# Spectral Measures: Spectral Spaces (I)

Problem

Given a Hilbert space $\mathcal{H}$.

Let the Lebesgue measure be $\lambda$.

Consider a Borel spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H})$$

Denote its probability measures by: $$\nu_\varphi(A):=\|E(A)\varphi\|^2$$

Introduce the spectral space: $$\mathcal{H}_\parallel:=\{\varphi:\nu_\varphi\ll\lambda\}$$ $$\mathcal{H}_\perp:=\{\varphi:\nu_\varphi\perp\lambda\}$$

Then they decompose: $$\mathcal{H}=\mathcal{H}_\parallel\oplus\mathcal{H}_\perp$$

How to prove this?

Attention

This thread has been split: Spectral Spaces (II)

• Can you find an operator approach for defining these subspaces for $(Nf)(\lambda)=\lambda f(\lambda)$ on $L^{2}_{\mu}(S)$, where $\mu$ is a finite positive Borel measure on $\mathbb{C}$ with support $S$? – DisintegratingByParts Oct 21 '14 at 17:49
• No not really I thought about it for some time what I ended up was something about decomposing rather the spectrum than the subspaces: $\lambda\in\sigma(T)\iff T\varphi_U-\lambda_U\varphi_U\to0\quad(\lambda_U\to\lambda)$ where the net is indexed over the neighborhoods $U\in\mathcal{N}(\lambda)$ and ordered by inclusion $U\subseteq U'$. Does this characterization hold true for noneigenvalues? – C-Star-W-Star Oct 27 '14 at 15:18
• The only way I've seen to decompose the space is using the Radon-Nikodym theorem. I don't know what operator property there could be to characterize the decomposition, especially when you think about decomposing the measures along an arbitrary fixed measure. – DisintegratingByParts Oct 27 '14 at 16:29
• Ok can you tell me what was the idea you had in mind w.r.t. the L^2 interpretation? – C-Star-W-Star Oct 27 '14 at 16:37

If you start with a unit vector $v \in \mathcal{H}$, then you can define a subspace $\mathcal{H}_{v}$ as the closure in $\mathcal{H}$ of all polynomials in $N$, $N^{\star}$ acting on $v$. $\mathcal{H}_{v}$ is invariant under $N$, $N^{\star}$. This defines a unitary map $$\mathcal{F}_{v} : \mathcal{H}_{v}\rightarrow L^{2}_{\mu_{v}}$$ where $\mu_{v}(S)=\|E(S)v\|^{2}$. And $\mathcal{F}_{v}N\mathcal{F}_{v}^{-1}=M_{\lambda}$ is a multiplication operator. Now decompose the measure $\mu_{v}$ into an atomic part, a part which is absoutely continuous with respect to Lebesgue measure and another which is concentrated on a Lebesgue measure $0$ set. This decomposition is achieved with multiplication operators on $L^{2}_{\mu_{v}}$ by characteristic functions, i.e., projections on $L^{2}_{\mu_{v}}$. The corresponding subspaces are invariant under $N$, $N^{\star}$ and, therefore, commute with the original spectral measure.

Next, find another unit vector $w$ in the orthogonal complement of $\mathcal{H}_{v}$. The resulting space $\mathcal{H}_{w}$ is orthogonal to $\mathcal{H}_{v}$, and you can do the same thing there. If $\mathcal{H}$ is separable, then there are--at nost--countably many of such spaces needed to span the original space, and you find projection operators onto the desired spaces, which you can sum to obtain the final projections required to decompose the space.

If the space is not separable, then there are complications in summing the projections, but nets probably work. However, if you want to decompose the spectral measure directly, and you still want to end up with measurable sets along which to decompose the spectral measure, then you may need to stick with separable spaces.

• Ok first of all thanks for the answer!! (+1) So you basically gave a proof that it really decomposes the space, right? However, I thought of a characterization directly related to the operator itself not relating to the probability measure. (But probably this results in a different decomposition.) Anyway, I hope you don't mind if I won't mark it as answered. – C-Star-W-Star Oct 27 '14 at 17:41
• @Freeze_S : As I mentioned in the comments, I seriously doubt that there's an operator theoretic way to decompose the measures along another measure. This question seems to be a measure theoretic one. – DisintegratingByParts Oct 27 '14 at 18:46
• Yeah you're right that is already the most natural way to do so. – C-Star-W-Star Oct 27 '14 at 18:49
• I'd like to split this thread into this one stressing on the orthogonal decomposition and another one stressing on the equivalent description. I have chosen so as your answer focuses on the orthogonal decomposition and therefore still fits quite well. Is this ok for you? – C-Star-W-Star Nov 24 '14 at 13:52

Preparation

For finite measures: $$\nu_\varphi(A^\complement)=0\iff\nu_\varphi(A)=\nu_\varphi(\Omega)$$

Check equivalences: $$\nu_\varphi(A)=0\iff E(A)\varphi=0$$ $$\nu_\varphi(A)=\nu_\varphi(\Omega)\iff E(A)\varphi=\varphi$$ Concluding preparation.

Decomposition

By preparation one has: $$\mathcal{H}_\parallel=\bigcap_{\lambda(N)=0}\mathcal{N}E(N)\quad\mathcal{H}_\perp=\bigcup_{\lambda(N)=0}\mathcal{R}E(N)$$

Remind projections: $$\mathcal{R}E(A)^\perp=\mathcal{N}E(A)\quad\mathcal{R}E(A)=\mathcal{N}E(A)^\perp$$

Orthogonal complement:* $$(\mathcal{H}_\perp)^\perp=\left(\bigcup_{\lambda(N)=0}\mathcal{R}E(N)\right)^\perp=\bigcap_{\lambda(N)=0}\mathcal{R}E(N)^\perp=\bigcap_{\lambda(N)=0}\mathcal{N}E(N)=\mathcal{H}_\parallel$$

It is closed subspace: $$\varphi^{(\prime)}\in\mathcal{H}_\perp\implies\varphi^{(\prime)}\in\mathcal{R}E(N^{(\prime)})\implies\varphi+\varphi'\in\mathcal{R}E(N\cup N')\implies\varphi+\varphi'\in\mathcal{H}_\perp$$ $$\varphi_n\in\mathcal{H}_\perp\implies\varphi_n\in\mathcal{R}E(N_n)\implies\lim_n\varphi_n\in\mathcal{R}E(\cup_n N_n)\implies\lim_n\varphi_n\in\mathcal{H}_\perp$$

Concluding decomposition.

*Reference: Hilbert vs. De Morgan