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Let $H$ be a Hilbert space and $T \in \mathcal{L}(H)$ be a normal operator. It is known, that there exists a spectral measure $E$ on $(\mathbb{C},\mathcal{B})$ (where $\mathcal{B}$ is the set of Borel measurable sets) such that $$ T = \int_{\mathbb{C}} z\,dE(z) $$ and that for every $\delta \in \mathcal{B}$ we have that $H(\delta):= E(\delta)H$ is a $T$-reducing subspace and that $\sigma(T\mid_{H(\delta)}) \subseteq \bar{\delta}$. However, for a decomposition $\sigma(T) = \delta \sqcup \delta'$ into two closed sets $\delta, \delta'$ we can also define $$ \mathcal{P}_{\delta} = \frac{1}{2\pi i}\int_{\Gamma} (\lambda I -T)^{-1}\, d\lambda $$ by the holomorphic functional calculus where $\Gamma$ encloses $\delta$ but not $\delta'$. One can easily show that $\mathcal{P}_{\delta}$ is a projection operator and since $\mathcal{P}_{\delta}$ commutes with $T$, the subspace $F(\delta) = \mathcal{P}_\delta H$ is a $T$-reducing subspace.\

Question: What is the relationship between these two concepts? What is the relationship between the subspaces $H(\delta)$ and $F(\delta)$? Is the spectral projection expressible through the spectral measure?

I would also appreciate any references where this is covered!

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The answer to your question about $\delta,\delta'$ is that $$ \frac{1}{2\pi i}\oint_{\Gamma}(\lambda I-T)^{-1}d\lambda = E(\delta), $$ provided that $\Gamma$ is a positively oriented contour that encloses only the component $\delta$ of the spectrum. It's hard to isolate the components of the spectrum for a general normal operator. However, if $A$ is selfadjoint, then Stone's formula gives you a way to construct the spectral measure from a vector limit of the resolvent:

$$ \frac{1}{2}(E(a,b)+E[a,b])x\\=\lim_{\epsilon\downarrow 0}\frac{1}{2\pi i}\int_{a}^{b}((t-i\epsilon)I-A)^{-1}x-((t+i\epsilon I-A)^{-1}xdt $$ This was one of the earliest constructions of the spectral measure. Functional Analysis by Peter Lax details this approach to construct the spectral measure for (un)bounded self-adjoint linear operators on a Hilbert space.

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  • $\begingroup$ Do you really mean $E(\delta) = 2\pi i \mathcal{P}_{\delta}$? Because then $2\pi i \mathcal{P}_{\delta} = E(\delta) = E(\delta)^2 = (2 \pi i)^2 \mathcal{P}_\delta^2 = (2 \pi i)^2 \mathcal{P}_\delta$ and thus $\mathcal{P}_\delta = 0$ always. Or I am missing something obvious here? $\endgroup$
    – Andrei Kh
    Commented Feb 19, 2019 at 17:33
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    $\begingroup$ @AndreiKh : I added the $1/2\pi i$. Thanks for catching that. $\endgroup$ Commented Feb 19, 2019 at 17:45
  • $\begingroup$ Ah, alright! So at the end, the spectral projection and the spectral measure are the same, but only the latter is defined on all Borel sets while the former is defined only on the connected components of the spectrum right? Is there any significant difference between the reducing subspace $H(\delta)$ when $\delta$ is an arbitrary Borel set and when it is a connected component? Has the latter some special properties? $\endgroup$
    – Andrei Kh
    Commented Feb 19, 2019 at 18:02
  • $\begingroup$ @AndreiKh : The spectral measure $E$ of a selfadjoint $A$ is determined on intervals. It's the same type of construction that takes you from a non-decreasing function on $\mathbb{R}$ to a Borel measure on $\mathbb{R}$. $\endgroup$ Commented Feb 19, 2019 at 20:19

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