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Given a Hilbert space $\mathcal{H}$.

Consider a Hamiltonian: $$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$

Denote resolvent set: $$\rho(H):=\{z\in\mathbb{C}:(z-H)^{-1}\in\mathcal{B}(\mathcal{H})\}$$

Define the ratio: $$\eta_z:\sigma(H)\to\mathbb{C}:\lambda\mapsto\tfrac{|z-\lambda|}{1+|\lambda|}$$

Then one has estimates: $$z\in\rho(H):\quad\delta_-(z)\leq\eta_z\leq\delta_+(z)$$

Moreover one has: $$z\in\rho(H):\quad\delta_\pm(z)=\delta_\pm(\overline{z})$$

How to prove these?

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  • $\begingroup$ I'm not sure what you are asking. Presumably $m_z$ and $M_z$ are constants that depend on $z$ but not on $\lambda$, but what is $\lambda$? When you ask "how to check these," are you asking about a proof? a computation? $\endgroup$
    – hardmath
    Commented Mar 26, 2015 at 18:40
  • $\begingroup$ @hardmath: Oh yes these are supposed to be constants w.r.t. $\lambda$ that still depend on $z$. $\lambda$ is meant to lie within the spectrum. (I've added that.) I'm asking for a quick 'proof'. Hope it is clearer now. Thanks for pointing me out on this!!! $\endgroup$
    – C-star-W-star
    Commented Mar 27, 2015 at 17:24

1 Answer 1

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This has very little to do with Hilbert spaces or Hamiltonians. If $\sigma(H)$ is a closed set in the complex plane and $z \notin \sigma(H)$, then $f(\lambda) = \dfrac{|z - \lambda|}{1 + |\lambda|}$ is a continuous positive function on $\mathbb C$, with limit $1$ as $|\lambda| \to \infty$, and its only zero is at $\lambda = z$. Therefore it is bounded away from $0$ on $\{\lambda: |\lambda - z| \ge d(z, \sigma(H))\}$.

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  • $\begingroup$ I great so that gives the bound, thanks!! :) (Yeah I thought it would be nicer to provide some context.) $\endgroup$
    – C-star-W-star
    Commented Mar 23, 2015 at 19:47
  • $\begingroup$ In fact, your check reveals more... Do you mind adding? (See update above.) $\endgroup$
    – C-star-W-star
    Commented Mar 24, 2015 at 18:54

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