Let $A$ be (densely-defined) self-adjoint operator on a (complex) Hilbert space $H$. Then, the Cayley transform $U=(A-i)(A+i)^{-1}$ is a unitary operator, and $(A\pm i)^{-1} \in B(H)$.

Using the fact that $U$ is unitary, from the spectral theorem for bounded normal operators, we can express $$ U = \int_\mathbb{T} \lambda \; dF(\lambda) $$ where $dF(\lambda)$ is a projection-valued measure, and $\mathbb{T}$ is the unit circle.

My question is: Using only these facts, how can I obtain the spectral measure for $A$, and prove that $$A = \int_\mathbb{R} \lambda \; dE(\lambda)$$

P.S: I am not looking for a proof via the other versions of the spectral theorem (multiplication / functional calculus) or a proof which uses Hergoltz's theorem, etc.


1 Answer 1


This requires only the inverse of the Cayley transform. Start with $$ (U-I)=(A-iI)(A+iI)^{-1}-(A+iI)(A+iI)^{-1}=-2i(A+iI)^{-1}. $$ It follows that $\mathcal{N}(U-I)=\{0\}$ and $\mathcal{R}(U-I)=\mathcal{D}(A)$. Similarly, $$ (U+I) = 2A(A+iI)^{-1} = iA(U-I). $$ Let $U=\int_{T}\lambda dF(\lambda)$, and, for each $0 < \delta < \pi$, define $G_{\delta}$ to be the characteristic function of the arc $\{ e^{i\theta} : \theta \in [\delta,2\pi-\delta]\}$. Then $$ P_{\delta} = \int G_{\delta}(\lambda)dF(\lambda) $$ is a projection with $P_{\delta}x \in \mathcal{D}(A)$ because $$ Q_{\delta}=\int_{T} G_{\delta}\frac{1}{\lambda-1}dF(\lambda) $$ is bounded and $(U-I)Q_{\delta}=P_{\delta}$ implies that the range of $P_{\delta}$ is in $\mathcal{R}(U-I)=\mathcal{D}(A)$. Furthermore, $$ iAP_{\delta} = iA(U-I)Q_{\delta}=(U+I)Q_{\delta}=\int_{T}G_{\delta}(\lambda)\frac{\lambda+1}{\lambda-1}dF(\lambda) \\ AP_{\delta} = \int_{T}i\frac{1+\lambda}{1-\lambda}G_{\delta}(\lambda)dF(\lambda). $$ Because $x \in \mathcal{D}(A)$ iff $x = (U-I)y$ for some $y$, then, for all $x \in \mathcal{D}(A)$, one has $$ \begin{align} P_{\delta}Ax & = P_{\delta}A(U-I)y \\ & =-iP_{\delta}(U+I)y\\ & =-i(U+I)P_{\delta}y \\ & = A(U-I)P_{\delta}y \\ & = AP_{\delta}(U-I)y = AP_{\delta}x \end{align} $$ If $x \in \mathcal{D}(A)$, then $$ Ax = \lim_{\delta\downarrow 0}P_{\delta}Ax=\lim_{\delta\downarrow 0}AP_{\delta}x = \lim_{\delta\downarrow 0}\int i\frac{1+\lambda}{1-\lambda}G_{\delta}(\lambda)dF(\lambda)x. $$ Therefore, by the monotone convergence theorem, if $x\in\mathcal{D}(A)$, then $$ \begin{align} \|Ax\|^{2} & =\lim_{\delta\downarrow 0}\|AP_{\delta}x\|^{2} \\ & = \lim_{\delta\downarrow 0}\int \left|\frac{1+\lambda}{1-\lambda}\right|^{2}|G_{\delta}(\lambda)|^{2}d\|F(\lambda)x\|^{2} \\ & = \int \left|\frac{1+\lambda}{1-\lambda}\right|^{2}d\|F(\lambda)x\|^{2} < \infty. \end{align} $$ Conversely if the last integral on the right is finite for some $x$, then the following limit exists in $X$: $$ y = \lim_{\delta\downarrow 0}\int i\frac{1+\lambda}{1-\lambda}G_{\delta}(\lambda)dF(\lambda)x = \lim_{\delta\downarrow 0}AP_{\delta}x. $$ Then, because $\lim_{\delta}P_{\delta}x=x$ exists, and because $A$ is closed, it follows that $x\in\mathcal{D}(A)$ and $Ax=y$. Finally, one concludes that $$ x \in \mathcal{D}(A) \iff \int \left|\frac{1+\lambda}{1-\lambda}\right|^{2}d\|F(\lambda)x\|^{2} < \infty. $$ And, in that case, $$ Ax = \lim_{\delta\downarrow 0}\int i \frac{1+\lambda}{1-\lambda}G_{\delta}(\lambda)dF(\lambda)x. $$ Change of variables: The final step is a change of variables. Define a new spectral measure $E$ on $\mathbb{R}$ by $E(S)=F(\{ \frac{t-i}{t+i} : t\in S\})$. It follows that $\frac{t-i}{t+i}=\lambda$ gives $t=i\frac{1+\lambda}{1-\lambda}$. So, $$ x \in \mathcal{D}(A) \iff \int_{-\infty}^{\infty}t^{2}d\|E(t)x\|^{2} < \infty, $$ and, for any such $x$, the following exists as an improper integral: $$ Ax = \int_{-\infty}^{\infty}tdE(t)x,\;\; x \in \mathcal{D}(A). $$


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