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Given a Hilbert space $\mathcal{H}$ and spectral a measure $E:\Sigma(\Omega)\to\mathcal{B}(\mathcal{H})$.

How to define the integral for unbounded measurable functions: $$f:\Omega\to\mathbb{C}:\quad\int_\Omega f\mathrm{d}E=?$$

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Define a linear operator $\int_{\Omega} f\,dE$ on the domain $$ \mathcal{D}\left(\int_{\Omega}f\,dE\right) = \left\{ x \in \mathcal{H} : \int_{\Omega}|f(t)|^{2}\,d\|E(t)x\|^{2} < \infty \right\}. $$ The linear operator is most easily defined as the unique vector $\int_{\Omega} f\,dEx$ such that $$ \int_{\Omega}f(t)\,d(E(t)x,y) = \left(\int_{\Omega}f\,dEx\,,\,y\right),\;\;\; y \in \mathcal{H}. $$ Justification: The measure $\mu_{x,y}(S)=(E(S)x,y)$ is a complex measure of finite total variation on $\Omega$, because $\mu_{w,w}$ is a finite positive measure for each $w \in X$, and $\mu_{x,y}$ can be written as a sum of four such measures by using the polarization identity for Hilbert space: $$\mu_{x,y}=\frac{1}{4}\sum_{n=0}^{3}i^{n}\mu_{x+i^{n}y,x+i^{n}y}.$$ For simple functions $f = \sum_{j=1}^{n}\alpha_{j}\chi_{S_{j}}$ where the $S_{j}$ are disjoint Borel subsets of $\Omega$, $$ \begin{align} \left|\int_{\Omega} f d(E(t)x,y)\right|^{2} & = \left| \sum_{j=1}^{n}\alpha_{j}(E(S_{j})x,y)\right|^{2} \\ & = \left|\sum_{j=1}^{n}\alpha_{j}(E(S_{j})x,E(S_{j})y)\right|^{2} \\ & \le \sum_{j=1}^{n}|\alpha_{j}|^{2}\|E(S_{j})x\|^{2}\sum_{j=1}^{n}\|E(S_{j})y \|^{2} \\ & = \int_{\Omega}|f|^{2}d\|E(t)x\|^{2}\sum_{j=1}^{n}\|E(S_{j})y\|^{2} \\ & \le \left(\int_{\Omega}|f|^{2}d\|E(t)x\|^{2}\right)\|y\|^{2} \end{align} $$ Therefore, for any $f \in \mathcal{D}\left(\int_{\Omega}fdE\right)$, $$ \left|\int_{\Omega}f(t)d(E(t)x,y)\right|\le |f|_{x}\|y\|, $$ where $|f|_{x}=\left(\int_{\Omega}|f(t)|^{2}\,d\|E(t)x\|^{2}\right)^{1/2}$, which, by the Riesz Representation Theorem, guarantees the existence of a unique vector $\int_{\Omega}f\,dEx$ such that $$ \int_{\Omega}f(t)d(E(t)x,y) = \left(\int_{\Omega} fdE x\,,\,y\right),\;\;\; y \in \mathcal{H}. $$ Norm Identities: It is easy to check that the domain of $\int_{\Omega}fdE$ is a linear subspace and $\int_{\Omega}fdE$ is linear on its domain. If $f(t)=\sum_{j=1}^{n}\alpha_{j}\chi_{S_{j}}(t)$ where the $S_{j}$ are disjoint Borel sets, then uniqueness of the Riesz Representation guarantees $$ \int_{\Omega} f dEx = \sum_{j=1}^{n}\alpha_{j}E(S_{j})x. $$ Therefore, for a simple function $f$, $$ \left\|\int_{\Omega}f\,dEx\right\|^{2} = \int_{\Omega}|f(t)|^{2}d\|E(t)x\|^{2}. $$ The Lebesgue dominated convergence theorem shows that the above equality continues to hold for all $x \in \mathcal{D}\left(\int_{\Omega}fdE\right)$, for any bounded or unbounded Borel function $f$.

Operator Properties: It should be noted that if the Borel function $f$ is uniformly bounded by $M$, then $\int_{\Omega}fdE$ is a bounded linear operator whose operator norm is bounded by $M$.

For a general Borel function $f$, the domain of $\int_{\Omega}fdE$ is invariant under $E(S)$ because $$ \int_{\Omega}|f(t)|^{2}d\|E(t)E(S)x\|^{2}=\int_{S}|f(t)|^{2}\,d\|E(t)x\|^{2} \le \int_{\Omega}|f(t)|^{2}d\|E(t)x\|^{2}. $$ Let $S_{N}=\{ t\in\Omega :|f(t)| \le N\}$. The range of $E(S_{N})$ is in the domain of $\int_{\Omega}f\,dE$ for any positive integer $N$. Furthermore, $x = \lim_{N}E(S_{N})x$ for all $x$ because $\bigcup_{N=1}^{\infty}S_{N}=\Omega$. Therefore, the domain of $\int_{\Omega}f\,dE$ is dense, even for an unbounded Borel function $f$.

Next it is shown that $\int_{\Omega}f\,dE$ is closed. To this end, suppose that $\{x_{n}\}_{n=1}^{\infty}\subseteq\mathcal{D}(\int_{\Omega}fdE)$ converges to some $x$, and suppose that $\int_{\Omega}fdEx_{n}$ converges to some $y$. Let $S_{N}=\{ t\in\Omega : |f(t)| \le N\}$. Then $(\int_{\Omega}fdE)E(S_{N})$ is a bounded linear operator, which gives $$ \begin{align} \int_{\Omega}fdE\cdot E(S_{N})x & = \lim_{n}\int_{\Omega}fdE\cdot E(S_{N})x_{n} \\ & = \lim_{n}E(S_{N})\int_{\Omega}fdE x_{n} \\ & = E(S_{N})y. \end{align} $$ Thus, $$ \int_{S_{N}}|f(t)|^{2}d\|E(t)x\|^{2} = \|E(S_{N})y\|^{2} \le \|y\|^{2}. $$ Because this holds for all $N \ge 1$, it follows that $x \in \mathcal{D}(\int_{\Omega}fdE)$ and the previous identity gives $$ E(S_{N})\left(\int_{\Omega}fdE x - y\right) = 0,\;\;\; N \ge 1 \\ \implies \int_{\Omega}fdE x = y. $$ Therefore, $\int_{\Omega}f dE$ is a closed densely-defined operator for a complex unbounded Borel function $f$.

It's not hard to show that the following adjoint equation holds: $$ \left(\int_{\Omega} fdE\right)^{\star} = \int_{\Omega}\overline{f}dE. $$

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  • $\begingroup$ Hey that seems kind of similar to your other answer: math.stackexchange.com/a/941481/79762 $\endgroup$ – C-Star-W-Star Sep 29 '14 at 0:50
  • $\begingroup$ Well using the polarization identity together with Riesz' representation might give not the best estimates as observed in the other thread: math.stackexchange.com/a/924505/79762 $\endgroup$ – C-Star-W-Star Sep 29 '14 at 0:54
  • $\begingroup$ I'll leave it to you to show $|\int_{\Omega}fd(E(t)x,y)|^{2}\le \int_{\Omega}|f|^{2}d(E(t)x,x)\|y\|^{2}$. Because all of the measures $\mu_{x,y}(S)=(E(S)x,y)$ are finite, you can also do interesting things with Radon-Nikodym once you realize that $\mu_{x,y} << \mu_{x,x}$ for all $y$, which follows from Cauchy-Schwarz. There are lots of options here. $\endgroup$ – DisintegratingByParts Sep 29 '14 at 1:04
  • $\begingroup$ @Freeze_S : For the first inequality, start with simple functions. $\endgroup$ – DisintegratingByParts Sep 29 '14 at 1:14
  • $\begingroup$ @Freeze_S : I decided it might be generally useful to cover this topic in some depth. So I've added a lot of detail for you. $\endgroup$ – DisintegratingByParts Sep 30 '14 at 4:29

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