# 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 $X$?

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.
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.
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.