According to Reed and Simon, Methods of Mathematical Physics II Section X.5 Example 1 (pg. 195), the operator $$A:= -\Delta+x^{2}-x^{4}$$ is not essentially self-adjoint on $C_{c}^{\infty}(\mathbb{R})$, the space of $C^{\infty}$ functions with compact support. I am trying to understand the flaw in the reasoning of the following argument for showing that $A$ is essentially self-adjoint on $C_{c}^{\infty}(\mathbb{R})$.
Since $A$ is symmetric, $A$ is closable and we denote its closure by $\bar{A}: D(\bar{A})\rightarrow L^{2}(\mathbb{R})$, where $$D(\bar{A}) = \{f\in L^{2}(\mathbb{R}) : ((-\Delta+x^{2}-x^{4})f)_{dist} \in L^{2}(\mathbb{R})\}$$ In words, $D(\bar{A})$ consists of all $L^{2}$ functions such that the distribution $Af$ is given by the integral action of a (unique) $L^{2}$ equivalence class.
To see that $\bar{A}:D(\bar{A}) \rightarrow L^{2}(\mathbb{R})$ is self-adjoint, we argue as follows. Suppose $g\in D(\bar{A}^{*})$. If $f\in C_{c}^{\infty}(\mathbb{R})\subset D(\bar{A})$, then by definition of $D(\bar{A}^{*})$, $$\langle{f,\bar{A}^{*}g}\rangle_{L^{2}} = \langle{\bar{A}f,g}\rangle_{L^{2}}=\langle{(-\Delta+x^{2}-x^{4})f,g}\rangle_{L^{2}}$$ $(-\Delta+x^{2}-x^{4}g)$ is a well-defined distribution in $\mathcal{D}'(\mathbb{R})$, and there by definition of differentiation and multiplication of distributions, we have that $$\langle{(-\Delta+x^{2}-x^{4})f,g}\rangle_{L^{2}} = \langle{f,(-\Delta+x^{2}-x^{4})g}\rangle_{\mathcal{D}-\mathcal{D}'},$$ where $\langle{\cdot,\cdot}\rangle_{\mathcal{D}-\mathcal{D}'}$ denotes the duality pairing. Since the preceding holds for all $f\in C_{c}^{\infty}(\mathbb{R})$, we conclude that $$\bar{A}^{*}g=(-\Delta +x^{2}-x^{4})g \in \mathcal{D}'(\mathbb{R}).$$ Sincere $\bar{A}^{*}g\in L^{2}(\mathbb{R})$, we have by definition of $D(\bar{A})$, that $\bar{A}^{*}g\in D(\bar{A})$.
Looking back to the preceding argument, I am guessing that the error in my reasoning is that $$D(\bar{A})\neq \{f\in L^{2}(\mathbb{R}) : ((-\Delta+x^{2}-x^{4})f)_{dist} \in L^{2}(\mathbb{R})\}$$ It seems clear that $D(\bar{A})$ is a subset of the set on the RHS, but it does not seem so obvious to me how to show the reverse inclusion.