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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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    First you say, that you look at unbounded $A$, then you write $\|A\| = 1$?! – 2017-01-06
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    What is the actual question being asked here? – 2017-01-06
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    @martini, Is it not true that self-adjoint operators have operator norm = 1? – 2017-01-06
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    Now, it is not, there are unbounded self-adjoint operators, which (by the very definition of being unbounded) do not have a finite norm. – 2017-01-06
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    @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? – 2017-01-06
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    Which book is "this book"? It is always better to include your sources in the question. It is easier to give fitting answers then ... – 2017-01-06
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    @martini GTM: Quantum Theory for Mathematicians, chapter 9 – 2017-01-06
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    As far as I know, Arzela-Ascoli only holds on compact spaces, the general Hilbert space does not have big compact sets. – 2017-01-06
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    It seems like the natural way to approach this requires me to figure out that somehow this sequence of sequences is uniformly convergent. I reached of a theorem I had seen about uniform convergence and erred optimistically. I wonder if it still turns out that I can get uniform convergence another way. – 2017-01-06

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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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    Shoot. Is it really that easy? – 2017-01-06
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    Yes, it is that easy. – 2017-01-06