Let $V \in C_{b}^{1}(\mathbb{R}, \mathbb{R})$ be a differentiable real-valued function defined on $\mathbb{R}$ bounded with its first derivative. Consider the Hamilton's operator $H$ such that:
$$(Hf)(x)=V(x)f(x)-\frac{1}{2}f''(x)$$ Operator $H$ has a closure and $\overline{(H, C_0^{\infty}(\mathbb{R}))} = (H, W_2^2(\mathbb{R}))$ (see 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).
The question. Is it possible to show this fact using Fourier transform and will it be an easier way than showing it ''by hands''?
What gives the idea. In the book ''Methods of modern mathematical physics'' by M.Reed and B.Simon, volume 2, the Sobolev space is defined using Fourier transform. After that they show that this definition is equivalent to a common definition of Sobolev space (using week derivatives). Adopting Reed&Simon's definition to our case we get that $W_2^2(\mathbb{R})$ is a subspace of $L^2(\mathbb{R})$ such that $f\in W_2^2(\mathbb{R})$ iff $f$ is measurable and $$ \int\limits_{\mathbb{R}}(1+x^2)^2|\hat{f}(x)|^2dx < \infty $$ In other words, function $f\in W_2^2(\mathbb{R})$ iff $x\mapsto (1+x^2)\hat{f}(x) \in L^2(\mathbb{R})$.
Notation. Here $C_0^\infty(\mathbb{R}) \subset L^2(\mathbb{R})$ is a linear space of smooth complex-valued functions with compact support defined on $\mathbb{R}$ and $W_2^2(\mathbb{R}) \subset L^2(\mathbb{R})$ is the Sobolev space, i.e. the linear space of all functions $f\in L^2(\mathbb{R})$ such that $f'\in L^2(\mathbb{R})$ and $f''\in L^2(\mathbb{R})$ where derivitives considered in a week sence.
Let $A=-\frac{d^{2}}{dx^{2}}$ on the domain $\mathcal{D}(A)=W^{2}_{2}(\mathbb{R})$. The restriction $A_{0}$ of $A$ to $\mathcal{C}_{0}^{\infty}(\mathbb{R})$ has a closure $\overline{A_{0}}=A$. In other words, the closure of the graph of $A_{0}$ is equal to the graph of $A$.
If $V$ is any bounded measurable function on $\mathbb{R}$, then $V$ defines a bounded linear operator on $L^{2}(\mathbb{R})$ defined by $$ (Vf)(x) = V(x)f(x). $$ This is bounded because, if $|V(x)| \le M$ for some constant $M$ and for all $x\in\mathbb{R}$, then $$ \|Vf\|_{L^{2}}^{2}=\int_{-\infty}^{\infty}|V(x)|^{2}|f(x)|^{2}dx \le M^{2}\int_{-\infty}^{\infty}|f(x)|^{2} = M^{2}\|f\|_{L^{2}}^{2}. $$ So $V \in \mathcal{L}(L^{2}(\mathbb{R}))$ and the operator norm is bounded by $M$.
The closure of $A_{0}+V$ is $\overline{A_{0}}+V=A+V$. This is a general fact about operators that follows because $V$ is a bounded operator on the underlying Hilbert space $X=L^{2}(\mathbb{R})$ and because the closure $\overline{A_{0}}$ of $A_{0}$ is $A$. The proof of this fact follows from the observation that if $f_n \in \mathcal{D}(A_0)$, then $$ \langle f_n , Af_n \rangle \in X\times X\rightarrow \langle f,g \rangle \\ \implies \langle f_n , Af_n + Vf_n \rangle \in X\times X \rightarrow \langle f, g+Vf\rangle. $$ and $$ \langle f_n , Af_n + Vf_n \rangle \in X\times X \rightarrow \langle f, h\rangle \\ \implies \langle f_n , Af_n \rangle \in X\times X\rightarrow \langle f,h-Vf \rangle $$