Projection valued measure of bounded self-adjoint operator.

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

• This is part of the Spectral Theorem. How you would prove it depends on which approach to the Spectral Theorem you are using, and which parts you have already seen proved. – Robert Israel May 14 '15 at 20:16

1 Answer

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.

• Is there a nice proof for the unbounded case? – user412674 Apr 11 '17 at 1:32