Closure of the Hamilton's operator $(Hf)(x)=\frac{1}{2}f''(x)-V(x)f(x)$ with $C_c^\infty(\mathbb{R}, \mathbb{C})$ domain Let $V \in C_{b}^{1}(\mathbb{R}, \mathbb{R})$ be a differentiable function bounded with its first derivative and $H$ be a Hamilton's operator such that:
$$(Hf)(x)=\frac{1}{2}f''(x)-V(x)f(x)$$
The quastion is how to show that operator $(H, C_c^\infty(\mathbb{R}, \mathbb{C}))$ has a closure $(H, W_2^2(\mathbb{R}, \mathbb{C}))$? Here $C_c^\infty(\mathbb{R}, \mathbb{C}) \subset L^2(\mathbb{R}, \mathbb{C})$ is a linear space of smooth functions with compact support and $W_2^2(\mathbb{R}, \mathbb{C}) \subset L^2(\mathbb{R}, \mathbb{C})$ is the Sobolev class, i.e. the linear space of all the functions $f\in L^2(\mathbb{R, \mathbb{C}})$ such that $f'\in L^2(\mathbb{R, \mathbb{C}})$ and $f''\in L^2(\mathbb{R, \mathbb{C}})$.
 A: Assume that $V \in L^{\infty}(\mathbb{R})$. Then $Lf = Vf$ is a bounded linear operator on $L^{2}$ with $\|Lf\|\le \|V\|_{\infty}\|f\|$. Therefore the closure of $Hf = \frac{1}{2}f''-Vf$ on a domain $\mathscr{D}$ has the same domain as the closure of $H_{0}f=\frac{1}{2}f''$ on $\mathscr{D}$. This is because if $\{f_{n}\}\subset \mathscr{D}$ converges to $f$, then $Hf_{n}$ converges to some $g$ iff $H_{0}f_n$ converges to $g+Vf$. So the domain of the closure of $H$ is the same as the domain of the closure of $H_{0}$. The domain of the closure of $H_{0}$ on $C_{c}^{\infty}$ is $W_{2}^{2}$, which proves your result.
Added because of a question in your comment: Let $H_{m}$ be the minimal operator defined by $H_{m}f = -f''$ on $\mathcal{C}_{c}^{\infty}(\mathbb{R})$. You question was about the closure of $H_{m}$. First, $H_{m}$ is symmetric on its domain, meaning that $(H_{m}f,g)=(f,H_{m}g)$ for $f,g\in\mathcal{D}(H_{m})=\mathcal{C}_{c}^{\infty}$. Therefore
$$
                      H_{m} \preceq H_{m}^{\star} \;\;\; (\preceq \mbox{ means graph inclusion.})
$$
On the other hand, $g\in \mathcal{D}(H_{m}^{\star})$ iff there exists $h \in L^{2}$ such that
$$
                     (H_{m}f,g)=(f,h),\;\;\; f \in \mathcal{C}_{c}^{\infty}
$$
In other words $h$ is the negative of the weak second derivative of $g$. Assuming you know some Sobolev theory, that means $g$ is twice absolutely continuous with $-g'' \in L^{2}$ and $-g''=h$. So $\mathcal{D}(H_{m}^{\star})$ consists of all twice absolutely continuous functions in $L^{2}$ whose second derivative is in $L^{2}$. It is easy to check that
$$
                  (H_{m}^{\star}f,g)=(f,H_{m}^{\star}g),\;\;\; f,g\in \mathcal{D}(H_{m}^{\star}).
$$
That is, $H_{m}^{\star}$ is also symmetric, which gives
$$
                H_{m}^{\star} \preceq (H_{m}^{\star})^{\star}=\overline{H_{m}},
$$
where $\overline{H_{m}}$ is the closure of $H_{m}$. But we alreay knew $H_{m}\preceq H_{m}^{\star}$ because $H_{m}$ is symmetric. Hence $\overline{H_{m}}\preceq H_{m}^{\star}$ because $H_{m}^{\star}$ is closed. Putting these pieces together gives
$$
                   \overline{H_{m}}=H_{m}^{\star}.
$$
In other words, the closure of $H_{m}$ from its domain $\mathcal{C}_{c}^{\infty}$ is selfadjoint, which is to say that $H_{m}$ is essentially selfadjoint. This domain is the same as $W_{2}^{2}$ because
$$
              \|H_{m}f\|=\|f''\|,\;\;(H_{m}f,f) = \|f'\|^{2},\;\;\; f \in \mathcal{D}(H_{m}).
$$
This is an outline of how you know that $H_{m}$ closes to a selfadjoint operator $H_{0}$ with domain $W_{2}^{2}$.
Your problem, however, is a general operator issue.

Theorem: Let $A : \mathcal{D}(A)\subseteq X\rightarrow X$ be a closable linear operator on a Hilbert space $X$, and let $B$ be a bounded linear operator on $X$. Then $A+B$ is closable and $\mathcal{D}(\overline{A})=\mathcal{D}(\overline{A+B})$ and $\overline{A+B}=\overline{A}+B$.

Proof: Suppose $\langle x_{n},Ax_{n}\rangle$ converges in $X\times X$ to some $\langle x,y \rangle$. Then $x_{n} \rightarrow x$ and $Ax_{n} \rightarrow y$. Because $B$ is bounded, then $Bx_n \rightarrow Bx$ and, hence, $\langle x_n, (A+B)x_n\rangle$ converges in $X\times X$ to $\langle x,y+Bx\rangle$. Conversely, if $\langle x_n, (A+B)x_n \rangle \rightarrow \langle x,y\rangle \in X\times X$ then $Ax_n = (A+B)x_n-Bx_n \rightarrow y-Bx$. So the domains of the closures of $A+B$ and of $A$ are the same, and $\overline{A+B}=\overline{A}+B$. There are a few details, but not many. $\;\;\Box$
