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

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?

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.

• Thank you for your answer. I think I have been confusing myself this whole time by thinking of projection valued measures as projecting onto eigenspaces, but this seems wrong. I'm guessing there is no interpretation of $V_E$ in terms of eigenspaces, and so nothing more can be said about $V_E$ than what you have written? Commented Jul 25, 2023 at 6:26
• The current intuition that I have is that if $E_\lambda$ is a projection-valued measure, then $E_\lambda$ projects onto the span of all the eigenspaces corresponding to eigenvalues less than or equal to $\lambda$. What has been confusing me is how points in the continuous spectrum are to be interpreted here as they have no eigenvectors and thus no eigenspaces. Commented Jul 25, 2023 at 6:31
• @CBBAM As you can see from this example your intuition is not correct. But we do have that if $\lambda$ is an eigenvalue, then $\mu(\{\lambda \})$ is the projection onto the eigenspace to that eigenvalue.
– jd27
Commented Jul 25, 2023 at 6:46
• Thanks I think my problem is understanding what happens in the continuous spectrum, so let me put that aside for the moment. Suppose $\Omega$ is a set consisting of only eigenvalues, then does $\mu(\Omega)$ match my intuition of it being the projection onto the span of all eigenspaces corresponding to the eigenvalues in $\Omega$? Commented Jul 25, 2023 at 6:49
• @CBBAM If $\Omega$ consists of finitely many eigenvalues, then $\mu (\Omega)$ is the projection onto the direct sum of the corresponding eigenspaces. This follows from the additivity of the spectral measure and the fact that projections for different eigenvalues have orthogonal range. I am not sure if the result also holds for countable many eigenvalues. I think it might be false for uncountable many (although that can only happen in non separable Hilbert spaces anyways).
– jd27
Commented Jul 25, 2023 at 7:34

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

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

• Thank you for your answer, but this is not quite what I am looking for. I know what $V_E$ is by definition and that all of this relates to the spectral theorem. What I am looking for is an explanation on what $V_E$ is explicitly, for example how does it contain generalized eigenvectors if they do not belong to the original Hilbert space. Sorry if my question was unclear. Commented Jul 25, 2023 at 6:13
• I think $V_E$ I defined are the generalized eigenvectors which are generalized eigenvectors for the spectrum of bounded self adjoint operator $A$. What you are trying to understand seems to be what happens to $V_{A \setminus \{\lambda \} }$ for multiplication operator and it looks like the vectors in it are $L^2$ functions which are non-zero only at the point $\lambda$ and are all equivalent to $0$ as $L^2$ function. But the authors for you are pointing out as a $\delta$ function at $\lambda$. This might be true in distributional sense but it is not part of generalized eigenvectors. Cheers ! Commented Jul 25, 2023 at 7:58
• It must be $V_{\{\lambda\}}$ and not $V_{A \setminus \{\lambda\}}$ in my previous comment. Thanks. Hope this helps ! Commented Jul 25, 2023 at 8:12