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 self-adjoint on $H^2(\mathbb{R}^3)$. My idea is to use Kato-Rellich 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 Gagliardo-Niremberg-Sobolev 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 Kato-Rellich theorem gives the thesis. Is there something worng in this reasoning?

  • $\begingroup$ 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

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