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Consider $[0,1] $ with $\mu$ Lebsgue measure on $[0,1]$ .Let $H = L^2([0,1], \mu)$. For $\phi \in L^{\infty}([0,1],\mu)$.

Define $M_{\phi} \in B(L^2([0,1], \mu))$ by $M_{\phi}(f) = \phi f, \ f\in L^2([0,1], \mu)$ .

I have shown that $\sigma(M_{\phi}) = \text{ess range} \ \phi$.

Now to find the spectral measure $E : M_{\phi} \to B(H)$ of $M_{\phi}$.

Required Hints to do the problem.

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  • $\begingroup$ Do you mean $\Omega=[0,1]$? $\endgroup$ Commented Mar 12, 2018 at 13:18
  • $\begingroup$ Hint. Try the case $\phi(x)=x$ and the case, when $\phi(x)$ has finitely many values. $\endgroup$ Commented Mar 12, 2018 at 13:19
  • $\begingroup$ The hint I can give is that this question is already answered on this site. $\endgroup$ Commented Mar 12, 2018 at 23:40
  • $\begingroup$ @MartinArgerami is this math.stackexchange.com/questions/256330/… $\endgroup$
    – User8976
    Commented Mar 13, 2018 at 0:02
  • $\begingroup$ That would be a particular case. I was thinking about this one. $\endgroup$ Commented Mar 13, 2018 at 0:28

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