# The Hardy-Weyl algebra

Jim Agler and John E. McCarthy Partially supported by National Science
Foundation Grant DMS 2054199

###### Abstract

We study the algebra ${\mathcal{A}}$ generated by the Hardy operator $H$ and
the operator $M_{x}$ of multiplication by $x$ on $L^{2}[0,1]$. We call
${\mathcal{A}}$ the Hardy-Weyl algebra. We show that its quotient by the
compact operators is isomorphic to the algebra of functions that are
continuous on $\Lambda$ and analytic on the interior of $\Lambda$ for a planar
set $\Lambda$ = $[-1,0]\cup\overline{\mathbb{D}(1,1)}$, which we call the
lollipop. We find a Toeplitz-like short exact sequence for the $C^{*}$-algebra
generated by ${\mathcal{A}}$.

We study the operator $Z=H-M_{x}$, show that its point spectrum is
$(-1,0]\cup\mathbb{D}(1,1)$, and that the eigenvalues grow in multiplicity as
the points move to $0$ from the left.

##  1 Introduction

The classical Weyl algebra is generated by the operators of multiplication by
$x$, denoted $M_{x}$, and differentiation, denoted by $D$. These operators
satisfy the commutator relation

$DM_{x}-M_{x}D\ =\ 1.$ (1.1)

The algebra generated by these relations has been studied extensively in both
algebra and operator theory—see e.g. the books [14, 12, 19, 9]. In operator
theory, the study is complicated by the fact that no bounded operators satisfy
(1.1). In this note, we shall study the associated algebras that arise when
one replaces the differentiation operator $D$ by a bounded integration
operator.

The Hardy operator $H$ is defined on $L^{2}[0,1]$ by

$Hf(x)\ =\ \frac{1}{x}\int_{0}^{x}f(t)dt.$

Let $V=M_{x}H$ denote the Volterra operator

$V:f\mapsto\int_{0}^{x}f(t)dt,$

which is a right inverse of $D$. These operators give rise to a new set of
relations instead of (1.1), namely

$\displaystyle VM_{x}-M_{x}V$ $\displaystyle\ =\ $ $\displaystyle-V^{2}$
$\displaystyle HM_{x}-M_{x}H$ $\displaystyle=$ $\displaystyle-HM_{x}H\ =\
-HV.$

We shall use $L^{2}$ to mean $L^{2}[0,1]$ throughout.

######  Definition 1.2.

Let ${\mathcal{A}}$ denote the closure of the unital algebra generated by $H$
and $M_{x}$ in the norm topology of $B(L^{2})$. We call ${\mathcal{A}}$ the
Hardy-Weyl algebra.

When is an operator in ${\mathcal{A}}$? Can all the elements be described?
Since the commutator of $H$ and $M_{x}$ is compact, we first describe the
quotient of ${\mathcal{A}}$ by the compact operators. Define $\Lambda$, a
subset of the plane, by

$\Lambda\ =\ [-1,0]\cup\overline{\mathbb{D}(1,1)}.$

We call $\Lambda$ the lollipop. The lollipop algebra $A(\Lambda)$ is the
Banach algebra of functions that are continuous on $\Lambda$ and analytic on
${\rm int}(\Lambda)$, equipped with the maximum modulus norm. Let
${\mathcal{K}}$ denote the compact operators on $L^{2}$, and let
${\mathcal{K}}_{{\mathcal{A}}}={\mathcal{K}}\cap{\mathcal{A}}$,

######  Theorem 1.3.

There is a Banach algebra isomorphism $\gamma$ from $A(\Lambda)$ onto
${\mathcal{A}}/{\mathcal{K}}_{{\mathcal{A}}}$. It is given by

$\gamma(f)\ =\ f|_{[-1,0]}(-M_{x})+f|_{\overline{\mathbb{D}(1,1)}}(H)-f(0).$

Let $\theta:{\mathcal{A}}\to A(\Lambda)$ be defined by
$\theta(T)=\gamma^{-1}([T])$, where $[T]$ is the projection of $T$ onto
${\mathcal{A}}/{\mathcal{K}}_{{\mathcal{A}}}$. As a corollary to Theorem 1.3
we obtain that every element $T$ of ${\mathcal{A}}$ can be written uniquely as

$T\ =\ M_{\phi}+g(H)+K,$ (1.4)

where $\phi\in C([0,1])$, $g$ is in $A(\overline{\mathbb{D}(1,1)})$,
$\phi(0)=g(0)$, and $K\in{\mathcal{K}}_{{\mathcal{A}}}$.

If we look at $C^{*}({\mathcal{A}})$, the $C^{*}$-algebra generated by
${\mathcal{A}}$, we get something similar to the Toeplitz algebra short exact
sequence. See [4, 8, 18] for some recent results on the Toeplitz algebra, and
[3, 10, 15, 16] for some applications. The lollipop algebra is replaced by the
functions that are continuous on the boundary.

######  Theorem 1.5.

There is a short exact sequence of $C^{*}$-algebras

$0\to{\mathcal{K}}\to C^{*}({\mathcal{A}})\to C(\partial\Lambda)\to 0.$

In addition to studying the algebra generated by $H$ and $M_{x}$, one may also
study the smaller algebra generated by $V$ and $M_{x}$.

######  Definition 1.6.

The algebra ${\mathcal{A}}_{0}$ is the norm-closed unital algebra generated by
$V$ and $M_{x}$.

One can see that ${\mathcal{A}}_{0}$ is a proper subalgebra of ${\mathcal{A}}$
by noting that its quotient by the compact operators is isomorphic to
$C([0,1])$.

The lattice of closed invariant subspaces of the Volterra operator $V$ was
shown independently by Brodskii and Donoghue [5, 6] to be

${\rm Lat}(V)\ =\ \Big{\\{}\\{f\in L^{2}:f=0{\rm\ on\
}[0,s]\\}:s\in[0,1]\Big{\\}}.$

We let ${\rm AlgLat}(V)$ denote the set of bounded operators on $L^{2}$ that
leave invariant every element of ${\rm LatV}$. This is a large algebra—it
includes all right translation operators for example. It is described by the
following theorem of Radjavi and Rosenthal [17, Example 9.26].

######  Theorem 1.7.

The algebra ${\rm AlgLat}(V)$ is the weak operator topology closure of the
algebra generated by $V$ and $M_{x}$.

It follows from Theorem 1.7 that $H$ is in the WOT closure of
${\mathcal{A}}_{0}$, and hence ${\mathcal{A}}$ and ${\mathcal{A}}_{0}$ have
the same WOT closure. However, no extra compact operators are added in the WOT
closure.

######  Theorem 1.8.

${\rm AlgLat}(V)\cap{\mathcal{K}}\ \subset\ {\mathcal{A}}_{0}.$

In Section 6 we consider the operator $Z=H-M_{x}$. It follows from Theorem 1.3
that $[Z]$ generates ${\mathcal{A}}/{\mathcal{K}}_{{\mathcal{A}}}$. We show
$Z$ has a surprisingly rich collection of eigenvectors.

######  Theorem 1.9.

Let $Z=H-M_{x}$. Then

$\sigma_{p}(Z)\ =\ (-1,0]\cup\mathbb{D}(1,1).$

The algebraic multiplicity of the eigenvalues of $Z$ on the stick $(-1,0]$
increases as $\lambda\to 0^{-}$, and hence operators $X$ in the closed algebra
generated by $Z$ have the property that $\theta(X)$ is not just in
$A(\Lambda)$, but is smoother. We prove:

######  Theorem 1.10.

Let $X$ be in the norm-closed algebra generated by $Z$. Then $\theta(X)$ is
$C^{m}$ on $(-\frac{2}{2m+1},0)$.

##  2 Preliminaries

######  Definition 2.1.

A monomial operator is a bounded linear operator $T:L^{2}[0,1]\to L^{2}[0,1]$
with the property that there exist constants $c_{n}$ and $p_{n}$ so that

$T:x^{n}\mapsto c_{n}\,x^{p_{n}}\qquad\forall n\in{\mathbb{N}}.$ (2.2)

We call it a flat monomial operator if there exists some $\tau$ so that
$p_{n}=n+\tau$ for all $n$.

In [2] we showed that every flat monomial operator is in ${\rm AlgLat}(V)$,
and hence by Theorem 1.7 in the weak closure of ${\mathcal{A}}_{0}$. The
Volterra operator $V$ is Hilbert-Schmidt, and hence compact. See e.g. [11] for
a proof.

Let $H^{2}$ denote the Hardy space of the unit disk, and
$k_{w}(z)=\frac{1}{1-\bar{w}z}$ the Szegő kernel. We shall let $S:f(z)\mapsto
zf(z)$ denote the unilateral shift on $H^{2}$. Let

$\beta(z)\ =\ \frac{1}{2-z},$

and let $C_{\beta}:f\mapsto f\circ\beta$ denote the composition operator of
composing with $\beta$.

There is a unitary $U:L^{2}\to H^{2}$ that is defined on monomials by

$U:x^{s}\mapsto\frac{1}{s+1}k_{\frac{\bar{s}}{\bar{s}+1}},$ (2.3)

and extended by linearity and continuity to the whole space. If $T\in
B(L^{2})$, we shall let $\widehat{T}$ denote $UTU^{*}$. It is easy to see that
$U$ is unitary, as it preserves inner products. In [1] we prove that $U$ is
given by the formula

$Uf(z)\ =\ \frac{1}{1-z}\int_{0}^{1}f(x)x^{\frac{z}{1-z}}dx,$ (2.4)

and show that

$\displaystyle\widehat{M_{x}}$ $\displaystyle\ =\ $ $\displaystyle
S^{*}C_{\beta}^{*}$ $\displaystyle\widehat{V}$ $\displaystyle\ =\ $
$\displaystyle(1-S^{*})C_{\beta}^{*}$ $\displaystyle\widehat{H}$
$\displaystyle=$ $\displaystyle 1-S^{*}.$

If $X$ is a compact subset of $\mathbb{C}$, we shall let $C(X)$ denote the
Banach algebra of functions that are continuous on $X$, with the maximum
modulus norm. We shall let $A(X)$ denote the subalgebra of functions that are
continuous on $X$ and analytic on the interior of $X$, and $P(X)$ denote the
closure of the polynomials in $C(X)$. A theorem of Mergelyan [13] says that if
the complement of $X$ is connected, then $A(X)=P(X)$.

##  3 The Calkin Hardy-Weyl Algebra

We let $\mathcal{K}$ denote the ideal of compact operators acting on $L^{2}$
and set

$\mathcal{K}_{0}={\mathcal{A}}\cap\mathcal{K}.$

Evidently, $\mathcal{K}_{0}$ is a 2-sided ideal in ${\mathcal{A}}$.
Consequently, we may define an algebra $\mathcal{C}$, the _Calkin Hardy-Weyl
algebra_ , by

$\mathcal{C}={\mathcal{A}}/\mathcal{K}_{0}$

If $T\in{\mathcal{A}}$ we let $[T]$ denote the coset of $T$ in $\mathcal{C}$,
i.e.,

$[T]=\\{T+K\,|\,K\in\mathcal{K}_{0}\\}.$

######  Proposition 3.1.

$\mathcal{C}$ is an abelian Banach algebra.

###### Proof.

That $\mathcal{C}$ is a Banach algebra follows from the fact that
$\mathcal{K}_{0}$ is closed in ${\mathcal{A}}$. To see that $\mathcal{C}$ is
abelian, observe that as $M_{x}$ and $H$ generate ${\mathcal{A}}$, $[M_{x}]$
and $[H]$ generate $\mathcal{C}$. Furthermore, as
$M_{x}H=V\in\mathcal{K}_{0}$,

$[M_{x}][H]=0.$ (3.2)

Likewise, as $HM_{x}=(1-H)V\in\mathcal{K}_{0}$,

$[H][M_{x}]=0,$ (3.3)

so that in particular we have that

$[M_{x}][H]=[H][M_{x}].$

As $[M_{x}]$ and $[H]$ commute and generate $\mathcal{C}$, $\mathcal{C}$ is
abelian. ∎

###  3.1 A Uniform Algebra Homeomorphically Isomorphic to $\mathcal{C}$

We begin by defining an algebra by gluing together two simpler algebras whose
maximal ideal spaces overlap at a single point. Let

${\mathcal{P}}\ =\ \\{\ f=(f_{-},f_{+})\ :\ f_{-}\in C([-1,0]),\ f_{+}\in
A(\overline{\mathbb{D}(1,1)})\text{ and }f_{-}(0)=f_{+}(0)\ \\}$

where we view ${\mathcal{P}}$ as an algebra with the operations

$cf=(cf_{-},cf_{+}),\ f+g=(f_{-}+g_{-},f_{+}+g_{+}),\text{and
}fg=(f_{-}g_{-},f_{+}g_{+}),$

and the norm

$\|f\|=\max\big{\\{}\max_{t\in[-1,0]}|f_{-}(t)|,\max_{z\in(\overline{\mathbb{D}(1,1)}}|f_{+}(z)|\
\big{\\}}.$

We abuse notation by letting

$f(0)=f_{-}(0)$

when $f\in{\mathcal{P}}$.

We note that if $f\in C([-1,0])$, then as $-M_{x}$ is self-adjoint and has
spectrum equal to $[-1,0]$, we may form the operator $f(-M_{x})$. Likewise, as
$H$ is cosubnormal and has spectrum equal to $\overline{\mathbb{D}(1,1)}$, if
$g\in A(\overline{\mathbb{D}(1,1)})$, then we may form the operator $g(H)$.
Concretely,

$f(-M_{x})=M_{f(-x)}\ \text{ and }\ \widehat{g(H)}=M_{h}^{*},$

where $h(z)=\overline{g(1-\bar{z})}$, and $M_{h}$ denotes multiplication by
$h$.

######  Lemma 3.4.

If $f\in C([-1,0])$ and $g\in A(\overline{\mathbb{D}(1,1)})$, then

$[f(-M_{x})][g(H)]=g(0)[f(-M_{x})]+f(0)[g(H)]-f(0)g(0).$

###### Proof.

Since $[-1,0]$ is a spectral set for $-M_{x}$ and $\overline{\mathbb{D}(1,1)}$
is a spectral set for $H$ it suffices to prove the lemma in the special case
when $f$ and $g$ are polynomials. Let $f(x)=f(0)+xf_{1}(x)$ and
$g(x)=g(0)+xg_{1}(x)$. Using (3.2) and (3.3) we see that

$\displaystyle[f(-M_{x})][g(H)]$
$\displaystyle=\big{(}[f(0)]+[-M_{x}][f_{1}(-M_{x})]\big{)}\big{(}[g(0)]+[H][g_{1}(H)]\big{)}$
$\displaystyle=f(0)g(0)+g(0)[-M_{x}][f_{1}(-M_{x})]+f(0)[H][g_{1}(H)]$
$\displaystyle=f(0)g(0)+g(0)[f(-M_{x})-f(0)]+f(0)[g(H)-g(0)]$
$\displaystyle=g(0)[f(-M_{x})]+f(0)[g(H)]-f(0)g(0).$

∎

If $f\in{\mathcal{P}}$ we define $\gamma(f)\in{\mathcal{A}}$ by the formula

$\gamma(f)=f_{-}(-M_{x})+f_{+}(H)-f(0).$

We also define $\Gamma:{\mathcal{P}}\to\mathcal{C}$ by the formula

$\Gamma(f)=[\gamma(f)]$

######  Proposition 3.5.

$\Gamma$ is a continuous unital homomorphism.

###### Proof.

$\gamma$ is linear and $\gamma(1)=1$. Therefore, $\Gamma$ is linear and
$\Gamma(1)=1$. Also,

$\displaystyle\|\Gamma(f)\|$ $\displaystyle=\|[\gamma(f)]\|$
$\displaystyle\leq\|\gamma(f)\|$
$\displaystyle=\|f_{-}(-M_{x})+f_{+}(H)-f(0)\|$
$\displaystyle\leq\|f_{-}(-M_{x})\|+\|f_{+}(H)\|+|f(0)|$
$\displaystyle=\max_{t\in[-1,0]}|f_{-}(t)|+\max_{z\in\overline{\mathbb{D}(1,1)}}|f_{+}(z)|+|f(0)|$
$\displaystyle\leq 3\|f\|,$

so $\Gamma$ is continuous.

Finally, to see that $\Gamma$ preserves products, fix $f,g\in{\mathcal{P}}$.

$\displaystyle\Gamma(f)\Gamma(g)$ $\displaystyle=[\gamma(f)]\ [\gamma(g)]$
$\displaystyle=[f_{-}(-M_{x})+f_{+}(H)-f(0)]\ [g_{-}(-M_{x})+g_{+}(H)-g(0)]$
$\displaystyle=\Big{(}[f_{-}(-M_{x})][g_{-}(-M_{x})]+[f_{+}(H)][g_{+}(H)]\Big{)}$
$\displaystyle\ \
+\Big{(}[f_{-}(-M_{x})][g_{+}(H)]+[g_{-}(-M_{x})][f_{+}(H)]\Big{)}$
$\displaystyle\ \
-\Big{(}f(0)[g_{-}(-M_{x})+g_{+}(H)]+g(0)[f_{-}(-M_{x})+f_{+}(H)]\Big{)}$
$\displaystyle\ \ +f(0)g(0)$ $\displaystyle=\ \ A+B-C+f(0)g(0).$

But

$\displaystyle A$
$\displaystyle=\Big{(}[f_{-}(-M_{x})][g_{-}(-M_{x})]+[f_{+}(H)][g_{+}(H)]\Big{)}$
$\displaystyle=\Big{(}[f_{-}g_{-}(-M_{x})]+[f_{+}g_{+}(H)]-f(0)g(0)\Big{)}+f(0)g(0)$
$\displaystyle=[\gamma(fg)]+f(0)g(0)$ $\displaystyle=\Gamma(fg)+f(0)g(0),$

and using Lemma 3.4, we see that

$\displaystyle B$
$\displaystyle=\Big{(}[f_{-}(-M_{x})][g_{+}(H)]\Big{)}+\Big{(}[g_{-}(-M_{x})][f_{+}(H)]\Big{)}$
$\displaystyle=\Big{(}g(0)[f_{-}(-M_{x})]+[f(0)g_{+}(H)]-f(0)g(0)\Big{)}+\Big{(}f(0)[g_{-}(-M_{x})]+[g(0)f_{+}(H)]-f(0)g(0)\Big{)}$
$\displaystyle=C-2f(0)g(0).$

Therefore,

$\displaystyle\Gamma(f)\Gamma(g)$ $\displaystyle=A+B-C+f(0)g(0)$
$\displaystyle=(\Gamma(fg)+f(0)g(0))+(C-2f(0)g(0))-C+f(0)g(0)$
$\displaystyle=\Gamma(fg).$

∎

######  Lemma 3.6.

If $p$ is a polynomial in two variables and we define $f\in{\mathcal{P}}$ by
letting

$f_{-}(t)=p(t,0),\ t\in[-1,0]\ \ \ \ \text{and}\ \ \ \ f_{+}(z)=p(0,z),\
z\in\overline{\mathbb{D}(1,1)},$

then

$p([-M_{x}],[H])=\Gamma(f).$

###### Proof.

If $p=p(x,y)$ is a polynomial in two variables and we let

$q(x,y)=p(x,y)-p(x,0)-p(0,y)+p(0,0),$

then

$p(x,y)=p(x,0)+p(0,y)-p(0,0)+q(x,y)$

and

$q([-M_{x}],[H])=0.$

Therefore,

$\displaystyle p([-M_{x}],[H])$ $\displaystyle=p([-M_{x}],0)+p(0,[H])-p(0,0)$
$\displaystyle=f_{-}([M_{x}])+f_{+}([h])-f(0)$ $\displaystyle=[\gamma(f)]$
$\displaystyle=\Gamma(f).$

∎

######  Corollary 3.7.

The range of $\Gamma$ is dense in $\mathcal{C}$.

###### Proof.

This follows immediately from Lemma 3.6 by recalling that $[-M_{x}]$ and $[H]$
generate $\mathcal{C}$ (cf. proof of Proposition 3.1). ∎

Lemma 3.6 suggests that we consider the subset ${\mathcal{P}}_{0}$ of
${\mathcal{P}}$ defined by

${\mathcal{P}}_{0}=\\{f\in{\mathcal{P}}\,|\,f_{-}\text{ and }f_{+}\text{ are
polynomials}\\}.$

We note that it follows from the facts that the polynomials are dense in both
$C([-1,0])$ and $A(\overline{\mathbb{D}(1,1)})$ that ${\mathcal{P}}_{0}$ is
dense in ${\mathcal{P}}$.

######  Lemma 3.8.

If $s\in[-1,0]$, then

$|f_{-}(s)|\leq\|\Gamma(f)\|$ (3.9)

for all $f\in{\mathcal{P}}$.

###### Proof.

As $f$ is continuous, it suffices to prove the lemma under the assumption that
$s\in(-1,0)$. For $n$ satisfying $1/n<\min\\{s,1-s\\}$ we define a unit vector
$\chi_{n}\in L^{2}$ by the formula

$\chi_{n}(t)=\left\\{\begin{array}[]{ll}\sqrt{\frac{n}{2}}&\mbox{if }|t-s|\leq
1/n\\\ 0&\mbox{if }|t-s|>1/n\end{array}\right.$

We observe that the mean value theorem for integrals implies that

$\lim_{n\to\infty}\langle g\,\chi_{n},\chi_{n}\rangle=g(s)$

whenever $g\in C([0,1])$. Also, as $\chi_{n}\to 0$ weakly,

$\lim_{n\to\infty}\|K\chi_{n}\|=0$

whenever $K$ is a compact operator acting on $L^{2}$. In particular, as $V$ is
compact and $V\chi_{n}(t)=0$ when $t\in[0,s-1/n)$,

$\lim_{n\to\infty}\|H\chi_{n}\|=\lim_{n\to\infty}\|M_{1/x}V\chi_{n}\|=0.$

More generally, if $q$ is a polynomial and $q(0)=0$, write $q(z)=zr(z)$, and
we get

$\lim_{n\to\infty}\|q(H)\chi_{n}\|=\lim_{n\to\infty}\|r(H)\ H\chi_{n}\|=0.$

Now fix $f\in{\mathcal{P}}_{0}$ and a compact operator $K$ acting on $L^{2}$.
Using the observations in the previous paragraph we have that

$\displaystyle\langle(\gamma(f)+K)\ \chi_{n},\chi_{n}\rangle$ $\displaystyle=\
$ $\displaystyle\langle(f_{-}(-M_{x})+f_{+}(H)-f(0)+K)\
\chi_{n},\chi_{n}\rangle$ $\displaystyle=\ $ $\displaystyle\langle
f_{-}(-x)\chi_{n},\chi_{n}\rangle+\langle(f_{+}-f_{+}(0))(H)\
\chi_{n},\chi_{n}\rangle+\langle K\ \chi_{n},\chi_{n}\rangle$
$\displaystyle\to\ $ $\displaystyle\ \ \ \ \
f_{-}(s)\qquad\qquad+\qquad\qquad\ \ 0\qquad\qquad\qquad\ +\qquad 0$
$\displaystyle=\ $ $\displaystyle\ \ f_{-}(s).$

Therefore, as $\|\chi_{n}\|=1$,

$|f_{-}(s)|\leq\|\gamma(f)+K\|$

for all $f\in{\mathcal{P}}_{0}$ and $K$ any compact operator acting on
$L^{2}$. Hence,

$|f_{-}(s)|\leq\inf_{K\in{\mathcal{K}}_{0}}\|\gamma(f)+K\|=\|\Gamma(f)\|$

for all $f\in{\mathcal{P}}_{0}$. As $\Gamma$ is continuous and
${\mathcal{P}}_{0}$ is dense in ${\mathcal{P}}$, it follows that (3.9) holds
for all $f\in{\mathcal{P}}$. ∎

######  Lemma 3.10.

If $z\in\overline{\mathbb{D}(1,1)}$, then

$|f_{+}(z)|\leq\|\Gamma(f)\|$ (3.11)

for all $f\in{\mathcal{P}}$.

###### Proof.

We first observe that as $f_{+}\in A(\overline{\mathbb{D}(1,1)})$, by the
Maximum Modulus Theorem it suffices to prove the lemma under the assumption
that $z=1+\tau$ where $\tau\in\mathbb{T}\setminus\\{-1\\}$. For
$\alpha\in\mathbb{D}$, let

$\Upsilon_{\alpha}=U^{*}\frac{k_{-\bar{\alpha}}}{\|k_{-\bar{\alpha}}\|},$

where $U$ is as in (2.3). Clearly, as
$k_{-\bar{\alpha}}/\|k_{-\bar{\alpha}}\|$ is a unit vector and $U^{*}$ is
unitary, $\Upsilon_{\alpha}$ is a unit vector. Also, as

$(1-S^{*})\frac{k_{-\bar{\alpha}}}{\|k_{-\bar{\alpha}}\|}=(1+\alpha)\frac{k_{-\bar{\alpha}}}{\|k_{-\bar{\alpha}}\|},$

it follows that $H\Upsilon_{\alpha}=(1+\alpha)\Upsilon_{\alpha}$, and more
generally,

$f_{+}(H)\Upsilon_{\alpha}\ =\ f_{+}(1+\alpha)\Upsilon_{\alpha}$ (3.12)

for all $f\in{\mathcal{P}}$.

Now notice that (2.3) implies that

$\Upsilon_{\alpha}\ =\
\frac{\sqrt{1-|\alpha|^{2}}}{1+\alpha}x^{-\frac{\alpha}{1+\alpha}}.$

######  Claim 3.13.

If $\rho>0$ and $\tau\in\mathbb{T}\setminus\\{-1\\}$, then

$\lim_{\alpha\to\tau}\langle
x^{\rho}\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle=0.$ (3.14)

###### Proof.

First note that

$\rho-(\frac{\alpha}{1+\alpha}+\frac{\bar{\alpha}}{1+\bar{\alpha}})+1=\rho+\frac{1-|\alpha|^{2}}{|1+\alpha|^{2}},$

so that

$\int_{0}^{1}x^{\rho-(\frac{\alpha}{1+\alpha}+\frac{\bar{\alpha}}{1+\bar{\alpha}})}dx\
=\ \Big{(}\rho+\frac{1-|\alpha|^{2}}{|1+\alpha|^{2}}\Big{)}^{-1}.$

Hence,

$\displaystyle\langle x^{\rho}\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle$
$\displaystyle=\frac{1-|\alpha|^{2}}{|1+\alpha|^{2}}\int_{0}^{1}x^{\rho-(\frac{\alpha}{1+\alpha}+\frac{\bar{\alpha}}{1+\bar{\alpha}})}dx$
$\displaystyle=\frac{1-|\alpha|^{2}}{|1+\alpha|^{2}}\Big{(}\rho+\frac{1-|\alpha|^{2}}{|1+\alpha|^{2}}\Big{)}^{-1}$
$\displaystyle=\frac{1-|\alpha|^{2}}{|1+\alpha|^{2}\rho\ +1-|\alpha|^{2}}.$

Therefore, if $\rho>0$ and $\tau\in\mathbb{T}\setminus\\{-1\\}$, (3.14) holds.
∎

Observe that if $q$ is a polynomial and $\tau\in\mathbb{T}\setminus\\{-1\\}$,
then Claim 3.13 implies that $\langle
q\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle\to q(0)$ as $\alpha\to\tau$. In
particular,

$\lim_{\alpha\to\tau}\langle
q(-M_{x})\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle=0$ (3.15)

whenever $\tau\in\mathbb{T}\setminus\\{-1\\}$ and $q$ is a polynomial
satisfying $q(0)=0$.

We now conclude the proof of the lemma. We need to show that if
$f\in{\mathcal{P}}$ and $\tau\in\mathbb{T}\setminus\\{-1\\}$ then (3.11) holds
with $z=1+\tau$. First assume that $f\in{\mathcal{P}}_{0}$ and fix
$K\in\mathcal{K}_{0}$. Since $\Upsilon_{\alpha}\to 0$ weakly as
$\alpha\to\tau$, using (3.12) and (3.14) we have

$\displaystyle\langle(\gamma(f)+K)\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle$
$\displaystyle\ =\ $
$\displaystyle\langle(f_{-}-f_{-}(0))(-M_{x})\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle+\langle
f_{+}(H)\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle+\langle
K\Upsilon_{\alpha},\Upsilon_{\alpha}\rangle$ $\displaystyle\to$
$\displaystyle\qquad\qquad\qquad 0\qquad\qquad\qquad+\ \ \ \ f_{+}(1+\tau)\ \
\ \ +\ \ 0$ $\displaystyle=$ $\displaystyle f_{+}(1+\tau).$

as $\alpha\to\tau$. Therefore, if $f\in{\mathcal{P}}_{0}$ and
$\tau\in\mathbb{T}\setminus\\{-1\\}$,

$|f_{+}(1+\tau)|\leq\|\gamma(f)+K\|.$

Hence, if $f\in{\mathcal{P}}_{0}$,

$|f_{+}(1+\tau)|\leq\inf_{K\in K_{0}}\|\gamma(f)+K\|=\|\Gamma(f)\|.$

As $\Gamma$ is continuous and ${\mathcal{P}}_{0}$ is dense in ${\mathcal{P}}$,
it follows that (3.11) holds with $z=1+\tau$ for all $f\in{\mathcal{P}}$. ∎

######  Lemma 3.16.

$\Gamma$ is a homeomorphism.

###### Proof.

In the proof of Proposition 3.5 we showed that

$\|\Gamma(f)\|\leq 3\|f\|$

for all $f\in{\mathcal{P}}$. On the other hand, Lemma 3.8 implies that

$\max_{t\in[-1,0]}|f_{-}(t)|\leq\|\Gamma(f)\|$

for all $f\in{\mathcal{P}}$ and Lemma 3.10 implies that

$\max_{z\in\overline{\mathbb{D}(1,1)}}|f_{+}(z)|\leq\|\Gamma(f)\|$

for all $f\in{\mathcal{P}}$. Therefore,

$\|f\|=\max\
\big{\\{}\max_{t\in[-1,0]}|f_{-}(t)|,\max_{z\in\overline{\mathbb{D}(1,1)}}|f_{+}(z)|\
\big{\\}}\leq\|\Gamma(f)\|$

for all $f\in{\mathcal{P}}$. ∎

Putting together the results of Subsection 3.1 we get the following theorem.

######  Theorem 3.17.

The map $\Gamma$ is a homeomorphic unital isomorphism from ${\mathcal{P}}$
onto $\mathcal{C}$.

###  3.2 Some Observations on the Gelfand Theory of $\mathcal{C}$

If $\Lambda=[-1,0]\cup\overline{\mathbb{D}(1,1)}$, then there is an isometric
isomorphism from ${\mathcal{P}}$ onto the lollipop algebra $A(\Lambda)$ given
by

$A(\Lambda)\ni f\mapsto\big{(}\ f|_{[-1,0]}\
,f|_{(\overline{\mathbb{D}(1,1)})}\ \big{)}.$

So one could just as well state Theorem 3.17 with ${\mathcal{P}}$ replaced by
$A(\Lambda)$ and $\Gamma$ replaced with the map
$\Gamma^{\sim}:A(\Lambda)\to\mathcal{C}$ defined by

$\ \Gamma^{\sim}(f)=\Big{[}\
f\big{|}_{[-1,0]}(-M_{x})+f\big{|}_{\overline{\mathbb{D}(1,1)}}(H)-f(0)\
\Big{]}.$

######  Definition 3.18.

Define $\theta:{\mathcal{A}}\to A(\Lambda)$ by

$\theta(X)\ =\ (\Gamma^{\sim})^{-1}([X]).$

Then Theorem 3.17 says that there is a short exact sequence

$0\to{\mathcal{K}}_{{\mathcal{A}}}\to{\mathcal{A}}\stackrel{{\scriptstyle\theta}}{{\to}}A(\Lambda)\to
0.$

######  Remark 3.19.

By Mergelyan’s theorem, $A(\Lambda)=P(\Lambda)$, and since $z$ generates
$P(\Lambda)$, it follows that

$\Gamma^{\sim}(z)=[H-M_{x}]$

generates $\mathcal{C}$. We shall examine $H-M_{x}$ in Section 6.

##  4 $C^{*}({\mathcal{A}})$

We shall let ${\mathcal{B}}=C^{*}({\mathcal{A}})$ denote the $C^{*}$-algebra
generated by ${\mathcal{A}}$. Since it is irreducible, ${\mathcal{B}}$
contains all the compact operators.

The Toeplitz $C^{*}$-algebra $\mathcal{T}$ is the $C^{*}$ algebra generated by
the shift $S$. There is a short exact sequence

$0\to{\mathcal{K}}\to{\mathcal{T}}\stackrel{{\scriptstyle\alpha}}{{\to}}\
A(\overline{\mathbb{D}})\to 0.$

(See e.g. [7, 7.23]). A cross-section of $\alpha$ is the map that sends a
function $m$ to the Toeplitz operator $T_{m}$ on $H^{2}$ with symbol $m$.

Since $\widehat{H}=S^{*}+1$, the $C^{*}$-algebra generated by $H$ is unitarily
equivalent to $\mathcal{T}$. We wish to think of it as living on
$\overline{\mathbb{D}(1,1)}$, so we must shift things over. Let $\tau(z)=z+1$.
For any function $f$ defined on some domain in $\mathbb{C}$, let
$f^{\cup}(z)=\overline{f(\bar{z})}$ be its reflection in the real axis.

######  Definition 4.1.

Let $\psi\in C(\partial\mathbb{D}(1,1))$. Let $H_{\psi}\in B(L^{2})$ be
defined by

$H_{\psi}\ =\ U^{*}T^{*}_{(\psi\circ\tau)^{\cup}}U.$

The map $\psi\mapsto H_{\psi}$ is unital and linear, but not multiplicative.
One checks that if $\psi(z)=z^{n}$, then $H_{z^{n}}=H^{n}$, and if
$\psi(z)=\bar{z}^{n}$, then $H_{\bar{z}^{n}}=(H^{*})^{n}$.

######  Theorem 4.2.

There is a short exact sequence

$0\to{\mathcal{K}}\to{\mathcal{B}}\stackrel{{\scriptstyle\pi}}{{\to}}C(\partial\Lambda)\to
0.$ (4.3)

For every $X$ in ${\mathcal{B}}$, its coset in ${\mathcal{B}}/{\mathcal{K}}$
can be written uniquely as

$[X]\ =\ [g(-M_{x})+H_{\psi}-g(0)]$ (4.4)

where $g\in C[-1,0]$, $\psi\in C(\partial\mathbb{D}(1,1))$, and
$g(0)=\psi(0)$. The essential spectrum of $X$ as in (4.4) is
$g([0,1])\cup\psi(\partial\mathbb{D}(1,1))$. If $\lambda\notin\sigma_{e}(X)$,
then the Fredhom index is given by the winding number of $\psi$ about
$\lambda$:

${\rm ind}(X-\lambda)\ =\ {\rm ind}_{\psi}(\lambda).$

Proof: For $X\in{\mathcal{B}}$, we shall let $[X]$ denote its equivalence
class in ${\mathcal{B}}/{\mathcal{K}}$. We have

$[HM_{x}]=[M_{x}H]=0,$

and

$[HH^{*}]=[H^{*}H]=H+H^{*}.$

So ${\mathcal{B}}/{\mathcal{K}}$ is abelian. Moreover, for any polynomial $q$
in 3 variables, there are polynomials $p_{1},p_{2},p_{3}$ in one variable so
that

$[q(M_{x},H,H^{*})]\ =\ [p_{1}(M_{x})+p_{2}(H)+p_{3}(H^{*})].$ (4.5)

Therefore operators of the form (4.5) are dense in
${\mathcal{B}}/{\mathcal{K}}$.

We wish to prove that ${\mathcal{B}}/{\mathcal{K}}$ is isomorphic to the
abelian $C^{*}$-algebra $C(\partial\Lambda)$. We will use a similar strategy
to the proof of Theorem 3.17. Let

${\mathcal{Q}}\ =\ \\{f=(f_{-},f_{+}):f_{-}\in C([-1,0]),f_{+}\in
C(\partial\mathbb{D}(1,1)),f_{-}(0)=f_{+}(0)\\}.$

The algebra ${\mathcal{Q}}$ is just $C(\partial\Lambda)$, but it is easier to
define the functional calculus on it. Define

$\displaystyle\delta:{\mathcal{Q}}$ $\displaystyle\ \to\ $
$\displaystyle{\mathcal{B}}$ $\displaystyle f$ $\displaystyle\mapsto$
$\displaystyle f_{-}(-M_{x})+H_{f_{+}}-f(0).$

$\Box$

Let $\Delta(f)=[\delta(f)]$. The following lemma is straightforward to prove.

######  Lemma 4.6.

(i) Let $\psi,\phi\in C(\partial\mathbb{D}(1,1))$. Then
$[H_{\psi}H_{\phi}]=[H_{\psi\phi}]$.

(ii) Let $g\in C([-1,0])$ and $\psi\in C(\partial\mathbb{D}(1,1))$. Then

$[g(-M_{x})][H_{\psi}]\ =\ [H_{\psi}][g(-M_{x})]\ =\
[g(0)H_{\psi}+\psi(0)g(-M_{x})-g(0)\psi(0)].$

Using Lemma 4.6, one can check that $\Delta$ is a unital *-homomorphism from
${\mathcal{Q}}$ into ${\mathcal{B}}/{\mathcal{K}}$. Its range is dense, so if
we can show it has no kernel, then it is a C*-isomorphism.

######  Lemma 4.7.

If $s\in[-1,0]$, then

$|f_{-}(s)|\leq\|\Delta(f)\|$ (4.8)

for all $f\in{\mathcal{Q}}$.

###### Proof.

As $f$ is continuous, it suffices to prove the lemma under the assumption that
$s\in(-1,0)$, and as $\Delta$ is continuous, we can assume that $f_{-}$ is a
polynomial, and that $f_{+}(z)=f(0)+zp_{2}(z)+\overline{zp_{3}(z)}$ where
$p_{2}$ and $p_{3}$ are polynomials.

As in Lemma 3.8, for $n$ satisfying $1/n<\min\\{s,1-s\\}$ we define a unit
vector $\chi_{n}\in L^{2}$ by the formula

$\chi_{n}(t)=\left\\{\begin{array}[]{ll}\sqrt{\frac{n}{2}}&\mbox{if }|t-s|\leq
1/n\\\ 0&\mbox{if }|t-s|>1/n\end{array}\right.$

Let $K$ be compact.

$\displaystyle\langle(\delta(f)+K)\ \chi_{n},\chi_{n}\rangle$ $\displaystyle=\
$ $\displaystyle\langle(f_{-}(-M_{x})+H_{zp_{2}}+H_{\bar{z}\bar{p}_{3}}+K)\
\chi_{n},\chi_{n}\rangle$ $\displaystyle=\ $ $\displaystyle\langle
f_{-}(-x)\chi_{n},\chi_{n}\rangle+\langle p_{2}(H)H\
\chi_{n},\chi_{n}\rangle+\langle\chi_{n},p_{3}(H)H\ \chi_{n}\rangle+\langle K\
\chi_{n},\chi_{n}\rangle$ $\displaystyle\to\ $ $\displaystyle\ \ \
f_{-}(s)\qquad\quad\ +\qquad\qquad 0\qquad\ \ \ +\qquad\ 0\qquad\ \ \ \ \ \ \
+\qquad 0$ $\displaystyle=\ $ $\displaystyle\ \ f_{-}(s).$

Therefore,

$|f_{-}(s)|\leq\inf_{K\in{\mathcal{K}}}\|\delta(f)+K\|=\|\Delta(f)\|.$

∎

If $\Delta(f)=0$, by Lemma 4.7 we must have $f_{-}=0$. So
$\delta(f)=H_{f_{+}}$ must be compact. But $H_{f_{+}}$ is unitarily equivalent
to a Toeplitz operator, and there are no non-zero compact Toeplitz operators.
Therefore $\Delta$ has a trivial kernel, and hence is a $*$-isomorphism.

The claim about the spectrum of $[x]$ now follows from the fact that the
spectrum of a function in $C(\partial\Lambda)$ equals its range. Finally, the
claim about the Fredholm index follows from the fact that the Fredholm index
at $\lambda$ will be unchanged under any homotopy of $f$ that keeps $\lambda$
outside its range. Then $f$ can be homotoped to $(f_{-},f_{+})$ where
$f_{+}(z)=(z-1-\lambda)^{n}$ and $f_{-}(x)=(-1-\lambda)^{n}$ for some integer
$n$, and the Fredholm index of $\delta(f)$ is $n$. $\Box$

##  5 Compact operators in the little algebra ${\mathcal{A}}_{0}$

Recall from Definition 1.6 that ${\mathcal{A}}_{0}$ is the norm-closed algebra
generated by $M_{x}$ and $V$. We shall prove that every compact operator in
${\rm AlgLat}(V)$ lies not just in ${\mathcal{A}}$ but in ${\mathcal{A}}_{0}$.

For $I$ an interval in $[0,1]$, let us write $L^{2}(I)$ for the subspace of
$L^{2}$ that vanishes a.e. off $I$, and let $P_{I}$ denote projection onto
$L^{2}(I)$. For $\phi,\psi\in L^{2}$ we write $\phi\otimes\psi$ to denote the
rank one operator

$\phi\otimes\psi:f\ \mapsto\ \langle f,\psi\rangle\phi.$

The key observation is the following:

######  Lemma 5.1.

Suppose $\phi\in L^{2}[t,1]$ and $\psi\in L^{2}[0,t]$. Then
$\phi\otimes\psi=M_{\phi}VM_{\psi}^{*}$.

Proof: We have

$M_{\phi}VM_{\psi}^{*}f(x)\ =\ \phi(x)\int_{0}^{x}f(s)\overline{\psi(s)}ds.$

The right-hand side is $0$ if $x<t$, and $\phi(x)\langle f,\psi\rangle$ if
$x>t$. $\Box$

######  Lemma 5.2.

Every finite rank operator on $L^{2}[0,1]$ can be written as an integral
operator whose kernel is in $L^{2}([0,1]\times[0,1])$.

Proof: Let $K=\sum_{j=1}^{n}\phi_{j}\otimes\psi_{j}$. Define

$k(x,s)\ =\ \sum_{j=1}^{n}\phi_{j}(x)\overline{\psi_{j}(s)}.$

Then $Kf(x)=\int_{0}^{1}k(x,s)f(s)ds$, and $k$ is in
$L^{2}([0,1]\times[0,1])$. $\Box$

######  Lemma 5.3.

Let $k$ be in $L^{2}([0,1]\times[0,1])$, and let
$Tf(x)=\int_{0}^{1}k(x,s)f(s)ds$. Then $T$ is in ${\rm AlgLat}(V)$ if and only
if $k(s,x)=0$ for $s>x$.

Proof: Sufficiency is clear. To prove necessity, assume that for some $0<t<1$,
the kernel

$k(s,x)\chi_{[0,t](x)}\chi_{[t,1]}(s)$

is not $0$ a.e. As an integral operator is zero if and only if the kernel is
$0$ a.e., this means that the corresponding integral operator is non-zero, and
hence $T$ maps a function in $L^{2}(t,1)$ to a function that is not $0$ a.e.
on $[0,t]$. $\Box$

######  Lemma 5.4.

Let $T=\phi\otimes\psi$ be a rank-one operator. Then $T$ is in ${\rm
AlgLat}(V)$ if and only if for some $0<t<1$, the support of $\phi$ is in
$[t,1]$ (i.e. $\phi=0$ a.e. on $[0,t]$) and the support of $\psi$ is in
$[0,t]$. In this case, $T\in{\mathcal{A}}_{0}$.

Proof: The first part follows from Lemma 5.3. For the second part, observe
that if the supports of $\phi,\psi$ are in $[t,1]$ and $[0,t]$ respectively,
then $\phi\otimes\psi=M_{\phi}VM_{\psi}^{*}$. If $\phi$ and $\psi$ are both in
$C([0,1])$, this proves that $\phi\otimes\psi\in{\mathcal{A}}_{0}$.

For the general case, choose continuous functions $f_{n}$ and $g_{n}$ that
converge to $\phi$ and $\psi$ respectively in $L^{2}$. It follows from Lemma
5.1 that $M_{f_{n}}VM_{g_{m}}^{*}$ converges to $M_{\phi}VM_{g_{m}}^{*}$ in
norm as $n\to\infty$, and that $M_{\phi}VM_{g_{m}}^{*}$ converges to
$M_{\phi}VM_{\psi}^{*}$. Therefore $\phi\otimes\psi\in{\mathcal{A}}_{0}$.
$\Box$

######  Theorem 5.5.

Let $K$ be a compact operator in ${\rm AlgLat}(V)$. Then
$K\in{\mathcal{A}}_{0}$, and can be approximated in norm by finite rank
operators in ${\mathcal{A}}_{0}$.

Proof: Note that $K\in{\rm AlgLat}(V)$ means that for all $0<s<1$, we have
$P_{[0,s]}KP_{[s,1]}=0$.

Let $\varepsilon>0$. First, consider $P_{[1/2,1]}KP_{[0,1/2]}$. This can be
approximated within $\varepsilon/2$ by a finite rank operator that is a sum of
rank one operators that map $L^{2}(0,1/2)$ to $L^{2}(1/2,1)$. By Lemma 5.4,
this means that this finite rank operator is in ${\mathcal{A}}_{0}$.

A similar argument shows that $P_{[1/4,1/2]}KP_{[0,1/4]}$ and
$P_{[3/4,1]}KP_{[1/2,3/4]}$ can both be approximated by finite rank operators
in ${\mathcal{A}}_{0}$ within $\varepsilon/8$. Iterating, we get that if $n$
is a power of $2$, we can approximate

$K-\sum_{j=1}^{n}P_{[(j-1)/n,j/n]}KP_{[(j-1)/n,j/n]}$

within $\varepsilon$ by a finite rank operator in ${\mathcal{A}}_{0}$.

Finally we observe that

$\|\sum_{j=1}^{n}P_{[(j-1)/n,j/n]}KP_{[(j-1)/n,j/n]}\|\ =\ \max_{1\leq j\leq
n}\|P_{[(j-1)/n,j/n]}KP_{[(j-1)/n,j/n]}\|.$ (5.6)

Since $K$ is compact,

$\lim_{n\to\infty}\sup_{|I|=1/n}\|KP_{I}\|=0,$

so (5.6) tends to $0$. $\Box$

##  6 The operator $H-M_{x}$

Let us write $Z$ for the operator $H-M_{x}$. We know that $[Z]$ generates the
Calkin Hardy-Weyl algebra $\mathcal{C}$. By Theorem 3.17 we know that the
spectrum of $[Z]$ in ${\mathcal{A}}/{\mathcal{K}}_{0}$ is $\Lambda$. It is not
surprising that $\mathbb{D}(1,1)$ are eigenvalues of $Z$, since they are
eigenvalues of $H$. It is perhaps surprising that every point in the stick,
except $-1$, is also an eigenvalue. Moreover as we move up the stick to the
bulb of the lollipop, the eigenvalues increase in multiplicity.

######  Theorem 6.1.

(i) $\sigma_{p}(Z)=\mathbb{D}(1,1)\cup(-1,0]$.

(ii) The point spectrum of $Z^{*}$ is empty.

(iii) The spectrum of $Z$ is $\Lambda$.

Proof: (i) Suppose $(Z-\lambda)f=0$. Let $F(x)=Vf(x)=\int_{0}^{x}f(t)dt$. Then
we have

$\frac{1}{x}F(x)\ =\ (x+\lambda)f(x).$

As $F^{\prime}(x)=f(x)$, we get the equation

$\frac{1}{x}F(x)=(x+\lambda)F^{\prime}(x),$ (6.2)

with the boundary condition

$\quad F(0)\ =\ 0.$ (6.3)

The function $F$ is continuous. Let $\Omega$ denote the relatively open subset
of $[0,1]$ on which it is non-zero.

We get that the solution of (6.2), with $\lambda\neq 0$, is

$F(x)\ =\ c\left(\frac{x}{x+\lambda}\right)^{1/\lambda}\chi_{\Omega}(x)\ ,$
(6.4)

where the constant $c$ can a priori be different on different components of
$\Omega$ (though we show below that $\Omega$ is actually connected). Hence

$f(x)\ =\
cx^{\frac{1}{\lambda}-1}(x+\lambda)^{-1-\frac{1}{\lambda}}\chi_{\Omega}(x).$
(6.5)

Case: $\Omega=[0,1]$.

For $f$ to be in $L^{2}$ with $c\neq 0$, we need

$\Re(\frac{1}{\lambda}-1)\ >\ -\frac{1}{2},$

which is the same as $\lambda\in\mathbb{D}(1,1)$. For
$\lambda\in\mathbb{D}(1,1)$, we get the eigenfunctions

$f(x)\ =\ x^{\frac{1}{\lambda}-1}(x+\lambda)^{-1-\frac{1}{\lambda}}.$ (6.6)

When $\lambda=0$, we get

$F(x)\ =\ ce^{-1/x},$

and

$f(x)\ =\ c\frac{1}{x^{2}}e^{-\frac{1}{x}}.$ (6.7)

Now suppose that $\Omega$ is not all of $[0,1]$. Decompose $\Omega$ as a union
of disjoint non-empty intervals. On each interval, we have that $f$ is given
by (6.5), with some constant $c$ that can depend on the interval. Choose such
an interval, $I$. Then $0$ cannot be an end-point of $I$, or we would have
that $F$ is given by (6.4), and this is discontinuous at the right-hand end-
point of $I$.

So assume that the left-hand end-point of $I$ is $t>0$. Then the boundary
condition (6.3) is replaced by $F(t)=0$. On $I$, we have

$F(x)\ =\ c\left(\frac{x}{x+\lambda}\right)^{1/\lambda},$

so to have $F(t)=0$ we need $\lambda=-t$. By continuity, $F$ cannot vanish
again, so we conclude that $I=(t,1]$ and that for $\lambda\in(-1,0)$ the
function

$f(x)\ =\
x^{\frac{1}{\lambda}-1}(x+\lambda)^{-1-\frac{1}{\lambda}}\chi_{[-\lambda,1]}(x)$
(6.8)

is an eigenvector of $Z$ with eigenvalue $\lambda$.

(ii) As $H^{*}f(x)=\int_{x}^{1}\frac{1}{t}f(t)dt$, the eigenvalue equation
becomes

$G(x)\ =\ (x+\lambda)f(x),$

where

$G(x)\ =\ \int_{x}^{1}\frac{1}{t}f(t)dt$

As $G^{\prime}(x)=-\frac{1}{x}f(x)$, we get the differential equation

$G(x)\ =\ -x(x+\lambda)G^{\prime}(x),\quad G(1)=0.$ (6.9)

Solving for $G$, we get

$G(x)\ =\
cx^{-\frac{1}{\lambda}}(x+\lambda)^{\frac{1}{\lambda}}\,\chi_{\\{G\neq 0\\}}.$

The only solution on an interval that vanishes on the right end-point is the
zero solution.

(iii) We know

$\sigma_{e}(Z)=\partial\Lambda\subseteq\Lambda=\overline{\sigma_{p}(Z)}\subseteq\sigma(Z).$

If $\lambda$ is a point in $\mathbb{C}\setminus\Lambda$, it must be a Fredholm
point. By (i), it is not an eigenvalue of $Z$, and by (ii) it is not an
eigenvalue of $Z^{*}$. Therefore $Z-\lambda$ has trivial kernel and cokernel,
and closed range. Therefore it is invertible, and $\lambda$ is in the
resolvent of $Z$. $\Box$

Not only are points on the stick of the lollipop eigenvalues, there is some
additional smoothness. By a generalized eigenvector of order $n$ we mean a
vector $f$ that satisfies $(Z-\lambda)^{n+1}f=0$ but $(Z-\lambda)^{n}f\neq 0$.

We shall prove the case $\lambda=0$ first.

######  Lemma 6.10.

At $0$, the operator $Z$ has generalized eigenvectors of all orders.

Proof: We want to show that if we let
$f_{0}(z)=\frac{1}{x^{2}}e^{-\frac{1}{x}}$ from (6.7), then for every
$n\in{\mathbb{N}}$ there exists $f\in L^{2}$ so that

$Zf_{n+1}\ =\ f_{n}.$ (6.11)

######  Claim 6.12.

For every $n\in{\mathbb{N}}$ there exists a polynomial $p_{n}$ of degree
$2n+2$, with lowest order term of degree $n+2$, so that the functions

$f_{n}(x)\ =\ p_{n}(\frac{1}{x})e^{-\frac{1}{x}}$

satisfy (6.11).

We prove this by induction on $n$. It is true when $n=0$. Assume we have
proved it up to level $n$, and we want to prove it for $n+1$. So we wish to
solve the equation

$Zf_{n+1}\ =\ f_{n}=p_{n}(\frac{1}{x})e^{-\frac{1}{x}}$ (6.13)

and show that the solution is of the form

$f_{n+1}(x)\ =\ p_{n+1}(\frac{1}{x})e^{-\frac{1}{x}}.$ (6.14)

Writing $F_{n+1}$ for $Vf_{n+1}$, equation (6.13) is

$\displaystyle(H-M_{x})f_{n+1}(x)$ $\displaystyle\ =\ $
$\displaystyle\frac{1}{x}\int_{0}^{x}f_{n+1}(t)dt-xf_{n+1}(x)$
$\displaystyle=$ $\displaystyle\frac{1}{x}F_{n+1}(x)-xF_{n+1}^{\prime}(x)$
$\displaystyle=$ $\displaystyle f_{n}(x).$

This gives us the linear differential equation

$F_{n+1}^{\prime}(x)-\frac{1}{x^{2}}F_{n+1}(x)\ =\ -\frac{1}{x}f_{n}(x).$

Multiply by the integrating factor $e^{\frac{1}{x}}$ to get

$\displaystyle\frac{d}{dx}\left[e^{\frac{1}{x}}F_{n+1}(x)\right]$
$\displaystyle\ =\ $ $\displaystyle-\frac{1}{x}e^{\frac{1}{x}}f_{n}(x)$
$\displaystyle=$ $\displaystyle-\frac{1}{x}p_{n}(\frac{1}{x}).$

Therefore

$e^{\frac{1}{x}}F_{n+1}(x)\ =\ q_{n}(\frac{1}{x}),$

where $q_{n}$ is a polynomial of degree $2n+2$ that may have a constant term,
and whose next lowest order term is of degree $n+2$. This gives

$\displaystyle f_{n+1}(x)$ $\displaystyle\ =\ $
$\displaystyle\frac{d}{dx}\left[e^{-\frac{1}{x}}q_{n}(\frac{1}{x})\right]$
$\displaystyle=$ $\displaystyle
e^{-\frac{1}{x}}\left[\frac{1}{x^{2}}q_{n}(\frac{1}{x})-\frac{1}{x^{2}}q_{n}^{\prime}(\frac{1}{x})\right]$

Let

$p_{n+1}(x)\ =\ x^{2}q_{n}(x)-x^{2}q_{n}^{\prime}(x).$

The degree of $p_{n+1}$ is two higher than $q_{n}$, so it is $2(n+1)+2$. There
may be a term of order $2$; the next lowest order term is $n+3=(n+1)+2$. But
as $Zf_{0}=0$, one can subtract a multiple of $f_{0}$ from $f_{n+1}$ without
changing (6.13), so we can assume that $p_{n+1}$ has no term of order $2$.

So we have proved Claim 6.12. As any function of the form (6.14) is in
$L^{2}$, we are done. $\Box$

For points in the stick, a similar method works, but there are restrictions
when requiring the generalized eigenvectors to be in $L^{2}$. Here is one
result.

######  Lemma 6.15.

Let $\lambda\in(-1,0)$. Then $Z$ has a generalized eigenvalue of order $1$ at
$\lambda$ if and only if $-\frac{2}{3}<\lambda<0$.

Proof: Let $s=-\lambda$.

$f_{0}(x)\ =\
\left(\frac{x-s}{x}\right)^{\frac{1}{s}}\frac{1}{x(x-s)}\chi_{[s,1]}.$ (6.16)

All the functions below are supported on $[s,1]$. We wish to find a function
$f_{1}$ that satisfies

$(Z+s)f_{1}\ =\ f_{0}.$ (6.17)

Writing $F_{1}$ for $Vf_{1}(x)=\int_{s}^{x}f_{1}(t)dt$, this becomes

$\frac{1}{x}F_{1}-(x-s)F_{1}^{\prime}\ =\ f_{0}.$

or

$F_{1}^{\prime}-\frac{1}{x(x-s)}F_{1}\ =\ -\frac{1}{x-s}f_{0}.$ (6.18)

An integrating factor for (6.18) is
$\left(\frac{x}{x-s}\right)^{\frac{1}{s}}$. This yields

$\displaystyle\frac{d}{dx}\left[\left(\frac{x}{x-s}\right)^{\frac{1}{s}}F_{1}\right]$
$\displaystyle\ =\ $
$\displaystyle-\left(\frac{x}{x-s}\right)^{\frac{1}{s}}\frac{1}{x-s}f_{0}$
$\displaystyle=$ $\displaystyle-\frac{1}{x(x-s)^{2}}.$

Integrating, we get

$\left(\frac{x}{x-s}\right)^{\frac{1}{s}}F_{1}\ =\
\frac{1}{s^{2}}\log\frac{x-s}{x}+\frac{1}{s}\frac{1}{x-s}+c.$

Dividing through by the integrating factor and differentiating, we get

$\displaystyle f_{1}(x)$ $\displaystyle\ =\ $
$\displaystyle\frac{1}{s^{2}}\left[\left(\frac{x-s}{x}\right)^{\frac{1}{s}-1}\frac{1}{x^{2}}\log\frac{x-s}{x}+\left(\frac{x-s}{x}\right)^{\frac{1}{s}}\frac{s}{x(x-s)}\right]$
$\displaystyle+\frac{1}{s}\left[-\frac{1}{(x-s)^{2}}\left(\frac{x-s}{x}\right)^{\frac{1}{s}}+\frac{1}{x-s}\left(\frac{x-s}{x}\right)^{\frac{1}{s}-1}\frac{1}{x^{2}}\right]$
$\displaystyle+c\left[\left(\frac{x-s}{x}\right)^{\frac{1}{s}-1}\frac{1}{x^{2}}\right].$

We can choose $c=0$, since it is the coefficient of $f_{0}$. This gives

$f_{1}(x)\ =\
\left(\frac{x-s}{x}\right)^{\frac{1}{s}}\left[\frac{1}{s^{2}}\frac{1}{x(x-s)}\log\frac{x-s}{x}+\frac{1-s}{s}\frac{1}{x(x-s)^{2}}\right].$
(6.19)

Examining this expression, we see that $f_{1}$ is smooth on $(s,1]$, and the
first term

$\frac{1}{s^{2}}\frac{(x-s)^{\frac{1}{s}-1}}{x^{\frac{1}{s}+1}}\log\frac{x-s}{x}$

vanishes at $s$ for every $s<1$. However the second term

$\frac{1-s}{s}\ \frac{(x-s)^{\frac{1}{s}-2}}{x^{\frac{1}{s}+1}}$

has a singularity that grows like $(x-s)^{\frac{1}{s}-2}$. This is integrable
for every $s<1$, but it is only in $L^{2}[s,1]$ for $s<\frac{2}{3}$. So we
have shown that (6.17) has a solution $f_{1}$ in $L^{2}$ if and only if
$\lambda>-\frac{2}{3}$. $\Box$

One can repeat the argument of Lemma 6.15 to get higher order generalized
eigenvectors, as $\lambda$ gets closer to $0$.

######  Lemma 6.20.

Let $m\geq 1$. Let $\lambda$ lie in the interval $(-\frac{2}{2m+1},0)$. Then
$Z$ has generalized eigenvectors up to order $m$ at $\lambda$.

Proof: We shall inductively find functions $f_{n}$ satisfying
$(Z-\lambda)f_{n+1}=f_{n}$, with $f_{0}$ as in (6.16). Let $s=-\lambda$, and
write

$\Phi(x)\ =\ \left(\frac{x-s}{x}\right)^{\frac{1}{s}}\chi_{[s,1]}(x).$

Then we have

$\displaystyle f_{0}(x)$ $\displaystyle\ =\ $
$\displaystyle\Phi(x)\frac{1}{x(x-s)}$ $\displaystyle f_{1}(x)$
$\displaystyle\ =\ $
$\displaystyle\Phi(x)\left[\frac{1}{s^{2}}\frac{1}{x(x-s)}\log\frac{x-s}{x}+\frac{1-s}{s}\frac{1}{x(x-s)^{2}}\right].$

Writing $F_{n+1}$ for $Vf_{n+1}=\int_{s}^{x}f_{n+1}(t)dt$, we want to solve

$\frac{1}{x}F_{n+1}(x)-(x-s)F_{n+1}^{\prime}(x)\ =\ f_{n}(x).$ (6.21)

After multiplying by the integrating factor $1/\Phi$, we have

$\frac{d}{dx}\left[\frac{1}{\Phi}F_{n+1}\right]\ =\
-\frac{1}{\Phi}\frac{1}{x-s}f_{n}(x).$ (6.22)

######  Claim 6.23.

There are constants $M_{n}$ such that the functions $f_{n}$ satisfy

$|f_{n}(x)|\ \leq\ M_{n}(x-s)^{\frac{1}{s}-n-1}\qquad\forall x\in(s,1].$
(6.24)

Proof of Claim 6.23: By induction on $n$. It is true when $n=0$. Assume it is
true up to $n$. From (6.22) we get for $x\in(s,1]$:

$\frac{1}{\Phi(x)}F_{n+1}(x)\ =\
\int_{x}^{1}\frac{1}{\Phi(t)}\frac{1}{t-s}f_{n}(t)dt+c_{n}.$ (6.25)

By the inductive hypothesis, the integrand in (6.25) is $O(t-s)^{-n-2}$, so
the integral is $O(x-s)^{-n-1}$. So $F_{n+1}$ satisfies

$F_{n+1}(x)\ =\ c_{n}\Phi(x)+O(x-s)^{\frac{1}{s}-n-1}.$ (6.26)

From (6.21) we have

$f_{n+1}(x)\ =\ \frac{1}{x(x-s)}F_{n+1}(x)-\frac{1}{x-s}f_{n}(x).$ (6.27)

When we use (6.26) for $F_{n+1}$, we get

$f_{n+1}(x)\ =\ c_{n}f_{0}(x)-\frac{1}{x-s}f_{n}(x)+O(x-s)^{\frac{1}{s}-n-2}.$

Now the claim follows from the inductive hypothesis on $f_{n}$. $\Box$

It follows from (6.27) that that $f_{m}$ is continuous on $(s,1]$, and Claim
6.23 shows that its singularity at $s$ is of order $(x-s)^{\frac{1}{s}-m-1}$.
This means $f_{m}$ is in $L^{2}$ provided $\frac{1}{s}-m-1\ >\ -\frac{1}{2},$
which is the same as $s<\frac{2}{2m+1}$. $\Box$

For later use, let us note that if you track the constants $M_{n}$ in Claim
6.23, you can show:

######  Lemma 6.28.

In Claim 6.23 one can take $M_{0}=\frac{1}{(s^{1+1/s})}$ and the consants
$M_{n}$ satisfy

$M_{n+1}\leq M_{n}\left(1+\frac{M_{0}}{n+1}\right).$

######  Lemma 6.29.

Let $m\geq 1$. On the interval $(-\frac{2}{2m+1},0)$ one can choose the
generalized eigenvectors of order $n$ of $Z-\lambda$ continuously in
$\lambda$, for every $n\leq m$, and satisfying
$(Z-\lambda)f_{\lambda,n}=f_{\lambda,n-1}$, for every $1\leq n\leq m$.

Proof: Let us write $f_{\lambda,n}$ for the choice of generalized eigenvector
of order $n$ at $\lambda$. Write

$\Phi(\lambda,x)\ =\
\left(\frac{x+\lambda}{x}\right)^{-\frac{1}{\lambda}}\chi_{[-\lambda,1]}(x).$

We have

$f_{\lambda,0}(x)\ =\ \frac{1}{x(x+\lambda)}\Phi(\lambda,x).$

On every compact subset $K$, of $(-1,0)$ the functions
$\\{f_{\lambda,0}:\lambda\in K\\}$ are uniformly bounded, and
$\lim_{\lambda^{\prime}\to\lambda}f_{\lambda^{\prime},0}(x)=f_{\lambda,0}(x)$
a.e., so the map $\lambda\mapsto f_{\lambda,0}$ is continuous as a map from
$(-\frac{2}{2m+1},0)$ into $L^{2}$.

For higher $n$, we find $f_{n+1}$ as in Lemma 6.20. At each stage, we take the
constant $c_{n}$ in (6.26) to be $0$. (We can do this because
$c_{n}f_{\lambda,0}$ will be in the kernel of $Z-\lambda$.) This gives us

$f_{\lambda,n+1}(x)\ =\
\frac{1}{x(x-s)}\Phi(\lambda,x)\int_{x}^{1}\frac{1}{\Phi(\lambda,t)}\frac{1}{t-s}f_{\lambda,n}(t)dt-\frac{1}{x-s}f_{\lambda,n}(x).$

Moreover we have $f_{\lambda,n}(x)=0$ if $x<-\lambda$ and

$|f_{\lambda,n}(x)|\ \leq\
M_{n,\lambda}(x+\lambda)^{-\frac{1}{\lambda}-n-1},\qquad x>-\lambda.$

By Lemma 6.28, we have that each $M_{n,\lambda}$ can be chosen uniformly in
$\lambda$ for $\lambda$ in $(-\frac{2}{2m+1},0)$. For any interval $I$ of
length $\delta$, we get

$\displaystyle\int_{I}|f_{\lambda,n}(x)|^{2}dx$ $\displaystyle\ \leq\ $
$\displaystyle M_{n}\int_{0}^{\delta}|x+\lambda|^{-\frac{2}{\lambda}-2n-2}$
$\displaystyle=$ $\displaystyle
M_{n}\frac{1}{-\frac{2}{\lambda}-2n-1}\delta^{-\frac{2}{\lambda}-2n-1}.$

Therefore, as $\lambda$ ranges over any compact subset of
$(-\frac{2}{2m+1},0)$, the functions $|f_{\lambda,n}(x)|^{2}$ are uniformly
integrable in $x$. So, by the Vitali convergence theorem, the map

$\displaystyle(-\frac{2}{2m+1},0)$ $\displaystyle\ \to\ $ $\displaystyle
L^{2}$ $\displaystyle\lambda$ $\displaystyle\mapsto$ $\displaystyle
f_{\lambda,n}$

is continuous. $\Box$

These lemmas say that operators in the closed algebra generated by $Z$ have
certain smoothness properties when mapped by $\sigma$ into $A(\Lambda)$. The
functions get smoother as we get closer to $0$.

######  Theorem 6.30.

Let $X$ be in the norm-closed algebra generated by $Z$. Then $\theta(X)$ is
$C^{m}$ on $(-\frac{2}{2m+1},0)$.

Proof: Let $\theta(X)=\phi\in A(\Lambda)$. Let $p_{j}$ be a sequence of
polynomials so that $\|p_{j}(Z)-X\|\to 0$. It follows from Theorem 3.17 that
$p_{j}$ converges to $\phi$ uniformly on $\Lambda$.

Case: $m=1$. Let $f_{\lambda,n}$ be as in Lemma 6.29. For any polynomial $p$,
we have

$\langle p(Z)f_{\lambda,0},f_{\lambda,0}\rangle\ =\
\|f_{\lambda,0}\|^{2}p(\lambda).$

Moreover,

$p(Z)f_{\lambda,1}\ =\
p(\lambda)f_{\lambda,1}+p^{\prime}(\lambda)f_{\lambda,0}.$

Let $g^{1}_{\lambda}$ be the linear combination of $f_{\lambda,0}$ and
$f_{\lambda,1}$ that satisfies $\langle
f_{\lambda,0},g^{1}_{\lambda}\rangle=1$ and $\langle
f_{\lambda,1},g^{1}_{\lambda}\rangle=0$. Then

$\langle p(Z)f_{\lambda,1},g^{1}_{\lambda}\rangle\ =\ p^{\prime}(\lambda).$

So as functions on $(-\frac{2}{3},0)$, we get that $p_{j}^{\prime}(\lambda)$
converges to some function $\psi(\lambda)=\langle
Xf_{\lambda,1},g^{1}_{\lambda}\rangle$.

######  Claim 6.31.

For all $x$ in $(-\frac{2}{3},0)$, we have $\phi^{\prime}(x)=\psi(x)$, and
$\psi$ is continuous.

We have
$g^{1}_{\lambda}=a_{\lambda,0}f_{\lambda,0}+a_{\lambda,1}f_{\lambda,1}$, where
the coefficients $a_{\lambda,0}$ and $a_{\lambda,1}$ solve the linear system

$\begin{pmatrix}\langle f_{\lambda,0},f_{\lambda,0}\rangle&\langle
f_{\lambda,0},f_{\lambda,1}\rangle\\\ \langle
f_{\lambda,1},f_{\lambda,0}\rangle&\langle
f_{\lambda,1},f_{\lambda,1}\rangle\end{pmatrix}\begin{pmatrix}a_{\lambda,0}\\\
a_{\lambda,1}\end{pmatrix}\ =\ \begin{pmatrix}1\\\ 0\end{pmatrix}.$

By Lemma 6.29, since $f_{\lambda,0}$ and $f_{\lambda,1}$ are continuous in
$\lambda$, so as they are linearly independent, we have that $g^{1}_{\lambda}$
is also continuous in $\lambda$. Therefore $\psi$ is continuous, and
$p_{j}^{\prime}$ converges to $\psi$ locally uniformly on $(-\frac{2}{3},0)$.
As $\int_{-\frac{1}{3}}^{x}p_{j}^{\prime}(t)dt=\phi(x)-\phi(-\frac{1}{3})$
converges to $\int_{-\frac{1}{3}}^{x}\psi(t)dt$, we get that
$\phi^{\prime}(x)=\psi(x)$. $\Box$

We have shown that $\phi$ is in $C^{1}(-\frac{2}{3},0)$. A similar argument
with higher derivatives proves that $\phi$ is in $C^{m}(-\frac{2}{2m+1},0)$.
$\Box$

Of course one can also find generalized eigenvectors for $Z$ of all orders at
points in $\mathbb{D}(1,1)$, but we already know that $\theta(X)$ is analytic
on $\mathbb{D}(1,1)$ for every $X\in{\mathcal{A}}$.

##  7 Open Questions

######  Question 7.1.

Is there a good description of ${\mathcal{K}}_{{\mathcal{A}}}$, the compact
operators in ${\mathcal{A}}$?

######  Question 7.2.

Is $C^{*}(Z)=C^{*}({\mathcal{A}})$? To prove this, it is sufficient to show
that $C^{*}(Z)$ is irreducible, since then it would contain all the compacts,
and its quotient by the compacts would be all of $C(\partial\Lambda)$.

######  Question 7.3.

Do the eigenvectors of $Z$ span $L^{2}$?

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