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There's an exercise in this book that asks us to prove the closure of the spectrum of self-adjoint (unbounded) operators on a Hilbert space, using a theorem that elements in such a spectrum are almost-eigenvalues (iff); that is,

$$ \lambda\in\sigma(A)\iff\lim_{n\to\infty}\frac{\|(A-\lambda I)\psi_n\|}{\|\psi_n\|}=0,\text{ for some }\psi_n\to\psi $$

I have a proof, but I'm not sure that it's valid because I'm uncomfortable with some of the theorems used. And given the ease of other problems in this book, it seems like I'm making this too complicated.

Suppose that $\lambda_n\to\lambda$, $\lambda_n\in\sigma(A)$. We know that all $\lambda_n\in\mathbb R$ for self-adjoint operators. I want to show that $\lambda\in\sigma(A)$, then I will be done.

Firstly, $\lambda_n\leq M$ and $1=\|A^*A\|=\|A\|^2\implies\|A\|=1$, so that $\|A-\lambda_n I\|\leq\|A\|+\|\lambda_n I\|\leq 1+M$. Therefore the family of operators $\|A-\lambda_n I\|$. Therefore this family is equicontinuous.

I also know that this sequence converges pointwise to $A-\lambda I$ because for any $\psi\in\mathbb H$ and $\varepsilon>0$, there's certainly some $i$ for which $\|(A-\lambda_i I)\psi-(A-\lambda I)\psi\|=\|(\lambda_i-\lambda)\psi\|\leq\|\lambda_i-\lambda\|\|\psi\|<\varepsilon$. Now equicontinuity and pointwise convergence give me uniform convergence by the Arzelà–Ascoli theorem.

Given $\varepsilon_1, \varepsilon_2>0$, I can fix a tail so that

$$ \frac{\|(A-\lambda I)\psi_{i,i}\|}{\|\psi_{i,i}\|}\leq\frac{\|(A-\lambda _iI)\psi_{i,i}\|}{\|\psi_{i,i}\|}+\frac{\|(\lambda_iI-\lambda I)\psi_{i,i}\|}{\|\psi_{i,i}\|}\leq\varepsilon_1+\varepsilon_2, $$

where $\psi_{i,i}$ is the $\psi$ in the $i$-th term of the sequence gauranteed from $\lambda_i$ being an eigenvalue. This implies that $\lambda$ is also an almost-eigenvalue. QED.

I appreciate any feedback.

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    $\begingroup$ First you say, that you look at unbounded $A$, then you write $\|A\| = 1$?! $\endgroup$ – martini Jan 6 '17 at 9:15
  • $\begingroup$ What is the actual question being asked here? $\endgroup$ – Aweygan Jan 6 '17 at 9:20
  • $\begingroup$ @martini, Is it not true that self-adjoint operators have operator norm = 1? $\endgroup$ – T.J. Gaffney Jan 6 '17 at 9:20
  • $\begingroup$ Now, it is not, there are unbounded self-adjoint operators, which (by the very definition of being unbounded) do not have a finite norm. $\endgroup$ – martini Jan 6 '17 at 9:22
  • $\begingroup$ @martini, without boundedness, can I get uniform convergence? Without uniform convergence, can I get convergence of that final sequence? Basically can any of this be salvaged? $\endgroup$ – T.J. Gaffney Jan 6 '17 at 9:25
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You can argue almost directly: Now let $\epsilon > 0$ be given. Choose $n \in \mathbf N$ with $\def\abs#1{\left|#1\right|}\abs{\lambda_n - \lambda} < \frac \epsilon 2$. By part (1) of the statement, there is a $\psi \in D(A)$ such that $$ \def\norm#1{\left\|#1\right\|} \norm{(A- \lambda_nI)\psi} < \frac\epsilon 2 \norm{ \psi}$$ We have: \begin{align*} \norm{(A- \lambda I)\psi} &= \norm{(A - \lambda_n I)\psi + (\lambda_n - \lambda)\psi}\\ &\le \norm{(A - \lambda_n I)\psi} + \abs{\lambda_n - \lambda}\norm\psi\\ &< \frac \epsilon 2 \norm\psi + \frac\epsilon 2 \norm\psi\\ &= \epsilon \norm\psi. \end{align*} That proves the statement.

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  • $\begingroup$ Shoot. Is it really that easy? $\endgroup$ – T.J. Gaffney Jan 6 '17 at 9:49
  • $\begingroup$ Yes, it is that easy. $\endgroup$ – martini Jan 6 '17 at 10:00

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