Definition The group of unitary operators $u(\theta)$ on $L^2(\mathbb R^3)$ given by $(u(\theta)\Phi)(r) =\Phi(\theta) \equiv e^{\frac{3\theta}{2}}\Phi(e^\theta r)$ is called the group of dilation operators on $\mathbb R^3.$

Let $H$ be a compact operator and $R(z) \equiv(H-z)^{-1}$ the resolvent. Define $R(z,\theta)$ by $R(z,\theta) \equiv u(\theta)R(z)u(\theta)^{-1}$. Let us define also $\mathcal O \equiv \{\theta \in \mathbb C: (u(\theta)\Phi)(r)$ for $\theta \in \mathbb R$ has an analytic continuation $ \}$

In this article A Class of Analytic Perturbations for One-body Schrδdinger Hamiltonians they were able to show that the function $\Psi_z(\theta)=(\phi(\theta) ,R(z,\theta)\phi(\theta))$ is meromorphic in $z$ for $z \in \mathcal C^{++} \equiv \{z \in \mathbb C : Im \ z > 0 ,\ Re \ z > 0 \} $ and $\theta \in \mathcal O^\epsilon \equiv \{\theta \in \mathcal O: Im \ \theta > \epsilon \}$

Now since for $\theta \in \mathbb R,\ u(\theta)$ is unitary we have that

$\Psi_z \equiv (\Phi,R(z)\Phi)=\Psi_z(\theta)$

from this they claim that $\Psi_z(\theta)$ that for fixed $z \in \mathcal C^{++}$ and $\theta \in \mathcal O^\epsilon$ the function $\Psi_z(\theta)$ is constant in $\theta$.

My question is why is $\Psi_z(\theta)$ constant for fixed $z \in \mathcal C^{++}$ and $\theta \in \mathcal O^\epsilon$ ?


1 Answer 1


As I understand it, they have shown:

  1. That $\Phi_z(\theta)$ is meromorphic in a region containing part of the real line.
  2. That $\Phi_z(\theta)$ is constant on the real line (using unitarity of $u(\theta)$ for real $\theta$).

Hence by analytic continuation, $\Phi_z(\theta)$ must be constant over the entire region connected to the real line where it is meromorphic.

  • $\begingroup$ So if a meromorphic function is constant in a interval than i will be constant in all its domain? $\endgroup$ Jan 12, 2020 at 3:07
  • 1
    $\begingroup$ @amiltonmoreira In all of the domain which is connected to the interval, yes. $\endgroup$
    – Yly
    Jan 12, 2020 at 3:08
  • $\begingroup$ I gave a proof as question here math.stackexchange.com/questions/3505785/… $\endgroup$ Jan 12, 2020 at 3:09
  • 1
    $\begingroup$ @amiltonmoreira Yes, your proof looks basically correct. It all follows from uniqueness of analytic continuation. $\endgroup$
    – Yly
    Jan 12, 2020 at 3:13
  • 1
    $\begingroup$ @amiltonmoreira Yes. The wikipedia page on meromorphic functions says so explicitly in the second paragraph: en.wikipedia.org/wiki/Meromorphic_function $\endgroup$
    – Yly
    Jan 12, 2020 at 3:18

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