How is a spectral subspace of a bounded linear operator defined?

Question

Let $A$ be a bounded linear operator on some Hilbert space. In a previous question (How to interpret spectral projections?) I learned that the spectral projectors, which are defined using the Borel functional calculus applied to the indicator function: $$P_\Omega = \chi_\Omega(A), \quad \Omega \subset \mathbb{R} \text{ is a Borel set,}$$ have the interpretation of projecting onto the closed linear span of the eigenvectors associated to eigenvalues in $\Omega$ (including generalized eigenvectors associated to points in the continuous spectrum). A similar statement is also mentioned in Hall's Quantum Theory for Mathematicians:

In cases where $A$ does not have a true orthonormal basis of eigenvectors, we would like the spectral theorem to provide a family of projection operators $P_E$, one for each Borel subset $E \subset \mathbb{R}$, which will allow us to define probabilities as in (6.2). We will call these projection operators spectral projections and the associated subspaces $V_E$ spectral subspaces. (Thus, $P_E$ is the orthogonal projection onto $V_E$.) Intuitively, $V_E$ may be thought of as the closed span of all the generalized eigenvectors with eigenvalues in $E$.

and

Given a bounded self-adjoint operator $A$, we hope to associate with each Borel set $E \subset \sigma(A)$ a closed subspace $V_E$ of $H$, where we think intuitively that $V_E$ is the closed span of the generalized eigenvectors for $A$ with eigenvalues in $E$.

I am having some trouble understanding the spaces $V_E$ and what exactly they consist of. If $E$ only contains points in the point spectrum then $V_E$ is simply the span of the associated eigenspaces, but what if we also have points in the continuous spectrum?

Taking an example from Hall's book, consider the multiplication operator $M$ acting on $L^2(\mathbb{R}$) by $$Mf(x) = xf(x).$$ Since $M$ is self-adjoint its residual spectrum is empty. It can be shown that the point spectrum of $M$ is also empty and that its continuous spectrum is $\mathbb{R}$. Thus, $$\sigma(M)= \sigma_c(M) = \mathbb{R}.$$ In this case the generalized/approximate eigenvectors are $\delta$-functions which clearly do not belong in $L^2(\mathbb{R})$, so what exactly do the spaces $V_E$ look like in this example? Hall says

If we think that the generalized eigenvectors for $M$ are the distributions $\delta(x-\lambda)$, $\lambda \in \mathbb{R}$, then we may make an educated guess that the spectral subspace $V_E$ should consist of those functions that are "supported" on $E$, that is, those that are zero almost everywhere on the complement of $E$. (A superposition of the "functions" $\delta(x-\lambda)$ with $\lambda \in E$, should be a function supported on $E$.)

The spectral projection $P_E$ is then the orthogonal projection onto $V_E$, which may be computed as $$P_E\varphi = 1_E \varphi,$$ where $1_E$ is the indicator function of $E$.

but I did not fully understand this passage. What do the $V_E$ look like, both in this example and in general?

Answer 1

For simplicity consider the multiplication operator $M$ on $L^2[0,1]$.

Define a projection valued measure $\mu$ by $\mu(A) \psi =\chi_{A} \psi$ for $ A \in \mathcal{B}([0,1])$ and $\psi \in L^2[0,1]$.

Let $\varphi , \psi \in L^2$ arbitrary. Then for any $ A \in \mathcal{B}([0,1])$ $$ \mu_\varphi^\psi (A) := \langle \varphi , \mu(A) \psi\rangle = \int_0^1 \chi_{A} (x)\bar{\varphi}(x) \psi(x) d x . $$ And so $ \mu_\varphi^\psi $ has the density $\bar{\varphi} \psi $ with respect to the Lebesgue measure on $[0,1]$.

Therefore $$ \int_{ [0,1]} \lambda d\mu_\varphi^\psi (\lambda) = \int_0^1 \lambda \bar{\varphi}(\lambda ) \psi(\lambda) d \lambda = \langle \varphi, M \psi \rangle $$ showing that $\mu$ is the spectral measure of $M$.

For $E \in \mathcal{B}([0,1])$ the spectral subspace $V_E$ is defined by $$V_E = \mathrm{ran} \, \mu (E) = \{ \psi \in L^2 : \exists \varphi \in L^2 : \psi = \chi_E \varphi \} = \{ \psi \in L^2 : \psi|_{[0,1]\setminus E } = 0 \},$$ which is exactly the expression in your post.

There is a big difference between "generalized eigenvectors" (in the distributional sense, which is what the author is reffering to here) and "approximate eigenvectors" (sequence of vectors in the Hilbert space with certain properties). They are not the same. Of course no distribution is in the spectral subspace.

The author is merely guessing from the distributional eigenvectors how the spectral subspaces might look. Perhaps this can also be justified by using the nuclear spectral theorem.

Answer 2

What you are looking for is the spectral theorem for bounded self adjoint linear operators.

See for example in: https://web.ma.utexas.edu/users/drp/files/Fall2020Projects/Perales%20DRP%20-%20Samuel%20Perales.pdf

$V_E$ are certain subspaces satisfying following properties (intuitively based on the following properties):

1. $V_E$ are invariant subspaces such that for $v_E \in V_E$, if $E = [\lambda-\epsilon,\lambda + \epsilon]$, we have $||(A-\lambda I) v_{E} || \leq \epsilon ||v_E||$
2. Further, $V_{\sigma(A)} = H$ (full space), $V_{\emptyset} = \{0\}$.
3. For $E \cap F = \emptyset$, $v \in V_E$, $u \in V_F$ $\implies \langle v,u \rangle = 0$.
4. $V_{E \cap F} = V_E \cap V_F$.

We can think of $V_E = \bigoplus_{\lambda \in E} \{v: ||(A-\lambda I)^{-1} v_n|| \rightarrow \infty,v_n \rightarrow v\}$.