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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 !

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    What's the meaning of the bar over the laplacian? I've never seen it.2012-07-08

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$-\bar{\Delta}$ stands for the operator closure of $(-\Delta, C^{\infty}_o(\mathbb{R}^3))$, I suppose. If this is the case then you should:

  1. Apply a Fourier transform to diagonalize $-\Delta$, so that it becomes a multiplication operator in Fourier space;
  2. Ascertain that this multiplication operator is essentially self-adjoint and determine its domain of self-adjointness;
  3. By means of an inverse Fourier transform, deduce from this that the domain of $-\bar{\Delta}$ is Sobolev space $H^2(\mathbb{R}^3)$;
  4. 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.

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    Thanks a lot ! I'll try to go over it on the weekend and try to complete the details myself. If I won't understand something, I'll reply here again. Thanks again !2012-07-08
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    Here is my try to complete all the details: (1) By denoting $U:= - \Delta$ we get from (1) (and from the fact that the unitary operator that diagonlize $U$ is the fourier transform) that $FUF^{-1} = x^2\psi$ (where $F$ stands for the Fourier Transform). (2) The operator above will be essentially self-adjoint if its closure is self-adjoint. But I have no idea how to prove it (or how to determine the domain of self-adjointness of this operator). Hope you can help. (3) I assume you meant here that I should move the "$F$" to the other side and get that $-\Delta$ is located in the Sobole2012-07-13
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    Hi, I've converted Joshua's not-really-answer into a comment. But it got cut-off. Here is the rest: "... get that $-\Delta$ is located in the Sobolev space. Can you please explain me this? (Since we need the closure of this operator...) (4) understandable!"2012-07-13
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    @Wilie Wong: Thanks a lot ! I hope Giuseppe Will be able to help2012-07-13