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I'm currently studying S. Albeverio's book "Solvable models in quantum mechanics" where some technical things are used that I don't fully understand. It is a general technical operator theory question, but I will introduce the setting.

General setting:

Looking at the Hamiltonian $-\Delta + V$ with the underlying Hilbert space $\mathrm{L}^2(\mathbb{R}^3)$, where $V$ is a real potential, the aim in the first chapter of the book is to approximate a $\delta$-Potential in 3D by scaling $V$. We denote $v:=|V|^{1/2}$, where the potential $V$ is an element of the Rollnik-class, i. e. real functions for which $$\int_{\mathbb{R}^6}\frac{|V(x)||V(y)|}{|x-y|^2}dxdy < \infty$$ The operator $vG_0 v: \mathrm{L}^2(\mathbb{R}^3)\rightarrow \mathrm{L}^2(\mathbb{R}^3)$ defined by the kernel $$(vG_0 v)(x,y)=\frac{v(x)v(y)}{4\pi|x-y|}$$ plays an important role in this setting (again $v:=|V|^{1/2}$). Note that the kernel is pointwise positive and the Rollnik-condition ensures that $vG_0v$ is Hilbert-Schmidt. Furthermore $G_0$ denotes the Operator given by convolution with the fundamental solution of the Laplace operator, i.e. $$G_0(x,y)=\frac{1}{4\pi|x-y|},$$ and $G_0(-\Delta \varphi)=\varphi$ for all $\varphi \in C_c^\infty$.

The (technical) problem:

On p.21 and p.22 he uses the fact, that $(f,vG_0vf)$, $f\in\mathrm{L}^2(\mathbb{R}^3)$, can be written as $$(f,vG_0vf)=\Vert G_0^{1/2}vf\Vert^2,$$ so in a sense he uses that $vG_0v$ is a positive Operator and can be written as $vG_0v=vG_0^{1/2}G_0^{1/2}v=(G_0^{1/2}v)^*(G_0^{1/2}v)$. Also he uses that $\Vert G_0^{1/2}vf\Vert^2=0$ implies $vf=0$.

I know that if an bounded linear operator $A$ is positive, it can be decomposed as $A=B^*B$ with $B$ also bounded. But I'm having great trouble in understanding how $G_0^{1/2}v$ is defined, as the operators $v$ (multiplication) and $G_0$, which $vG_0v$ is composed of, are unbounded. Why does $G_0^{1/2}v$ even exist and why is it bounded? How can I conclude $(f,vG_0vf)=\Vert G_0^{1/2}vf\Vert^2=0$ implies $vf=0$ and in what sense? Why is $vG_0v$ (obviously?) positive?

I thought about this a long time but couldn't come up with a satisfying solution. I would appreciate your help and comments!

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  • $\begingroup$ Kato pp. 281-334 discusses roots of unbounded operators. $\endgroup$ Dec 17, 2018 at 17:49

1 Answer 1

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Theorem 9.8 in

Elliott H. Lieb und Michael Loss. Analysis. Second Ed. Bd. 14. Graduate Studies in Mathematics. American Mathematical Society, 2001.

gives a important result regarding this issue. It resolved all my doubts.

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