Tridiagonal shifts as compact + isometry

Abstract

Let $$\{a_n\}_{n\ge 0}$$ and $$\{b_n\}_{n\ge 0}$$ be sequences of scalars. Suppose $$a_n \ne 0$$ for all $$n \ge 0$$ . We consider the tridiagonal kernel (also known as band kernel with bandwidth one) as $$\begin{aligned} k(z, w) = \sum _{n=0}^\infty ((a_n + b_n z)z^n) \overline{(({a}_n + {b}_n {w}) {w}^n)} \qquad (z, w \in \mathbb {D}), \end{aligned}$$ where $$\mathbb {D} = \{z \in \mathbb {C}: |z| < 1\}$$ . Denote by $$M_z$$ the multiplication operator on the reproducing kernel Hilbert space corresponding to the kernel k. Assume that $$M_z$$ is left-invertible. We prove that $$M_z =$$ compact $$+$$ isometry if and only if $$|\frac{b_n}{a_n}-\frac{b_{n+1}}{a_{n+1}}|\rightarrow 0$$ and $$|\frac{a_n}{a_{n+1}}| \rightarrow 1$$ .