I asked this question on math.stackexchange.com two weeks ago but got no answers so far and I got no clues from literature, so maybe someone here knows a reference. I hope it is ok to ask this question in this forum again.

I'm currently studying S. Albeverio's book "Solvable models in quantum mechanics" where one technical things is used that I don't fully understand. I will introduce the setting first:

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 $\varphi \in \mathcal{S}$.

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 $vf=0$.

So if I understand correctly he uses that the unbounded convolution operator $G_0$ is exactly the same as the inverse laplacian $(-\Delta)^{-1}=\mathcal{F}^{-1} 1/|p|^2 \mathcal{F}$ defined by functional calculus, but this requires $$D(G_0)=D((-\Delta)^{-1})$$ and $$G_0\phi=(-\Delta)^{-1} \phi, \quad \forall \phi \in D(G_0)=D((-\Delta)^{-1}).$$ Then obviously $G_0^{1/2}=\mathcal{F}^{-1} 1/|p| \mathcal{F}$. I found this to be commonly used in various papers and textbooks, but never explained in detail. I see that both operators conincide on the schwartz functions, but as $\mathcal{S}\subsetneq D((-\Delta)^{-1}) $ why does this equality of convolution operator and fourier multiplier extend?

By Michael Cwikel, "Weak type estimates for singular values and the number of bound states of Schrödinger operators" (1977) it follows that $\mathcal{F}^{-1} 1/|p| \mathcal{F} v = (-\Delta)^{-1/2}v$ is bounded, so the remaining parts of the statement should follow then.

  • $\begingroup$ I can't say anything for sure without checking the original source, but I suspect two things here: (1) The authors did not think about these issues as carefully as you did; (2) $G_0$ may just be notation for $(-\Delta)^{-1}$, and we do know it's convolution on nice functions. Then these issues disappear, if you use the fact (that you already alluded to) that the Fourier transform gives a spectral representation of the free Laplacian. $\endgroup$ – Christian Remling Jan 3 at 17:18
  • $\begingroup$ In the book they define G_k=(-\Delta-k^2)^{-1} which can be written as a convolution operator with kernel e^{ik|x|}/(4pi|x|) on L^2 for Im k>0, as (|p|^2-k^2)^{-1} is bounded and L^2. As far as I can see, they never explicitly define G_0 on the few pages before they suddenly use the notation vG_0v for the operator with the kernel I provided above. $\endgroup$ – MrMatzetoni Jan 4 at 11:27

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.