Unilateral Shift

A symmetry is a unitary involution, i.e., an operator $Q$ such that $Q^{*}Q = QQ^{*} = Q^{2} = 1$. It may be pertinent to recall that if an operator possesses any two of the properties "unitary," "involutory," and "Hermitian," then it possesses the third; the proof is completely elementary algebraic manipulation.

The commutant of an operator (or of a set of operators) is the set of all operators that commute with it (or with each operator in the set). The bilateral shift $W$ can be viewed as multiplication by $e_1$ on $L^{2}$ of the unit circle. Here $e_n(z) = z^n$ ($n = 0, \pm 1, \pm 2, \ldots$) whenever $\left|z\right| = 1$, and $L^2$ is formed with normalized Lebesgue measure.

The commutant of an operator (or of a set of operators) is the set of all operators that commute with it (or with each operator in the set). The unilateral shift is the restriction of the bilateral shift $W$ to $\mathbf{H}^2$, where $\mathbf{H}^2$ is the closed span in $L^2$ of the unit circle of the functions $e_n$, $n = 0, 1, 2, \ldots$ (with $e_n(z) = z^n$), and $\mathbf{H}^{\infty }$ is the set of essentially bounded functions in $\mathbf{H}^2$. If the bilateral shift is regarded as a multiplication (by $e_1$ on $L^2$), then its commutant is the set of all multiplications on the same $L^2$ (Problem 146). If, however, the multiplier itself is in $\mathbf{H}^2$ (and hence in $\mathbf{H}^{\infty }$), then $\mathbf{H}^2$ is invariant under the induced multiplication (cf. Problem 34), and the restriction of that multiplication to $\mathbf{H}^2$ makes sense.

A direct sum (finite or infinite) of isometries is an isometry. An isometry for which the unitary direct summand is absent is called pure.

Consider an isometry $U$ on a Hilbert space $\mathbf{H}$ such that the subspaces $(U\mathbf{H})^{\perp }$, $U(U\mathbf{H})^{\perp }$, $U^2(U\mathbf{H})^{\perp }, \ldots$ are pairwise orthogonal and span $\mathbf{H}$, but make no demands on the value of the co-rank (the co-rank being $\dim (U\mathbf{H})^{\perp }$). Every such isometry may be called a shift (a unilateral shift). The co-rank of a shift (also called its multiplicity) constitutes a complete set of unitary invariants for it; the original unilateral shift is determined (to within unitary equivalence) as the shift of multiplicity $1$ (the simple unilateral shift). For the purpose of this problem every infinite cardinal number $m$ is even; recall that $m = 2m$.

A part of an operator is a restriction of it to an invariant subspace.

A quasinilpotent operator is one whose spectrum is the single point $0$, equivalently one whose spectral radius $r(A)$ is $0$.

If $A$ is an operator on a Hilbert space $\mathbf{H}$, a subspace $\mathbf{N}$ of $\mathbf{H}$ is called wandering for $A$ if it is orthogonal to all its images under the positive powers of $A$.

Let $U$ be the unilateral shift, with orthonormal basis $\left\{ e_n\right\}$, $n = 0, 1, 2, \ldots$, for the underlying space $\mathbf{H}^2$ so that $Ue_n = e_{n+1}$. Each complex number $\lambda$ of modulus less than $1$ is a simple eigenvalue of $U^{*}$, with corresponding eigenvector $f_{\lambda } = \sum_{n=0}^{\infty } \lambda^n e_n$.

Regard $U$ as the restriction to $\mathbf{H}^2$ of the multiplication induced by $e_1$ on $L^2$ of the unit circle. A function $\varphi$ in $\mathbf{H}^{\infty }$ of constant modulus $1$ almost everywhere is called an inner function; informally, $\mathbf{M} = \varphi \cdot \mathbf{H}^2$ can be described as the set of all multiples of $\varphi$ by functions in $\mathbf{H}^2$.