I'm trying to prove the next:
Consider the multiplication operator $T:L^{2}[0,1]\rightarrow L^{2}[0,1]$ defined by $y(t)=Tx(t)=tx(t).$ Then $\sigma(T)=\sigma_{c}(T)=[0,1]$ and the corresponding spectral family is defined by $$E_{\lambda}x= \left\{ \begin{array}{lcc} 0 & if & \lambda<0 \\ \\ v_{\lambda} & if & 0\leq\lambda\leq 1 \\ \\ x & if & \lambda \geq 1 \end{array} \right.$$ where $$v_{\lambda}(t)= \left\{ \begin{array}{lcc} x(t) & if & 0\leq t \leq \lambda \\ \\ 0 & if & \lambda<t\leq 1. \end{array} \right.$$
First, I've proved that operator multiplication $Tx(t)=tx(t)$ is linear, bounded and self-adjoint. Even more, such operator has no eigenvalues. Then residual spectrum $\sigma_{r}(T)=\emptyset$ and $\sigma_{c}(T)=\sigma(T).$
Computing $m=\displaystyle\inf_{||x||=1}\langle Tx,x\rangle=t=\displaystyle\sup_{||x||=1}\langle Tx,x\rangle=M$ and since $T$ is self-adjoint we can conclude $\sigma(T)=[0,1]$ because $t\in[0,1].$
I'm stuck proving that $\{E_{\lambda}\}_{\lambda}$ is the spectral family, defined as above, of such operator.
I've seen in some cases this can be computed directly depending of the operator: if $T$ is the null operator or the identity such spectral family is easy to find, but in this case I can't see it.
Any kind of help is thanked in advanced.
The spectral family $E_{\lambda}$ is multiplication by a characteristic function: $$ E_{\lambda}f = 0,\;\; \lambda \le 0 \\ E_{\lambda}f = \chi_{[0,\lambda]}f,\;\;\; 0 \le \lambda \le 1, \\ E_{\lambda}f = f,\;\; 1 \le \lambda. $$ If you unravel your definition, you'll have the same thing. Notice that $$ E_{\lambda+\Delta\lambda}f-E_{\lambda}f = \chi{[\lambda,\lambda+\Delta\lambda]}f $$ $$ T(E_{\lambda+\Delta\lambda}f-E_{\lambda}f) \approx \lambda(E_{\lambda+\Delta\lambda}f-E_{\lambda}f) $$ $T$ has approximate eigenvalues $[0,1]$, but no actual ones.