Core for the Laplace operator in a bounded domain Let $X$ be a bounded connected open subset of the $n$-dimensional real Euclidean space. Consider the Laplace operator defined on the space of infinitely differentiable functions with compact support in $X$. 
Does the closure of this operator generate a strongly continuous semigroup on $C_0(X)$ endowed with the supremum norm?
I think it is equivalent to the following question: 

Is the space of infinitely differentiable functions with compact support in $X$ a core for the Dirichlet Laplacian on $X$?

 A: The closure of the Laplacian on $C^\infty_0(X)$ 
cannot generate a strongly continuous semigroup on $C(X)$,
since there are two distinct (closed!) generators that extend $(\Delta, C^\infty_0(X))$; namely
the Dirichlet Laplacian and the Neumann Laplacian. Probabilistically, 
these correspond to absorbing and reflecting Brownian motion, respectively. 
This follows from the basic fact that generators of semigroups cannot strictly extend one another. 
Lemma: 
Suppose that $(T_t)_{t\geq 0}$ is a strongly continuous semigroup with generator $(L,D(L))$, and that
$(S_t)_{t\geq 0}$ is a strongly continuous semigroup with generator $(K,D(K))$.
Suppose also that $L$ extends $K$, that is, $D(K)\subseteq D(L)$ and $Ku=Lu$ for all
$u\in D(K)$. Then $T_t=S_t$ for all $t\geq 0$. 
Proof. Fix $u\in D(K)$ and $t>0$, and consider the  continuous map from $[0,t]$ to $C(X)$ given by $s\to T_{t-s} S_su$. This map is differentiable on $(0,t)$ with 
$$\biggl({d\over{ds}}\biggr) T_{t-s} S_su= -L T_{t-s} S_su+ T_{t-s} K S_su=
T_{t-s} (K-L)(S_su)=0.$$
It follows that the value of this map at $s=0$ and at $s=t$ must be equal,
that is $T_tu=S_tu$.
Since this equation is true on the dense set $D(K)$ and since
$T_t$ and $S_t$ are bounded operators, it follows that $T_t=S_t$.
Therefore, the extension 
$D(K)\subseteq D(L)$ is not genuine; that is, $D(L)=D(K)$. 
So if any two generators extend one another, they in fact must coincide. 
A: I completely agree with the previous answer but I would like to add that in general there are even infinitely many extensions of the Laplacian - not only Dirichlet and Neumann. E.g., the Laplacian with all Robin-type boundary conditions
$$
\frac{\partial u}{\partial n}=pu_{|\partial X}
$$
will do the job.
