# Invariant subspaces of the Cesàro operator

Eva A. Gallardo-Gutiérrez  Eva A. Gallardo-Gutiérrez
Departamento de Análisis Matemático y Matemática Aplicada,
Facultad de Matemáticas,
Universidad Complutense de Madrid,
Plaza de Ciencias 3, 28040 Madrid, Spain
and
Instituto de Ciencias Matemáticas ICMAT (CSIC-UAM-UC3M-UCM),
Madrid, Spain<EMAIL_ADDRESS>,  Jonathan R. Partington  Jonathan R.
Partington,
School of Mathematics,
University of Leeds,
Leeds LS2 9JT, United Kingdom<EMAIL_ADDRESS>and  William T. Ross
William T. Ross,
Department of Mathematics and Statistics,
University of Richmond,
Richmond, VA 23173, USA<EMAIL_ADDRESS>
###### Abstract.

This paper explores various classes of invariant subspaces of the classical
Cesàro operator $C$ on the Hardy space $H^{2}$. We provide a new
characterization of the finite co-dimensional $C$-invariant subspaces, based
on earlier work of the first two authors, and determine exactly which model
spaces are $C$-invariant subspaces. We also describe the $C$-invariant
subspaces contained in model spaces and establish that they are all cyclic.
Along the way, we re-examine an associated Hilbert space of analytic functions
on the unit disk developed by Kriete and Trutt. We also make a connection
between the adjoint of the Cesàro operator and certain composition operators
on $H^{2}$ which have universal translates in the sense of Rota.

###### Key words and phrases:

Cesàro operator, Hardy space, invariant subspace

###### 2010 Mathematics Subject Classification:

Primary 47A15, 47A55, 47B15

First author is partially supported by Plan Nacional I+D grant no.
PID2019-105979GB-I00 (Spain), the Spanish Ministry of Science and Innovation,
through the “Severo Ochoa Programme for Centres of Excellence in R&D”
(CEX2019-000904-S) and from the Spanish National Research Council, through the
“Ayuda extraordinaria a Centros de Excelencia Severo Ochoa” (20205CEX001).

##  1\. Introduction

This paper examines the invariant subspaces of the Cesàro operator

$(Cf)(z):=\frac{1}{z}\int_{0}^{z}\frac{f(\xi)}{1-\xi}d\xi$

on the Hardy space $H^{2}$ [19] of the unit disk
$\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$. Recall that $H^{2}$ is the Hilbert
space of analytic functions $f$ on $\mathbb{D}$ with

$\|f\|_{H^{2}}:=\Big{(}\sup_{0<r<1}\int_{\mathbb{T}}|f(r\xi)|^{2}dm(\xi)\Big{)}^{\frac{1}{2}}<\infty.$

In the above, $m$ denotes Lebesgue measure on the unit circle
$\mathbb{T}:=\partial\mathbb{D}$, normalized so that $m(\mathbb{T})=1$. Note
also that $\|f\|_{H^{2}}^{2}=\sum_{n\geqslant 0}|a_{n}|^{2}$, where
$\\{a_{n}\\}_{n\geqslant 0}$ is the sequence of Taylor coefficients of $f$.

Hardy’s inequality [26] (see also [27]) says that $C$ is a bounded operator on
$H^{2}$. Moreover, a calculation shows that the matrix representation of $C$
with respect to the orthonormal basis $\\{z^{n}\\}_{n\geqslant 0}$ for
$H^{2}$, i.e., the infinite matrix whose entries are $\langle
Cz^{n},z^{m}\rangle_{H^{2}}$, $m,n\geqslant 0$, is the well known Cesàro
matrix

(1.1)
$C:=\begin{bmatrix}1&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\cdots\\\\[3.0pt]
\frac{1}{2}&\frac{1}{2}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\cdots\\\\[3.0pt]
\frac{1}{3}&\frac{1}{3}&\frac{1}{3}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\cdots\\\\[3.0pt]
\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&\frac{1}{4}&{\color[rgb]{.75,.75,.75}0}&\cdots\\\\[3.0pt]
\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\frac{1}{5}&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix}.$

In this matrix setting, one thinks of $C$ as acting on column vectors ${\bf
a}=(a_{0},a_{1},a_{2},\ldots)^{T}$ in $\ell^{2}$ via matrix multiplication

$C{\bf
a}=\big{(}a_{0},\frac{a_{0}+a_{1}}{2},\frac{a_{0}+a_{1}+a_{2}}{3},\ldots\big{)}^{T}.$

As was known for quite some time, the Cesàro matrix is the basis for an
important summability method for Fourier series [41]. A more detailed analysis
from a paper of Brown, Halmos, and Shields establishes the following basic
operator theory facts about $C$ and its adjoint $C^{*}$ (which we record here
for later use).

######  Proposition 1.2 (Brown–Halmos–Shields [4]).

For the Cesàro operator $C$ on $H^{2}$,

  1. (a)

$\|I-C\|=1$;

  2. (b)

$\|C\|=2$;

  3. (c)

$\sigma(C)=\\{z:|z-1|\leqslant 1\\}$;

  4. (d)

$\sigma_{p}(C)=\varnothing$;

  5. (e)

$\sigma_{p}(C^{*})=\\{z:|z-1|<1\\}$.

Here, $\sigma(C)$ denotes the spectrum of $C$ and $\sigma_{p}(C)$ the point
spectrum (eigenvalues of $C$).

The Cesàro operator is hyponormal, namely, $C^{*}C-CC^{*}\geqslant 0$ [4, Thm.
3] (see also [35]). One of the gems in the study of the Cesàro operator, and
to be explored further in this paper, is a theorem of Kriete and Trutt [31]
which extends the above hyponormality result and says that $I-C$ is unitarily
equivalent to the operator $M_{z}f=zf$ (multiplication by the independent
variable $z$) on $\mathcal{H}^{2}(\mu)$, where $\mu$ is a certain positive
finite Borel measure on $\overline{\mathbb{D}}$, the closure of $\mathbb{D}$,
and $\mathcal{H}^{2}(\mu)$ denotes the closure of the analytic polynomials
$\mathbb{C}[z]$ in $L^{2}(\mu)$. This shows that the Cesàro operator is
subnormal (a normal operator restricted to one of its invariant subspaces).

A follow up paper of Kriete and Trutt [32] began a discussion of some of the
complexities of the invariant subspaces of the Cesàro operator via a
discussion of the $M_{z}$-invariant subspaces of $\mathcal{H}^{2}(\mu)$. We
find this space quite interesting and this paper will continue this line of
inquiry. A recent paper [22] of the first two authors of this current paper
rekindled the discussion of the complexities of the Cesàro invariant subspaces
from the vantage point of semigroups of composition operators and of
translation operators $f(x)\mapsto f(x-t)$ on a certain weighted $L^{2}$ space
of the real line $\mathbb{R}$.

The purpose of this paper is to continue this invariant subspace discussion.
One particularly interesting class of invariant subspaces of the Cesàro
operator will be the model spaces $(u_{\alpha}H^{2})^{\perp}$, where
$\alpha>0$ and $u_{\alpha}$ is the atomic inner function

$u_{\alpha}(z)=\exp\Big{(}\alpha\frac{z+1}{z-1}\Big{)}.$

Using semigroup ideas from [22], along with a Beurling–Lax theorem developed
in that same paper, we will show in Theorem 7.2 that indeed the model space
$(u_{\alpha}H^{2})^{\perp}$, where $\alpha>0$, is an invariant subspace of the
Cesàro operator and that the vector

$\frac{1-u_{\alpha}(z)}{1+z}$

belongs to $(u_{\alpha}H^{2})^{\perp}$ and is a cyclic vector for the Cesàro
operator when restricted to $(u_{\alpha}H^{2})^{\perp}$ (Proposition 9.4).
Moreover, in Theorem 7.7 we will show that $(u_{\alpha}H^{2})^{\perp}$,
$\alpha>0$, are the only model spaces which are Cesàro invariant. In
Proposition 8.2 we prove that these model spaces $(u_{\alpha}H^{2})^{\perp}$
correspond to a curious class of $M_{z}$-invariant subspaces of the associated
Kriete–Trutt space $\mathcal{H}^{2}(\mu)$ that are different from the
“standard” Beurling-type ones consisting of the closure of $uH^{2}$ in
$\mathcal{H}^{2}(\mu)$ (as explored in [32]), where $u$ is an inner function.

An integral calculation will verify that the adjoint $C^{*}$ of the Cesàro
operator on $H^{2}$ is given by the integral formula

(1.3) $(C^{*}f)(z)=\frac{1}{1-z}\int_{z}^{1}f(\xi)d\xi.$

A semigroup discussion in [22] characterized the finite dimensional
$C^{*}$-invariant subspaces (equivalently the finite co-dimensional
$C$-invariant subspaces) in the setting of translation invariant subspaces of
a certain weighted $L^{2}$ space on $\mathbb{R}$. In Theorem 3.2 we use this
discussion to recast this characterization in the $H^{2}$ setting and show
that every finite dimensional $C^{*}$-invariant subspace is the span of finite
unions of the functions

$(1-z)^{\mu}(\log(1-z))^{j},\quad 0\leqslant j\leqslant k,$

where $\Re\mu>-\tfrac{1}{2}$ and $k\in\mathbb{N}_{0}$ is fixed. We will give
two proofs of this fact. The first, proved in §3, will use some semigroup
techniques from [22] while the second, shown in §5, will be obtained by
describing the finite dimensional $M_{z}^{*}$-invariant subspaces of
$\mathcal{H}^{2}(\mu)$ (which is an interesting result in itself) through
ideas developed in [1].

In §10 we establish a connection between the Cesàro operator and the concept
of a universal operator in the sense of Rota. In Theorem 10.1 we show that
although $C^{*}$ is not universal, there is a bounded analytic function $F$ on
the disk $\\{z:|z-1|<1\\}$ such that $F(C^{*})$ is universal. We prove this by
using an interesting result from [8] which says that certain linear translates
$C_{\varphi}-\lambda I$ of a class of composition operators $C_{\varphi}$ on
$H^{2}$ are universal in the sense of Rota.

Through the results in this paper, we hope to make the case that the
Kriete–Trutt space $\mathcal{H}^{2}(\mu)$ is an important Hilbert space of
analytic functions on $\mathbb{D}$ that is very much worthy of further study.
It has a rich variety of “nonstandard” $M_{z}$-invariant subspaces and the
overall complexity of these invariant subspaces is yet unknown.

##  2\. Semigroups of operators

For each $t\geqslant 0$ let

$\varphi_{t}(z):=e^{-t}z+1-e^{-t},\quad z\in\mathbb{D}.$

These are analytic self maps of the open unit disk $\mathbb{D}$ and
$\varphi_{t}(\mathbb{D})$ is an internally tangent disk at $\xi=1$ (see Figure
1). This collection of maps $\\{\varphi_{t}\\}_{t\geqslant 0}$ defines a
holomorphic flow in that

  1. (i)

$\varphi_{0}(z)=z$ for all $z\in\mathbb{D}$;

  2. (ii)

$\varphi_{t+s}=\varphi_{t}\circ\varphi_{s}$ for all $s,t\geqslant 0$ and all
$z\in\mathbb{D}$; and

  3. (iii)

for any fixed $s\geqslant 0$ and any $z\in\mathbb{D}$,
${\displaystyle\lim_{t\to s}\varphi_{t}(z)=\varphi_{s}(z)}$.

Figure 1. The circles $\varphi_{t}(\mathbb{T})$ for the maps
$\varphi_{t}(z)=e^{-t}z+1-e^{-t}.$ The circles get smaller as $t\to\infty$.

The Littlewood subordination theorem [37, p. 13] implies that the composition
operator

$C_{\varphi_{t}}:H^{2}\to H^{2},\quad C_{\varphi_{t}}f=f\circ\varphi_{t},$

defines a bounded linear operator which satisfies
$\|C_{\varphi_{t}}\|=e^{t/2}$ [12, Thm. 9.4] and
$\sigma(C_{\varphi_{t}})=\\{z:|z|\leqslant e^{t/2}\\}$ [12, Thm. 7.26]. In
fact, one can quickly verify that the collection
$\\{C_{\varphi_{t}}\\}_{t\geqslant 0}$ defines a strongly continuous (or
$C_{0}$) semigroup of operators on $H^{2}$ in that

  1. (i)

$C_{\varphi_{0}}=I$;

  2. (ii)

$C_{\varphi_{t}}C_{\varphi_{s}}=C_{\varphi_{s+t}}$ for all $s,t\geqslant 0$;
and

  3. (ii)

${\displaystyle\lim_{t\to 0^{+}}C_{\varphi_{t}}f=f}$ for all $f\in H^{2}$.

Using an analysis of the generator and co-generator of the $C_{0}$ semigroup
$\\{C_{\varphi_{t}}\\}_{t\geqslant 0}$, the authors in [22, Cor. 2.3] connect
the above composition operators with the adjoint $C^{*}$ of the Cesàro
operator from (1.3) by means of the following formula:

$(C^{*}f)(z)=-\int_{0}^{\infty}e^{-t}(C_{\varphi_{t}}f)(z)dt.$

The formula above has the following important consequence which is key to some
of our results concerning the invariant subspaces of the Cesàro operator.

######  Theorem 2.1.

A closed subspace $\mathcal{M}$ of $H^{2}$ satisfies
$C\mathcal{M}\subseteq\mathcal{M}$ if and only if
$C_{\varphi_{t}}\mathcal{M}^{\perp}\subseteq\mathcal{M}^{\perp}$ for all
$t\geqslant 0$.

We pause to mention the papers [13, 39, 40] that also use semigroups to obtain
information about the Cesàro operator.

In a way, Theorem 2.1 is a Beurling–Lax theorem for the Cesàro operator,
reminiscent of the Beurling–Lax theorem for the invariant subspaces for the
semigroup of shifts $f(x)\mapsto f(x-t)$, $t\in\mathbb{R}$, on
$L^{2}(\mathbb{R})$ [34, p. 204].

There is another associated $C_{0}$ semigroup of shift operators from [22]
which will play an important role in our discussion. Let $\mathbb{C}^{+}$
denote the right half plane

$\mathbb{C}^{+}:=\\{z:\Re z>0\\}$

and let $H^{2}(\mathbb{C}^{+})$ denote the Hardy space of $\mathbb{C}^{+}$
[25, 34]. These are the analytic functions $F$ on $\mathbb{C}^{+}$ for which

$\|F\|_{H^{2}(\mathbb{C}^{+})}:=\Big{(}\sup_{0<x<\infty}\int_{-\infty}^{\infty}|F(x+iy)|^{2}dy\Big{)}^{\frac{1}{2}}<\infty.$

######  Remark 2.2.

As to not be overly pedantic, we will use the notation $H^{2}$ (and not
$H^{2}(\mathbb{D})$) to denote the Hardy space of the disk $\mathbb{D}$.

By a version of the Paley–Wiener theorem [34, p. 203], every $F\in
H^{2}(\mathbb{C}^{+})$ can be realized as

(2.3)
$F(s)=(\mathcal{L}f)(s):=\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}f(x)e^{-sx}dx,\quad
s\in\mathbb{C}^{+},$

(the normalized Laplace transform) where $f\in L^{2}(\mathbb{R}_{+})$ (and
$\mathbb{R}_{+}=(0,\infty)$). Furthermore,

$\|F\|_{H^{2}(\mathbb{C}^{+})}^{2}=\int_{0}^{\infty}|f(x)|^{2}dx.$

Conversely, $\mathcal{L}f\in H^{2}(\mathbb{C}^{+})$ whenever $f\in
L^{2}(\mathbb{R}_{+})$. In other words, the normalized Laplace transform
$\mathcal{L}$ is a unitary operator from $L^{2}(\mathbb{R}_{+})$ onto
$H^{2}(\mathbb{C}^{+})$.

The function

(2.4) $\gamma(z)=\frac{1+z}{1-z},\quad z\in\mathbb{D},$

is a conformal map from $\mathbb{D}$ onto $\mathbb{C}^{+}$ which induces the
unitary operator $\mathcal{U}:H^{2}\to H^{2}(\mathbb{C}^{+})$ given by

(2.5)
$(\mathcal{U}g)(s)=\frac{1}{\sqrt{\pi}(1-s)}g\big{(}\frac{1+s}{1-s}\big{)},\quad
g\in H^{2},$

with

(2.6)
$(\mathcal{U}^{-1}G)(z)=\frac{2\sqrt{\pi}}{1-z}G\big{(}\frac{z-1}{z+1}\big{)},\quad
G\in H^{2}(\mathbb{C}^{+}).$

Moreover, as discussed in [9, Lemma 2.1], if $\psi$ is an analytic self map of
$\mathbb{C}^{+}$ such that $C_{\psi}F=F\circ\psi$ defines a bounded
composition operator on $H^{2}(\mathbb{C}^{+})$ (and this happens if and only
if $\psi(\infty)=\infty$ and

$\lim_{z\to\infty}\frac{z}{\psi(z)}$

exists and is finite) [21, 33]), then with

$\Phi(z):=\gamma^{-1}\circ\psi\circ\gamma,$

which defines an analytic self map of $\mathbb{D}$, the linear transformation

$(\mathcal{L}_{\Phi}f)(z):=\frac{1-\Phi(z)}{1-z}f(\Phi(z))$

defines a weighted composition operator on $H^{2}$ and

$\mathcal{U}^{-1}C_{\psi}\mathcal{U}=\mathcal{L}_{\Phi}.$

Applying this discussion to the family of functions defined on
$\mathbb{C}^{+}$ by

$\psi_{t}(s)=e^{t}s+(e^{t}-1),\quad t\geqslant 0,$

which are analytic self maps of $\mathbb{C}^{+}$ (see Figure 2) that induce
bounded composition operators $C_{\psi_{t}}$ on $H^{2}(\mathbb{C}^{+})$, a
discussion in [8] shows that

$\mathcal{U}C_{\varphi_{t}}\mathcal{U}^{-1}=e^{t}C_{\psi_{t}}\quad\mbox{for
all $t\geqslant 0$}.$

Figure 2. The lines $\psi_{t}(\\{\Re s=0\\})$ for the maps
$\psi_{t}(s)=e^{t}s+(e^{t}-1)$. These lines drift off to infinity as
$t\to\infty$.

Now consider the semigroup $\\{V_{t}\\}_{t\geqslant 0}$ of operators on
$L^{2}(\mathbb{R}_{+})$ given by

$(V_{t}g)(x):=e^{-t}e^{-(1-e^{-t})x}g(e^{-t}x),\quad t\geqslant 0.$

Using the unitary operator

(2.7) $T:L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R}_{+}),\quad(Th)(x):=\frac{1}{\sqrt{x}}h(\log x),\quad x>0,$

(for which $T^{-1}g(y)=e^{y/2}g(e^{y})$), one sees that

$(T^{-1}V_{t}Th)(y)=e^{-t/2}e^{-(1-e^{-t})e^{y}}h(y-t),\quad y\in\mathbb{R}.$

Finally, with the weight function

(2.8) $w(y):=e^{-2(e^{y}-1)}$

(see Figure 3), one can show that the operator

(2.9) $W:L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R},w(y)dy),\quad(Wh)(y):=\frac{h(y)}{\sqrt{w(y)}},$

is unitary and a calculation from [22, Prop. 2.4] shows that

$(W\sigma_{t}W^{-1}f)(y)=f(y-t),\quad f\in L^{2}(\mathbb{R},w(y)dy),\quad
t>0.$

This gives us the following result from [22, Prop. 2.4].

######  Proposition 2.10.

The semigroup $\\{\sigma_{t}\\}_{t\geqslant 0}$ on $L^{2}(\mathbb{R})$ given
by

$(\sigma_{t}h)(y):=e^{-(1-e^{-t})e^{y}}h(y-t),\quad y\in\mathbb{R},$

is unitarily equivalent, via

$\mathfrak{F}:=WT^{-1}\mathcal{L}^{-1}\mathcal{U}:H^{2}\to
L^{2}(\mathbb{R},w(y)dy),$

to the semigroup

$(S_{t}f)(y):=f(y-t)$

acting on $L^{2}(\mathbb{R},w(y)dy)$.

Putting this all together, yields the following result from [22, Thm. 2.5].

######  Theorem 2.11.

A closed subspace $\mathcal{M}$ of $H^{2}$ is invariant for the Cesàro
operator if and only if the closed subspace $\mathfrak{F}\mathcal{M}^{\perp}$
of $L^{2}(\mathbb{R},w(y)dy)$ satisfies
$S_{t}(\mathfrak{F}\mathcal{M}^{\perp})\subseteq\mathfrak{F}\mathcal{M}^{\perp}$
for all $t\geqslant 0$.

For the unweighted $L^{2}(\mathbb{R})$ space, the closed subspaces
$\mathcal{F}$ of $L^{2}(\mathbb{R})$ for which
$S_{t}\mathcal{F}\subseteq\mathcal{F}$ for all $t\geqslant 0$ are described by
the Beurling–Lax theorem (see [34, p. 204]). For a wide class of weights $v$
on $\mathbb{R}$, the same result is true for the $\\{S_{t}\\}_{t\geqslant
0}$-invariant subspaces of $L^{2}(\mathbb{R},v(y)dy)$. However, for the weight
$w$ from (2.8), there is a different lattice of $\\{S_{t}\\}_{t\geqslant 0}$\-
invariant subspaces. Some restrictions on these subspaces were known by Domar
[16, Eqn. 8] and the paper [22] gives some specific examples. The complexity
of the $\\{S_{t}\\}_{t\geqslant 0}$-invariant subspaces stems from the fact
that the weight $w$ is uniformly bounded above and below on every interval
$(-\infty,a)$ but decreases rapidly on $(a,\infty)$. Thus, the space
$L^{2}((-\infty,a),w(y)dy)$ can be seen as a renormed version of
$L^{2}(-\infty,a)$ whereas $L^{2}((a,\infty),w(y)dy)$ is a much larger space
than $L^{2}(a,\infty)$.

Figure 3. The weight function $w(y)=e^{-2(e^{y}-1)}$ from (2.8).

##  3\. Finite co-dimensional invariant subspaces – first proof

An integral computation, using the fact that the adjoint $C^{*}$ of the Cesàro
operator $C$ on $H^{2}$ given by

(3.1) $(C^{*}f)(z)=\frac{1}{1-z}\int_{z}^{1}f(\xi)d\xi,$

shows that the linear span of the functions

$(1-z)^{\mu}(\log(1-z))^{j},\quad 0\leqslant j\leqslant k,$

where $\Re\mu>-\tfrac{1}{2}$ and $k\in\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$
is fixed, is a finite dimensional invariant subspace for $C^{*}$. For example,
the one dimensional space

$\operatorname{span}(1-z)^{\mu}$

is an eigenspace for $C^{*}$, with eigenvalue $(\mu+1)^{-1}$, while the spaces

$\operatorname{span}\\{(1-z)^{\mu},(1-z)^{\nu}\\}\quad\mbox{and}\quad\operatorname{span}\\{(1-z)^{\mu},(1-z)^{\mu}\log(1-z)\\}$

are two dimensional invariant subspaces for $C^{*}$. Theorem 3.2 below
describes all of the finite dimensional $C^{*}$-invariant subspaces. We will
give two proofs of this result. The first, given in this section, uses the
semigroup techniques developed in §2 while the second proof, given in §5, uses
an associated Hilbert space of analytic functions $\mathcal{H}^{2}(\mu)$
developed by Kriete and Trutt in [31, 32].

######  Theorem 3.2.

The finite dimensional $C^{*}$-invariant subspaces of $H^{2}$ are spans of
finite unions of sets of the form

$\\{(1-z)^{\mu}(\log(1-z))^{j},\quad 0\leqslant j\leqslant k\\},$

where $\Re\mu>-\tfrac{1}{2}$ and $k\in\mathbb{N}_{0}$.

###### Proof.

Our starting point is the result in [22, Thm. 2.9] which says that the finite
dimensional $\\{S_{t}\\}_{t\geqslant 0}$-invariant subspaces of
$L^{2}(\mathbb{R},w(y)dy)$ (recall the translations $S_{t}$ from Proposition
2.10) are all spanned by finite unions of sets of the form

$\\{e^{\lambda y},ye^{\lambda y},\ldots y^{n}e^{\lambda y}\\}$

for some $n\in\mathbb{N}_{0}$ and $\Re\lambda>0$. We now translate this into
information about invariant subspaces of the Cesàro operator via our semigroup
discussion from §2.

Under the unitary operator $W^{-1}:L^{2}(\mathbb{R},w(y)dy)\to
L^{2}(\mathbb{R})$ from (2.9), the function $y^{k}e^{\lambda y}$ is sent to

$y^{k}e^{\lambda y}e^{-(e^{y}-1)}.$

Next, under the unitary operator $T:L^{2}(\mathbb{R})\to
L^{2}(\mathbb{R}_{+})$ from (2.7), this last function is sent to

$(\log x)^{k}x^{\lambda-1/2}e^{-(x-1)}.$

We may multiply by a meaningless constant and instead consider the function

$(\log x)^{k}x^{\lambda-1/2}e^{-x}.$

The next step is to use the normalized Laplace transform
$\mathcal{L}:L^{2}(\mathbb{R}_{+})\to H^{2}(\mathbb{C}_{+})$ from (2.3).
Observe that

$\int_{0}^{\infty}e^{-sx}x^{\lambda-1/2}e^{-x}\,dx=(s+1)^{-\lambda-1/2}\Gamma(\lambda+\tfrac{1}{2})$

(where $\Gamma$ is the standard gamma function) and, differentiating with
respect to $\lambda$, we deduce that

$\int_{0}^{\infty}e^{-sx}(\log x)^{k}x^{\lambda-1/2}e^{-x}\,dx$

is equal to

(3.3)
$(s+1)^{-\lambda-1/2}\Big{\\{}a_{0}+a_{1}\log(s+1)+\ldots+a_{k}(\log(s+1))^{k}\Big{\\}},$

for constants $a_{0},\ldots,a_{k}$ (depending on $\lambda$ but not $s$).

Finally, with the unitary operator $\mathcal{U}^{-1}:H^{2}(\mathbb{C}^{+})\to
H^{2}$ from (2.6), the functions from (3.3) are mapped to functions of the
form

$(1-z)^{\lambda-1/2}\Big{\\{}b_{0}+b_{1}\log(1-z)+\ldots+b_{k}(\log(1-z))^{k}\Big{\\}}.$

Note that $\Re(\lambda-\tfrac{1}{2})>-\tfrac{1}{2}$ and we pick up the (only)
eigenvectors $(1-z)^{\mu}$ with $\Re\mu>-\tfrac{1}{2}$.

Next, two-dimensional invariant subspaces are spanned either by
$(1-z)^{\mu_{1}}$ and $(1-z)^{\mu_{2}}$ with $\mu_{1}\neq\mu_{2}$, or by
$(1-z)^{\mu}$ and $(1-z)^{\mu}(\log(1-z))$. Since $1-z\mapsto
1-e^{-t}z-1+e^{-t}=e^{-t}(1-z)$ under the composition semigroup, it is easy to
see that this is an invariant subspace. And so on, for higher dimensions. ∎

######  Corollary 3.4.

The finite co-dimensional closed invariant subspaces of the Cesàro operator on
$H^{2}$ are orthogonal complements of spans of finite unions of sets of the
form

$\\{(1-z)^{\mu}(\log(1-z))^{j},\quad 0\leqslant j\leqslant k\\},$

where $\Re\mu>-\tfrac{1}{2}$ and $k\in\mathbb{N}_{0}$.

##  4\. The Kriete–Trutt transform

A result of Kriete and Trutt [31], used to prove that the Cesàro operator on
$H^{2}$ is subnormal, involves the following transform. Observe that

$\Re\big{(}\frac{w}{1-w}\big{)}>-\tfrac{1}{2}\quad\mbox{for all
$w\in\mathbb{D}$}$

and so the family of functions

(4.1) $q_{w}(z):=(1-z)^{\frac{w}{1-w}},\quad w\in\mathbb{D},$

belong to $H^{2}$. Since

$q_{\frac{n}{n+1}}(z)=(1-z)^{n}\quad\mbox{for all $n\geqslant 0$},$

and the linear span of $(1-z)^{n},n\geqslant 0$, is dense in $H^{2}$, we see
that

(4.2) $\overline{\operatorname{span}}\\{q_{w}:w\in\mathbb{D}\\}=H^{2}.$

The integral formula from (3.1) verifies the eigenvalue identity

$C^{*}q_{w}=(1-w)q_{w}\quad\mbox{for all $w\in\mathbb{D}$}.$

For $f\in H^{2}$, define the function $Kf$ by

(4.3) $(Kf)(w):=\langle f,q_{\overline{w}}\rangle_{H^{2}},\quad
w\in\mathbb{D}.$

Note that $Kf$ is analytic on $\mathbb{D}$ and (4.2) enables us to create a
Hilbert space $\mathcal{H}$ of analytic functions on $\mathbb{D}$, we will
call the Kriete–Trutt space, as the range of $\mathcal{H}$, i.e.,

$\mathcal{H}:=\\{Kf:f\in H^{2}\\}.$

The Hilbert space structure on $\mathcal{H}$ comes from the range norm

$\|Kf\|_{\mathcal{H}}:=\|f\|_{H^{2}}.$

With this norm, the Kriete–Trutt transform $K$ defined in (4.3) is a unitary
map from $H^{2}$ onto $\mathcal{H}$.

For $f\in H^{2}$ and $z\in\mathbb{D}$, use the eigenvalue identity from (3.1)
to see that

$\displaystyle(KCf)(z)$ $\displaystyle=\langle
Cf,q_{\overline{z}}\rangle_{H^{2}}$ $\displaystyle=\langle
f,C^{*}q_{\overline{z}}\rangle_{H^{2}}$ $\displaystyle=\langle
f,(1-\overline{z})q_{\overline{z}}\rangle_{H^{2}}$ $\displaystyle=(1-z)\langle
f,q_{\overline{z}}\rangle_{H^{2}}$ $\displaystyle=(1-z)(Kf)(z).$

This shows that the multiplication operator $M_{z}f=zf$ on $\mathcal{H}$ is
well defined, bounded, and $\|M_{z}\|=1$ (since $\|I-C\\}=1$ from Proposition
1.2). We summarize the discussion above with the following.

######  Proposition 4.4 (Kriete–Trutt [31]).

The operator $K:H^{2}\to\mathcal{H}$ is unitary and $K(I-C)K^{*}=M_{z}$,
multiplication by $z$, on $\mathcal{H}$.

A further analysis of Kriete and Trutt from [31] shows that $\mathcal{H}$
contains the analytic polynomials $\mathbb{C}[z]$ as a dense set and, more
importantly (and quite difficult to prove), there is a positive finite Borel
measure $\mu$ on $\overline{\mathbb{D}}$ such that

(4.5) $\int_{\mathbb{D}}|p|^{2}d\mu=\|p\|^{2}_{\mathcal{H}}\quad\mbox{for all
$p\in\mathbb{C}[z]$}.$

This measure $\mu$ is supported on the sequence of circles

$\gamma_{n}=\Big{\\{}z:\Big{|}z-\frac{n}{n+1}\Big{|}=\frac{1}{n+1}\Big{\\}}\quad\mbox{for
$n\geqslant 0$};$

(see Figure 4).

Figure 4. The circles (from left to right)
$\gamma_{0},\gamma_{1},\gamma_{2},\gamma_{3},\gamma_{4},\gamma_{5}$.

Furthermore, $\mu(\gamma_{n})=2^{-n-1}$ and $\mu|_{\gamma_{n}}$ is mutually
absolutely continuous with respect to arc length measure on $\gamma_{n}$. This
says that $\mathcal{H}=\mathcal{H}^{2}(\mu)$ (the closure of $\mathbb{C}[z]$
in $L^{2}(\mu)$) and thus, $C$ is unitarily equivalent to $I-M_{z}$ on
$\mathcal{H}^{2}(\mu)$. This last operator is subnormal (with $M_{1-z}$ on
$L^{2}(\mu)$ being a normal extension) and hence $C$ is subnormal.

##  5\. Finite co-codimensional invariant subspaces – second proof

Results from [32] show that the invariant subspaces for $M_{z}$,
multiplication by $z$ on $\mathcal{H}^{2}(\mu)$, can be quite complicated and
are not well understood. We will say more about this later on. For now though,
we show that the finite co-dimensional $M_{z}$-invariant subspaces are known.
This is in line with a series of analogous results for some other classical
Banach spaces of analytic functions on $\mathbb{D}$ such as the Bergman and
Hardy spaces. Our result is the following.

######  Theorem 5.1.

A closed subspace $\mathcal{M}$ of $\mathcal{H}^{2}(\mu)$ is a finite co-
dimensional $M_{z}$-invariant subspace if and only if there is a polynomial
$p$ all of whose zeros lie in $\mathbb{D}$ such that
$\mathcal{M}=p\mathcal{H}^{2}(\mu)$.

The proof of this result follows an analogue of an argument from [1] (see also
[11, Ch. 10]) and will be briefly sketched below. To make those arguments
work, we need to first establish that $\mathcal{H}^{2}(\mu)$ has the so-called
division property: if $\lambda\in\mathbb{D}$ and $f\in\mathcal{H}^{2}(\mu)$
with $f(\lambda)=0$, then $f(z)/(z-\lambda)\in\mathcal{H}^{2}{\mu)}$. See a
discussion from [36] for more on the division property for general Banach
spaces of analytic functions.

To prove this, we proceed as follows. The measure $\mu$ from
$\mathcal{H}^{2}(\mu)$ is a finite positive Borel measure on
$\overline{\mathbb{D}}$ with no point masses in $\mathbb{D}$ (recall the
discussion at the end of §4). Moreover, using the fact that
$\mathcal{H}^{2}(\mu)$ is a reproducing kernel Hilbert space of analytic
functions on $\mathbb{D}$ [31, Lemma 3], it follows that for each
$\lambda\in\mathbb{D}$, there is a positive constant $C_{\lambda}$ such that

(5.2) $|f(\lambda)|^{2}\leqslant C_{\lambda}\int|f|^{2}d\mu\quad\mbox{for all
$f\in\mathcal{H}^{2}(\mu)$}.$

Fix $\lambda\in\mathbb{D}$ and, for $f\in\mathcal{H}^{2}(\mu)$, consider the
difference quotient

(5.3) $(Q_{\lambda}f)(z):=\frac{f(z)-f(\lambda)}{z-\lambda},\quad
z\in\mathbb{D}.$

Note that $Q_{\lambda}f$ is analytic on $\mathbb{D}$ (since $f$ is analytic on
$\mathbb{D}$).

Let us first show that $Q_{\lambda}f\in L^{2}(\mu)$. Fix $r>0$ small enough so
that $\overline{D(\lambda,r)}\subseteq\mathbb{D}$ (where
$D(\lambda,r)=\\{z:|z-\lambda|<r\\}$). Then

$\int|Q_{\lambda}f|^{2}d\mu=\int_{\overline{D(\lambda,r)}}|Q_{\lambda}f|^{2}d\mu+\int_{\overline{\mathbb{D}}\setminus\overline{D(\lambda,r)}}|Q_{\lambda}f|^{2}d\mu.$

The first integral above is bounded above by

$\mu(\overline{D(\lambda,r)})\cdot\sup_{z\in
D(\lambda,r)}\big{|}\frac{f(z)-f(\lambda)}{z-\lambda}\big{|}^{2}$

which is finite. The second integral is bounded above by

$\frac{1}{r^{2}}\int_{\overline{\mathbb{D}}}|f-f(\lambda)|^{2}d\mu$

which is also finite.

Next we show that $Q_{\lambda}f\in\mathcal{H}^{2}(\mu)$. Let
$\\{p_{n}\\}_{n\geqslant 1}$ be a sequence polynomials which approximate
$f\in\mathcal{H}^{2}(\mu)$ in $L^{2}(\mu)$ norm. From (5.2) one can see that
$p_{n}\to f$ pointwise in $\mathbb{D}$. By [32, Lemma 2], there is a positive
constant $C_{w}$ such that

(5.4) $\|(z-\lambda)q\|_{\mathcal{H}^{2}(\mu)}\geqslant
C_{w}\|q\|_{\mathcal{H}^{2}(\mu)}\quad\mbox{for all $q\in\mathbb{C}[z]$}.$

Apply this inequality to the sequence of polynomials

$q_{n}(z)=\frac{p_{n}(z)-p_{n}(\lambda)}{z-\lambda},$

to see that $q_{n}\to f$ pointwise in $\mathbb{D}$ and are norm bounded. By
[5, Prop. 2], $q_{n}\to Q_{\lambda}f$ weakly in $\mathcal{H}^{2}(\mu)$. This
means that $Q_{\lambda}f$ belongs to the weak closure of the polynomials.
However, the weak closure and norm closure are the same, since the polynomials
form a convex set (this is sometimes known as Mazur’s theorem [18, Thm.
V.3.13]). Moreover, by the closed graph theorem, the map $f\mapsto
Q_{\lambda}f$ defines a bounded linear operator on $\mathcal{H}^{2}(\mu)$.

The final ingredient needed to use the argument from [1] (see also [11, Ch.
10]) is the following fact.

######  Lemma 5.5.

The vector space $(z-\xi)\mathcal{H}^{2}(\mu)$ is dense in
$\mathcal{H}^{2}(\mu)$ whenever $\xi\in\mathbb{T}$.

###### Proof.

By [32, Theorem 4] the inclusion map $i:H^{2}\to\mathcal{H}^{2}(\mu)$ is
continuous. Thus, there is a positive constant $C$ such that

(5.6) $\|p(z)(z-\xi)-1\|_{H^{2}}\geqslant
C\|p(z)(z-\xi)-1\|_{\mathcal{H}^{2}(\mu)}\quad\mbox{for all
$p\in\mathbb{C}[z]$}.$

Since $(z-\xi)$ is an outer function, Beurling’s theorem [19, Thm. 7.4] shows
that the left hand side of the above can be made as small as desired by
choosing an appropriate $p\in\mathbb{C}[z]$.

The previous paragraph shows that
$1\in\mathcal{S}:=\overline{(z-\xi)\mathcal{H}^{2}(\mu)}$ and the
$M_{z}$-invariance of $\mathcal{S}$ will show that
$\mathcal{S}=\mathcal{H}^{2}(\mu)$. ∎

Below is a rough sketch of the proof of Theorem 5.1. We refer the reader to
[11, Ch. 10] for the full details. Let $q\in\mathbb{C}[z]$ with simple roots
$w_{1},\ldots,w_{d}$ in $\mathbb{D}$. If these roots are not simple, the
argument below can be adjusted by taking appropriate derivatives. First note
that

$q\mathcal{H}^{2}(\mu)=\\{f\in\mathcal{H}^{2}(\mu):f(w_{j})=0\mbox{ for all
$1\leqslant j\leqslant d$}\\}.$

The $\subseteq$ containment follows from inspection. For the $\supseteq$
containment, note that if $f(w_{j})=0$ for all $1\leqslant j\leqslant d$, then

$Q_{w_{1}}Q_{w_{2}}\ldots Q_{w_{d}}f=\frac{c}{q}f$

for some $c\in\mathbb{C}\setminus\\{0\\}$ from which $\supseteq$ holds.

One can argue that
$\operatorname{dim}(\mathcal{H}^{2}(\mu)/q\mathcal{H}^{2}(\mu))=d$, by
observing that the cosets $z^{j}+q\mathcal{H}^{2}(\mu)$, $0\leqslant
j\leqslant d-1$, are linearly independent.

If $\mathcal{M}$ is an $M_{z}$-invariant subspace of $\mathcal{H}^{2}(\mu)$
with $\operatorname{dim}(\mathcal{H}^{2}(\mu)/\mathcal{M})=d$, one can prove
that

$\mathfrak{T}:\mathcal{H}^{2}(\mu)/\mathcal{M}\to\mathcal{H}^{2}(\mu)/\mathcal{M},\quad\mathfrak{T}(f+\mathcal{M})=zf+\mathcal{M}$

is a well defined linear transformation. For any $Q\in\mathbb{C}[z]$, the
$M_{z}$-invariance of $\mathcal{M}$ will show that
$Q(\mathfrak{T})(f+\mathcal{M})=Qf+\mathcal{M}$. Since
$\operatorname{dim}(\mathcal{H}^{2}(\mu)/\mathcal{M})=d$, the cosets
$z^{j}+\mathcal{M}$, $0\leqslant j\leqslant d$, are linearly dependent and so
there is a $Q_{0}\in\mathbb{C}[z]\setminus\\{0\\}$ so that
$Q_{0}\in\mathcal{M}$. Thus, $Q_{0}(\mathfrak{T})$ is zero on
$\mathcal{H}^{2}(\mu)/\mathcal{M}$.

Now factor $Q_{0}=qh$, where $q\in\mathbb{C}[z]$ has its zeros in $\mathbb{D}$
and $h\in\mathbb{C}[z]$ has its roots in $\mathbb{D}^{c}$. Use the facts that
$(z-w)\mathcal{H}^{2}(\mu)=\mathcal{H}^{2}(\mu)$ whenever $|w|>1$ and that
$(z-w)\mathcal{H}^{2}(\mu)$ is dense in $\mathcal{H}^{2}(\mu)$ whenever
$|w|=1$ (Lemma 5.5) to prove that

$q\mathcal{H}^{2}(\mu)\subseteq\overline{Q_{0}\mathcal{H}^{2}(\mu)}\subseteq\mathcal{M}.$

A dimension argument shows that $\mathcal{M}=q\mathcal{H}^{2}(\mu)$ – which
completes the proof.

Since the Kriete–Trutt transform $K$ maps a finite co-dimensional
$C$-invariant subspace of $H^{2}$ to a finite co-dimensional $M_{z}$-invariant
subspace of $\mathcal{H}^{2}(\mu)$, we obtain the following result.

######  Theorem 5.7.

The finite co-dimensional $C$-invariant subspaces of $H^{2}$ are orthogonal
complements of spans of finite unions of sets of the form

$\\{(1-z)^{\mu}(\log(1-z))^{j},\quad 0\leqslant j\leqslant k\\},$

where $\Re\mu>-1/2$ and $k\in\mathbb{N}_{0}$.

###### Proof.

Note that

$\frac{d}{dw}q_{w}(z)=(1-w)\log(1-z)q_{w}(z)$

(and versions of this formula with higher order derivatives) and thus the
finite co-dimensional subspaces of $C$ correspond via the operator $K$ from
(4.3) to spaces of $\mathcal{H}^{2}(\mu)$ of functions which vanish, with an
appropriate number of derivatives, at a finite set of points of $\mathbb{D}$.
∎

##  6\. Other interesting invariant subspaces

Although the finite co-dimensional $M_{z}$-invariant subspaces of the
Kriete–Trutt space $\mathcal{H}^{2}(\mu)$ are thoroughly understood (Theorem
5.1), a general $M_{z}$-invariant subspace of $\mathcal{H}^{2}(\mu)$ can be
quite complicated. Recall the family of $H^{2}$ functions $q_{w}$,
$w\in\mathbb{D}$, from (4.1).

######  Proposition 6.1.

If $A\subseteq\mathbb{D}$ is a zero sequence for $\mathcal{H}^{2}(\mu)$,
meaning there is an $f\in\mathcal{H}^{2}(\mu)\setminus\\{0\\}$ which vanishes
on $A$, then

$\overline{\operatorname{span}}\\{q_{a}:a\in A\\}$

is a nontrivial invariant subspace of the adjoint of the Cesàro operator.

###### Proof.

Notice that

$\displaystyle(\overline{\operatorname{span}}\\{q_{a}:a\in A\\})^{\perp}$
$\displaystyle=\\{f\in H^{2}:\langle f,q_{a}\rangle=0\;\forall a\in A\\}$
$\displaystyle=\\{f\in H^{2}:(Kf)(a)=0\;\forall a\in A\\}$
$\displaystyle=\\{g\in\mathcal{H}^{2}(\mu):g|_{A}\equiv 0\\}\not=0.$

This shows that $\overline{\operatorname{span}}\\{q_{a}:a\in A\\}\not=H^{2}$.
From the fact that

$C^{*}q_{a}=(1-a)q_{a},$

it follows that $\overline{\operatorname{span}}\\{q_{a}:a\in A\\}$ is
$C^{*}$-invariant. ∎

######  Corollary 6.2.

If $A\subseteq\mathbb{D}$ is a zero sequence for $\mathcal{H}^{2}(\mu)$, then

$(\overline{\operatorname{span}}\\{q_{a}:a\in A\\})^{\perp}$

is a nontrivial invariant subspace of the Cesàro operator.

The zero sequences of $\mathcal{H}^{2}(\mu)$ can be quite complicated (see
§11) and do not have a complete characterization.

One can also create interesting “chains” of invariant subspaces for the Cesàro
operator. Since the functions $(1-z)^{\alpha}$, $\Re\alpha>-\tfrac{1}{2}$, are
eigenfunctions for $C^{*}$, we may construct invariant subspaces by joining
together eigenspaces. The following result [32, Cor. 7] will be of use here.

######  Corollary 6.3.

Let $\\{\lambda_{n}\\}_{n\geqslant 1}$ be a sequence of positive numbers
satisfying $\lambda_{n+1}-\lambda_{n}\geqslant\delta>0$ for all
$n\in\mathbb{N}$ and consider the quantity

$b_{r}:=\sum_{\lambda_{n}<r}\frac{1}{\lambda_{n}}-a\log r$

for $r>0$, where $a>0$ is independent of $r$. If $b_{r}$ is unbounded above as
$r\to\infty$ for some $a\geqslant\tfrac{1}{2}$ then

$\operatorname{span}\\{(1-z)^{\lambda_{n}}:n\geqslant 1\\}$

is a dense subset of $H^{2}$. However, if $b_{r}$ is bounded above as
$r\to\infty$ for some $a<\tfrac{1}{2}$ then

$\operatorname{span}\\{(1-z)^{\lambda_{n}}:n\geqslant 1\\}$

is not a dense subset of $H^{2}$.

For example, for each $k\in\mathbb{N}$, the set

$\Lambda_{k}:=\\{k+3n:n\geqslant 1\\}$

yields a proper closed $C^{*}$-invariant subspace $V_{k}$ of $H^{2}$ defined
by

$V_{k}:=\overline{\operatorname{span}}\\{(1-z)^{\lambda_{n}}:n\geqslant 1\\}.$

(take $\tfrac{1}{3}<a<\tfrac{1}{2}$ in the above corollary). However, the join
(the closure of $V_{k}+V_{\ell}$) of any two such subspaces $V_{k}$ and
$V_{\ell}$ with $k\not\equiv\ell$ mod 3 will be all of $H^{2}$ (take
$\tfrac{1}{2}<a<\tfrac{2}{3}$).

We can actually say a bit more about these $C^{*}$-invariant subspaces
$V_{k}$. Translating this into the Kriete–Trutt space $\mathcal{H}^{2}(\mu)$
determined by $f\mapsto Kf$, upon which $C$ is unitarily equivalent to
$M_{1-z}$ (recall Proposition 4.4), the images (under $K$) of the orthogonal
complements of the subspaces $V_{k}$ above (which are clearly invariant under
$M_{1-z}$) are subspaces of $\mathcal{H}^{2}(\mu)$ consisting of functions
that vanish at $w=j/(j+1)$, $j\geqslant 1$.

Now the division property for $\mathcal{H}^{2}(\mu)$ (recall that the
difference quotient $Q_{\lambda}f$ belongs to $\mathcal{H}^{2}(\mu)$ whenever
$f\in\mathcal{H}^{2}(\mu)$ and $\lambda\in\mathbb{D}$, from (5.3)) can be used
to show that, for example,

$V_{1}\supsetneq V_{4}\supsetneq V_{7}\supsetneq\ldots$

(or equivalently, that $(1-z)\not\in V_{4}$, etc.). Indeed, by the division
property, it is possible to construct functions in $\mathcal{H}^{2}(\mu)$ that
vanish at $j/(j+1)$ for $j=4,7,10,\ldots$ but not for $j=1$.

##  7\. The role of model spaces

This section contains a very curious class of invariant subspaces for the
Cesàro operator coming from the well studied theory of model spaces. For
$\alpha>0$ let $u_{\alpha}$ be the atomic inner function

(7.1) $u_{\alpha}(z):=\exp\Big{(}\alpha\frac{z+1}{z-1}\Big{)},\quad
z\in\mathbb{D}.$

Notice that $u_{\alpha}$ is a bounded analytic function on $\mathbb{D}$ and
$|u_{\alpha}(e^{i\theta})|=1$ for all $0<\theta<2\pi$. Thus, $u_{\alpha}H^{2}$
is a closed subspace of $H^{2}$. From here one can consider the model space

$(u_{\alpha}H^{2})^{\perp}:=\\{f\in H^{2}:f\perp u_{\alpha}H^{2}\\}.$

We refer the reader to [23] for the basics of model spaces.

Let us denote $H^{\infty}(\mathbb{D})$ to be the set of bounded analytic
functions on $\mathbb{D}$ and for $g\in H^{\infty}(\mathbb{D})$, define

$\|g\|_{\infty}:=\sup_{z\in\mathbb{D}}|g(z)|.$

######  Theorem 7.2.

For each $\alpha>0$, the model space $(u_{\alpha}H^{2})^{\perp}$ is a closed
invariant subspace of the Cesàro operator.

###### Proof.

By Theorem 2.1 it suffices to show that

$C_{\varphi_{t}}(u_{\alpha}H^{2})\subseteq u_{\alpha}H^{2}\quad\mbox{for all
$t>0$}.$

This comes from a special case of a result of Cowen and Wahl [14, Lemma 5].
For the sake of completeness, we outline the proof here. Observe that

(7.3)
$\Re\Big{(}\frac{\varphi_{t}(z)+1}{\varphi_{t}(z)-1}-\frac{z+1}{z-1}\Big{)}\leqslant
0\quad\mbox{for all $z\in\mathbb{D}$}.$

This follows from the simple fact that $\varphi_{t}(\mathbb{T})$ is a circle
inside $\overline{\mathbb{D}}$ that is internally tangent to $\mathbb{T}$ at
$\\{1\\}$ and that the linear fractional transformation

$z\mapsto\frac{z+1}{z-1}$

maps $\mathbb{D}$ onto the left half plane $-\mathbb{C}^{+}$ (One can also see
(7.3) via Julia’s Lemma). For any $g\in H^{2}$ we have

$\displaystyle
C_{\varphi_{t}}(u_{\alpha}(z)g(z))=\exp\Big{(}\alpha\frac{\varphi_{t}(z)+1}{\varphi_{t}(z)-1}\Big{)}g(\varphi_{t}(z))$
$\displaystyle=\Bigg{(}\exp\Big{(}\alpha\frac{z+1}{z-1}\Big{)}\exp\Big{(}\alpha\Big{(}\frac{\varphi_{t}(z)+1}{\varphi_{t}(z)-1}-\frac{z+1}{z-1}\Big{)}\Big{)}\Bigg{)}g(\varphi_{t}(z))$
(7.4)
$\displaystyle=u_{\alpha}(z)\exp\Big{(}\alpha\Big{(}\frac{\varphi_{t}(z)+1}{\varphi_{t}(z)-1}-\frac{z+1}{z-1}\Big{)}\Big{)}g(\varphi_{t}(z)).$

From (7.3) it follows that the function

$\exp\Big{(}\alpha\Big{(}\frac{\varphi_{t}(z)+1}{\varphi_{t}(z)-1}-\frac{z+1}{z-1}\Big{)}\Big{)}$

belongs to $H^{\infty}(\mathbb{D})$. Since $g(\varphi_{t})\in H^{2}$, (7.4)
shows that

$C_{\varphi_{t}}(u_{\alpha}g)\in H^{2}\quad\mbox{for all $t\geqslant 0$},$

which completes the proof. ∎

As it turns out (see Theorem 7.7 below), the family of model spaces

$\\{(u_{\alpha}H^{2})^{\perp}:\alpha>0\\}$

are the only model spaces that are invariant under the Cesàro operator. To
prove this, we need a few reminders about inner functions.

Recall [19, Ch. 2] that an analytic function $u$ on $\mathbb{D}$ is inner if
$u\in H^{\infty}(\mathbb{D})$ and the radial boundary function

$u(\xi):=\lim_{r\to 1^{-}}u(r\xi),$

which exists for almost every $\xi\in\mathbb{T}$ by Fatou’s theorem [19, Thm.
1.3], satisfies $|u(\xi)|=1$ for almost every $\xi\in\mathbb{T}$. Any inner
function can be factored (uniquely up to a unimodular constant factor) as
$u=Bs_{\mu}$, where $B$ is the Blaschke product

$B(z):=z^{N}\prod_{n=1}^{\infty}\frac{\overline{z_{n}}}{|z_{n}|}\frac{z_{n}-z}{1-\overline{z_{n}}z},\quad
z\in\mathbb{D},$

with zeros at $z=0$ and at $z_{n}\in\mathbb{D}\setminus\\{0\\}$ satisfying the
Blaschke condition $\sum_{n\geqslant 1}(1-|z_{n}|)<\infty$, and $s_{\mu}$ is
the singular inner function

$s_{\mu}(z):=\exp\Big{(}-\int_{\mathbb{T}}\frac{\xi+z}{\xi-z}d\mu(\xi)\Big{)},\quad
z\in\mathbb{D},$

with associated singular measure (with respect to Lebesgue measure $m$) on
$\mathbb{T}$. Note that $u_{\alpha}=s_{\alpha\delta_{1}}$, where $\alpha>0$.
Both $B$ and $s_{\mu}$ are inner. An important tool needed below is boundary
spectrum

(7.5)
$\Sigma(u):=\Big{\\{}\xi\in\mathbb{T}:\liminf_{z\in\mathbb{D},z\to\xi}|u(z)|=0\Big{\\}}$

of $u$. It is known (see [23, Thm. 7.18 and Prop. 7.19]) that $\Sigma(u)$ is a
closed subset of $\mathbb{T}$, $u$ has an analytic continuation across
$\mathbb{T}\setminus\Sigma(u)$, and $\Sigma(u)$ consists of the accumulation
points of the zeros of $B$ and the support of the singular measure $\mu$.

There is a corresponding notion of inner function for $H^{2}(\mathbb{C}^{+})$.
A bounded analytic function $\Theta$ on $\mathbb{C}^{+}$ is inner if
$|\Theta(iy)|=1$ for almost every $y\in\mathbb{R}$. Note that $\Theta$ is
inner for $\mathbb{C}^{+}$ if and only if $u=\Theta\circ\gamma$ is inner for
$\mathbb{D}$, where $\gamma$ is the conformal map from $\mathbb{D}$ to
$\mathbb{C}^{+}$ from (2.4). Though there is an analogous factorization
theorem for inner functions $\Theta$ on $\mathbb{C}^{+}$ as there was for
inner functions $u$ on $\mathbb{D}$ (see [25, Ch. 2] or [34, Ch. 6]), we will
only need the fact that the function

(7.6) $\Theta(s)=e^{-as},\quad s\in\mathbb{C}^{+},$

where $a>0$, is inner on $\mathbb{C}^{+}$.

As we did for the atomic inner functions $u_{\alpha}$, we can define the model
space $(uH^{2})^{\perp}$ for any inner function $u$.

######  Theorem 7.7.

If $u$ is a nonconstant inner function and $(uH^{2})^{\perp}$ is an invariant
subspace for the Cesàro operator, then $u=u_{\alpha}$ for some $\alpha>0$.

###### Proof.

We use Theorem 2.1. For the closed subspace $uH^{2}$ to be invariant under
each composition operator $C_{\varphi_{t}}$, $t\geqslant 0$, it is necessary
for $u$ to divide $u\circ\varphi_{t}$ for each $t$ (meaning that
$(u\circ\varphi_{t})/u\in H^{\infty}(\mathbb{D})$). Thus, if $u$ has a zero
$z_{0}\in\mathbb{D}$ then $u(\varphi_{t}(z_{0}))=0$ for every $t\geqslant 0$.
As $t\to 0^{+}$, these zeros accumulate at $z=0$ which is impossible by the
isolated zeros theorem for analytic functions.

By the classical factorization theorem for inner functions mentioned earlier
(and the argument in the previous paragraph), we may now suppose that $u$ is a
singular inner function $s_{\mu}$ and consider what happens if a point
$\xi_{0}\in\mathbb{T}\setminus\\{1\\}$ lies in its boundary spectrum
$\Sigma(u)$ from (7.5). Let us examine what happens to
$C_{\varphi_{t}}(uH^{2})$ when $t=-\log 2$. Assuming that $u(\varphi_{-\log
2})\in uH^{2}$, we see that $u(z)$ divides the bounded analytic function
$u((1+z)/2)$, say $u((1+z)/2)=u(z)g(z)$ where $g\in H^{\infty}(\mathbb{D})$
(removing inner factors does not change the $H^{\infty}$ norm).

Now observe that

(7.8) $\liminf_{z\in\mathbb{D},z\to\xi_{0}}|u((1+z)/2)|=|u((1+\xi_{0})/2)|\neq
0$

(since $u$ is a singular inner function and thus has no zeros in
$\mathbb{D}$). However, by the definition of boundary spectrum from (7.5),

(7.9) $\liminf_{z\in\mathbb{D},z\to\xi_{0}}|u(z)|=0.$

Since $u((1+z)/2)=u(z)g(z)$, the facts from (7.8) and (7.9) contradict the
fact that $g$ must be bounded near $\xi_{0}$. ∎

From Theorem 7.2, the Cesàro operator has the model spaces
$(u_{\alpha}H^{2})^{\perp}$ as invariant subspaces for $\alpha>0$ (and no
other model spaces are invariant under $C$). Thus, its adjoint $C^{*}$ has the
invariant subspaces $u_{\alpha}H^{2}$ and no other Beurling subspaces $uH^{2}$
are invariant.

With the notation of [22], we can see that these Beurling spaces
$u_{\alpha}H^{2}$ are equivalent to the standard $\\{S_{t}\\}_{t\geqslant
0}$-invariant subspaces

$L^{2}((a,\infty),w(y)dy)\quad\mbox{ with $a=\log\alpha$}.$

For under

$TW^{-1}:L^{2}(\mathbb{R},w(y)dy)\to L^{2}(0,\infty)$

(recall the operators $T$ and $W$ from §2) such a subspace is sent to
$L^{2}(e^{a},\infty)$. Now, under the normalized Laplace transform
$\mathcal{L}$, it maps to the space $\Theta_{a}H^{2}(\mathbb{C}^{+})$, where
$\Theta_{a}$ is the inner function $\Theta_{a}(s)=\exp(-e^{a}s)$ from (7.6).
Finally, under the unitary operator $\mathcal{U}^{-1}:H^{2}(\mathbb{C}^{+})\to
H^{2}$ from (2.6) we arrive at $u_{\alpha}H^{2}$ with $\alpha=e^{a}$.

The model spaces $(u_{\alpha}H^{2})^{\perp}$, $\alpha>0$, also have the
following interesting property.

######  Theorem 7.10.

If $\mathcal{M}$ is a closed invariant subspace for the Cesàro operator such
that $(u_{\alpha}H^{2})^{\perp}\subseteq\mathcal{M}$ for some $\alpha>0$, then
$\mathcal{M}=(u_{\beta}H^{2})^{\perp}$ for some $\beta\geqslant\alpha$.

###### Proof.

As we have just seen, the transformation
$\mathfrak{F}=WT^{-1}\mathcal{L}^{-1}\mathcal{U}$ maps the subspace
$u_{\alpha}H^{2}$ to $L^{2}((a,\infty),w(y)dy)$, with $a=\log\alpha$. This has
the remarkable property that the only translation-invariant subspaces
contained in it are standard, that is, of the form $L^{2}((b,\infty),w(y)dy)$,
for $b\geqslant a$ (this follows from Domar’s work and appears in the
discussion preceding Theorem 2.9 in [22]). On applying $\mathfrak{F}^{-1}$ the
result follows. ∎

We now examine the $C$-invariant subspaces which are contained in a fixed
model space $(u_{\alpha}H^{2})^{\perp}$.

######  Theorem 7.11.

Fix $\alpha>0$. Up to isomorphism, the restriction of $I-C$ to the model space
$(u_{\alpha}H^{2})^{\perp}$ is a shift of multiplicity one and its invariant
subspaces are parameterized by Beurling spaces of the form $\Theta
H^{2}(\mathbb{C}^{+})$ where $\Theta\in H^{\infty}(\mathbb{C}^{+})$ is inner.

###### Proof.

To examine the $C$-invariant subspaces contained in a model space
$(u_{\alpha}H^{2})^{\perp}$, let us return to the translation model on
$L^{2}(\mathbb{R},w(y)dy)$ from §2 and note that the adjoint of the weighted
translation $e^{t/2}S_{t}$ is given by

$(T_{t}g)(y)=e^{t/2}g(y+t)\frac{w(y+t)}{w(y)},\quad y\in\mathbb{R}.$

We seek the common invariant subspaces for $\\{T_{t}\\}_{t\geqslant 0}$ that
are contained in the space $L^{2}((-\infty,a),w(y)\,dy)$.

Note that the weight $w$ is uniformly bounded above and below on the interval
$(-\infty,a)$ (see Figure 3), and thus the operator

$V:L^{2}((-\infty,a),w(y)\,dy)\to
L^{2}(-\infty,a),\quad(Vg)(y):=\frac{g(y)}{w(y)}$

is an isomorphism (but not an isometry).

The question therefore reduces to parameterizing the common invariant
subspaces of the semigroup $\\{VT_{t}V^{-1}\\}_{t\geqslant 0}$ on
$L^{2}(-\infty,a)$, and we see that

$(VT_{t}V^{-1}g)(y)=e^{t/2}g(y+t),$

is a backward translation. Now we can use the Beurling–Lax theorem to find the
lattice of common invariant subspaces, but there is an alternative method that
gives us more information about the action of $C$ on
$(u_{\alpha}H^{2})^{\perp}$.

Namely, we use a reversal operator

$R:L^{2}(-\infty,a)\to L^{2}(0,\infty),\quad(Rf)(y)=f(a-y)$

and then, applying the normalized Laplace transform $\mathcal{L}$, we have the
isomorphism

$\mathcal{L}R:L^{2}(-\infty,a)\to H^{2}(\mathbb{C}^{+}).$

Under this, the weighted left translation $e^{t/2}S_{t}$ becomes

$(\mathcal{L}R)VT_{t}V^{-1}(\mathcal{L}R)^{-1}G(s)=e^{t/2}e^{-st}G(s),\quad
s\in\mathbb{C}^{+}.$

Let $A$ be the infinitesimal generator of the composition semigroup
$\\{C_{\varphi_{t}}\\}_{t\geqslant 0}$, namely

$(Af)(z)=(1-z)f^{\prime}(z),\quad z\in\mathbb{D},$

(see the pioneering work by Berkson and Porta [3], for instance). Then we may
represent the resolvent $(I-A^{*})^{-1}$ of the adjoint semigroup – which is
the restriction of the Cesàro operator – equivalently as multiplication by the
function

$F(s)=\int_{0}^{\infty}e^{-t}e^{t/2}e^{-st}\,dt=\frac{1}{s+\tfrac{1}{2}}.$

It is more productive now to use $I-C$ in which case we have the operator of
multiplication by the function

$\dfrac{s-\tfrac{1}{2}}{s+\tfrac{1}{2}}.$

That is, up to isomorphism, the restriction of $I-C$ is a shift of
multiplicity one and its invariant subspaces are parameterized by Beurling
spaces of the form $\Theta H^{2}(\mathbb{C}^{+})$ where $\Theta\in
H^{\infty}(\mathbb{C}^{+})$ is inner. ∎

From Proposition 1.2 we know that

$\sigma(I-C)=\overline{\mathbb{D}}.$

The construction above yields the following.

######  Corollary 7.12.

For $\alpha>0$, the spectrum of $I-C$ restricted to the model space
$(u_{\alpha}H^{2})^{\perp}$, or any of its invariant subspaces, is
$\overline{\mathbb{D}}$ and

$\operatorname{dim}(\mathcal{M}/(I-C)\mathcal{M})=1$

for all $C$-invariant subspaces
$\mathcal{M}\subseteq(u_{\alpha}H^{2})^{\perp}$.

###### Proof.

From the proof of Theorem 7.11, the operator $I-C$, when restricted to a
closed $C$-invariant subspace of a model space $(u_{\alpha}H^{2})^{\perp}$
will be the same as the spectrum of the operator of multiplication by

$b(s)=\frac{s-\frac{1}{2}}{s+\frac{1}{2}}$

on $\Theta H^{2}(\mathbb{C}^{+})$ for an appropriate inner function $\Theta$
on $\mathbb{C}^{+}$. This is independent of $\Theta$ as we can solve

$(b-\lambda)\Theta f=\Theta g$

for $f,g\in H^{2}(\mathbb{C}^{+})$ precisely when we can solve
$(b-\lambda)f=g$. The spectrum of the analytic Toeplitz operator $M_{b}$ on
$H^{2}(\mathbb{C}^{+})$ is easily seen to be
$\overline{b(\mathbb{C}^{+})}=\overline{\mathbb{D}}$ (this is a special case
of what is sometimes known as Wintner’s theorem [24, p. 365]).

For the second part, notice, by what we have shown above,

$\operatorname{dim}\Big{(}(u_{\alpha}H^{2})^{\perp}/(I-C)(u_{\alpha}H^{2})^{\perp}\Big{)}$

is equal to

$\operatorname{dim}(\Theta H^{2}(\mathbb{C}^{+})/b\Theta
H^{2}(\mathbb{C}^{+})),$

and this last quantity is equal to one. ∎

######  Remark 7.13.

Suppose that $\mathcal{M}$ is a closed $C$-invariant subspace of $H^{2}$. For
fixed $\alpha>0$ consider

$\overline{\mathcal{M}+(u_{\alpha}H^{2})^{\perp}}.$

Notice how this is a $C$-invariant subspace that contains
$(u_{\alpha}H^{2})^{\perp}$. Thus, by Theorem 7.10, either

$\overline{\mathcal{M}+(u_{\alpha}H^{2})^{\perp}}=(u_{\beta}H^{2})^{\perp}$

for some $\beta\geqslant\alpha$ or

$\overline{\mathcal{M}+(u_{\alpha}H^{2})^{\perp}}=H^{2}.$

In the first case we have $\mathcal{M}\subseteq(u_{\beta}H^{2})^{\perp}$ and,
by Theorem 7.11, we can, in a sense, describe $\mathcal{M}$. Notice how the
second possibility above says that $\mathcal{M}^{\perp}\cap
u_{\alpha}H^{2}=\\{0\\}$. An an example of when this occurs, one can consider
a finite co-dimensional invariant subspace $\mathcal{M}$ consisting of
eigenfunctions of $C$. Then $\mathcal{M}^{\perp}\cap u_{\alpha}H^{2}=\\{0\\}$
and so $\overline{\mathcal{M}+(u_{\alpha}H^{2})^{\perp}}=H^{2}$.

##  8\. Model spaces and the Kriete–Trutt space

Results from the previous section bring up further interesting things one can
say about the lattice of invariant subspaces for $M_{z}$ on the Kriete–Trutt
space $\mathcal{H}^{2}(\mu)$. The Kriete–Trutt transform

$K:H^{2}\to\mathcal{H}^{2}(\mu),\quad(Kf)(z)=\langle
f,q_{\overline{w}}\rangle_{H^{2}},$

where

$q_{w}(z)=(1-z)^{\frac{w}{1-w}}$

from (4.3) is a unitary operator which maps $C$-invariant subspaces of $H^{2}$
to $M_{z}$-invariant subspaces of $\mathcal{H}^{2}(\mu)$. What type of
$M_{z}$-invariant subspace is $K((u_{\alpha}H^{2})^{\perp})$?

To begin to unpack this, we require a few further detains about
$\mathcal{H}^{2}(\mu)$ from [32]. The first [32, Thm. 4] (and mentioned
earlier in §5) is that the inclusion map

$i:H^{2}\to\mathcal{H}^{2}(\mu),\quad i(f)=f,$

is bounded. The second [32, Lemma 4] is that if $u$ is an inner function whose
boundary spectrum $\Sigma(u)$ from (7.5) does not contain the point $1$, then
the multiplication operator $f\mapsto uf$ is bounded below on
$\mathcal{H}^{2}(\mu)$ and so $u\mathcal{H}^{2}(\mu)$ is a closed subspace of
$\mathcal{H}^{2}(\mu)$. Third [32, Cor. 1], for inner functions $A$ and $B$,

$\overline{A\mathcal{H}^{2}(\mu)}=\overline{B\mathcal{H}^{2}(\mu)}\quad\mbox{if
and only if $u_{s}A=cu_{t}B$}$

for some $|c|=1$ and some $s,t\geqslant 0$, where $u_{t}$ is the standard
atomic inner function from (7.1). Thus, an inner function $B$ is cyclic for
$M_{z}$ on $\mathcal{H}^{2}(\mu)$ if and only if $B=cu_{t}$. They also note
that

$K\frac{1}{1-(1-a)z}=e^{t}u_{t},\quad t=-(\log a)/2,\quad 0<a<1.$

This shows that

$\frac{1}{1-(1-a)z}$

is a cyclic vector for $C$, meaning

$\overline{\operatorname{span}}\Big{\\{}C^{n}\frac{1}{1-(1-a)z}:n\geqslant
0\Big{\\}}=H^{2}.$

Fourth [32, Thm. 5], for a closed $M_{z}$-invariant subspace $\mathcal{M}$ of
$\mathcal{H}^{2}(\mu)$, we have that
$\mathcal{M}=\overline{u\mathcal{H}^{2}(\mu)}$ for some inner function $u$ if
and only if $\mathcal{M}\cap i(H^{2})$ is dense in $\mathcal{M}$.

One can describe $K((u_{\alpha}H^{2})^{\perp})$ as a “liminf” space as
follows. A theorem of Tumarkin [17, Thm. 4.3.1] says that for a given
$\alpha>0$, there is a sequence $\\{B_{n}\\}_{n\geqslant 1}$ of finite
Blaschke products with simple zeros such that

$(u_{\alpha}H^{2})^{\perp}=\varliminf(B_{n}H^{2})^{\perp},$

meaning that given any $f\in(u_{\alpha}H^{2})^{\perp}$ there are
$f_{n}\in(B_{n}H^{2})^{\perp}$ such that $f_{n}\to f$ in $H^{2}$ norm. Note
that

$(B_{n}H^{2})^{\perp}=\operatorname{span}\Big{\\{}\frac{1}{1-\overline{\lambda}z}:\lambda\in
B_{n}^{-1}(\\{0\\})\Big{\\}}$

is a convenient space of rational functions [23, Prop. 5.6].

If $\lambda\in\mathbb{D}$ and $k_{\lambda}(z)=(1-\overline{\lambda}z)^{-1}$,
then [32, p. 203]

$(Kk_{\lambda})(z)=(1-\lambda)^{\frac{z}{1-z}}.$

When $0<\lambda<1$,

(8.1)
$(Kk_{\lambda})(z)=c_{\lambda}\exp\Big{(}-a_{\lambda}\frac{1+z}{1-z}\Big{)}$

for some positive constants $c_{\lambda}$ and $a_{\lambda}$. As mentioned
earlier, $k_{\lambda}$ is cyclic for $C$ and so $Kk_{\lambda}$ is cyclic for
$M_{z}$ on $\mathcal{H}^{2}(\mu)$.

The discussion above says that

$K((u_{\alpha}H^{2})^{\perp})=\varliminf\Big{\\{}\operatorname{span}\big{\\{}(1-\lambda)^{\frac{z}{1-z}}:\lambda\in
B_{n}^{-1}(\\{0\\})\big{\\}}\Big{\\}},$

meaning that given any $g\in K((u_{\alpha}H^{2})^{\perp})$, there are

$g_{n}\in\operatorname{span}\big{\\{}(1-\lambda)^{\frac{z}{1-z}}:\lambda\in
B_{n}^{-1}(\\{0\\})\big{\\}}$

with $g_{n}\to g$ in $\mathcal{H}^{2}(\mu)$. But what does
$K((u_{\alpha}H^{2})^{\perp})$ really contain? Is

$K((u_{\alpha}H^{2})^{\perp})=\overline{u\mathcal{H}^{2}(\mu)}$

for some inner function $u$? The answer is no .

######  Proposition 8.2.

For any $\alpha>0$, the $M_{z}$-invariant subspace
$K((u_{\alpha}H^{2})^{\perp})$ of $\mathcal{H}^{2}(\mu)$ is not equal to
$\overline{u\mathcal{H}^{2}(\mu)}$ for any inner function $u$.

###### Proof.

First note that $\overline{uH^{2}}=\overline{u\mathcal{H}^{2}(\mu)}$, since
$H^{2}$ is a dense subspace of $\mathcal{H}^{2}(\mu)$ and multiplication by
$u$ is a bounded operator on both $H^{2}$ and $\mathcal{H}^{2}(\mu)$.

Certainly

$K((u_{\alpha}H^{2})^{\perp})\not=\overline{u_{t}H^{2}}$

for any $t>0$ since, by (8.1), $u_{t}$ corresponds to a Cauchy kernel via $K$
and these are cyclic vectors for $C$.

Is $K((u_{\alpha}H^{2})^{\perp})=\overline{uH^{2}}$ where the inner function
$u$ has a Blaschke factor? No, since otherwise, $u(\lambda_{0})=0$ for some
$\lambda_{0}\in\mathbb{D}$ and thus

$0=(Kf)(\lambda_{0})=\langle
f(w),(1-w)^{\frac{\lambda_{0}}{1-\lambda_{0}}}\rangle_{H^{2}}\quad\mbox{for
all $f\in(u_{\alpha}H^{2})^{\perp}$}.$

This would mean that

$(1-w)^{\frac{\lambda_{0}}{1-\lambda_{0}}}\in u_{\alpha}H^{2}.$

But this cannot be since $(1-w)^{\lambda_{0}/(1-\lambda_{0})}$ is outer
(Beurling’s theorem).

Thus, if
$K((u_{\alpha}H^{2})^{\perp})=\overline{uH^{2}}=\overline{u\mathcal{H}^{2}(\mu)}$,
then the inner function $u$ must be a singular inner function $s_{\nu}$. Now,
unless the measure $\nu$ consists of a single atom, $u$ has nonconstant inner
factors $J_{1}$ and $J_{2}$ with associated singular measures with disjoint
supports. In this case, it follows that the $C$-invariant subspaces containing
$\overline{uH^{2}}$ are not totally ordered by inclusion, and so, by Theorem
7.7, we cannot have $K((u_{\alpha}H^{2})^{\perp})=\overline{uH^{2}}$.

Finally, suppose that
$K((u_{\alpha}H^{2})^{\perp})=\overline{uH^{2}}=u\mathcal{H}^{2}(\mu)$ where
$u$ is a singular inner function corresponding to a point mass at
$p\in\mathbb{T}\setminus\\{1\\}$. From our discussion above, note that
$u\mathcal{H}^{2}(\mu)$ is closed in $\mathcal{H}^{2}(\mu)$. Then there is a
function $g\in(u_{\alpha}H^{2})^{\perp}$ such that $Kg=u$. Now, and this is
interesting on its own, if $f\in H^{2}$, then

$\langle S^{*}f,(1-w)^{\frac{\bar{z}}{1-\bar{z}}}\rangle_{H^{2}}=\langle
f,(w-1+1)(1-w)^{\frac{\bar{z}}{1-\bar{z}}}\rangle_{H^{2}}.$

So that $K(S^{*}f-f)(z)$ is $-Kf(\varphi(z))$ where

$\frac{\varphi(z)}{1-\varphi(z)}=\frac{z}{1-z}+1;$

that is, $\varphi(z)=1/(2-z)$. But now $S^{*}g-g$ belongs to
$(u_{\alpha}H^{2})^{\perp}$ and so $K(S^{*}g-g)\in u\mathcal{H}^{2}(\mu)$.
However, $-Kg(\varphi(z))$ is analytic in an open neighborhood of $p$, which
is a contradiction unless $S^{*}g-g=0$, which is impossible. ∎

######  Remark 8.3.

The $M_{z}$-invariant subspaces
$\mathcal{M}_{\alpha}:=K((u_{\alpha}H^{2})^{\perp})$ are not the “standard”
Beurling-type $M_{z}$-invariant subspaces $\overline{u\mathcal{H}^{2}(\mu)}$.
Moreover, via Theorem 7.10, they also inherit the interesting property that if
$\mathcal{M}$ is an $M_{z}$-invariant subspace that contains
$\mathcal{M}_{\alpha}$ for some $\alpha$, then
$\mathcal{M}=\mathcal{M}_{\beta}$ for some $\beta\geqslant\alpha$.

##  9\. Cyclic subspaces

Recall that $I-C$ is unitarily equivalent to multiplication by the function
$z$ on the Kriete–Trutt space $\mathcal{H}^{2}(\mu)$ and is bounded below
(recall (5.4)). Thus, we may discuss the index

$\operatorname{dim}(\mathcal{M}/z\mathcal{M})$

of an invariant subspace $\mathcal{M}$. See [36] for some general facts about
the index of an invariant subspace of $M_{z}$ on Banach spaces of analytic
functions. We start with some preliminary observations about $C$ itself, which
is not bounded below.

######  Proposition 9.1.

Let $\mathcal{M}$ be a closed subspace of $\mathcal{H}^{2}(\mu)$ that is
invariant under $M_{1-z}$, and hence $M_{z}$. Then $(1-z)\mathcal{M}$ is dense
in $\mathcal{M}$.

###### Proof.

Let

$p_{n}(z)=1-(z+z^{2}+...+z^{n})/n.$

Since $p_{n}(1)=0$, we have that $p_{n}(z)=(1-z)q_{n}(z)$ for some
$q_{n}\in\mathbb{C}[z]$.

By Parseval’s theorem, observe that $\|p_{n}h-h\|_{H^{2}}\to 0$ for every
monomial $h(z)=z^{k}$ (hence, by (5.6), in $\mathcal{H}^{2}(\mu)$ norm as
well). Use this, along with the facts that $|p_{n}(z)|\leqslant 2$ for all
$z\in\mathbb{D}$ and the density of $\mathbb{C}[z]$ in $\mathcal{H}^{2}(\mu)$,
to see that

$(1-z)q_{n}h=p_{n}h\to h$

in $\mathcal{H}^{2}(\mu)$ norm for all $h\in\mathcal{H}^{2}(\mu)$. Now, if
$h\in\mathcal{M}$ then

$(1-z)q_{n}(z)h\in(1-z)\mathcal{M}$

and the result follows. ∎

######  Corollary 9.2.

Let $\mathcal{M}\subseteq H^{2}$ be a closed invariant subspace for the Cesàro
operator $C$. Then $C\mathcal{M}$ is dense in $\mathcal{M}$.

Recall that a vector ${\bf x}$ in a Hilbert space $\mathcal{H}$ is a cyclic
vector for a bounded operator $T$ on $\mathcal{H}$ if

$\overline{\operatorname{span}}\\{T^{n}{\bf x}:n\geqslant 0\\}=\mathcal{H}.$

From Theorem 7.11 we know that the operator $I-C$, when restricted to
$(u_{\alpha}H^{2})^{\perp}$, is similar to the operator $M_{b}$ of
multiplication by $b(s)=(s-1/2)/(s+1/2)$ on $H^{2}(\mathbb{C}^{+})$. From this
we can determine some cyclic vectors for $C$ when restricted to the model
spaces $(u_{\alpha}H^{2})^{\perp}$, $\alpha>0$.

######  Proposition 9.3.

For every $\lambda\in\mathbb{C}^{+}$ the function

$\frac{1}{s+\lambda}$

is a cyclic vector for $M_{b}$ on $H^{2}(\mathbb{C}^{+})$. Moreover, for every
inner function $\Theta\in H^{\infty}(\mathbb{C}^{+})$ the function

$\frac{\Theta(s)}{s+\lambda}$

is cyclic for the restriction of $M_{b}$ to $\Theta H^{2}(\mathbb{C}^{+})$.

###### Proof.

There is an orthonormal basis of $H^{2}(\mathbb{C}^{+})$ given by

$\frac{1}{\sqrt{2\pi}}\frac{(s-\frac{1}{2})^{n}}{(s+\frac{1}{2})^{n+1}},\quad
n\geqslant 0.$

This can be seen most easily by transforming to $\mathbb{D}$ using the
conformal mapping

$s\mapsto\frac{s-\frac{1}{2}}{s+\frac{1}{2}}.$

With respect to this basis, $M_{b}$ is is a unilateral shift (so we see again
that its spectrum is $\overline{\mathbb{D}}$). This tells us that $1/(s+1/2)$
is cyclic, as a nonzero multiple of the first basis vector. Now a vector
remains cyclic for $M_{b}$ when multiplied by an invertible function in
$H^{\infty}(\mathbb{C}^{+})$, and in particular,

$\frac{s+\frac{1}{2}}{s+\lambda}.$

(We have many other choices but these ones lead to simpler expressions.) The
result for $\Theta H^{2}(\mathbb{C}^{+})$ follows easily. ∎

We now track the vectors from Proposition 9.3 back to the original $H^{2}$
setting. For $\alpha>0$ write $a=\log\alpha$. There is a surjective
isomorphism $L_{a}$ between the spaces $L^{2}((-\infty,a),w(y)\,dy)$ and
$H^{2}(\mathbb{C}^{+})$ given by

$(L_{a}f)(s)=\int_{-\infty}^{a}f(y)e^{-s(a-y)}\,dy.$

Taking $f(y)=e^{\lambda y}$, we conclude from Proposition 9.3 that this is a
cyclic vector for the operator $L_{a}^{-1}M_{b}L_{a}$ on $L^{2}(-\infty,a)$
corresponding to $I-C$ (and hence for the operator corresponding to $C$) as in
Theorem 7.11. Note that, since we are restricting $w$ to $(-\infty,a)$, on
which it is bounded above and below (Figure 3), we can drop all references to
$w$ from now on.

We now recall the operator $(Th)(x)=x^{-1/2}h(\log x)$ from (2.7) to transfer
this to $L^{2}(0,\alpha)$ and obtain the cyclic vectors

$f_{\lambda}(x)=x^{-\frac{1}{2}}x^{\lambda}.$

Let us calculate the Laplace transform of $f_{\lambda}\in L^{2}(0,\alpha)$,
i.e.,

$\int_{0}^{\alpha}x^{-1/2}x^{\lambda}e^{-sx}\,dx$

to obtain a cyclic vector for the unitarily equivalent form of $C$ on the
model space $(e^{-\alpha s}H^{2}(\mathbb{C}^{+}))^{\perp}$. The answer is too
complicated to be usable for most values of $\lambda$ but if we take
$\lambda=\tfrac{1}{2}$ we obtain

$\dfrac{1-e^{-\alpha s}}{s}.$

Finally, using $\mathcal{U}^{-1}$ defined in (2.6), we obtain a cyclic vector
for $C$ on $(u_{\alpha}H^{2})^{\perp}$, namely

$g_{\alpha}(z)=\frac{1-u_{\alpha}(z)}{1+z}.$

The singularity at $-1$ for $g_{\alpha}$ is removable. Also note that
$g_{\alpha}$ belongs to $(u_{\alpha}H^{2})^{\perp}$ since if
$h=(z+1)k\in(z+1)H^{2}$ then

$\langle g_{\alpha},u_{\alpha}h\rangle_{H^{2}}=\langle
1-u_{\alpha},u_{\alpha}zk\rangle_{H^{2}}=0,$

and this is true for a dense set of $h$ hence for all $h\in H^{2}$. We
summarize this discussion with the following.

######  Proposition 9.4.

For $\alpha>0$, the vector

$\frac{1-u_{\alpha}(z)}{1+z}$

belongs to $(u_{\alpha}H^{2})^{\perp}$ and is a cyclic vector for the
restriction of $C$ to $(u_{\alpha}H^{2})^{\perp}$.

Even though $C|_{\mathcal{M}}$, where $\mathcal{M}$ is a closed $C$-invariant
subspace contained in $(u_{\alpha}H^{2})^{\perp}$, is cyclic, it seems that
finding cyclic vectors cannot be done explicitly, except perhaps for some very
simple inner functions $\Theta$ in Proposition 9.3.

For $w\in\mathbb{D}$ and

$\lambda=\frac{\bar{w}}{1-\bar{w}}\in\\{z\in\mathbb{C}:\mathop{\rm
Re}\nolimits z>-\tfrac{1}{2}\\},$

we want an expression for

$\langle g_{\alpha},(1-z)^{\lambda}\rangle_{H^{2}},$

the cyclic vector for $M_{z}$ when restricted to
$K((u_{\alpha}H^{2})^{\perp})$. Since $\mathcal{U}$ from (2.5) is a unitary
operator, and thus preserves inner products, we may do the calculation in
$H^{2}(\mathbb{C}^{+})$. Thus, to within irrelevant constants, we obtain

$\left\langle\frac{1-e^{-\alpha
s}}{s},\frac{2^{\lambda}}{(s+1)^{\lambda+1}}\right\rangle_{H^{2}(\mathbb{C}^{+})}.$

The first term in the inner product is the Laplace transform of the
characteristic function of $(0,\alpha)$, as we have seen already, while the
second is the Laplace transform of the function

$\frac{2^{\lambda}e^{-t}t^{\lambda}}{\Gamma(\lambda+1)},$

where $\Gamma$ is the standard gamma function.

Thus, since the normalized Laplace transform is unitary, we obtain the
Kriete–Trutt cyclic vectors given by the function

$U_{\alpha}(w):=\frac{2^{w/(1-w)}}{\Gamma(1/(1-w))}\int_{0}^{\alpha}e^{-t}t^{w/(1-w)}\,dt,\quad
w\in\mathbb{D}.$

This is summarized with the following.

######  Proposition 9.5.

For $\alpha>0$, the function $U_{\alpha}$ is a cyclic vector for $M_{z}$ when
restricted to $K((u_{\alpha}H^{2})^{\perp})$.

The function $U_{\alpha}$ does not have a usable formula. However, we can say
a few things.

######  Proposition 9.6.

The function $U_{\alpha}$ has no zeros on $\mathbb{D}$.

###### Proof.

This will follow from the general fact that if $f\in(u_{\alpha}H^{2})^{\perp}$
and

$\overline{\operatorname{span}}\\{C^{n}f:n\geqslant
0\\}=(u_{\alpha}H^{2})^{\perp},$

then $Kf$, which will be a generator for the $M_{z}$-invariant subspace
$K((u_{\alpha}H^{2})^{\perp})$, has no zeros in $\mathbb{D}$.

Indeed, suppose $(Kf)(\lambda)=0$ for some $\lambda\in\mathbb{D}$. Since
$KCf(z)=(1-z)Kf(z)$ (Proposition 4.4), every function in

$\overline{\operatorname{span}}\\{(1-z)^{n}Kf(z):n\geqslant 0\\}$

would have a zero at $\lambda$ (recall (5.2)). The previous identity would say
that every function in

$\overline{\operatorname{span}}\\{KC^{n}f:n\geqslant 0\\}$

vanishes at $\lambda$. However, $f$ is a cyclic vector for
$C|_{(u_{\alpha}H^{2})^{\perp}}$ and so every function from
$K((u_{\alpha}H^{2})^{\perp})$ would vanish at $\lambda$. It follows that

$\langle
f,(1-w)^{\frac{\overline{\lambda}}{1-\overline{\lambda}}}\rangle_{H^{2}}=0\quad\mbox{for
all $f\in(u_{\alpha}H^{2})^{\perp}$}$

which, in turn, implies that

$(1-z)^{\frac{\overline{\lambda}}{1-\overline{\lambda}}}\in u_{\alpha}H^{2}.$

This last statement is impossible since
$(1-z)^{\frac{\overline{\lambda}}{1-\overline{\lambda}}}$ is an outer function
and this can not belong to any Beurling subspace $u_{\alpha}H^{2}$. ∎

######  Proposition 9.7.

The function $U_{\alpha}$ does not belong to $H^{2}$.

###### Proof.

From Proposition 8.2, we know that $K((u_{\alpha}H^{2})^{\perp})$ is never
equal to $\overline{u\mathcal{H}^{2}(\mu)}$ for any inner function $u$. If
$U_{\alpha}\in H^{2}$, then, since $\\{pU_{\alpha}:p\in\mathbb{C}[z]\\}$ is
dense in $K((u_{\alpha}H^{2})^{\perp})$ (recall Proposition 9.5), then [32,
Thm. 5] (mentioned above) would say that
$K((u_{\alpha}H^{2})^{\perp})=\overline{u\mathcal{H}^{2}(\mu)}$, which we know
is not the case. ∎

##  10\. A connection to universal operators

A bounded operator $U$ on a Hilbert space $\mathcal{H}$ is universal for
$\mathcal{H}$ if given any bounded operator $T$ on $\mathcal{H}$, there exists
a constant $a\not=0$ and an closed invariant subspace $\mathcal{M}$ for $U$
such that $U|_{\mathcal{M}}$ is similar to $aT$. Universal operators thus have
an extraordinarily rich class of invariant subspaces. Though universal
operators might not seem to exist at all, a theorem of Caradus [7] (see also
[10, Ch. 8]) shows they exist in abundance. Indeed, if $\mathcal{H}$ is an
infinite dimensional separable Hilbert space and $U$ is a bounded operator on
$\mathcal{H}$ such that $\ker(U)$ is infinite dimensional and $U$ is
surjective, then $U$ is universal.

For example, if $u$ is an inner function that is not a finite Blaschke
product, then the co-analytic Toeplitz operator $T_{\overline{u}}$ on $H^{2}$
satisfies Caradus’ criterion and is thus universal [24, Prop. 16.7.1].
Although neither $C$ nor $C^{*}$ is universal (because they are injective), we
do have the following.

######  Theorem 10.1.

There exists a bounded analytic function $F$ on the disk $D(1,1)$ such that
$F(C^{*})$ is universal.

Note that $C^{*}$ has an $H^{\infty}$ functional calculus (see the remarks in
§11) and so the operator $F(C^{*})$ makes sense. Our proof of Theorem 10.1
uses a universality result for composition operators from [8], an analysis
from [15], and a commutant result from [38].

The following theorem is from Carmo and Noor [8, Theorem 4.3].

######  Theorem 10.2.

Let $\varphi$ be a hyperbolic non-automorphism of $\mathbb{D}$ with a fixed
point $\zeta\in\mathbb{T}$ and the other outside the closed unit disk
$\overline{\mathbb{D}}$, possibly at $\infty$. If
$a:=\varphi^{\prime}(\zeta)\in(0,1)$, then for each $\lambda$ with
$0<|\lambda|<a^{-1/2}$ the operator $C_{\varphi}-\lambda I$ is universal on
the Hardy space $H^{2}$.

In our case, each of the symbols in the semigroup
$\varphi_{t}(z)=e^{-t}z+1-e^{-t},t>0,$ from §2 induces a composition operator
with a universal translate. Here $a_{t}=\varphi_{t}^{\prime}(1)=e^{-t}$ and
the fixed points are $1\in\mathbb{T}$ and at $\infty$. Clearly, the lattice of
invariant subspaces of $C_{\varphi_{t}}-\lambda I$ and $C_{\varphi_{t}}$ are
the same.

###### Proof of Theorem 10.1.

To help with the typesetting, fix $t>0$ and let $\alpha=e^{-t}$. The
composition operator induced by $\varphi_{t}$ becomes $C_{\alpha
z+(1-\alpha)}$ and, with respect to the usual orthonormal basis
$\\{z^{n}\\}_{n\geqslant 0}$ for $H^{2}$, has the matrix representation

$\begin{bmatrix}1&\alpha&\alpha^{2}&\alpha^{3}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&(1-\alpha)&2\alpha(1-\alpha)&3\alpha^{2}(1-\alpha)&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&(1-\alpha)^{2}&3\alpha(1-\alpha)^{2}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&(1-\alpha)^{3}&\cdots\\\
\vdots&\vdots\ &\vdots&\vdots&\ddots\end{bmatrix}.$

See a paper of Deddens [15, p. 862] for the details of this. Note that $C^{*}$
has the matrix representation

$\begin{bmatrix}1&\frac{1}{2}&\frac{1}{3}&\frac{1}{4}&\frac{1}{5}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&\frac{1}{2}&\frac{1}{3}&\frac{1}{4}&\frac{1}{5}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\frac{1}{3}&\frac{1}{4}&\frac{1}{5}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\frac{1}{4}&\frac{1}{5}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\frac{1}{5}&\cdots\\\\[-3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix}.$

One can check, as Deddens did, that the two operators above commute. By
Theorem 10.2 there exists a constant $\beta_{\alpha}$ for which

$C_{\alpha z+(1-\alpha)}-\beta_{\alpha}I$

is universal. Moreover, this operator also commutes with $C^{*}$ and has the
matrix representation

$\begin{bmatrix}1-\beta_{\alpha}&\alpha&\alpha^{2}&\alpha^{3}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&(1-\alpha)-\beta_{\alpha}&2\alpha(1-\alpha)&3\alpha^{2}(1-\alpha)&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&(1-\alpha)^{2}-\beta_{\alpha}&3\alpha(1-\alpha)^{2}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&(1-\alpha)^{3}-\beta_{\alpha}&\cdots\\\
\vdots&\vdots\ &\vdots&\vdots&\ddots\end{bmatrix}.$

By a result of Shields and Wallen [38] (which describes the commutant of $C$ –
and thus $C^{*}$), there is a bounded analytic function $F$ on $D(1,1)$ such
that

$F(C^{*})=C_{\alpha z+(1-\alpha)}-\beta_{\alpha}I.$

This says that $F(C^{*})$ is universal. ∎

By a trick from [15, p. 863], we can actually compute $F$. Note that

$F(C^{*})=\begin{bmatrix}F(1)&*&*&*&*&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&F(\frac{1}{2})&*&*&*&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&F(\frac{1}{3})&*&*&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&F(\frac{1}{4})&*&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&F(\frac{1}{5})&\cdots\\\\[-3.0pt]
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix},$

where the $*$ entries in the matrix above are not important. Comparing the
diagonal entries of the matrix representations of $C_{\alpha
z+(1-\alpha)}-\beta_{\alpha}I$ and $F(C^{*})$, we see that

$F(\tfrac{1}{n})=(1-\alpha)^{n-1}-\beta_{\alpha}\quad\mbox{for all $n\geqslant
1,$}$

and thus, since $\\{\tfrac{1}{n}\\}_{n\geqslant 1}$ is not a Blaschke sequence
for the disk $D(1,1)$, we see that

$F(z)=(1-\alpha)^{1/z-1}-\beta_{\alpha},\quad z\in D(1,1).$

######  Remark 10.3.

The lattice of invariant subspaces for $C^{*}$ is strictly contained in the
corresponding lattice for $F(C^{*})$. For the function $F$ is not injective on
$\sigma_{p}(C^{*})=D(1,1)$ (Proposition 1.2), and so there exist distinct
eigenvalues $\lambda$, $\mu$ for $C^{*}$, with corresponding eigenvectors
$g,h\in H^{2}$, such that $F(\lambda)=F(\mu)$. Now the one-dimensional space
$\mathcal{E}$ spanned by $g+h$ is not invariant under $C^{*}$, but
$F(C^{*})(g+h)=F(\lambda)g+F(\mu)h$ and so $\mathcal{E}$ is an invariant
subspace for $F(C^{*})$.

##  11\. Some final remarks

So how complicated is the invariant subspace structure for the Cesàro
operator? Since the invariant subspaces for the Cesàro operator are in one-to-
one (and order preserving) correspondence with the invariant subspaces for the
multiplication operator $M_{z}$ on the Kriete–Trutt space
$\mathcal{H}^{2}(\mu)$, the complexity of the invariant subspaces of the the
Cesàro operator is reflected in the complexity of the $M_{z}$-invariant
subspaces of $\mathcal{H}^{2}(\mu)$.

So what kind of space is $\mathcal{H}^{2}(\mu)$? Is it more like the Hardy
space $H^{2}$, where $\operatorname{dim}(\mathcal{M}/z\mathcal{M})=1$ for any
nonzero $M_{z}$-invariant subspace $\mathcal{M}$ of $H^{2}$ and these
invariant subspaces are cyclic? From Kriete and Trutt, and from §9, there are
$M_{z}$-invariant subspaces which are Beurling-like, and of the form
$\overline{u\mathcal{H}^{2}(\mu)}$, and those which are not.

Or, possibly, is $\mathcal{H}^{2}(\mu)$ more like the Bergman space $A^{2}$ of
analytic functions on $\mathbb{D}$ which are square integrable with respect to
area measure [20]. Here the $M_{z}$–invariant subspaces of $A^{2}$ are known
to be very complicated and will probably never have a suitable description. A
salient indication of the complexity of the $M_{z}$–invariant subspaces of
$A^{2}$ comes from the following fact: given any
$n\in\mathbb{N}\cup\\{\infty\\}$, there is an $M_{z}$-invariant subspace
$\mathcal{M}$ of $A^{2}$ for which the index
$\operatorname{dim}(\mathcal{M}/z\mathcal{M})$ is equal to $n$. Does the
Kriete–Trutt space $\mathcal{H}^{2}(\mu)$ have this property? One might be
tempted to use the trick from [29] and use the notions of sampling and
interpolating for $A^{2}$ to construct two “zero-based” invariant subspaces

$\mathcal{M}_{Z_{j}}:=\\{f\in A^{2}:f|_{Z_{j}}\equiv 0\\},\quad j=1,2,$

(where $Z_{j}$ are sequences in $\mathbb{D}$) such that
$\mathcal{M}_{Z_{1}}+\mathcal{M}_{Z_{2}}$ is closed in $A^{2}$. Then the
invariant subspace $\mathcal{M}=\mathcal{M}_{Z_{1}}+\mathcal{M}_{Z_{2}}$ will
satisfy $\operatorname{dim}(\mathcal{M}/z\mathcal{M})=2$ [28]. However, one
can show there is no corresponding notion of “sampling” in
$\mathcal{H}^{2}(\mu)$ as there is in $A^{2}$.

Kriete and Trutt [32, Cor. 6] showed that there are two sequences $A$ and $B$
in $\mathbb{D}$ for which $\mathcal{M}_{A}$ and $\mathcal{M}_{B}$ are both
nonzero but

(11.1) $\mathcal{M}_{A}\cap\mathcal{M}_{B}=\\{0\\}.$

In other words, the union of the two zero sequences $A$ and $B$ does not form
a zero sequence (which is what happens in $A^{2}$ [30]). Can one use this to
produce an index $2$ invariant subspace of $\mathcal{H}^{2}(\mu)$ as one did
for the Bergman space?

Some further parallels between the Hardy shift, the Bergman shift, and the
operator $T=I-C$ can be observed as follows:

  * •

$\|T\|=\|T^{*}\|=1$ and $\sigma(T)=\sigma(T^{*})=\overline{\mathbb{D}}$
(Proposition 1.2);

  * •

Using the matrix representations of $C$ and $C^{*}$ from (1.1), a calculation
shows that

$TT^{*}=\begin{bmatrix}0&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&\frac{1}{2}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\frac{2}{3}&{\color[rgb]{.75,.75,.75}0}&\cdots\\\\[3.0pt]
{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&{\color[rgb]{.75,.75,.75}0}&\frac{3}{4}&\cdots\\\\[3.0pt]
\vdots&\vdots&\vdots&\vdots&\ddots\\\ \end{bmatrix}$

and so $\|T^{*}{\bf x}\|<\|{\bf x}\|$ for all ${\bf x}\in\ell^{2}$. Hence
$T^{*}$ is completely non-unitary.

  * •

By [2, Prop. 4.6], the properties above imply that $T$ (and hence $T^{*}$) has
an isometric $H^{\infty}$ functional calculus. That is, the operators lie in
the class $\mathbb{A}$.

  * •

As shown in [6], every operator of class $\mathbb{A}$ is reflexive, which
means that every operator fixing all $T$’s invariant subspaces can be
approximated in the weak operator topology by polynomials in $T$. This is a
property that guarantees a rich subspace lattice.

There are other curious facts from [32] which only add to the mystery of
whether or not $\mathcal{H}^{2}(\mu)$ is closer to a Hardy space or a Bergman
space. If $f\in\mathcal{H}^{2}(\mu)\setminus\\{0\\}$ vanishes on a sequence
$\\{z_{n}\\}_{n\geqslant 1}$ in $\mathbb{D}$ which does not accumulate at the
point $1$, then $\\{z_{n}\\}_{n\geqslant 1}$ must satisfy the Blaschke
condition $\sum_{n\geqslant 1}(1-|z_{n}|)<\infty$ (which seems to point
towards $\mathcal{H}^{2}(\mu)$ being closer to $H^{2}$). Moreover, if $J$ is a
closed arc of $\mathbb{T}$ which does not contain $1$ and
$f\in\mathcal{H}^{2}(\mu)\setminus\\{0\\}$, then $\int_{J}\log|f|dm>-\infty$
[19, Thm. 2.2] (again pointing towards $\mathcal{H}^{2}(\mu)$ being closer to
$H^{2}$). However, there are $f\in\mathcal{H}^{2}(\mu)\setminus\\{0\\}$ which
vanish on sequences $\\{z_{n}\\}_{n\geqslant 1}$ which do not satisfy the
Blaschke condition (pointing towards $\mathcal{H}^{2}(\mu)$ being closer to
$A^{2}$). In addition, there is the property (11.1) mentioned earlier, again
pointing towards $\mathcal{H}^{2}(\mu)$ being closer to the $A^{2}$.

Furthermore [31, Lemma 2], the functions

$g_{0}(z)=1;$
$g_{n}(z)=\frac{1}{(1-z)^{n}}z(z-\tfrac{1}{2})\cdots(z-\tfrac{n}{n-1}),\quad
n\geqslant 1$

form an orthonormal basis for $\mathcal{H}^{2}(\mu)$. In particular,

$\frac{1}{(1-z)^{n}}\in\mathcal{H}^{2}(\mu)\quad\mbox{for all $n\geqslant 1$}$

which shows that $\mathcal{H}^{2}(\mu)$ is a space much bigger than $H^{2}$
(and even bigger than $A^{2}$). It is often the case that “large” Hilbert
spaces of analytic functions on $\mathbb{D}$ have a rich class of
$M_{z}$-invariant subspaces.

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