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: $$f:\Omega\to\mathbb{C}:\quad\int_\Omega f\mathrm{d}E=?$$


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: $$ \left(\int_{\Omega} fdE\right)^{\star} = \int_{\Omega}\overline{f}dE. $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.