Let $V_1$ and $V_2\in L^2(\mathbb{R}^3)$ be real-valued functions, then $V_1(x_1)$ and $V_2(x_2)$ can be viewed as multiplicative operators in $L^2(\mathbb{R}^6)$ by $V_i(u) = V_i(x_i)u(x_1,x_2)$. Prove that $-\Delta+V_1(x_1)+V_2(x_2)$ is essentially self-adjoint in $C_0^\infty(\mathbb{R}^6)$.
I want to use fourier transforms to establish that for any $a>0$, then there exists $b>0$ s.t. $|\!|u|\!|_\infty\leq a|\!|\Delta u|\!|_2+b|\!|u|\!|_2$ first, then I can use some theorem to solve it. But I found that for $\mathbb{R}^n$, when $n>3$, then $\frac{1}{1+|\xi|^2}$ is not in $L^2(\mathbb{R}^n)$. Then I cannot establish the inequality I want. Do you have some ideas?
