Weighted shift operators

A weighted shift is the product of a shift (one-sided or two) and a compatible diagonal operator. More explicitly, suppose that $\left\{ e_n\right\}$ is an orthonormal basis ($n = 0, 1, 2, \ldots$, or else $n = 0, \pm 1, \pm 2, \ldots$), and suppose that $\left\{ \alpha_n\right\}$ is a bounded sequence of complex numbers (the set of $n$'s being the same as before). A weighted shift is an operator of the form $SP$, where $S$ is a shift ($Se_n = e_{n+1}$) and $P$ is a diagonal operator with diagonal $\left\{ \alpha_n\right\}$ ($Pe_n = \alpha_n e_n$). It is sometimes convenient to use the symbol $\operatorname {shift}\left\langle \alpha_0, \alpha_1, \alpha_2, \ldots \right\rangle$ to denote the weighted shift with the indicated weights.

For each operator $A$ there exists a complex number of modulus $r(A)$ in the approximate point spectrum of $A$ (Problem 78). If $A$ is a weighted shift, then the circular symmetry of each part of the spectrum (Problem 89) implies that every complex number of modulus $r(A)$ belongs to the approximate point spectrum of $A$. Consequence: the approximate point spectrum of a weighted shift always includes a circle (possibly degenerate).

Let $p = \left\{ p_0, p_1, p_2, \ldots \right\}$ be a sequence of strictly positive numbers, and let $l^2(p)$ be the set of complex sequences $\left\langle \xi_0, \xi_1, \xi_2, \ldots \right\rangle$ with $\sum_{n=0}^{\infty } p_n \left|\xi_n\right|^2 < \infty$. With respect to the coordinatewise linear operations and the inner product defined by

the set $l^2(p)$ is a Hilbert space; it may be called a weighted sequence space.

An operator is nilpotent if some positive integral power of it is zero (and the least such power is the index of nilpotence); an operator $A$ is quasinilpotent if $\lim_n \left\| A^n\right\|^{1/n} = 0$. Nilpotence implies quasinilpotence. The spectral mapping theorem implies that if $A$ is nilpotent, then $\operatorname {spec} A = \left\{ 0\right\}$. The expression for the spectral radius in terms of norms implies that if $A$ is quasinilpotent, then $\operatorname {spec} A = \left\{ 0\right\}$, and, moreover, the converse is true. A nilpotent operator always has a non-trivial kernel, and hence a non-empty point spectrum; for quasinilpotent operators that is not so. (Observe that on finite-dimensional spaces such a construction is clearly impossible.)

A subspace is invariant under an operator if the operator maps it into itself; "non-trivial" means different from both $0$ and the whole space.

Suppose that $\left\{ \alpha_n\right\}$ is a monotone sequence ($\alpha_n \geq \alpha_{n+1}$, $n = 0, 1, 2, \ldots$) of positive numbers ($\alpha_n > 0$) such that $\sum_{n=0}^{\infty } \alpha_n^2 < \infty$. The unilateral weighted shift with the weight sequence $\left\{ \alpha_n\right\}$ is called a monotone $l^2$ shift. The span of the basis vectors $e_n, e_{n+1}, e_{n+2}, \ldots$ is invariant under such a shift, $n = 0, 1, 2, \ldots$. The orthogonal complement, i.e., the span $\mathbf{M}_n$ of $e_0, \ldots , e_{n-1}$, is invariant under the adjoint, $n = 1, 2, 3, \ldots$; the principal result is that every invariant subspace of that adjoint is one of these orthogonal complements.