The Smirnov class for spaces with the complete Pick property
Alexandru Aleman Michael Hartz John E. McCarthy Stefan Richter



  We show that every function in a reproducing kernel Hilbert space with a normalized
  complete Pick kernel is the quotient of a multiplier and a cyclic multiplier.
  This extends a theorem of Alpay, Bolotnikov and Kaptanoğlu.
  We explore various consequences of this result regarding zero sets, spaces on compact sets and
  Gleason parts. In particular, using a construction of Salas, we exhibit a
  rotationally invariant complete Pick space of analytic functions on the unit disc
  for which the corona theorem fails.


Ballistic, diffusive, and arrested transport in disordered momentum-space lattices
    Bryce Gadway
    December 30, 2023
==================================================================================





§ INTRODUCTION



The Smirnov class in the unit disc  can be characterized as the space

    N^+  =  {φ/ψ :  φ, ψ∈ H^∞,  ψ outer},

where H^∞ denotes the algebra of bounded analytic functions on .
Alternatively, it is the space of analytic functions on  for which the functions

    θ↦log ( 1+  | f(re^iθ) | )

are uniformly integrable over [0,2 π] for all 0 < r < 1.
The class occurs frequently in function theory <cit.>, and has interesting connections to operator theory — see, e.g. <cit.>. It contains H^p for all p > 0.

The algebra H^∞ is the multiplier algebra of the Hardy space H^2,
which is the  Hilbert function space on   which has the kernel

    s(z,w)  = 1/1- z w

as its reproducing kernel.
Many interesting properties of H^∞ and H^2 can be shown to follow from the fact that the kernel is a complete Pick kernel — see the book <cit.> for a comprehensive account on this topic.

We shall give a definition of complete Pick spaces in Section <ref> below. For now,
we simply mention that in addition to the Hardy space H^2, the Dirichlet space and the
Drury-Arveson space, which has been studied by many different authors over the years
<cit.>,
satisfy this property.

Suppose that  is a Hilbert
function space on a set X whose kernel k is normalized at x_0 (this means that k(x,x_0) = 1 for every x).
Let () denote the multiplier algebra of . A multiplier ψ is said to be cyclic if ψ
is dense in .
We define

    N^+() = {φ/ψ: φ, ψ∈()  and ψ is cyclic}.

(Taking = H^2, we recover the classical Smirnov class.)
Observe that a cyclic multiplier does not vanish anywhere, so the quotient φ/ψ is defined on all of X.
Moreover, since the product of any two cyclic multipliers is cyclic (see Subsection <ref>),
N^+() is an algebra.

Using the inner-outer factorization of functions in H^2, one can show that H^2 ⊂ N^+.
On the other hand, there are functions in the Bergman space L^2_a
on  which do not possess boundary radial
limits anywhere on ∂, and hence are not
quotients of two functions in H^∞ = (L^2_a).
Thus, L^2_a is not contained in N^+(L^2_a).
The following result shows that the inclusion ⊂ N^+() is valid
for every complete Pick space  with a normalized kernel.
It was shown in the case of the Drury-Arveson space on a finite dimensional ball
by Alpay, Bolotnikov and Kaptanoğlu
<cit.>, but their proof can be extended to general complete Pick spaces.


  
  Let  be a
   complete Pick space on X whose kernel is normalized at x_0 ∈ X and let f ∈ with ||f||_≤ 1.
  Then
  there are φ,ψ∈() of multiplier norm at most 1 with ψ(x_0) = 0
  such that
  
    f = φ/1 - ψ.

  In particular, ⊂ N^+().


We will provide a proof of Theorem <ref> in Section <ref>.
This theorem applies in particular to the Dirichlet space,
which answers a question posed in <cit.> and at the end of <cit.>.

The majority of this note is devoted to exploring consequences of this theorem.
Our first application concerns zero sets.
If S is a class of functions on a set X, then a subset Z of X is called
a zero set for S if there exists a function in S that vanishes exactly on Z.
It is well known that the zero sets for
H^2 and for H^∞ agree and are precisely the Blaschke
sequences, along with the entire disc  (see <cit.>).
Zero sets for functions in L^2_a can be much more complicated, for instance,
the union of two zero sets for L^2_a need not be a zero set for L^2_a <cit.>.
On the other hand, it was shown by Marshall and Sundberg <cit.>, see also <cit.>,
that the zero sets for functions and for multipliers in the Dirichlet space agree.
We show that this fact extends to all complete Pick spaces, thereby providing a positive answer to
<cit.> in the case of complete Pick kernels.
The case of Pick kernels which are not necessarily complete Pick kernels remains open.
We say that a reproducing kernel Hilbert space on X is normalized if its kernel is normalized
at some point in X.


  
  Let  be a normalized complete Pick space. Then the zero sets for , ()
  and N^+() agree. In particular, the union of two zero sets for  is a zero set for .


If  is a Hilbert function space on X, then it may happen that
all functions in  extend uniquely to a larger set.
For instance, we could start with a space of analytic functions on ,
let X = 1/2 and let  be the restriction of  to X,
so that every function in  extends uniquely to an analytic function on .
The notion of partially multiplicative functional (or generalized kernel function) allows
one to find the largest possible domain of definition for the functions in . A non-zero
bounded functional ρ on  is said to be partially multiplicative
if ρ(f g) = ρ(f) ρ(g) whenever f,g ∈ such that f g ∈.
Clearly, evaluation at each point of X gives rise to a partially multiplicative
functional on . We say that X is a maximal domain for  if conversely, every partially
multiplicative functional is given by evaluation at a point in X (this is called  is algebraically
consistent by Cowen and MacCluer, see <cit.>).
It was shown in <cit.> that every Hilbert function space can be considered as a space of functions on the
set of partially multiplicative
functionals, and this set is a maximal domain for .
For more discussion, the reader is referred to
<cit.>, <cit.> and <cit.>.
In the context of normalized complete Pick spaces, there is another notion of partially multiplicative functional,
and we show that the two notions agree. This answers a question that was left open in <cit.>.



  Let  be a normalized complete Pick space and let ρ be a bounded non-zero functional on .
  Then the following are equivalent.
  
    
  * ρ(φ g) = ρ(φ) ρ(g) for all φ∈() and all g ∈.
    
  * ρ(f g) = ρ(f) ρ(g) whenever f,g ∈ such that f g ∈.
  


A classical notion in uniform algebras is that of a Gleason part in the maximal ideal space,
see <cit.> and <cit.>. For instance, Gleason parts play an important
role in the study of the maximal ideal space of H^∞, see for example <cit.>.
This notion can be generalized to multiplier algebras.
Given two characters ρ_1,ρ_2 on a multiplier
algebra, we write ρ_1 ∼ρ_2 if ||ρ_1 - ρ_2|| < 2. This defines
an equivalence relation on the maximal ideal space of the multiplier algebra (see Lemma <ref>),
and the equivalence classes are called Gleason parts.
In this context, we prove in Section <ref> the following result.



  Let  be a normalized complete Pick space on a set X. Then the following are equivalent.
  
    
  * X is a maximal domain for .
    
  * Every weak-* continuous character on () is given by evaluation at a point in X.
    
  * The characters of evaluation at points in X form a Gleason part in the maximal
      ideal space of X.
  


The equivalence of (i) and (ii) in Proposition <ref> was already observed in <cit.> (using
(i) in Corollary <ref> as the definition of partially multiplicative functional and under the assumption
that  separates the points in X).

In Section <ref>
we study complete Pick spaces of continuous functions on compact sets — for example, spaces
of analytic functions on the disc that extend to be continuous on .
In particular, we investigate the validity of a corona theorem in this context.
Carleson's famous corona theorem for H^∞ <cit.> asserts that the open unit disc is dense
in the maximal ideal space of H^∞. This was extended to multiplier algebras
of certain Besov-Sobolev spaces on the unit ball in finite dimensions,
including the multiplier algebra of the Drury-Arveson space, by Costea, Sawyer and Wick <cit.>.
If () consists of continuous functions on a compact set X, then
a corona theorem for () would assert that the maximal ideal space of ()
is equal to the characters of evaluation at points of X.


  
  Let  be a
  normalized complete Pick space of continuous functions
  on a compact set X such that X is a maximal domain for  and
  such that () separates the points of X. Then the following are equivalent.
  
    
  * () = as vector spaces.
    
  * The corona theorem holds for (), that is,
      the maximal ideal space of ()
      is X.
    
  * The one-function corona theorem holds for (), that is, if
      φ∈() is non-vanishing, then 1/ φ∈().
  


As a consequence, using a very interesting example of Salas <cit.>, we exhibit a rotationally invariant
complete Pick space on  for which the one-function corona theorem (and hence the full corona theorem) fails.


  
  There exists a complete Pick space  on  such that 
  is a maximal domain for  with a reproducing kernel
  of the form
  
    k(z,w) = ∑_n=0^∞ a_n (z w)^n

  and such that the one-function corona theorem for () fails. In particular, 
  is a proper compact subset of the maximal ideal space of ().


In the terminology of Nikolski, the fact that the one-function corona theorem fails for ()
means that the spectrum of () is not 1-visible, see Definition 0.2.1 and Lemma 0.2.2
in <cit.>. We also remark that an example of a multiplier algebra
of functions on a subset of  for which the one-function corona theorem fails was already
obtained by Trent <cit.>, but the multiplier algebra of Theorem <ref> is quite different.
Indeed, Trent remarks that the functions in his multiplier algebra
“have no smoothness properties in general” (see the discussion preceding <cit.>),
whereas the multipliers of Theorem <ref> are analytic functions in the unit disc.



§ PRELIMINARIES





 §.§ Kernels, multipliers and normalization

Let  be a reproducing kernel Hilbert space on a set X with reproducing kernel k.
For background material on reproducing kernel Hilbert spaces, the reader is referred to
<cit.> and <cit.>. We will assume throughout that k does not vanish on the diagonal.
Let

    () = {φ: X →: φ f ∈ for all  f ∈}

denote the multiplier algebra of . The closed graph theorem shows that
every φ∈() determines a bounded multiplication operator M_φ on ,
and we set ||φ||_() = ||M_φ||.
We say that k (or ) is normalized at x_0 ∈ X
if k(x,x_0) = 1 for all x ∈ X, and we say that k (or ) is normalized
if it is normalized at some point in X.
If k(z,w) ≠ 0 for all z,w ∈ X, then it is always
possible to normalize the kernel at a point (see <cit.>). This procedure multiplies
all functions in  by a fixed non-vanishing function and leaves () unchanged.
If k is normalized, then  contains in particular the constant function 1 and
||1||_ = 1, so that
() ⊂, and the inclusion is contractive.



 §.§ The Pick property

We say that  is a complete Pick space
if for every r ∈ and every finite collection of points z_1,…,z_n ∈ X
and matrices W_1,…,W_n ∈ M_r(),
positivity of the nr × nr-block matrix

    [ k(z_i,z_j) (I_^r - W_i W_j^*) ]_i,j=1^n

implies that there exists Φ∈ M_r(()) of norm at most 1 such that

    Φ(z_i) = W_i    (i=1,…,n).

In this setting, we also say that k is a complete Pick kernel.
If  is a normalized complete Pick space, then k(z,w) ≠ 0 for all z,w ∈ X
by <cit.>.
Complete Pick spaces were characterized
by a theorem of Quiggin <cit.> and McCullough <cit.>. We require the following
characterization of Agler and McCarthy <cit.>.


  
  Let  be a reproducing kernel Hilbert space on X with kernel k which is normalized at x_0 ∈ X.
  Then  is a complete Pick space if and only if there exists an auxiliary Hilbert space 
  and a function b: X → with ||b(w)|| < 1 for all w ∈ X, b(x_0) = 0 and
  
    k(z,w) = 1/1 - ⟨ b(z),b(w) ⟩_.

  If  is separable, then  can be chosen to be separable.


A comprehensive treatment of complete Pick spaces, as well as examples besides the ones already mentioned
in the introduction, can be found in <cit.>.



 §.§ Cyclic multipliers



We say that a multiplier φ of a reproducing kernel Hilbert space 
is cyclic if
the multiplication operator M_φ on  has dense range.
Observe that φ is cyclic if and only if
(M_φ^*) = {0}.
In particular, the product of two cyclic multipliers is cyclic.

We require the following version of the maximum modulus principle
for Hilbert function spaces with non-vanishing kernels.

  
  Let  be a reproducing kernel Hilbert space on X with kernel k such that k(z,w) ≠ 0 for all
  z,w ∈ X and let φ∈() with ||φ||_()≤ 1.
  If there exists z ∈ X with |φ(z)| = 1, then φ is constant.



  By multiplying φ by a complex number number of modulus 1, we may assume that φ(z) = 1. Let
  w ∈ X be arbitrary. Since ||φ||_()≤ 1, the Pick matrix at {z,w}, which is
  
    [                      0      k(z,w) (1 - φ(w));     k(w,z) ( 1 - φ(w)) k(y,y) ( 1 - |φ(w)|^2) ],

  is positive semidefinite. Taking its determinant, we see that
  
    0 ≤ - |k(z,w)|^2  |1 - φ(w)|^2.

  Since k(z,w) ≠ 0 by assumption, this inequality implies that φ(w) = 1.
  Consequently, φ is identically 1.


The following lemma gives a sufficient condition for cyclicity of multipliers, which is probably known.


  
  Let  be a reproducing kernel Hilbert space on X with kernel k such that k(z,w) ≠ 0
  for all z,w ∈ X.
  Let φ∈() with ||φ||_()≤ 1.
  If φ≠ 1, then 1 - φ is cyclic.



  It suffices to show that M_1 - φ^* = 1 - M_φ^* is injective.
  Thus, let h ∈( 1 - M_φ^*). Then M_φ^* h = h. Since M_φ is a contraction,
  it follows that M_φ h = h and hence that
  (φ - 1) h = 0.
  Since φ≠ 1, Lemma <ref> implies that φ - 1 has no zeros on X. Consequently,
  h = 0, so that 1 - φ is cyclic.





§ PROOF OF THEOREM <REF> AND FIRST CONSEQUENCES



We now present the proof of Theorem <ref>. For the convenience of the reader,
we restate the result.

  
  Let  be a complete Pick space on X whose kernel is normalized at x_0 ∈ X and let f ∈ with ||f||_≤ 1.
  Then
  there are φ,ψ∈() of multiplier norm at most 1 with ψ(x_0) = 0
  such that
  
    f = φ/1 - ψ.

  In particular, ⊂ N^+().



  By Theorem <ref>, there exists
  a Hilbert space  and a function b: X →(,) with b(x_0) = 0 such that
  
    k(z,w) = 1/1- b(z) b(w)^* .

  Note that b ∈(⊗, ) has multiplier norm at most 1,
  as k(z,w) (1 - b(z) b(w)^*) = 1 is positive definite.
  Define Φ: X →( ⊕,) by
  
    Φ(z) = (1, f(z) b(z)).

  Then
  
    k(z,w) Φ(z) Φ(w)^*    = k(z,w) + f(z) f(w) k(z,w) b(z) b(w)^* 
       = k(z,w) + f(z) f(w) k(z,w) - f(z) f(w)

  and consequently
  
    k(z,w) ( Φ(z) Φ(w)^* - f(z) f(w)) = k(z,w) - f(z) f(w).

  Since ||f||_≤ 1 this function is positive definite (see <cit.>). In this setting,
  a version of Leech's theorem <cit.>, or more precisely the implication
  (i) ⇒ (ii) of <cit.>, shows that there exists
  a vector valued multiplier Ψ∈(, ⊗ (⊕)) of norm
  at most 1 such that
  
    Φ(z) Ψ(z) = f(z)    (z ∈ X).

  In the statement of <cit.>, it is assumed that k satisfies an irreducibility
  condition that implies that  separates the points of X, but the proof shows that it suffices to assume that k is normalized.
  Write
  
    Ψ(z) =
        [ φ(z); Ψ(z) ],

  where φ∈() and Ψ(z) ∈(, ⊗)
  both have norm at most one.
  Then
  
    φ(z) + f(z) b(z) Ψ(z) = f(z),

  so that
  
    f(z) ( 1 - b(z) Ψ(z)) = φ(z).

  Defining ψ(z) = b(z) Ψ(z), we obtain the desired representation of f.
  Lemma <ref> implies that 1 - ψ is a cyclic multiplier, which shows
  that f ∈ N^+() and hence proves the additional assertion.


Corollary <ref> is now an immediate consequence of the preceding theorem.


  
  Let  be a normalized complete Pick space.
  Then the zero sets for , ()
  and N^+() agree. In particular, the union of two zero sets for  is a zero set for .




  Observe that
  
    () ⊂⊂ N^+(),

  where the first inclusion holds since  is normalized, so that 1 ∈,
  and the second inclusion follows from Theorem <ref>.
  It is immediate from the definition of N^+()
  that every zero set for N^+() is a zero set for (),
  so that the zero sets for all three spaces agree. Since ()
  is an algebra, the union of two zero sets is a zero set.



  To obtain equality of the zero sets for  and for (), it suffices
  to assume that  is a complete Pick space whose kernel does not vanish anywhere.
  Indeed, in this case, the kernel can be normalized at a point (see <cit.>),
  which does not affect the zero sets.


It is also not hard to deduce Corollary <ref> from Theorem <ref>.


  
  Let  be a normalized complete Pick space and let ρ be a bounded non-zero functional on .
  Then the following are equivalent.
  
    
  * ρ(φ g) = ρ(φ) ρ(g) for all φ∈() and all g ∈.
    
  * ρ(f g) = ρ(f) ρ(g) whenever f,g ∈ such that f g ∈.
  



  The implication (ii) ⇒ (i) is trivial since () ⊂.
  Conversely, assume that (i) holds  and let f,g ∈
  such that f g ∈. By Theorem <ref>, we may write f = φ /ψ, where φ , ψ∈()
  and ψ is cyclic.
  We claim that ρ(ψ) ≠ 0.
  Indeed, if ρ(ψ) = 0, then
  ρ(ψ g) = 0 for all g ∈ and hence ρ = 0, as ψ is cyclic.
  Therefore, ρ(ψ) ≠ 0.
  Moreover,
  
    ρ(ψ) ρ(f g) = ρ(ψ f g) = ρ( φ g) = ρ(φ) ρ(g) = ρ(ψ f) ρ(g)
        = ρ(ψ) ρ(f) ρ(g).

  Since ρ(ψ) ≠ 0, it follows that ρ(f g) = ρ(f) ρ(g).





§ GLEASON PARTS



Gleason parts were originally introduced in the context of uniform algebras,
but it is possible to generalize this notion to multiplier algebras of
reproducing kernel Hilbert spaces, and indeed to arbitrary unital operator algebras of functions.
This was observed by Rochberg <cit.> for multiplier algebras of
certain weighted Dirichlet spaces on ,
but his arguments readily generalize. For completeness, we provide the proof of the lemma below.
It is an easy adaptation of <cit.>.


  
  Let  be a reproducing kernel Hilbert space and let ρ_1,ρ_2 be characters
  on (). Then the following are equivalent.
  
    
  * ||ρ_1 - ρ_2 || < 2.
    
  * || ρ_1 |_(ρ_2)|| < 1.
    
  * Whenever (φ_n)_n is a sequence in the unit ball of () such that
      lim_n →∞ |ρ_1(φ_n)| = 1, then lim_n →∞ |ρ_2(φ_n)| = 1.
  
  In particular, the relation ρ_1 ∼ρ_2 if and only if ||ρ_1 - ρ_2|| < 2
  is an equivalence relation.



  For a ∈, let θ_a denote the conformal automorphism of  defined by
  
    θ_a(z) = a - z/1 - a z.

  Thus, θ_a is an involution which interchanges 0 and a.
  We will make repeated use of the following consequence of von Neumann's inequality:
  if φ∈() with ||φ||_()≤ 1, then
  θ_a ∘φ∈()
  with ||θ_a ∘φ||_()≤ 1 for all a ∈.

  (i) ⇒ (ii) Suppose that ||ρ_1 |_(ρ_2)|| =1. Then
  there exists a sequence (φ_n) in the unit ball of (ρ_2) such that if t_n = ρ_1(φ_n), then
  0 < t_n < 1 for all n ∈ and lim_n →∞ t_n = 1. Let a_n = (1 - √(1 - t_n^2))/t_n. It is straightforward
  to check that a_n ∈ (0,1) and that θ_a_n(t_n) = - θ_a_n(0) for all n ∈. Thus,
  
    ||ρ_1 - ρ_2||    ≥ |ρ_1 (θ_a_n∘φ_n) - ρ_2( θ_a_n∘φ_n)| =
        |θ_a_n (t_n) - θ_a_n (0)| 
       = 2 |a_n|  2.

  Consequently, ||ρ_1 - ρ_2|| =2.

  (ii) ⇒ (iii) Suppose that there exists a sequence (φ_n) in the unit ball
  of () such that |ρ_1(φ_n)| tends to 1, but a_n = ρ_2(φ_n) is bounded away
  from 1 in modulus. Then θ_a_n∘φ_n belongs to the unit ball of (ρ_2)
  and it is easy to see that |ρ_1(θ_a_n∘φ_n)| tends to 1. So ||ρ_1 |_(ρ_2)|| = 1.

  (iii) ⇒ (ii) Suppose that ||ρ_1 |_(ρ_2)|| = 1. Then there exists a sequence
  (φ_n) in the unit ball of (ρ_2) such that lim_n →∞ |ρ_1(φ_n)| = 1.
  Therefore, (iii) fails.

  (ii) ⇒ (i) Suppose that ||ρ_1 - ρ_2|| = 2. Then there exists a sequence (φ_n) in the unit
  ball of () such that a_n = ρ_1(φ_n) and b_n = ρ_2(φ_n) both belong to  and |a_n - b_n|
  tends to 2. Then θ_b_n∘φ_n belongs to the unit ball of (ρ_2) and
  
    |ρ_1(θ_b_n∘φ_n)| = |θ_b_n (a_n)| ≥|a_n - b_n|/2 1.

  Thus, ||ρ_1 |_(ρ_2)|| = 1.

  Finally, it follows from (i) that ∼ is reflexive and symmetric, and it follows from (iii) that ∼ is transitive.



Let  be a reproducing kernel Hilbert space on X.
Then the assignment φ↦ M_φ identifies ()
with a weak-* closed subalgebra of B(), hence () is a dual space in its own right,
and we endow it with this weak-* topology.
It is straightforward to check that on bounded subsets of (), the weak-* topology
coincides with the topology of pointwise convergence on X.
We are now in the position to prove a more detailed version of Proposition <ref>.


  
  Let  be a complete Pick space on a set X whose kernel is normalized at x_0 ∈ X
  and let δ_0 denote the character of evaluation at x_0.
  For a character ρ on (), the following assertions are equivalent.
  
    
  * ρ extends to a bounded partially multiplicative functional on .
    
  * ρ is weak-* continuous.
    
  * ρ belongs to the Gleason part of δ_0.
  
  It follows that the following are equivalent.
  
    
  * X is a maximal domain for .
    
  * Every weak-* continuous character on () is given by evaluation at a point in X.
    
  * The characters of evaluation at points in X form a Gleason part in the maximal
      ideal space of X.
  



  (i) ⇒ (ii) If ρ is a bounded partially multiplicative functional
  on  which extends ρ, then ρ(φ) = ρ( M_φ 1) for φ∈(),
  which shows that ρ is WOT continuous and thus weak-* continuous.

  (ii) ⇒ (iii) Suppose that ρ is weak-* continuous and assume
  toward a contradiction that ||ρ|_(δ_0)|| = 1.
  Since δ_0 is weak-* continuous,
  the unit ball of (δ_0) is weak-* compact, so there exists a multiplier
  ψ of norm 1 such that ψ(x_0) = 0 and ρ(ψ)=1. If follows
  from Lemma <ref> that |ψ(x)| < 1 for all x ∈ X, so the sequence
  (ψ^n) converges to zero pointwise, and hence in the weak-* topology.
  But ρ(ψ^n) = 1 for all n ∈, contradicting the fact that ρ is weak-* continuous.
  Therefore, ||ρ|_(δ_0)|| < 1, so (iii) holds by Lemma <ref>.

  (iii) ⇒ (i) Suppose that ρ belongs to the Gleason part of δ_0,
  so that α = ||ρ|_(δ_0)|| < 1 by Lemma <ref>.
  If f ∈ has a representation f = φ_1 / φ_2, where φ_1 and φ_2
  are multipliers and ρ(φ_2) ≠ 0, we define
  
    ρ(f) = ρ(φ_1)/ρ(φ_2).

  By Theorem <ref>, every f ∈ can be written
  as f = φ / (1 - ψ) with ||φ||_()≤ ||f||_ and
  ψ belonging to the unit ball of (δ_0).
  Since |ρ(ψ)| ≤α < 1 by assumption, we see in particular that every f ∈
  admits a representation as above.
  Since ρ is a character, ρ is well defined, linear and satisfies
  ρ(φ f)
  = ρ(φ) ρ(f) for φ∈() and f ∈.
  Moreover, writing f = φ/(1 - ψ) as above,
  we obtain the estimate
  
    |ρ(f)| ≤|ρ(φ)|/1 - |ρ(ψ)|≤1/1 - α ||f||_,

  so that ρ is a bounded functional that extends ρ, and ρ
  is partially multiplicative by Corollary <ref>.

  Finally, to deduce the equivalence of (i'), (ii') and (iii') from the equivalence of (i), (ii) and (iii),
  it only remains to show that a bounded partially multiplicative functional on 
  is uniquely determined by its values on (). This can be seen as in the proof of Corollary
  <ref>, or alternatively, if follows from the fact that
  () is dense in .



  Suppose that  is a normalized complete Pick space on X and assume for convenience
  that  separates the points of X. If X is not a maximal
  domain for , then Proposition <ref> provides three equivalent ways
  of enlarging X to a maximal domain for . There is a fourth equivalent way, which we will now
  briefly describe.

  For a cardinal number d, let _d denote the open unit ball
  _d in a Hilbert space of dimension d. The Drury-Arveson space H^2_d is the reproducing kernel
  Hilbert space on _d with kernel
  
    1/1 - ⟨ z,w ⟩.

  Let b: X →_d be the map of Theorem <ref> and let S = b(X). Then
  
    H^2_d |_S →,    f ↦ f ∘ b,

  is a unitary operator. Let
  
    I = {φ∈(H^2_d): φ|_S = 0}

  and let
  
    V = {z ∈_d: φ(z) = 0  for all φ∈ I }.

  (V is an analogue of the Zariski closure from algebraic geometry.)
  Tautologically, S ⊂ V, and it was observed by Davidson, Ramsey and Shalit <cit.>
  that H^2_d |_S can be identified with H^2_d |_V. More precisely,
  every function in H^2_d |_V extends uniquely to a function in H^2_d |_S.
  Moreover, <cit.>
  and its proof show that the partially multiplicative functionals on H^2_d |_S (and hence on )
  precisely correspond to points in V.

  In particular, this last description of a maximal domain for  shows that  remains
  a complete Pick space after enlarging the domain X to a maximal one.




§ SPACES ON COMPACT SETS AND THE CORONA THEOREM



In this section, we study spaces of continuous functions on compact sets and prove
Proposition <ref> and Theorem <ref>. We begin with a few preliminaries about
corona theorems.

Let  be a reproducing kernel Hilbert space on a set X such that () separates the points of X
and let (()) denote the maximal ideal space of ().
Then X can be identified with a subset of (()) via point evaluations.
We say that the corona theorem holds for ()
if X is dense in (()) in the Gelfand topology.

Gelfand theory shows that the following two statements are equivalent.

  
  * The corona theorem holds for ().

  * If φ_1,…,φ_n ∈() such that

    inf_x ∈ X∑_j=1^n |φ_j(x)| > 0,

then the ideal generated by φ_1,…,φ_n inside () is all of ().

We say that the one-function corona theorem holds for ()
if whenever φ∈() and inf_x ∈ X |φ(x)| > 0, then 1/φ∈().
Thus, the corona theorem for () implies the one-function corona theorem for ().


  It is not hard to see that in the setting above, the one-function corona
  theorem holds for () if and only if for every φ∈(),
  we have
  
    σ_() (φ) = {φ(x): x ∈ X }.

  Here, σ_() denotes the spectrum in the unital Banach algebra ().
  Similarly, the corona theorem holds for () if and only if for every n ∈,
  we have
  
    σ_() (φ_1,…,φ_n) = { (φ_1(x), …, φ_n(x)): x ∈ X  }⊂^n.

  If  is a normalized complete Pick space, then
  we may replace σ_() with other notions of joint spectrum. Indeed,
  the Toeplitz corona theorem
  (see <cit.>) shows that in this case,
  
    σ_() (φ_1,…,φ_n) = σ_r (M_φ_1,…,M_φ_n)
        = σ_T(M_φ_1,…,M_φ_n),

  where σ_r and σ_T denote the right spectrum and the Taylor spectrum, respectively
  (see <cit.> for a definition and discussion of these notions).


For normalized complete Pick spaces, it is easy to tell from the kernel whether the multiplier algebra
separates the points of the underlying set.


  
  Let  be a normalized complete Pick space on X with kernel k. Then the following are equivalent.
  
    
  * () separates the points of X.
    
  *  separates the points of X.
    
  * If z ≠ w, then k(·,z) ≠ k(·,w).
    
  * If z ≠ w, then k(·,z) and  k(·,w) are linearly independent.
  



  (i) ⇒ (ii) ⇒ (iii) is trivial, and (iii) ⇒ (iv) follows
  from the fact that k is normalized at a point.

  Finally, if (iv) holds, then the Cauchy-Schwarz inequality and the Pick property, applied to the points z,w,
  show that there exists φ∈() with φ(z) = 0 and φ(w) ≠ 0.

Thus, () separates the points of X if and only if the kernel is irreducible in the strong
sense of <cit.>.

Suppose now that  is a normalized complete Pick space which separates the points of X.
When investigating whether the corona theorem holds for (), we wish to exclude
constructions such as the restriction of the Hardy space H^2 to 1/2,
which is really a space on  in disguise. Thus, we will typically assume that X
is a maximal domain for .

In this section, we are interested in the case when X is a compact topological space
and the functions in  are continuous on X.
Such spaces are easy to construct, as the following class of examples shows.


  
  Let d ∈ and let  be a complete Pick space on _d
  with reproducing kernel of the form
 
    k(z,w) = ∑_n=0^∞ a_n ⟨ z,w ⟩^n,

  where a_0 = 1 and a_n > 0 for n ≥ 1. If ∑_n=0^∞ a_n < ∞,
  but the power series ∑_n=0^∞ a_n t^n has radius of convergence 1,
  then k extends to a continuous function on _d×_d,
  and  thus becomes a space of continuous functions on _d in a natural way.
  It is not hard to see that _d is a maximal domain for such a space
  (see, for example, <cit.>). Moreover,
  since a_1 > 0,  contains the coordinate functions, so that  separates the points of _d.

  More concretely, for s ∈, let
  
    k(z,w) = ∑_n=0^∞ (n+1)^s (z w)^n    (z,w ∈)

  and let _s be the reproducing kernel Hilbert space on  with kernel k.
  This scale of spaces contains in particular the Bergman space (s=1), the Hardy space (s=0)
  and the Dirichlet space (s=-1). If s ≤ 0, then  is a normalized complete Pick space (see,
  for example, <cit.>). If s < -1, then k satisfies the conditions
  in the preceding paragraph, hence _s becomes normalized complete Pick
  space on  and  is a maximal domain for .


Let  be a normalized complete Pick space of continuous functions on a compact
set X which separates the points of X such that X is a maximal domain for .
Then the embedding of X into (()) via point evaluations is a homeomorphism
onto its image, hence X can be identified with a compact
subset of (()). Thus,
the corona theorem holds for () if and only if X = (()).
Moreover, a multiplier
φ∈() is bounded below on X if and only if it is non-vanishing.
Consequently, the one-function corona theorem for () holds if and only if
every non-vanishing multiplier on  is invertible.

We are now in the position to prove Proposition <ref>.


  
  Let  be a
  normalized complete Pick space of continuous functions
  on a compact set X which separates the points of X such that X is a maximal domain for .
  Then the following are equivalent.
  
    
  * () = as vector spaces.
    
  * The corona theorem holds for ().
    
  * The one-function corona theorem holds for ().
  



  (i) ⇒ (ii) Suppose that () = as vector spaces. Since the multiplier norm
  dominates the norm of , an application of the open mapping theorem shows that
  these two norms are in fact equivalent. Thus, if ρ is a character on (),
  then ρ is a bounded functional on  which is partially multiplicative.
  Since X is a maximal domain for , the functional ρ equals evaluation
  at a point in X.

  (ii) ⇒ (iii) follows from Gelfand theory.

  (iii) ⇒ (i) Let f ∈. By Theorem <ref>,
  there are φ,ψ∈() with ψ non-vanishing such that
  f = φ / ψ. The assumption (iii) implies that 1 / ψ∈(),
  thus f ∈(). The reverse inclusion always holds, hence
  = () as vector spaces.


The spaces _s of Example <ref>, where s < -1, satisfy
condition (i) of the preceding proposition (see Proposition 31 and Example 1 on page 99 in <cit.>).
Hence, ((_s)) = (see also Corollary 1 on page 95 in <cit.>).
We now use an example of Salas <cit.>, which answered <cit.>,
and Proposition <ref> to exhibit a complete Pick space
on  for which the one-function corona theorem fails, thereby proving Theorem <ref>.


  
  There exists a complete Pick space  on  with a reproducing kernel
  of the form
  
    k(z,w) = ∑_n=0^∞ a_n (z w)^n,

  where a_0 = 1, a_n > 0 for all n ∈, lim_n →∞ a_n / a_n+1 = 1 and ∑_n a_n <∞
  such that  is a maximal domain for  and such that
  the one-function corona theorem for () fails.



  In <cit.>, Salas constructs a weighted shift T on ℓ^2
  with weight sequence (w_n) which satisfies
  
    
  * w_n is decreasing and lim_n →∞ w_n = 1.
    
  * ∑_n=0^∞β(n)^-2 < ∞, where β(n) = ∏_j=0^n-1 w_j.
    
  * T is not strictly cyclic.
  
  The original definition of strict cyclicity can be found in <cit.>;
we shall give an equivalent one.
Define a Hilbert space  by
  
    = { f(z) = ∑_n=0^∞f(n) z^n : ||f||^2 = ∑_n=0^∞ |f(n)|^2
      β(n)^2 < ∞}.

Property (2) implies that  is in fact
  a reproducing kernel Hilbert space on  whose reproducing kernel is given by
  
    k(z,w) = ∑_n=0^∞ a_n (z w)^n,

  where a_n = β(n)^-2, see <cit.>.
Property (3) is equivalent to saying that () ⊊,
see <cit.>.

We have a_0 = 1,
  so  k is normalized at 0.
  Moreover,
  
    a_n/a_n+1 = β(n+1)^2/β(n)^2 = w_n^2,

  hence Property (1) implies that a_n / a_n+1 decreases to 1. An application
  of a lemma of Kaluza (see Lemma 7.31 and 7.38 in <cit.>) shows that  is a
  complete Pick space.
  Moreover, we see that the radius of convergence of the power series
  ∑_n=0^∞ a_n t^n is 1, hence  is a
  space of continuous functions on , which is a maximal domain for ,
  and  separates the points of
   (see Example <ref>).
  Since () ⊊,
  the implication (iii) ⇒ (i) of Proposition <ref> shows
  that the one-function corona theorem fails for ().


We call a space  on  as in the statement of Theorem <ref> a Salas space.


  
  We can say slightly more about the maximal ideal space of the multiplier algebra of a Salas space .
  It is easy to check (see, for example, <cit.>) that there exists a continuous map
  
    π : (()) →,   ρ↦ρ(z).

  Since lim_n →∞ a_n / a_n+1 = 1,
  it follows from <cit.> that if λ∈, then
  π^-1(λ) is the singleton containing the character of evaluation at λ. In particular,
  every character on () which is not given by evaluation at a point in 
  is contained in π^-1(∂).
  Since there exists at least one such character by Theorem <ref>,
  and since  is rotationally invariant, we deduce
  that π^-1(λ) contains at least one such character for every λ∈∂.

  Moreover, since () has no non-trivial idempotent elements, the Šilov idempotent theorem (see <cit.>) shows that (()) is connected.



  Let  be a unital Banach algebra of continuous functions on  such that the polynomials
  form a dense subspace of  and such that the maximal ideal space of  is equal to .
  In <cit.>, the following problem is studied: given δ > 0, does there exist
  a constant C(δ) > 0 such that for all f ∈ with ||f||_≤ 1 and
  inf_z ∈ |f(z)| ≥δ, the estimate
  
    ||f^-1||_≤ C(δ)

  holds? The authors of <cit.> obtain a positive answer for rotationally
  invariant algebras  which satisfy some additional assumptions.

  We can use a Salas space to obtain a rotationally invariant algebra for which the question above has
  a negative answer. To this end, let  be a Salas space and let A() denote the norm
  closure of the polynomials inside of (). The map π of Remark <ref>
  shows that the maximal ideal space of A() is . By Theorem <ref>,
  there exists a multiplier φ in the unit ball of ()
  such that δ = inf_z ∈ |φ(z)| > 0, but such that φ is not invertible
  inside of (). For r ∈ (0,1), define φ_r(z) = φ(r z). Then each φ_r
  is analytic in an open neighborhood of , so since σ_A()(z) =, we conclude that
  φ_r ∈ A() for all r ∈ (0,1).
  Clearly, |φ_r| is bounded below by δ for all r ∈ (0,1), so
  each φ_r is invertible inside of A() by Gelfand theory. Moreover,
  a routine application of the Poisson kernel, combined with rotational invariance of ,
  shows that ||φ_r||_A()≤ ||φ||_()≤ 1 for all r ∈ (0,1).

  We claim that ||φ_r^-1||_A() is not bounded as r → 1. Indeed, suppose otherwise.
  Then by weak-* compactness of the closed unit ball of (), the net (φ_r^-1)_r < 1
  has a weak-* cluster point ψ∈(). In particular, φ_r^-1 converges to ψ
  pointwise on , so that ψ  = φ^-1, contradicting the fact that φ is not invertible
  inside of ().


If  is a Salas space, then the polynomials are not norm dense in (), since the corona
theorem fails for (). However, besides this and Remark <ref>, we know very little
about the size of () or of its maximal ideal space.


  Let  be a Salas space on .
  
    
  * Is () separable?
    
  * Is (()) metrizable?
    
  * Is the cardinality of (()) equal to that of the continuum?
  


Observe that a positive answer to any of these questions implies
a positive answer to the questions below it.

Recall that H^2_d denotes the Drury-Arveson space on _d, the
open unit ball in a Hilbert space of dimension d.
If d = ℵ_0, we simply write _∞ and H^2_∞.
In the case d < ∞, Costea, Sawyer and Wick <cit.> showed that the corona theorem
holds for (H^2_d). Fiang and Xia <cit.> provide
a more elementary proof of the one-function corona theorem in this case; an even shorter
proof was found by Richter and Sunkes <cit.>. But none of these proofs
extend to infinite d in a straightforward manner.
Let  be a Salas space.
It follows from Theorem <ref> of Agler and McCarthy that ()
can be identified with

    {φ|_V: φ∈(H^2_∞) }

for a set V ⊂_∞ (indeed, we can choose V= b(), where b
is the map from Theorem <ref>).
We are not aware of an argument which would show that the failure of the corona
theorem for () implies the failure of the corona theorem for (H^2_∞).
We therefore ask:


  Does the corona theorem hold for (H^2_∞)? Does the one-function corona theorem hold for (H^2_∞)?



amsplain


