Let $H =\ \mathcal{l}^{2}(\mathbb{Z},\mathbb{C})$ and $A = R + L$, where $L$ is the left-shift operator (and $R$ is the right-shift $(Ra)_{n}=a_{n-1}$). Set
$$U : \mathcal{l}^{2}(\mathbb{Z},\mathbb{C}) \longrightarrow L^{2}([0,1]) ,\quad U(a_n)\ =\ \sum_{n=-\infty}^{\infty}a_n\mathrm e^{2\pi \mathrm int}\text{.}$$ Then $ULU^{-1}$ is multiplication by $\mathrm e^{-2\pi \mathrm it}$ and $URU^{-1}$ is multiplication by $e^{2\pi \mathrm it}$. Therefore $UAU^{-1}$ is multiplication by $2\cos(2\pi t).\\$ Show that $A$ can be represented as a multiplication operator by $t$ on $L^{2}(\mathbb{R},\mathrm d\mu_{1})\oplus L^{2}(\mathbb{R},\mathrm d\mu_2)$, where $\mu_1$, $\mu_2$ have support in $[-2,2]$.
I started to proof, that $UAU^{-1}$ is multiplication by $2\cos(2\pi t)$: \begin{align} UAU^{-1} &= U(L+R)U^{-1}\\ &=U(LU^{-1}+RU^{-1})\\ &=ULU^{-1}+URU^{-1}\\ &=( .\ )\ \cdot\ \mathrm e^{-2\pi \mathrm it}\ +\ (\ .\ )\ \cdot\ \mathrm e^{2\pi \mathrm it}\\ &=(\ .\ )\ \cdot\ 2\cos(2\pi t) \end{align}