Let's consider a potential $V(x)\in L^3(\mathbb{R}^3)$. I want to know if the following Hamiltonian $$\Delta+V(x)$$ is selfadjoint on $H^2(\mathbb{R}^3)$. My idea is to use KatoRellich theorem; so for $f\in D(\Delta)=H^2(\mathbb{R}^3)$ we have $$\Vert Vf\Vert_{L^2(\mathbb{R}^3)}\leq\Vert V\Vert_{L^3(\mathbb{R}^3)}\Vert f\Vert_{L^6(\mathbb{R}^3)}$$ Now I can use GagliardoNirembergSobolev inequality to get $$\Vert f\Vert_{L^6(\mathbb{R}^3)}\leq C\Vert\Delta f\Vert_{L^2(\mathbb{R}^3)}^{\frac{1}{2}}\Vert f\Vert_{L^2(\mathbb{R}^3)}^{\frac{1}{2}}$$ Now I can use Young inequality $$\Vert f\Vert_{L^6(\mathbb{R}^3)}\leq C(\epsilon\Vert\Delta f\Vert_{L^2(\mathbb{R}^3)}+\frac{1}{\epsilon}\Vert f\Vert_{L^2(\mathbb{R}^3)})$$ So we have that $V$ is bounded with respect to the Laplacian and so KatoRellich theorem gives the thesis. Is there something worng in this reasoning?
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$\begingroup$ Nope, every step is as legit as it can be. Plus, you might be interested by this post on MSE. $\endgroup$ – Hachino Feb 3 '15 at 12:58

$\begingroup$ Thank you for sharing the question, I do not know how you got the implication from Gagliardo Nuremberg Sobolev inequality, the way I do your thing is following $$f_6^2\leq C\nabla f_2^2$$ then I use young inequality in Fourier space, for R.H.S. as $\nabla f_2^2=\xi \hat f(\xi)^2\leq C(\epsilon \xi^2 \hat f(\xi)^2+\frac{1}{\epsilon}\hat f (\xi)^2) $. RHS gives you laplacian when you go back to Original space. I would be happy to see if you could elaborate how you got your conclusion. $\endgroup$ – Harish Mar 20 '15 at 11:47