Definition 43.8.6 (Unbounded Integration Against Spectral Measure).label Let $X$ be a compact Hausdorff space, $H$ be a complex Hilbert space, $E: \cb_{X} \to B(H)$ be a spectral measure relative to $H$. For each Borel measurable function $f: X \to \complex$, let

\[D\paren{\int_X f dE}= \bracs{\xi \in H \bigg | \int_X |f|^2 dE_{\xi, \xi} < \infty}\]

and

\[\int_{X} f dE: D\paren{\int_X f dE}\to H \quad \angles{\int_X fdE \xi, \eta}_{H} = \int_{X} f dE_{\xi, \eta}\]

Then,

  1. (1)

    $\int_{X} f dE: D(\int_{X}f dE) \to H$ is a densely defined linear operator.

  2. (2)

    For any Borel measurable functions $f, g: X \to \complex$,

    \[D\paren{\int_X f dE \int_X g dE}= D\paren{\int_X g dE}\cap D\paren{\int_X fg dE}\]

    and $\int_{X} f dE \int_{X} g dE = \int_{X} fg dE$.

  3. (3)

    For any Borel measurable function $f: X \to \complex$, $(\int_{X} f dE)^{*} = \int_{X} \ol f dE$. In particular, $\int_{X} f dE$ is always closed.

Proof. (1): For each $n \in \natp$, $\one_{\bracsn{f \le n}}f \in B^{\infty}(X; \complex)$. Thus for each $\xi \in H$, by a monotone class argument,

\begin{align*}\int_{X} |f|^{2} dE_{E(\bracsn{f \le n})\xi, E(\bracsn{f \le n})\xi}&= \int_{\bracsn{f \le n}}|f|^{2} dE_{\xi, \xi}< \infty \\ \normn{\xi - E(\bracsn{f \le n})\xi}_{H}^{2}&= \normn{E(\bracsn{f > n})\xi}_{H}^{2}\end{align*}

where by continuity from above, $E(\bracsn{f > n})\xi \to 0$ as $n \to \infty$. Therefore $D(\int_{X} f dE)$ is dense in $H$.

It remains to show that $\int_{X} f dE$ is defined on the claimed domain. Let $\xi \in D(\int_{X} f dE)$, $\eta \in H$, and $B \in \cb_{X}$, then by (1) and (3) of Lemma 43.6.2,

\[\int_{X} \one_{B} d|E_{\xi, \eta}| \le E_{\xi, \xi}(B)^{1/2}E_{\eta, \eta}(B)^{1/2}\]

For any $\phi \in \Sigma(X; \complex)$, by the Cauchy-Schwarz inequality,

\begin{align*}\int_{X} |\phi| d|E_{\xi, \eta}|&\le \norm{\eta}_{H}\sum_{\alpha \in \phi(X)}|\alpha| E_{\xi, \xi}(\bracsn{\phi = \alpha})^{1/2}E_{\eta, \eta}(\bracsn{\phi = \alpha})^{1/2}\\&\le \norm{\eta}_{H}\braks{\int_X |\phi|^2 dE_{\xi, \xi}}^{1/2}\end{align*}

By the Monotone Convergence Theorem,

\[\int_{X} |f| d|E_{\xi, \eta}| \le \norm{\eta}_{H} \braks{\int_X |f|^2 dE_{\xi, \xi}}^{1/2}\]

so $\int_{X} f dE \xi \in H$ is well-defined.

(2): For each $\xi \in D(\int_{X} g dE)$, by a monotone class argument,

\[\int_{X} |fg|^{2} dE_{\xi, \xi}= \int_{X} |f|^{2} |g|^{2} dE_{\xi, \xi}= \int_{X} |f|^{2} dE_{\int_X gdE\xi, \int_X gdE\xi}\]

so $\int_{X} g dE \xi \in D(\int_{X} f dE)$ if and only if $\xi \in D(\int_{X} fg dE)$, and

\[D\paren{\int_X f dE \int_X g dE}= D\paren{\int_X g dE}\cap D\paren{\int_X fg dE}\]

Let $\xi \in D(\int_{X} g dE) \cap D(\int_{X} fg dE)$ and $\eta \in H$, then

\begin{align*}\angles{\int_X f dE \int_X g dE \xi, \eta}_{H}&= \int_{X} f dE_{\int_X g dE \xi, \eta}= \int_{X} fg dE_{\xi, \eta}\\&= \angles{\int_X f g dE \xi, \eta}_{H}\end{align*}

(3): For any $\xi, \eta \in D(\int_{X} f dE) = D(\int_{X} \ol f dE)$,

\begin{align*}\angles{\int_X f dE\xi, \eta}_{H}&= \int_{X} f dE_{\xi, \eta}= \ol{\int_X \ol f dE_{\eta, \xi}}\\&= \ol{\dpn{\int_X \ol f dE\eta, \xi}H}= \dpn{\xi, \int_X \ol f dE \eta}{H}\end{align*}

so $\int_{X} \ol f dE \subset (\int_{X} f dE)^{*}$. On the other hand, for each $\xi \in H$ and $\eta \in D[(\int_{X} f dE)^{*}]$, by the bounded $L^{\infty}$ functional calculus and (2),

\begin{align*}\angles{\xi, \int_X \frac{\ol f}{1 + |f|} dE\eta}_{H}&= \angles{\int_X \frac{f}{1 + |f|} dE\xi, \eta}_{H} \\&= \angles{\int_X f dE \int_X (1 + |f|)^{-1}dE\xi, \eta}_{H} \\&= \angles{ \int_X (1 + |f|)^{-1}dE\xi, \paren{\int_X f dE}^{*}\eta}_{H} \\&= \angles{\xi, \int_X (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*}\eta}_{H} \\\end{align*}

so

\[\int_{X} \frac{\ol f}{1 + |f|}dE = \int_{X} (1 + |f|)^{-1}dE\paren{\int_X f dE}^{*}\]

By (2) and multiplication of $\int_{X} (1 + |f|) dE$ from the left, $\int_{X} \ol f dE = \paren{\int_X f dE}^{*}$.$\square$

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