Spectral Radius

Suppose that $A$ is an operator on a Hilbert space $\mathbf{H}$. If $\lambda$ does not belong to the spectrum of $A$, then the operator $A - \lambda$ is invertible; write $\rho (\lambda ) = (A-\lambda )^{-1}$. (When it is necessary to indicate the dependence of the function $\rho$ on the operator $A$, write $\rho = \rho_A$.) The function $\rho$ is called the resolvent of $A$. The domain of $\rho$ is the complement of $\operatorname {spec} A$; its values are operators on $\mathbf{H}$.

Consider, quite generally, functions $\varphi$ whose domains are open sets in the complex plane and whose values are operators on $\mathbf{H}$. Such a function $\varphi$ is called analytic if, for each $f$ and $g$ in $\mathbf{H}$, the numerical function $\lambda \mapsto \left\langle \varphi (\lambda )f, g \right\rangle$ (with the same domain as $\varphi$) is analytic in the usual sense. (This concept is sometimes called weak analyticity.) In case the function $\psi$ defined by $\psi (\lambda ) = \varphi (1/\lambda )$ can be assigned a value at the origin so that it becomes analytic there, then (just as for numerical functions) $\varphi$ is called analytic at $\infty$, and $\varphi$ is assigned at $\infty$ the value of $\psi$ at $0$.

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.

The power norms of an operator $A$, that is the numbers $\left\| A^k\right\|$, constitute an interesting sequence.

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.)

If $A$ is a quasinilpotent operator, and if $f$ is a function that is analytic in a neighborhood of $0$, then $f$ has a power series expansion, $f(z) = \sum_{n=0}^{\infty } \alpha_n z^n$, convergent when, say, $\left|z\right| < \varepsilon$; in that case write $f(A) = \sum_{n=0}^{\infty } \alpha_n A^n$. This operator series converges (in the norm), since $\left\| A^n\right\|^{1/n} < \varepsilon$ for $n$ sufficiently large, so that $\left\| A^n\right\| < \varepsilon^n$ for $n$ sufficiently large. The correspondence $f \mapsto f(A)$ between functions analytic at $0$ and operators $f(A)$ (for a fixed quasinilpotent operator $A$) is called a functional calculus; it is an extension of the trivial functional calculus for polynomials.

An operator $A$ is called algebraic if there exists a non-zero polynomial $p$ such that $p(A) = 0$; by a natural extension of language it seems proper to say that $A$ is analytic if there exists a non-zero function $f$ analytic at $0$ such that $f(A) = 0$. Every nilpotent operator is algebraic.

The spectrum of the direct sum of two operators is the union of their spectra, and the same is true of the point spectrum, the approximate point spectrum, and the compression spectrum. The extension of this result from two direct summands to any finite number is a trivial induction.