Show minus Laplacian operator is densely defined on $L^2$

Question

I am trying to solve the following problem:

1. Let $U \subset \mathbb R^n$ be a bounded and smooth domain, and $L$ be the minus Lapacian operator with zero boundary condition. Prove that $$L : L^2(U) \rightarrow L^2(U)$$ is a densely defined, unbounded linear operator with domain $D(L) = H^2(U) \cap H^1_0(U)$.

2. Give an example showing that the difference between $H^2_0(U)$ and $H^2(U)\cap H^1_0(U)$.

Here $H^2(U)$ is the Sobolev space $W^{2,2}(U)$ and $H^1_0(U)$ denotes the closure of $C^{\infty}(U)$ in $H^1(U)$.

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I really have no clue. Should we use Green formula to define how does the minus Laplacian operator apply to a function from the above Sobolev space?

Any hints or answers are welcomed!