# Proving the spectral theorem for unbounded self-adjoint operators

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.

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: $$Ax = \int_{-\infty}^{\infty}tdE(t)x,\;\; x \in \mathcal{D}(A).$$