Notation: Let $a,b,m,M\in\mathbb{R}$, such that $a<m\leq M<b$, and let $A$ be an operator satisfying $mI\leq A\leq MI$. Let $K[a,b]$ is the set of all piecewise continuous bounded functions which are monotone decreasing limits of continuous functions.
Currently I'm self studying functional analysis, namely spectral (integral) decomposition of self-adjoint bounded operators. In the text, the author gives the following theorem:
Theorem A(Hilbert): For every self-adjoint bounded operator $A$ on a Hilbert space $H$ such that $mI\leq A\leq MI$ ($m,M\in\mathbb{R}$), there exists a family $E_\lambda$, for $\lambda\in\mathbb{R}$, of orthoprojections such that the following are true:
- $E_\lambda=0$ for $\lambda<m$ and $E_\lambda=I$ for $\lambda\geq M$.
- $E_{\lambda+0}=E_\lambda$ (continuity from the right).
- $E_{\lambda_1}\leq E_{\lambda_2}$ for $\lambda_1<\lambda_2$ or equivalently $E_{\lambda_1}\circ E_{\lambda_2}=E_{\lambda_1}$ for $\lambda_1<\lambda_2$.
(These first three properties mean that $\{E_\lambda\}$ is a spectral family.)
- $A=\int_{-\infty}^{\infty}\lambda dE_{\lambda}$, with the integral converging in the operator norm.
- $E_\lambda$ are strong limits of polynormal of $A$ and therefore they commute with any operator $B$ which commutes with $A$.
- $||Ax||^2=\int\lambda^2d\langle E_\lambda x,x \rangle$.
- A family $\{E_\lambda\}$ that satisfies all these properties is unique.
Note here that $E_\lambda=e_\lambda(A)$, with $$ e_\lambda(t):= \begin{cases} 1, & t\leq\lambda \\ 0, & t>\lambda. \tag{1} \end{cases} $$
I'd like to use the fourth bullet point for the operator $Ax(t)=t\cdot x(t)$ on $L^2[0,1]$, but I don't see how to. I suppose the first thing we have to do is find the spectral family $\{E_\lambda\}$ of $A$. First, we obviously have $A^kx=t^kx$ and hence for any polynomial $P(A)x=p(t)x$. Therefore, for any continuous function we obtain $\varphi(A)x=\varphi(t)x$ and the same for any $\varphi$ from the class $K$ (see notation); that is, $\varphi(A)x=\varphi(t)\cdot x(t)$ and $E_\lambda x(t)=e_\lambda(t)\cdot x(t)$. However, I don't see where to go from here. My guess would be to obtain $dE_\lambda$.
To make my question a bit more clear: Since $Ax(t)=t\cdot x(t)$ is a self-adjoint bounded operator between Hilbert spaces, and there are $m,M\in\mathbb{R}$ for which $mx(t)\leq t\cdot x(t)\leq M x(t)$, I should be able to use Hilbert's theorem; that is, write $A$ as some integral. However, I don't see how to write it as such. I have $E_\lambda x(t)=e_\lambda(t)\cdot x(t)$...but how can this help?