3\}$ vertices, $n_2$ of which
are of degree $2$. Upper bounds on $\iota_k(G)$ have been obtained for $k = 0$
and $k = 1$, namely $\iota_{0}(G) \le
\min\{\frac{n}{4},\frac{n+n_2}{5},\frac{n-n_2}{3}\}$ and $\iota_1(G) \le
\min\{\frac{n}{5},\frac{n+n_2}{6},\frac{n-n_2}{3}\}$. We prove that
$\iota_{k}(G) \le \min\{\frac{n}{k+4},\frac{n+n_2}{k+5},\frac{n-n_2}{k+2}\}$
for any $k \ge 0$. For the original setting of the Art Gallery Theorem, the
argument presented yields that if an art gallery has exactly $n$ corners and at
least one of every $k + 2$ consecutive corners must be visible to at least one
guard, then the number of guards needed is at most $n/(k+4)$. We also prove
that $\gamma(G) \le \frac{n - n_2}{2}$ unless $n = 2n_2$, $n_2$ is odd, and
$\gamma(G) = \frac{n - n_2 + 1}{2}$. Together with the inequality $\gamma(G)
\le \frac{n+n_2}{4}$, obtained by Campos and Wakabayashi and independently by
Tokunaga, this improves Chv\'{a}tal's bound. The bounds are sharp.

Combinatorics