Let $\mathbb{H}$ be a Hilbert space, $A$ a self-adjoint operator with domain $D_{A}$, $R_{A}$ the resolvent of $A$, and $z$ a point in the resolvent set $\rho(A)$. How could you prove the inequality \begin{equation} ||R_{A}(z)|| \leq 1/ d(z,\sigma(A)), \end{equation} where $\sigma(A)$ is the spectrum of $A$, and $d(z,\sigma(A))$ the distance of $z$ from $\sigma(A)$? I found this inequality in Hislop & Sigal, Introduction to Spectral Theory, Sect 5.2, where they reference Reed and Simon, Methods of Modern Mathematical Physics, vol. I, but I could not find a proof in that book. Thank you very much in advance.

PS I just note here that since for any closed operator \begin{equation} ||R_{A}(z)|| \geq 1/ d(z,\sigma(A)), \end{equation} (just note that all the point $w$ such that $|z-w| < ||R_{A}(z)||$ belong to $\rho(A)$), the above inequality must actually hold with equality.


1 Answer 1


For a bounded normal operator $N$, the norm and spectral radius of $N$ are the same. That is, $\|N\|=\sup_{\lambda\in\sigma(N)}|\lambda|$.

Let $\lambda \notin \sigma(A)$. Assume $A$ is unbounded. Then $(A-\lambda I)^{-1}$ is bounded and normal, with $$ \sigma((A-\lambda I)^{-1})=\frac{1}{\sigma(A)-\lambda}\cup\{0\}=\left\{ \frac{1}{\mu-\lambda} : \mu\in\sigma(A) \right\}\cup\{0\}. $$ Therefore, $$ \|(A-\lambda I)^{-1}\|=\sup_{\xi\in\sigma((A-\lambda I)^{-1})}|\xi| = \sup_{\mu\in\sigma(A)}\frac{1}{|\mu-\lambda|} = \frac{1}{\mbox{dist}(\lambda,\sigma(A))}. $$

  • $\begingroup$ It looks much more simple than I could ever imagine! I have only a doubt about the first line. I only know that $\mu \in \mathbb{C}$ belongs to $\sigma(A)$ if and only if $(\mu - \lambda)^{-1}$ belongs to $\sigma(R_{A}(\lambda))$. But still $\sigma(R_{A}(\lambda))$ could contain zero. Or am I wrong? This in any case shouldn't change the proof. Thank you very much for your invaluable help. $\endgroup$ Commented Jul 10, 2016 at 21:43
  • $\begingroup$ The result I quoted in my previous comment is from Schmudgen, Unbounded Self-Adjoint Operators on Hilbert Space, Proposition (2.10). $\endgroup$ Commented Jul 10, 2016 at 22:20
  • $\begingroup$ @MauryBarbato : If $A$ is unbounded, then $0$ is in the spectrum of $(A-\lambda I)^{-1}$, which is the only possible point being left out of $\frac{1}{\sigma(A)-\lambda}$, but $0$ is in the closure of the set $\frac{1}{\sigma(A)-\lambda}$ if $A$ is unbounded. I invoked closure to avoid a lengthy discussion of that one point $0$. $\endgroup$ Commented Jul 10, 2016 at 22:28
  • $\begingroup$ @MauryBarbato : Actually I think you have equality of the closure and the spectrum, which also gives equality for the norm in the final line. So that point was worth you mentioning it. I changed everything to equality (both of spectrum and of norm.) $\endgroup$ Commented Jul 10, 2016 at 22:34
  • $\begingroup$ If $A$ is bounded, then $0$ is not in $\sigma(R_{A}(\lambda))$, and since $\sigma(R_{A}(\lambda))$ is closed, there is a neighborhood $V$ of $0$ such that $\sigma(R_{A}(\lambda)) \cap V = \emptyset$. Consider now $A$ undbounded. Then $0$ is in $\sigma(R_{A}(\lambda))$. So to conclude that $\sigma(R_{A}(\lambda)) = \text{closure} \left( \frac{1}{\sigma(A) - \lambda} \right)$, it is enough to show that if $A$ is unbounded, then $0$ is not an isolated point of $\sigma(R_{A}(\lambda))$, as you claim. Could you give a proof or a reference for this statement? Thank you very much. $\endgroup$ Commented Jul 11, 2016 at 12:56

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