Show that operator is diagonalizable 
Let $T$ be the self-adjoint operator defined by
$$(T x)_{n} = x_{n + 1} + x_{n - 1} \text{ for } n \in \mathbb{Z}, (x_{n})_{n \in \mathbb{Z}}$$
on $\ell^{2}(\mathbb{Z})$.
Show that $T$ is diagonalizable by giving an explicit unitary operator $B : L^{2}(\mathbb{T}, \frac{d \theta}{2 \pi}) \rightarrow \ell^{2}(\mathbb{Z})$ such that $B M_{f} = TB$, where $f : \mathbb{T} \rightarrow \mathbb{R}$ is $f(e^{i \theta}) = 2 \cos(\theta)$. Here $M_f$ is multiplication by $f$ such that $(M_f g)(\xi) = f(\xi)g(\xi)$.
Note also that $\mathbb{T}$ here means the unit circle.

I am not sure what $B$ should be. Usually, to show diagonalizability I use the Spectral Theorem which states that every normal operator (acting on a separable
Hilbert space) is diagonalizable. But here one asks to explicitly give a unitary operator.
 A: Let $\{\epsilon_n:n\in\mathbb{Z}\}$ denote the standard ONB of $\ell^2(\mathbb{Z})$, so $\epsilon_n=(\dots,0,0,\dots,0,1,0,\dots,0,0,\dots)$ where $1$ lies in the $n$-th position. Also, let $e_n(z)=z^n$ for $n\in\mathbb{Z}$, so $\{e_n: n\in\mathbb{Z}\}$ is the standard orthonormal basis for $L^2(\mathbb{T})$.
Note that $T\epsilon_n=\epsilon_{n-1}+\epsilon_{n+1}$ for all $n\in\mathbb{Z}$, since
$$T(\epsilon_n)=(\dots,0,0,\dots,0,1,0,1,0,\dots,0,0,\dots)$$
where the aces lie on the $n-1$ and the $n+1$ positions.
Also, observe that if $z=e^{i\theta}$ then $$f(z)=2\cos(\theta)=e^{i\theta}+e^{-i\theta}=z^1+z^{-1}=e_1(z)+e_{-1}(z).$$
Now observe that $M_f(e_n)(z)=((e_1+e_{-1})\cdot e_n)(z)=(z+z^{-1})\cdot z^n=z^{n+1}+z^{n-1}=e_{n+1}(z)+e_{n-1}(z)$, so $M_f(e_n)=e_{n+1}+e_{n-1}$.
Define $B: L^2(\mathbb{T})\to \ell^2(\mathbb{Z})$ on the orthonormal basis and extend to the entire space: simply set $Be_n=\epsilon_n$. Note that $B$ maps the ONB onto an ONB, so $B$ is a unitary operator.
The relation $BM_f=TB$ is easily verified: it suffices to do so on the vectors of the ONB, and then continuity and linearity will yield the result on the entire space. Let $n\in\mathbb{Z}$. We have
$$BM_f(e_n)=B(e_{n+1}+e_{n-1})=\epsilon_{n+1}+\epsilon_{n-1}=T\epsilon_n=TB(e_n).$$
