Fourier Transforms and the Laplacian I need your help in the following question:
"Use Fourier transforms to prove that the domain of $H_0 := - \bar{\Delta} $ on $ L^2 (\mathbb{R}^3)$ consists entirely of continous bounded functions. If $V$ is a non-negative potential which is not $L^2 $ when restricted to any non-empty bounded open subset of $\mathbb{R}^3$ , prove that $Dom(H_0) \cap Dom(V) = \{0 \} $.
Help in any one of the two parts will be greatfuly acknowledged! 
Thanks in advance !  
 A: $-\bar{\Delta}$ stands for the operator closure of $(-\Delta, C^{\infty}_o(\mathbb{R}^3))$, I suppose. If this is the case then you should: 


*

*Apply a Fourier transform to diagonalize $-\Delta$, so that it becomes a multiplication operator in Fourier space; 

*Ascertain that this multiplication operator is essentially self-adjoint and determine its domain of self-adjointness;

*By means of an inverse Fourier transform, deduce from this that the domain of $-\bar{\Delta}$ is Sobolev space $H^2(\mathbb{R}^3)$;

*Apply the Sobolev imbedding theorem to conclude that this space is imbedded into a space of bounded and continuous functions (Hölder continuous, actually).


P.S.: I had not seen the second part of the question. For this you need to show that, if $\psi \in H^2(\mathbb{R}^3)$ is such that $V(x)\psi(x)\in L^2(\mathbb{R}^3)$, then $\psi\equiv 0$. Again, use the fact that $\psi$ is continuous and argue by contradiction: if $\psi \ne 0$ then there exists a bounded open subset $\Omega$ of $\mathbb{R}^3$ such that $\lvert \psi(x)\rvert \ge m>0$ for every $x \in\Omega$. From the fact that $V(x)\psi(x)\in L^2$ you infer from this $V\in L^2(\Omega)$, a contradiction.
