Understanding spectral measures Let $e_1=(1,0,0,....) \in \ell ^2 (\mathbb{N})$, and $A$ be the infinite matrix 
\begin{pmatrix}0 & 1 & 0\\1 & 0 & 1 & 0 & ...\\0 & 1 & 0 & 1 & 0\\0 & 0 & 1 & 0 & ...\\.. & ... &  & ... & ...\end{pmatrix}
I would like to calculate explicitly the functional calculus $f(A)$ for functions that are not polynomials, such as continuous functions and indicator functions., and/or to understand the spectral measure $\mu_{e_1}$, i.e. being able to calculate the integral $\int f d\mu_{e_1}$ for the functions mentioned above.
Is such a thing possible? 

This defines a right shift operator $S$:
$$
            Se_n = e_{n+1},\;\; n \ge 1.
$$
This operator has a standard representation on $H^2(\mathbb{T})$, consisting of holomorphic functions in the unit disk with square integrable boundary functions. On this space, $S$ is represented in function form by
$$
                      (Sf)(z) = zf(z).
$$
Here $1$ is identified with $e_1$, and $z^{n}$ is identified with $e_n$. The adjoint $S^*$ of $S$ is given by
$$
       (S^*f)(z)=\frac{f(z)-f(0)}{z}.
$$
Your operator is $S+S^*$, which is given by
$$
           (S+S^*)f = zf(z)+\frac{f(z)-f(0)}{z}.
$$
The resolvent of this operator $R(\lambda)=(S+S^*-\lambda I)^{-1}$ is defined at least for $\lambda\notin\mathbb{R}$, and $g=R(\lambda)f$ is determined by the equation:
$$
             (S+S^*-\lambda I)g=f.
$$
So the resolvent equation becomes
$$
         \left(z+\frac{1}{z}-\lambda\right)g(z)-\frac{g(0)}{z}=f(z).
$$
The singular terms on the left cancel, leaving the limit as $z\rightarrow 0$ of both sides to be
$$
             -\lambda g(0) = f(0), \\
                 g(0) = -\frac{1}{\lambda}f(0).
$$
Then you can obtain an expression for the resolvent $g=R(\lambda)f$.
