Theorem 43.8.7: Spectral Theorem I (Unbounded)

Theorem 43.8.7 (Spectral Theorem I (Unbounded)).label Let $H$ be a complex Hilbert space and $T: D(T) \to H$ be a densely defined, self-adjoint operator, then there exists a unique spectral measure $E: \cb_{\real} \to B(H)$ such that $T = \int_{\real}\lambda E(d\lambda)$.

Proof. Let $U(T) = (T - i)(T + i)^{-1}\in U(B(H))$ be the Cayley transform of $T$, then $U(T)$ is unitary. By Spectral Theorem I, there exists a unique spectral measure $E: \cb_{\partial B_\complex(0, 1)}\to B(H)$ such that $U(T) = \int_{\partial B_\complex(0, 1)}\lambda E(d\lambda)$. Moreover,

\[(I - U) = (T+i)(T+i)^{-1}- (T - i)(T+i)^{-1}= 2i(T + i)^{-1}\]

is injective, so $1$ is not an eigenvalue of $U$, and $E(\bracsn{1}) = 0$. Let

\[\phi: \partial B_{\complex}(0, 1) \setminus \bracsn{1}\to \real \quad \lambda \mapsto i\frac{1 + \lambda}{1 - \lambda}\]

then the pushforward $F = \phi_{*} E$ defined by $F(B) = E(\phi^{-1}(B))$ is also a spectral measure, with

\begin{align*}\int_{\real} \lambda F(d\lambda)&= \int_{\partial B_\complex(0, 1) \setminus \bracsn{1}}i\frac{1 + \lambda}{1 - \lambda}E(d\lambda) = i(1 + U)(1 - U)^{-1}\\&= i[(T + i)(T+i)^{-1}+ (T-i)(T+i)^{-1}] \cdot (2i(T + i)^{-1})^{-1}\\&= \frac{1}{2}[(T + i)(T+i)^{-1}(T+i) + (T-i)(T+i)^{-1}(T+i)] = T\end{align*}

$\square$

Direct References

circleTheorem 43.6.4: Spectral Theorem I

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circle43.8: Unbounded Functional Calculus
circleDefinition 43.8.10: Absolute Value

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