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We define $H=l^2(\mathbb{Z})$, $S\in L(H)$ to be the left (bilateral) shift and we look at $T=S+S^*$ ($S^*$ is actually the right shift). We need to prove that the spectrum of $T$ is $[-2, 2]$ and more importantly to find a unitary operator and 2 measures $\mu_1$ and $\mu_2$ in order to represent $T$ as multiplication by $x$ on $L^2(\mathbb{R}, \mu_1)\bigoplus L^2(\mathbb{R},\mu_2)$. Help would be greatly appreciated.

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We put $U\colon (L^2(\mathbb T),\frac{d\theta}{2\pi})\to H$ given by $U(g)=\left\{\int_0^{2\pi}g(e^{i\theta})e^{in\theta}d\frac{\theta}{2\pi}\right\}$. Then thanks to Parseval equality, $U$ is unitary. We have $UM_f=TU$, where $M_f\colon(L^2(\mathbb T),\frac{d\theta}{2\pi})\to (L^2(\mathbb T),\frac{d\theta}{2\pi})$ is given by $M_f(g)=fg$ and $f(e^{i\theta})=2\cos \theta$. Indeed, we have \begin{align*} (TU)_n&=\frac 1{2\pi}\int_0^{2\pi}g(e^{i\theta})e^{(n-1)\theta}d\theta+\frac 1{2\pi}\int_0^{2\pi}g(e^{i\theta})e^{(n+1)\theta}d\theta\\ &=\frac 1{2\pi}\int_0^{2\pi}g(e^{i\theta})e^{n\theta}\left(e^{-i\theta}+e^{i\theta}\right)d\theta\\ &=\frac 1{2\pi}\int_0^{2\pi}g(e^{i\theta})2\cos \theta e^{in\theta}d\theta\\ &=(UM_f)_n. \end{align*} This partially answers to your question, since now you have to represent the multiplication by $2\cos \theta$ by a multiplication by $x$ on $L^2(\mathbb R,\mu_1)\bigoplus L^2(\mathbb R,\mu_2)$. I chose to work with $L^2(\mathbb T)$ since using the fact that for a function $f\in L^{\infty}$, the spectrum of $M_f$ is the essential range of $f$, i.e. the set of $\lambda$ such that for all $\delta>0$, $\mu(\{x\in X: |f(x)-\lambda|<\delta\})=0$.

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Davide, first of all thank you very much for your answer. I have to admit I already came to the conclusions that you wrote here (perhaps I should have written them down). At least I'm happy with the fact that I was in the right direction and it is worth trying to continue. However, this is still the point that I got stuck at. I would greatly appreciate a bit more guidance, if that's possible. –  Jan Feb 6 '12 at 18:14

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