Projection valued measure of bounded self-adjoint operator.

Question

Let $A$ be a bounded self-adjoint operator with $P_E=\chi_E(A)$ as its projection valued measure on set $E\subset \mathbb{R}$, then $f(A)=\int f(\lambda)dP_\lambda$ and $A=\int \lambda dP_\lambda$. How to start this proof? Intuitively I can see why this is true.

Thanks

Answer 1

The easiest proof I have seen relies on knowing the dual space $C[a,b]^{\star}$ of $C[a,b]$. The characterization of this dual was one of the oldest results in functional Analysis, due to one of the Riesz brothers and probably proved in a fairly horrible way. Here's a simple way using Hahn-Banach:

Theorem: Let $\Phi$ be a continuous linear functional on $C[a,b]$. Then there exists a function $\rho$ of bounded variation on $[a,b]$ such that $\Phi$ is represented by the Riemann-Stieltjes integral $$ \Phi(f) = \int_{a}^{b}f(t)d\rho(t). $$ Furthermore, $\rho$ is unique if it is normalized to be continuous from the left (or right) at points of discontinuity and $\|\Phi\|=V_{a}^{b}(\rho)$ is the variation of $\rho$ on $[a,b]$.

Proof: A really elegant proof of this fact is obtained by using the Hahn-Banach theorem to extend $\Phi$ to a continuous linear functional $\tilde{\Phi}$ on $L^{\infty}[a,b]$. Then $\rho(t)=\tilde{\Phi}(\chi_{[a,t]})$, where $\chi_{[a,t]}$ is the $L^{\infty}$ characteristic function of the interval $[a,t]$. Now divide $[a,b]$ into $N$ equal pieces of length $(b-a)/N$ and let $t_{k} = a+k(b-a)/N$. By the uniform continuity of $f$ on $[a,b]$, $$ \lim_{N}\left\| f-\sum_{k=1}^{N}f(t_{k})\chi_{(t_{k-1},t_{k}]} \right\|_{\infty} = 0. $$ Hence, \begin{align} \Phi(f) & =\lim_{N}\tilde{\Phi}\left(\sum_{k=1}^{N}f(t_{k})\chi_{(t_{k-1},t_{k}]}\right) \\ & = \lim_{N}\sum_{k=1}^{N}f(t_{k})\{\rho(t_{k})-\rho(t_{k-1})\} = \int_{a}^{b}f(t)d\rho(t). \end{align} Everything else is just mop up. $\;\;\Box$

Now, if you have a bounded selfadjoint operator $A$ on a Hilbert space $X$, you can form a polynomial function $p$ of $A$ and show that $$ \|p(A)\| \le \|p\|_{C[a,b]} $$ if $\sigma(A)\subseteq[a,b]$. From this, you can represent functionals $$ \Phi_{x,y}(p) = (p(A)x,y) $$ as $$ (p(A)x,y) = \int_{a}^{b}p(t)d\rho_{x,y}(t). $$ Using the uniqueness of $\rho$ (after normalization) you obtain \begin{align} \rho_{\alpha x,y}(t) & = \alpha\rho_{x,y}(t), \\ \rho_{x+x',y}(t) & = \rho_{x,y}(t)+\rho_{x',y}(t) \\ \overline{\rho_{x,y}(t)} & = \rho_{y,x}(t), \\ |\rho_{x,y}(t)| & \le \|x\|\|y\|. \end{align} This now lifts to operators through representations of continuous bilinear forms: $$ \rho_{x,y}(t) = (E(t)x,y),\\ E(t)^{\star}=E(t),\\ E(a)=0,\;\; E(b)=I,\\ \int_{a}^{b}p(t)dE(t) = p(A). $$ After that it's a matter of isolating all of the properties of $E$ using properties of the representation.