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
and
Then,
- (1)
$\int_{X} f dE: D(\int_{X}f dE) \to H$ is a densely defined linear operator.
- (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)
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,
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,
For any $\phi \in \Sigma(X; \complex)$, by the Cauchy-Schwarz inequality,
By the Monotone Convergence Theorem,
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,
so $\int_{X} g dE \xi \in D(\int_{X} f dE)$ if and only if $\xi \in D(\int_{X} fg dE)$, and
Let $\xi \in D(\int_{X} g dE) \cap D(\int_{X} fg dE)$ and $\eta \in H$, then
(3): For any $\xi, \eta \in D(\int_{X} f dE) = D(\int_{X} \ol f dE)$,
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),
so
By (2) and multiplication of $\int_{X} (1 + |f|) dE$ from the left, $\int_{X} \ol f dE = \paren{\int_X f dE}^{*}$.$\square$
Post a Comment