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I am trying to fill in the details of the following (already discussed in these posts for example and also in Rudin, Reed and Simon, etc) fact: If $\lambda \in \sigma(A)$ is isolated for $A$ self-adjoint, then $\lambda$ is an eigenvalue. I have read the previous posts I linked above but am still seeming to have a fundamental misunderstanding.

The pieces I am trying to put together are as follows:

By the spectral theorem, we have $$ A = QM_\alpha Q^{-1} $$ where \begin{align*} Q: L^2(X,d\mu) &\to \mathcal{H} \ \text{is unitary} \\ X &= \bigcup_k (\mathbb{R}, \mu_k) \\ \mu_k &\text{are Borel measures on $\mathbb{R}$} \\ M_\alpha &= \text{multiplication operator on $L_2(X,d\mu)$} \\ \alpha(x) &= x \ \text{on each copy of $\mathbb{R}$}. \end{align*} (This is how the spectral theorem for unbounded, self-adjoint operators is stated in Spectral Theory by David Borthwick, for example).

We also have functional calculus: for a Borel function $f$ we define $$ f(A) := QM_{f\circ \alpha}Q^{-1} $$ where $Q$ and $\alpha$ are defined as in the above.

This allows us to define projectors: For $E$ a Borel set we define $$ \Pi_E := \chi_E(A) $$ where $\chi_E$ is the characteristic function for the Borel set $E$.

Finally, using the characterization of the eigenvalues of multiplication operators, one can prove that $$ \Pi_{\{\lambda\}} \neq 0 \implies \lambda \ \text{is an eigenvalue}. $$

I have been trying to put these pieces together to prove that an isolated point in the spectrum is an eigenvalue.

The route I have been trying to use is to prove that, if $\lambda$ is an isolated point of the spectrum, then $\Pi_{\{\lambda\}} \neq 0$.

I have tried many things to prove this. Some of them include the following:

  1. playing around with limits of spectral projectors (using the fact that i $f_n\to f$ pointwise and $f_n$ are uniformly bounded, then $f_n(A)v \to f(A)v$ for all $v \in \mathcal{H}$. So in particular since $$ \lim_{n\to \infty} \chi_{[\lambda - \frac{1}{n}, \lambda + \frac{1}{n}]} = \chi_{\{\lambda\}} \quad \text{p.w.} $$ we get that $$ \chi_{[\lambda - \frac{1}{n}, \lambda + \frac{1}{n}]}u \to \chi_{\{\lambda\}}u \quad \text{for all $u \in \mathcal{H}$}. $$ However, this didn't seem to work as the $u \in \mathcal{H}$ that yields a nonzero value of the projector could be different for each $n$. I also don't see how we are using the isolation of $\lambda$ with this approach.

  2. I have tried proving this more directly using the spectral theorem: \begin{align*} (QM_\alpha Q^{-1} - \lambda)u &= (QM_\alpha Q^{-1} - Q\lambda Q^{-1})u \\ &= Q(M_\alpha - \lambda)Q^{-1}u \\ &= Q(M_{\alpha - \lambda})Q^{-1}u \end{align*} We then must show, by the characterization of the eigenvalues for multiplication operators, that $$ \mu(\{x \ : \ \alpha(x) = \lambda x\}) > 0. $$ However, I have not been able to do this and again I don't see how we are using that $\lambda$ is isolated here.

Any hints or advice on how to proceed here would be greatly appreciated.

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2 Answers 2

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The point is that if $\lambda$ is an isolated point of the spectrum, then $\Pi_{\lambda} = \Pi_{(\lambda-\epsilon, \lambda+\epsilon)}$ for $\epsilon>0$ sufficiently small. That $\lambda$ is isolated implies that $\sigma (A) \cap (\lambda - \epsilon , \lambda + \epsilon ) = \{\lambda\}$. Thus by the addivitiy of the PVM, $\Pi_{((\lambda - \epsilon , \lambda + \epsilon ))} = \Pi_{((\lambda - \epsilon , \lambda + \epsilon ) \setminus \{\lambda\})} + \Pi_{(\{\lambda\})} = \Pi_{(\{\lambda\})}$.

So you do not actually need to take any sort of limit. The fact that $\lambda\in \sigma(A)$ implies that $\Pi_{(\lambda-\epsilon, \lambda+\epsilon)}\neq 0$ is enough to prove that $\Pi_\lambda\neq 0$. This can be seen by the multiplication operator representation for $A$. $UAU^{-1}= M_F$, and $\sigma(A) = \text{essran}F$ where $F$ is a multiplication operator on $L^2(M,\mathcal{M},\mu)$. So for every $\epsilon>0$, $\mu(F^{-1}(\lambda-\epsilon, \lambda+\epsilon))>0$. So $\Pi_{(\lambda-\epsilon,\lambda+\epsilon)}=\chi_{F^{-1}(\lambda-\epsilon,\lambda+\epsilon)}\neq 0$.

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    $\begingroup$ That $\lambda$ is isolated implies that $\sigma(A)\cap (\lambda-\epsilon, \lambda+\epsilon) = \{\lambda\}$. Thus by additivity of the PVM, $\Pi((\lambda-\epsilon,\lambda+\epsilon)) = \Pi((\lambda-\epsilon, \lambda+\epsilon)\setminus \{\lambda\}) + \Pi(\{\lambda\}) = \Pi(\{\lambda\})$. $\endgroup$ Commented Dec 8 at 1:01
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    $\begingroup$ No problem! PVM stands for Projection-Valued Measure. It refers to the spectral projections $\Pi(E)$. $\endgroup$ Commented Dec 8 at 1:13
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    $\begingroup$ If $\Pi_{(\lambda-\epsilon, \lambda+\epsilon)\setminus \{\lambda\}}\neq 0$, then there exists $\mu\neq \lambda\in \sigma(A)\cap (\lambda-\epsilon, \lambda+\epsilon)$. This contradicts the assumption that $\lambda$ is the only element of the spectrum in this interval. $\endgroup$ Commented Dec 8 at 1:28
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    $\begingroup$ Suppose that $\sigma(A)\cap (\lambda-\epsilon,\lambda+\epsilon) \setminus\{\lambda\}\neq \varnothing$. Then there exists $\mu\in \sigma(A)\cap (\lambda-\epsilon,\lambda+\epsilon) $, so $\mu$ is in the essential range of $F$. Then the indicator on $F^{-1}(\mu-\epsilon',\mu+\epsilon')$ is nonzero. It follows by the inclusion of sets $(\mu-\epsilon', \mu+\epsilon')\subset (\lambda-\epsilon, \lambda+\epsilon)\setminus\{\lambda\}$ that the associated projection on the latter set is nonzero as well. A good intuition to have is that in general the spectrum is the same as the support of $\Pi$. $\endgroup$ Commented Dec 8 at 2:40
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    $\begingroup$ Shoot sorry yes that was the wrong direction. In the other, use separability. The spectrum being disjoint from that open set means that every point in it is disjoint from the essential range of $F$. So there is some $\epsilon>0$ such that a neighborhood around that point is such that $F^{-1}(\mu\pm \epsilon)$ is a null set. Separability then bounds the measure of the open set by the sum of the measures of such balls which is zero, so the projector must vanish. $\endgroup$ Commented Dec 8 at 4:50
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First, note that the formulation of the spectral theorem for self-adjoint operators you stated is formulated slightly imprecisely and does not apply to nonseparable Hilbert spaces. Another version is formulated below which is sufficient for our purposes.

Fix a measure space $(X, \mathcal{A}, \mu )$ and $\mathbb{K}$ as one of $\mathbb{R}$ or $\mathbb{C}$. We need two definitions.

  • Let $h\colon X\to \mathbb{K}$ be an $\mathcal{A}$-measurable function. We define the essential range of $h$, denoted by ${\rm ess \, ran} \, h$, as the unique closed subset of $\mathbb{K}$ such that for each open subset $U$ of $\mathbb{K}$ we have \begin{equation} \mu (\{x\in X : h(x) \in U \}) > 0 \end{equation} if and only if $U\cap {\rm ess \, ran} \, h \neq \emptyset$. To see that such a closed subset exists, we may take \begin{equation} {\rm ess \, ran} \, h = \{\lambda \in \mathbb{K} : \mu (\{x\in X : |h(x) - \lambda | < \varepsilon \}) > 0 \text{ for all } \varepsilon > 0 \} \end{equation} and verify that it has the desired properties.

  • We say the measure space $(X, \mathcal{A}, \mu )$ is locally finite if for every $E\in \mathcal{A}$ with $\mu (E) > 0$ there is some $F\in \mathcal{A}$ with $F\subseteq E$ such that $0 < \mu (F) < \infty$. Note that every $\sigma$-finite measure space is locally finite. The reason for introducing locally finite measure spaces is to ensure relevant results about multiplication operators and the multiplication operator spectral theorem are also true for nonseparable Hilbert spaces. If the reader is not concerned about this, the locally finite condition may be replaced with $\sigma$-finite everywhere in the rest of this answer.

We now state the results that will be needed. Most of these results can be found in Lectures on Functional Calculus by Haase (link). More precise references, or even proofs, can be provided if needed. We always take $\mathbb{K}$ as a field that is fixed as one of $\mathbb{R}$ or $\mathbb{C}$.

Multiplication operators. Let $(X, \mathcal{A}, \mu )$ be a locally finite measure space and $h \colon X\to \mathbb{K}$ an $\mathcal{A}$-measurable function. We define a linear operator $M_{h}$ on $L_{2}(X, \mu )$ as follows. Let $u, v\in L_{2}(X, \mu )$. Then $u \in D(M_{h})$ and $M_{h}u = v$ if and only if $hu \in L_{2}(X, \mu )$ and $hu = v$ in $L_{2}(X, \mu )$. The operator $M_{h}$ has the following properties.

$({\rm a})$ We have $\sigma (M_{h}) = {\rm ess \, ran} \, h$.

$({\rm b})$ For each $\lambda\in \mathbb{K}$ we have that $\lambda$ is an eigenvalue of $M_{h}$ if and only if \begin{equation} \mu (\{x\in X : h(x) = \lambda \}) > 0 . \end{equation} Furthermore, if $\lambda\in \mathbb{K}$ is an eigenvalue of $M_{h}$, the multiplication operator $M_{1_{[h = \lambda ]}}$ is an orthogonal projection on $H$ onto the eigenspace of $M_{h}$ associated with $\lambda$, where $[h = \lambda] = \{x\in X : h(x) = \lambda \}$.

Multiplication operator spectral theorem. Let $H$ be a Hilbert space over $\mathbb{K}$ and let $A$ be a self-adjoint operator on $H$. There exists a triple $((X, \mathcal{A}, \mu ), \varphi , W)$, known as a multiplication operator representation of $A$, where $(X, \mathcal{A}, \mu )$ is a locally finite measure space, $\varphi \colon X\to \mathbb{R}$ is an $\mathcal{A}$-measurable function and $W\colon L_{2}(X, \mu ) \to H$ is a unitary operator such that \begin{equation} A = W M_{\varphi} W^{-1} . \end{equation} If $H$ is separable, the measure space in the multiplication operator representation of $A$ may be taken as a $\sigma$-finite (or even finite) measure space.

Spectral projections. Let $H$ be a Hilbert space over $\mathbb{K}$, let $A$ be a self-adjoint operator on $H$ and let $((X, \mathcal{A}, \mu ), \varphi , W)$ be a multiplication operator representation of $H$. For every $E\in \mathcal{A}$ we define a bounded orthogonal projection operator $\Pi_{E}$ on $H$ by \begin{equation} \Pi_{E} := W M_{1_{E}\circ \varphi} W^{-1} \end{equation} with the following properties.

$({\rm c})$ For each $E\in\mathcal{A}$ we have $\Pi_{E} \neq 0$ if and only if $\mu (\{x\in X : \varphi (x) \in E \}) > 0$.

$({\rm d})$ For each $\lambda \in \mathbb{K}$ we have that $\lambda$ is an eigenvalue of $A$ if and only if $\Pi_{\{\lambda\}} \neq 0$. Furthermore, if $\lambda\in \mathbb{K}$ is an eigenvalue of $A$ then $\Pi_{\{\lambda\}}$ is the orthogonal projection of $H$ onto the eigenspace of $A$ associated with $\lambda$.

We now combine the above results to show the result you are interested in.

Proposition. Let $A$ be a self-adjoint operator on a Hilbert space $H$ over $\mathbb{K}$. Let $((X, \mathcal{A}, \mu ), \varphi , W)$ be a multiplication operator representation of $H$. Let $\lambda\in\mathbb{K}$ and suppose $\lambda$ is an isolated point of $\sigma (A)$. Then $\lambda$ is an eigenvalue of $A$. Moreover, we have that $\Pi_{\{\lambda\}}$ is orthogonal projection on $H$ onto the eigenspace of $A$ associated with eigenvalue $\lambda$.

Proof. First note that for any $t \in \mathbb{K}$ we have \begin{equation} t I_{H} - A = W (t I_{L_{2}(\mu )} - M_{\varphi}) W^{-1} . \tag{1} \end{equation} As a consequence of $(1)$ combined with $({\rm a})$, we have $\sigma (A) = \sigma (M_{\varphi}) = {\rm ess \, ran} \, \varphi$. Now because $\lambda$ is an isolated point of $\sigma (A)$, there is some $\varepsilon > 0$ such that \begin{equation} \{t \in \mathbb{K} : |t - \lambda | < \varepsilon \} \cap \sigma (A) = \{\lambda \} . \end{equation} Define $U := \{t \in \mathbb{K} : |t - \lambda | < \varepsilon \}$ and $V := \{t \in \mathbb{K} : |t - \lambda | < \varepsilon \} \setminus \{\lambda\}$. Then $U$ and $V$ are open sets in $\mathbb{K}$ with $U\cap \sigma (A) \neq \emptyset$ and $V\cap \sigma (A) = \emptyset$. As $\sigma (A) = {\rm ess \, ran} \, \varphi$, it follows from the definition of the essential range of $\varphi$ that \begin{equation} \mu (\{x\in X : \varphi (x) \in U \}) > 0 \quad \text{and} \quad \mu (\{x\in X : \varphi (x) \in V \}) = 0 . \tag{2} \end{equation} Because $U = V \cup \{\lambda\}$ as a disjoint union, we deduce from $(2)$ that \begin{align*} \mu (\{x\in X : \varphi (x) = \lambda \}) &= \mu (\{x\in X : \varphi (x) \in \{\lambda\} \}) + \mu (\{x\in X : \varphi (x) \in V \}) \\ &= \mu (\{x\in X : \varphi (x) \in U \}) > 0 . \end{align*} Hence $\mu (\{x\in X : \varphi (x) = \lambda \}) > 0$ and consequently by $({\rm c})$ we see that $\Pi_{\{\lambda\}} \neq 0$. It now follows from $({\rm d})$ that $\lambda$ is an eigenvalue of $A$ and that $\Pi_{\{\lambda\}}$ is the orthogonal projection on $H$ onto the eigenspace of $A$ associated with eigenvalue $\lambda$.

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