# On the Gleason-Kahane-Żelazko theorem for associative algebras

Moshe Roitman  Department of Mathematics
University of Haifa 199 Abba Khoushy Avenue
Mount Carmel, Haifa 3498838
Israel<EMAIL_ADDRESS>and  Amol Sasane  Department of Mathematics
London School of Economics
Houghton Street
London WC2A 2AE
United Kingdom<EMAIL_ADDRESS>
###### Abstract.

The classical Gleason-Kahane-Żelazko Theorem states that a linear functional
on a complex Banach algebra not vanishing on units, and such that
$\Lambda(\mathbf{1})=1$, is multiplicative, that is,
$\Lambda(ab)=\Lambda(a)\Lambda(b)$ for all $a,b\in A$. We study the GKŻ
property for associative unital algebras, especially for function algebras. In
a GKŻ algebra $A$ over a field of at least $3$ elements, and having an ideal
of codimension $1$, every element is a finite sum of units. A real or complex
algebra with just countably many maximal left (right) ideals, is a GKŻ
algebra. If $A$ is a commutative algebra, then the localisation $A_{P}$ is a
GKŻ-algebra for every prime ideal $P$ of $A$. Hence the GKŻ property is not a
local-global property. The class of GKŻ algebras is closed under homomorphic
images. If a function algebra $A\subseteq\mathbb{F}^{X}$ over a subfield
$\mathbb{F}$ of $\mathbb{C}$, contains all the bounded functions in
$\mathbb{F}^{X}$, then each element of $A$ is a sum of two units. If $A$
contains also a discrete function, then $A$ is a GKŻ algebra. We prove that
the algebra of periodic distributions, and the unitisation of the algebra of
distributions with support in $(0,\infty)$ satisfy the GKŻ property, while the
algebra of compactly supported distributions does not.

###### Key words and phrases:

Gleason-Kahane-Żelazko Theorem, function algebras, generated by units,
convolution rings of distributions

###### 2010 Mathematics Subject Classification:

Primary 46H05; Secondary 46F10, 15A86, 47B49

##  1\. Introduction

We study the Gleason-Kahane-Żelazko property (Definition 1.1 below) of
associative unital algebras, especially of function algebras. We will also
investigate the validity of the GKŻ theorem in some natural convolution
algebras of distributions.

###  1.1. Background

The classical Gleason-Kahane-Żelazko Theorem, which was proved independently
by Gleason [4], and by Kahane and Żelazko [8], provides a characterisation of
the maximal ideals of a commutative complex Banach algebra $A$: a subspace of
$A$ is a maximal ideal if and only if it has codimension $1$ and contains no
units. Equivalently, a linear functional $\Lambda:A\to\mathbb{C}$ not
vanishing on units, and such that $\Lambda(\mathbf{1}_{A})=1$, is
multiplicative, that is, $\Lambda(ab)=\Lambda(a)\Lambda(b)$ for all $a,b\in
A$. The formulation in terms of linear functionals was extended by Żelazko to
any complex Banach algebra [25], thus providing a characterisation of the
ideals of codimension $1$ (that are necessarily maximal) in a complex Banach
algebra.

######  Definition 1.1.

Let $A$ be a unital algebra over a field $\mathbb{F}$. The algebra $A$ is said
to have the GKŻ property (or to be a GKŻ algebra) if every linear functional
$\Lambda:A\rightarrow\mathbb{F}$ not vanishing on units, and such that
$\Lambda(\mathbf{1}_{A})=1$ is multiplicative.

In Definition 1.1 we allow the case that $\mathbf{1}_{A}=\mathbf{0}_{A}$, that
is the case that $A$ is the zero algebra because, although we always start
with a nonzero algebra, we may obtain sometimes zero algebras. By the above
definition, a zero algebra is a GKŻ algebra.

The converse of the GKŻ property is obvious: if $\Lambda$ is a multiplicative
nonzero linear functional on $A$, then $\Lambda$ does not vanish on units and
$\Lambda(\mathbf{1}_{A})=1$.

######  Remark 1.2.

Let $A$ be an algebra over a field $\mathbb{F}$. Every subspace of $A$ is
contained in a subspace of codimension $1$. Thus the maximal subspaces of $A$
are the subspaces of codimension $1$.

An algebra $A$ satisfies the GKŻ-property if and only if each subspace of $A$
of codimension $1$ not containing units is multiplicatively closed,
equivalently it is an ideal (necessarily maximal).

There is an extensive literature on extensions of the Gleason-Kahane-Żelazko
Theorem. See, e.g., the surveys [6] and [13].

######  Definition 1.3.

A function algebra over a field $\mathbb{F}$ on a non-empty set $X$, is a
unital subalgebra of the $\mathbb{F}$-algebra $\mathbb{F}^{X}$ (the set of all
functions $X\to\mathbb{F}$, with addition and multiplication defined
componentwise) such that $\mathbf{1}_{\mathbb{F}^{X}}\in A$.

###  1.2. Summary

In section 2 we deal with the GKŻ property for algebras, in section 3 for
function algebras, and in section 4 for distribution algebras.

The main result of §2.1 (The GKŻ property of algebras and generation by
units), is that a GKŻ algebra $A$ over a field of at least $3$ elements and
having an ideal of codimension $1$, is generated by units, that is, every
element of $A$ is a finite sum of units (Theorem 2.1 (1)), equivalently,
$A=\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$. A GKŻ function
algebra over a field $\mathbb{F}$ with at least $3$ elements is unit
generated.

In §2.2 (Linear coverings), we prove that if the field $\mathbb{F}$ is
infinite, and $A\setminus\operatorname{{\mathcal{U}}}(A)$ is contained in a
union of less than $|\mathbb{F}|$ proper one-sided ideals, then the algebra
$A$ satisfies the GKŻ property. (Proposition 2.15). Thus a real or complex
algebra with just countably many maximal left ideals is a GKŻ algebra. We also
show that if $A$ is a commutative algebra, then the localisation $A_{P}$ is a
GKŻ-algebra for every prime ideal $P$ of $A$. Hence the GKŻ property is not a
local-global property.

In §2.3 (The GKŻ property for homomorphic images) we show that the class of
GKŻ algebras is closed under homomorphic images (Proposition 2.18), and
provide a sufficient condition for the GKŻ property of $A$ assuming that $A/I$
is a GKŻ algebra for an ideal $I$ of $A$ (Proposition 2.19). Naturally, $A$
satisfies the GKŻ property if and only if $A/\operatorname{Jac}(A)$ does
(Proposition 2.21), where $\operatorname{Jac}(A)$ is the Jacobson radical of
$A$.

In §4.2 (Unitisation), we show that the unitisation of a radical algebra is a
GKŻ algebra, and provide a topological sufficient condition for an algebra to
be radical (Proposition 2.26).

In section 3 (On the GKŻ property of function algebras), we prove that if a
function algebra $A\subseteq\mathbb{F}^{X}$ over a subfield $\mathbb{F}$ of
$\mathbb{C}$ contains all the bounded functions in $\mathbb{F}^{X}$, then each
element of $A$ is a sum of two units. If $A$ contains also a discrete
function, then $A$ is a GKŻ algebra.

In section 4 (Algebra of periodic distributions), we prove that the algebra of
periodic distributions, and the unitisation of the algebra of distributions
with support in $(0,\infty)$ satisfy the GKŻ property, while the algebra of
compactly supported distributions does not. Actually, we started our paper
with the study of the GKŻ property for some distribution algebras, and this
study led us to more general results.

###  1.3. Conventions and notations

Unless otherwise stated, all the algebras in this paper are nonzero unital
associative algebras, not necessarily commutative, over a field $\mathbb{F}$.
We use the notation $\mathbf{1}_{A}$ for the unity of the algebra $A$, and
$\operatorname{{\mathcal{U}}}(A)$ for the group of units (invertible elements)
of $A$. Using the $\mathbb{F}$-algebra monomorphism $\mathbb{F}\to A,\
f\mapsto f\mathbf{1}_{A}$, we may assume that $\mathbb{F}\subseteq A$. In the
whole paper, we denote by $\mathbb{F}$ a field, by $A$ a nonzero associative
unital algebra over $\mathbb{F}$, and by $X$ a nonempty set. Thus a function
algebra is a subalgebra of $\mathbb{F}^{X}$ with the same unity. We let
$\mathbb{F}^{\bullet}=\mathbb{F}\setminus\\{0\\}$. The cardinality of a set
$X$ is denoted by $|X|$. If $Y$ is a subset of $X$, we denote by $I_{Y}$ the
indicator (characteristic function) of $Y$. If $A\subseteq\mathbb{F}^{X}$ is a
function algebra, and $x\in X$, the projection $p_{x}$ is defined as the
function $p_{x}:A\to X\ (f\mapsto f(x))$. The field of two elements is denoted
by $\mathbb{F}_{2}$. A proper subspace of an algebra $A$ is a subspace of $A$
different from $A$, possibly zero. Similarly, we use proper ideals, etc. If
$A$ is an algebra and $a\in A$, then $\rho(a)$ (the resolvent of $a$) is the
set of all scalars $\lambda\in\mathbb{F}$ such that $a-\lambda\mathbf{1}$ is
invertible, and $\sigma(a)$ (the spectrum) of $a$ is the set
$\mathbb{F}\setminus\rho(a)$.

##  2\. The GKŻ property of general algebras

###  2.1. The GKŻ property of algebras and generation by units

There are several equivalent ways to define the property of an algebra to be
generated by units: as an algebra, as a ring, as a vector space (that is,
$A=\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$, and as an additive
group (which means that every element of $A$ is a finite sum of units). For
two surveys on rings generated by units see [18] and [19, Section 1].

######  Theorem 2.1.

Let $A$ be a GKŻ algebra over a field $\mathbb{F}$ of at least $3$ elements.
If $A$ has an ideal of codimension $1$, (equivalently, if there exists a
linear functional on $A$ not vanishing on units), then $A$ is generated by
units.

###### Proof.

Let $\Lambda$ be a linear functional that does not vanish on units. Replacing
$\Lambda$ by $\frac{\Lambda}{\Lambda(\mathbf{1})}$, we may assume that
$\Lambda(\mathbf{1})=1$. Assume that $A$ is not generated by units, that is,
$A\neq\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$. There exists a
subspace $V$ of $A$ of codimension $1$ containing
$\operatorname{{\mathcal{U}}}(A)$. Let $a\in A\setminus V$. Thus
$A=\mathbb{F}a\oplus V$. Let $a^{2}=\lambda a+w$, where
$\lambda\in\mathbb{F}$, and $w\in V$. Since $\mathbb{F}$ contains at least
three elements, we may choose $c\in\mathbb{F}$ such that

$c^{2}-\lambda c-\Lambda(w)\neq 0.$

There exists a unique linear functional $\Psi:A\rightarrow\mathbb{F}$ such
that $\Psi(v)\\!=\\!\Lambda(v)$ for all $v\in V$, and $\Psi(a)=c$. Hence

$\Psi(a^{2})=\Psi(\lambda a+w)=\lambda c+\Lambda(w)\neq c^{2}=(\Psi(a))^{2},$

so $\Psi$ is not multiplicative. On the other hand, $\Psi(u)=\Lambda(u)\neq 0$
for all $u\in\operatorname{{\mathcal{U}}}(A)$, and
$\Psi(\mathbf{1})=\Lambda(\mathbf{1})=1$, contradicting the assumption that
$A$ has the GKŻ property. ∎

######  Corollary 2.2.

A GKŻ function algebra over a field $\mathbb{F}$ with at least $3$ elements is
unit generated.

###### Proof.

Let $A\subseteq F^{X}$ be a nonzero function algebra. For all $x\in X$, the
projection $p_{x}$ is a multiplicative functional, and $p_{x}(I_{X})=1$. By
Theorem 2.1, $A$ is generated by units. ∎

An algebra $A$ is said to satisfy the GKŻ property vacuously if each linear
functional on $A$ vanishes on some unit, equivalently, if each maximal
subspace of $A$ contains a unit. Clearly, if $A$ is a GKŻ algebra vacuously,
then $A$ is a GKŻ algebra.

######  Remark 2.3.

In Theorem 2.1, the condition on the existence of an ideal of codimension $1$
is essential, but not necessary.

Indeed, consider the $\mathbb{C}$-algebras $\mathbb{C}(x)$ and
$\mathbb{C}(x)[y]$, where $x$ and $y$ are two independent indeterminates over
$\mathbb{C}$. Both these algebras satisfy the GKŻ property vacuously by the
next Lemma 2.4. However,
$\operatorname{span}(\operatorname{{\mathcal{U}}}(\mathbb{C}(x))=\mathbb{C}(x)$,
so the condition is not necessary, and
$\operatorname{span}(\operatorname{{\mathcal{U}}}(\mathbb{C}(x)[y]))=\mathbb{C}(x)\neq\mathbb{C}(x)[y]$,
so the condition is essential. ∎

######  Lemma 2.4.

Let $\mathbb{F}\subsetneqq\mathbb{K}$ be two fields, and let $A$ be an algebra
over $\mathbb{K}$. Then $A$ is a GKŻ algebra vacuously over $\mathbb{F}$.

###### Proof.

Let $V$ be an $\mathbb{F}$-subspace of $A$ of codimension $1$. Since
$\mathbb{F}\subsetneqq\mathbb{K}$, we have $[\mathbb{K}:\mathbb{F}]>1$, so $V$
contains a nonzero scalar $k\in\mathbb{K}$, and
$k\in\operatorname{{\mathcal{U}}}(A)$. ∎

Theorem 2.1 implies:

######  Corollary 2.5.

Let $A$ be an algebra over a field of cardinality $\geq 3$, that is not
generated by units. Then $A$ is a GKŻ algebra $\Leftrightarrow A$ is a GKŻ
algebra vacuously.

######  Remark 2.6.

If $A$ is an algebra, then
$\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$ is a subalgebra of $A$
with the same units.

######  Proposition 2.7.

Let $\mathbb{F}$ be a field of at least $3$ elements.

  1. (1)

The following three conditions are equivalent:

    1. (a)

$A$ is a GKŻ algebra.

    2. (b)

$A$ is a GKŻ algebra vacuously.

    3. (c)

$\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$ is a GKŻ algebra
vacuously.

  2. (2)

If $C\subsetneqq A$ are two algebras with the same units, then either both $A$
and $C$ are GKŻ algebras vacuously, or both of them are not GKŻ algebras.

###### Proof.

  1. (1)

$(a)\Rightarrow(b)$

If $A$ is not a GKŻ algebra vacuously, then by Theorem 2.1,
$A=\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$, a contradiction.

$(b)\Rightarrow(a)$ Clear.

$(b)\Rightarrow(c)$

If condition (c) does not hold, then there exists a linear functional
$\Lambda$ on $C$ not vanishing on units and such that
$\Lambda(\mathbf{1}_{C})=1$. Thus $\Lambda$ can be extended to a linear
functional $\widetilde{\Lambda}$ on $A$. We see that $\widetilde{\Lambda}$
does not vanish on units, since
$\operatorname{{\mathcal{U}}}(A)=\operatorname{{\mathcal{U}}}(C)$, and
$\widetilde{\Lambda}(\mathbf{1}_{A})=1$, contradicting $(b)$.

$(c)\Rightarrow(b)$

If (b) does not hold, then there exist a linear functional on $A$ not
vanishing on units and preserving unity. Its restriction to $C$ has the same
properties, contradicting (c).

  2. (2)

Clearly, (1) implies (2).

∎

######  Corollary 2.8.

If $A$ is a GKŻ algebra, then all proper subalgebras of $A$ with the same
units, are GKŻ algebras vacuously.

For group rings, see [14].

######  Remark 2.9.

If $A$ is an algebra, then
$\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$ is a homomorphic image
of the group ring $\mathbb{F}[\operatorname{{\mathcal{U}}}(A)]$.

######  Corollary 2.10.

No proper algebra extension of a group ring $\mathbb{F}[G]$ with the same
units satisfies the GKŻ property.

###### Proof.

This follows from Theorem 2.1, since the augmentation homomorphism of
$\mathbb{F}[G]$ is a multiplicative linear functional. ∎

For an application of Corollary 2.10 to distribution algebras, see §4.3.

######  Proposition 2.11.

  1. (1)

$\mathbb{F}_{2}$ is the only field for which all algebras generated by units
satisfy the GKŻ property.

  2. (2)

$\mathbb{F}_{2}$ is the only field for which the only unit generated function
algebra is the field itself (up to isomorphism).

###### Proof.

  1. (1)

Let $A$ be a unit generated algebra over $\mathbb{F}_{2}$. If
$\Lambda:A\to\mathbb{F}_{2}$ is a linear functional not vanishing on units,
then $\Lambda(u)=1$ for every unit $u$ in $A$. Since
$A=\operatorname{span}(\operatorname{{\mathcal{U}}}(A))$, we obtain that
$\Lambda$ is multiplicative, implying that $A$ is a GKŻ algebra.

On the other hand, if $|\mathbb{F}|>2$, let
$c\in\mathbb{F}\setminus\\{0,1\\}$. Consider the Laurent polynomial ring
$\mathbb{F}[x,x^{-1}]$. Let $\Lambda:\mathbb{F}[x,x^{-1}]\to\mathbb{F}$ be the
unique linear functional over $\mathbb{F}$, such that $\Lambda(x^{n})=1$ for
all integers $n\neq 2$, and $\Lambda(x^{2})=c$. We see that $\Lambda$ is not
multiplicative since $\Lambda(x^{2})=c\neq 1=(\Lambda(x))^{2}$. Since
$\operatorname{{\mathcal{U}}}(\mathbb{F}[x,x^{-1}])=\mathbb{F}^{\bullet}\\{x^{n}\mid
n\in\mathbb{Z}\\}$ (a direct product of two groups), we see that $\Lambda$
does not vanish on units. Hence the $\mathbb{F}$-algebra
$\mathbb{F}[x,x^{-1}]$ is unit generated, but it is not a GKŻ algebra.

  2. (2)

Let $A\subseteq\mathbb{F}_{2}^{X}$ be a function algebra. The only unit of
$\mathbb{F}_{2}^{X}$ is $I_{X}$. Hence
$\operatorname{span}(\operatorname{{\mathcal{U}}}(A))=\mathbb{F}_{2}I_{X}$. It
follows that
$A=\operatorname{span}(\operatorname{{\mathcal{U}}}(A))\Leftrightarrow
A=\mathbb{F}_{2}I_{X}$.

On the other hand, if $|\mathbb{F}|>2$, then every GKŻ algebra over
$\mathbb{F}$ is unit generated by Theorem 2.1, and by (1), there are GKŻ
algebras $A$ that are not isomorphic to $\mathbb{F}$.

∎

As shown by Vamos [21, page 418], each element of a real or complex Banach
algebra is a sum of two units, as an immediate consequence of the spectral
theorem. Analogously, we have:

######  Remark 2.12.

Let $A$ be an $\mathbb{F}$-algebra. Let $a\in A$ such that
$\rho(a)\neq\emptyset$. Then $a$ is a sum of at most $2$ units. Hence, if
every element of $A$ has a non-empty resolvent, then each element of $A$ is a
sum of at most $2$ units.

Indeed, let $\lambda\in\rho(a)$. If $\lambda=0$, then $a$ is invertible. If
$\lambda\neq 0$, then $a=(a-\lambda\mathbf{1})+\lambda\mathbf{1}$, is a sum of
two units. ∎

###  2.2. Linear coverings

######  Remark 2.13.

A one-sided ideal $V$ of codimension $1$ of the algebra $A$ is a maximal
ideal.

Indeed, $V$ is multiplicatively closed, so $V$ is an ideal. ∎

######  Lemma 2.14.

Let $V$ be a subspace $A$ of codimension $1$ that is not an ideal. Then there
exists a finitely generated subspace $V_{0}$ of $V$ such that
$AV_{0}=V_{0}A=A$.

###### Proof.

By Remark 2.13, $V$ is not a left ideal, so $V\subsetneqq AV$, implying that
$AV=A$. Similarly, $VA=A$. Hence there exist finitely generated subspace
$V_{1}$ and $V_{2}$ of $V$ such that $AV_{1}=V_{2}A=A$. Set
$V_{0}=V_{1}\cup~{}V_{2}$. ∎

######  Proposition 2.15.

Let $A$ be an $\mathbb{F}$-algebra. Then $A$ is a GKŻ algebra under each of
the following conditions:

  1. (1)

$A\setminus\operatorname{{\mathcal{U}}}(A)$ is a contained in a union of at
most two proper one-sided ideals.

  2. (2)

$\mathbb{F}$ is infinite, and $A\setminus\operatorname{{\mathcal{U}}}(A)$ is
contained in a union of less than $|\mathbb{F}|$ proper one-sided ideals.

  3. (3)

$\mathbb{F}$ is infinite, and $A\setminus\operatorname{{\mathcal{U}}}(A)$ is
contained in a union of finitely many proper one-sided ideals.

###### Proof.

Let $V$ be a subspace of $A$ of codimension $1$ that contains no units.

  1. (1)

We have $V\subseteq A\setminus\operatorname{{\mathcal{U}}}(A)\subseteq
I_{1}\cup I_{2}$, where $I_{1},I_{2}$ are two one-sided ideals of $A$, not
necessarily distinct. Viewing $A$ as an additive group, we obtain that
$V\subseteq I_{k}$ for some $k=1,2$. Since $\operatorname{codim}V=1$, we see
that $V=I_{k}$, so $V$ is an ideal by Remark 2.13. Hence $A$ is a GKŻ algebra.

  2. (2)

By Lemma 2.14, there exists a finitely generated subspace $V_{0}$ of $V$ such
that $AV_{0}=V_{0}A=A$. Since $V_{0}$ is contained in a union of less than
$|\mathbb{F}|$ proper one-sided ideals, it follows from[10, Theorem 1.2] or
from [1, Main Theorem], that $V_{0}$ is contained in one of the one-sided
ideals $I$ in this union. Hence $V=I$, so $V$ is an ideal by Remark 2.13, and
$A$ is a GKŻ algebra.

  3. (3)

is a particular case of (2).

∎

######  Corollary 2.16.

Let $A\subseteq\mathbb{F}^{X}$ be a function algebra. Assume that
$|X|<|\mathbb{F}|$, and that each $f\in A$ such that $f(x)\neq 0$ for all
$x\in X$ is invertible in $A$. Then $A$, is a GKŻ-algebra. In particular, if
$|X|<|\mathbb{F}|$, then $\mathbb{F}^{X}$ is a GKŻ algebra. More particularly,
if $\mathbb{F}=\mathbb{R},\mathbb{C}$, and $X$ is countable, then $A$ is a
GKŻ-algebra.

###### Proof.

We have $A\setminus\operatorname{{\mathcal{U}}}(A)\subseteq\bigcup_{x\in
X}p_{x}^{-1}(0)$. Hence $A$ is a GKŻ algebra by Proposition 2.15. ∎

######  Proposition 2.17.

Let $A$ be an algebra over a field $\mathbb{F}$.

  1. (1)

If $A$ has at most two maximal left (or at most two maximal right) ideals,
then $A$ satisfies the GKŻ property.

  2. (2)

Let $A$ be an algebra over an infinite field $\mathbb{F}$. If the cardinality
of the set of left ideals, or of the right ideals, is $<|\mathbb{F}|$, then
$A$ is a GKŻ algebra.

  3. (3)

If the field $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, and $A$ has just
countably many left ideals, then $A$ is a GKŻ algebra.

  4. (4)

If $A$ is commutative, then the localisation $A_{P}$ is a GKŻ-algebra for
every prime ideal $P$ of $A$. Hence the GKŻ property is not a local-global
property.

  5. (5)

Assume that $A$ is commutative and that the field $\mathbb{F}$ is infinite.
Let $\mathcal{P}$ be a set of less that $|\mathbb{F}|$ prime ideals. Then the
algebra $B=\bigcap_{P\in\mathcal{P}}A_{P}$ satisfies the GKŻ property.

###### Proof.

Items (1), (2) and (3) immediately follow from Proposition 2.15.

  1. (4)

$A_{P}$ has just one maximal ideal, so it a GKŻ-algebra by (1).

  2. (5)

This follows from Proposition 2.15 since the set of non-units of the algebra
$B$ is contained in $\bigcup_{P\in\mathcal{P}}PA_{P}$.

∎

###  2.3. The GKŻ property for homomorphic images

We show first that the class of GKŻ algebras is closed under homomorphic
images (Proposition 2.18). Equivalently, if $I$ is an ideal of a GKŻ algebra
$A$, then also $A/I$ is a GKŻ algebra. Then we deal with the converse, and
show that under suitable assumptions, if $A/I$ is a GKŻ algebra for a certain
ideal $I$, then $A$ is a GKŻ algebra.

######  Proposition 2.18.

A homomorphic image of a GKŻ algebra is a GKŻ algebra.

###### Proof.

We have to show that if $I$ is an ideal of a GKŻ algebra $A$, then $A/I$ is a
GKŻ algebra. Let $W$ be a subspace of $A/I$ of codimension $1$ that contains
no units. Hence $W=V/I$, where $V$ is a subspace of $A$ of codimension $1$,
that contains $I$, but no units modulo $I$. We infer that $V$ contains no
units of $A$, implying that $V$ is an ideal of $A$, since $A$ is a GKŻ
algebra. Hence $W=V/I$ is an ideal of $A/I$. It follows that $A/I$ is a GKŻ
algebra. ∎

For a partial converse of Proposition 2.18, see the next Proposition 2.19.
Recall that if $I$ is an ideal of $A$, an element $t\in A/I$ is liftable to a
unit in $A$ if $t$ contains a unit in $A$, that is, $t+I=u+I$ for some
$u\in\operatorname{{\mathcal{U}}}(A)$.

######  Proposition 2.19.

Let $A$ be an algebra over $\mathbb{F}$.

  1. (1)

Let $I$ be an ideal of $A$. Then $A/I$ is a GKŻ algebra $\Leftrightarrow$ each
maximal subspace of $A$ containing $I$, but no units modulo $I$, is an ideal.

  2. (2)

If each unit of $A/I$ is liftable to $A$, then $A$ is a GKŻ algebra
$\Leftrightarrow A/I$ is a GKŻ algebra.

###### Proof.

  1. (1)

$\Rightarrow$

See Proposition 2.18.

$\Leftarrow$

Let $V$ be a maximal subspace of $A$ containing $I$, but no units modulo $I$.
Thus $V/I$ is a maximal subspace of $A/I$ that contains no units. Hence $V/I$
is an ideal of $A/I$, implying that $V$ is an ideal of $A$.

  2. (2)

follows from (1) since in this case a subspace of $A$ containing $I$, contains
no units in $A$ if and only if it contains no units modulo $I$.

∎

By Proposition 2.17 (1), if $A$ is a unital algebra with a unique maximal left
ideal $J$, then $A$ satisfies the GKŻ property. For another generalisation,
see Proposition 2.21 (2) below. Indeed, in this case
$J=\operatorname{Jac}(A)$, the Jacobson radical of $A$. Recall that
$\operatorname{Jac}(A)$ is the intersection of all maximal left ideals of $A$,
and that in this definition ‘left ’ can be changed to ‘right’, so
$\operatorname{Jac}(A)$ is a two-sided ideal. For the Jacobson radical theory,
see e.g., [5, section 13B], [9, Part II, §1], and [11, Chapter 2.4].

######  Proposition 2.20.

Each maximal subspace $V$ of $A$ without units contains
$\operatorname{Jac}(A)$, so $\operatorname{Jac}(A)$ is the intersection of all
maximal subspaces of $A$ without units.

###### Proof.

There exists a unique linear functional $\Lambda$ on $A$ such that
$\ker\Lambda=V$ and $\Lambda(\mathbf{1})=1$. Let $c\in\operatorname{Jac}(A)$.
We have $\Lambda(c)\in\sigma(c)=\\{0\\}$. Hence $\Lambda(c)=0$, so $c\in V$. ∎

######  Proposition 2.21.

If $I$ is an ideal contained in $\operatorname{Jac}(A)$, then $A$ is a GKŻ
algebra $\Leftrightarrow A/I$ is a GKŻ algebra.

###### Proof.

This follows from Proposition 2.19, since $I\subseteq\operatorname{Jac}(A)$,
so an element of $A$ is a unit in $A$ if and only if it is a unit modulo $I$.
∎

######  Proposition 2.22.

Let $I$ be an ideal of the algebra $A$. Assume that each maximal subspace of
$A$ that does not contain units and does not contain $I$, is an ideal in $A$.
Equivalently, assume that if $\Lambda$ is a linear functional such that
$\Lambda(u)\neq 0$ for all $u\in\operatorname{{\mathcal{U}}}(A)$,
$\Lambda(\mathbf{1}_{A})=1$ and $\Lambda(I)\neq(0)$, then $\Lambda$ is
multiplicative. Then $A$ is a GKŻ algebra $\Leftrightarrow A/I$ is a GKŻ
algebra.

###### Proof.

In view of Proposition 2.18, we have to prove just the implication
$\Leftarrow$. Let $V$ be a maximal subspace of $A$ that does not contain units
and does not contain $I$. If $V$ is not an ideal, then $I\subseteq V$ and
$V/I$ is not an ideal in $A/I$, a contradiction. Thus every maximal subspace
of $A$ is an ideal, so $A$ is a GKŻ algebra. ∎

######  Remark 2.23.

If $A$ contains an element $a$ with empty spectrum, then $A$ is GKŻ algebra
vacuously.

Indeed, if $\Lambda$ is a linear functional that does vanish on units such
that $\Lambda(\mathbf{1})=1$, then $\Lambda(a)\in\sigma(a)=\emptyset$, a
contradiction. ∎

######  Proposition 2.24.

Let $I$ be an ideal of a generated by units algebra $A$ satisfying the
following two conditions:

  1. (1)

There exists an element $c\in A$ such that $c+\lambda{\mathbf{1}}$ is
invertible modulo $I$ for all $\lambda\in\mathbb{F}$.

  2. (2)

Each unit in $A/I$ is liftable to a unit in $A$.

Then $A$ is a GKŻ algebra.

###### Proof.

By (1), the element $c+I$ has an empty spectrum in $A/I$. Hence, $A/I$ is a
GKŻ algebra by Remark 2.23. By Proposition 2.19, $A$ is a GKŻ algebra. ∎

###  2.4. Unitisation

Recall that the unitisation of an $\mathbb{F}$-algebra $A$ (not necessarily
unital) is the algebra $B=\mathbb{F}\oplus A$ (a direct sum of additive
groups) with multiplication defined by $(\lambda+a)(\mu+b)=\lambda\mu+\lambda
b+\mu a+ab$ for $\lambda,\mu\in\mathbb{F}$, and $a,b\in A$. We identify
$\mathbb{F}$ with the subfield $\mathbb{F}\mathbf{1}_{B}$ of $B$.

For the next proposition recall that if $I$ is an ideal of an algebra $A$, not
necessarily unital, then $J\subseteq\operatorname{Jac}(A)$ if and only if each
element $a\in I$ is left quasiregular, that is, $1+a$ is left invertible in
the unitisation of $A$, equivalently, there exists an element $b\in A$ such
that $a+b+ab=0$ (see e.g., [22]). Moreover, an ideal $I$ of $A$ is contained
in $\operatorname{Jac}(A)$ if and only if each element of $I$ is left
invertible. Here, ‘left’ can be replaced by ‘right’ or omitted ([9] and [5]).
Hence, a non zero non-unital algebra $A$ is radical (that is,
$A=\operatorname{Jac}(A)$), if and only if each element of $A$ is
quasiregular.

######  Lemma 2.25.

Let $A$ be a nonzero $\mathbb{F}$-algebra, not necessarily unital, and let $B$
the unitisation of $A$. The following conditions are equivalent:

  1. (1)

$A$ is radical algebra.

  2. (2)

$A=\operatorname{Jac}(B)$.

  3. (3)

$1+A\subseteq\operatorname{{\mathcal{U}}}(B)$ (here $1\in\mathbb{F}\subseteq
B$, so $1\notin A$).

  4. (4)

$\operatorname{{\mathcal{U}}}(B)=\mathbb{F}^{\bullet}(1+A)$ (a direct product
of multiplicative groups).

  5. (5)

$\operatorname{{\mathcal{U}}}(B)=B\setminus A$.

###### Proof.

We have:
$A=\operatorname{Jac}(A)\Leftrightarrow$ each element of $A$ is quasiregular
$\Leftrightarrow A=\operatorname{Jac}(B)\Leftrightarrow(3)$.

Thus the first three conditions are equivalent. Since $B=\mathbb{F}+A$, we see
that the last three conditions are also equivalent.∎

######  Proposition 2.26.

The unitisation of a radical algebra is a GKŻ algebra. Moreover, $A$ is a
radical algebra if there exists a Hausdorff topology on $A$ such that
$\lim_{n\to\infty}a^{n}=0$ in this topology for all $a\in A$.

###### Proof.

Assume that $A$ is a radical algebra, and $B$ is the unitisation of $A$. By
Lemma 2.25, $A=\operatorname{Jac}(B)$. Since
$B/\operatorname{Jac}(B)=B/A\cong\mathbb{F}$, it follows that $B$ satisfies
the GKŻ property by Proposition 2.21. Alternatively, it follows from Lemma
2.25 that $\operatorname{{\mathcal{U}}}(B)=B\setminus A$, so $A$ is the unique
maximal one-sided ideal of $A$. By Proposition 2.15, $B$ is a GKŻ algebra.

Assume that $A$ has a Hausdorff topology as described. Let $a\in A$. Since
$\lim_{n\to\infty}a^{n}=0$, the geometric series $\sum_{n=0}^{\infty}a^{n}$
converges in $A$, and its sum $s$ satisfies in $B$: $s(1-a)=(1-a)s=1$, so
$1-a$ is invertible in $B$. By Lemma 2.25, $B$ is a GKŻ-algebra. ∎

For an application of Proposition 2.26 to distribution algebras see §4.2.

##  3\. On the GKŻ property of function algebras

In this section we investigate the GKŻ property for function algebras, using
ideas related to connections between ideals of a commutative ring and
ultrafilters, although we do not use ultrafilters explicitly. Instead of
ideals, we use maximal subspaces with no units. For ideals and ultrafilters
see the classical book [3], for later results see [12], and for the basics of
ultrafilter theory see [23].

Let $A$ be an algebra over a subfield $\mathbb{F}$ of $\mathbb{C}$. We prove
that if $\mathbb{F}$ contains all the bounded functions in $\mathbb{F}^{X}$,
then each element of $A$ is a sum of two units (Proposition 3.4). If $A$
contains also a discrete function, then $A$ is a GKŻ algebra (Corollary 3.12).
The condition that $A$ contains all bounded functions is implied by the
following condition: if $f\in A$, $g\in\mathbb{F}^{X}$, and $|g|\leq|f|$ (that
is, $|g(x)|\leq|f(x)|$ for all $x\in X$), then $g\in A$.

For the next Lemma 3.1, see [24] and especially Eric Wofsey’s answer. In this
lemma, the implications $(1)\Rightarrow(2)\Rightarrow(3)$ are easily proved.
Nevertheless, we indicate all the relevant implications.

######  Lemma 3.1.

Let $A$ be a commutative algebra, and let $\Lambda$ be a linear functional on
$A$ such that $\Lambda(u)\neq 0$ for all $u\in\operatorname{{\mathcal{U}}}(A)$
and $\Lambda(\mathbf{1})=1$. Let $e$ be an idempotent in $A$. We have:

  1. (1)

Let $u,v\in\operatorname{{\mathcal{U}}}(A)$. Then

    1. (a)

$ue+v(1-e)\in\operatorname{{\mathcal{U}}}(A)$.

    2. (b)

Exactly one of the two scalars $\Lambda(ue)$ and $\Lambda(v(1-e))$ vanishes.

  2. (2)

    1. (a)

$\Lambda(e)\in\\{0,1\\}$.

    2. (b)

For all $u\in\operatorname{{\mathcal{U}}}(A)$, $\Lambda(ue)=0$ if
$\Lambda(e)=0$, and $\Lambda(u)=\Lambda(ue)\neq 0$ if $\Lambda(e)=1$.

    3. (c)

$\Lambda(ue)=\Lambda(u)\Lambda(e)$ for all
$u\in\operatorname{{\mathcal{U}}}(A)$ and idempotents $e\in A$.

  3. (3)

If $A$ is generated by units, then

    1. (a)

$\Lambda(ae)=0$ for all $a\in A$ and idempotents $e\ in\ker\Lambda$, so
$\ker\Lambda$ contains the ideal generated by the idempotents it contains.

    2. (b)

$\Lambda(ae)=\Lambda(a)\Lambda(e)$ for all $a\in A$ and idempotents $e\in A$.

###### Proof.

  1. (1)

    1. (a)

$(ue+v(1-e))(u^{-1}e+v^{-1}(1-e))=1$, so
$ue+v(1-e)\in\operatorname{{\mathcal{U}}}(A)$.

    2. (b)

Since $\Lambda(e+(1-e))=1$, the scalars $\Lambda(e)$ and $\Lambda(1-e)$ cannot
both vanish.

Suppose that both of these scalars are nonzero. Then
$\Lambda\left(\Lambda(v(1-e))u(e)-\Lambda(ue)v(1-e)\right)=0$, although
$\Lambda(v(1-e))u(e)-\Lambda(ue)v(1-e)\in\operatorname{{\mathcal{U}}}(A)$ by
(1) (a), a contradiction.

  2. (2)

    1. (a)

follows from (1)(b) applied to $u=v=1$.

    2. (b)

follows from (1)(b) applied to $u$ and to $v=1$.

    3. (c)

follows from (2)(b).

  3. (3)

    1. (a)

follows from (2)(b).

    2. (b)

follows from (2)(c).

∎

######  Proposition 3.2.

Let $A$ be a function algebra that is generated by units. Suppose that $A$
contains the indicator functions of all subsets of $X$. Let $\Lambda$ be a
linear functional such that $\Lambda(u)\neq 0$ for all
$u\in\operatorname{{\mathcal{U}}}(A)$ and $\Lambda(\mathbf{1})=1$. If
$\Lambda(I_{\\{x_{0}\\}})\neq 0$ for some $x_{0}\in X$, then
$\Lambda=p_{x_{0}}$. Hence $\Lambda(I_{x})=0$ for all $x\neq x_{0}$ in $X$.

###### Proof.

Suppose that $\Lambda(I_{\\{x_{0}\\}})\neq 0$ for some $x_{0}\in X$. Since
$\Lambda(I_{\\{x_{0}\\}})=1$ by Lemma 3.1(2)(a), we obtain by Lemma 3.1 (3)
(b) for every $f\in A$:
$\Lambda(f)\Lambda(I_{\\{x_{0}\\}})=\Lambda(f\;\\!I_{\\{x_{0}\\}})=\Lambda(f(x_{0})I_{\\{x_{0}\\}})=f(x_{0})\Lambda(I_{\\{x_{0}\\}})=f(x_{0})=p_{x_{0}}(f)$.
∎

######  Remark 3.3.

If a function algebra $A$ on a set $X$ contains the indicators of all subsets
of $X$, then we may use definition by cases to obtain a function in $A$.

More precisely, given a partition of $X$ into $n$ disjoint subsets
$S_{1},\dots,S_{n}$, and functions $f_{1},\cdots,f_{n}$ in $A$, then the
function $f=\sum_{i=1}^{n}f_{i}I_{S_{i}}$ belongs to $A$. ∎

######  Proposition 3.4.

Let $A$ be a function algebra over a subfield $\mathbb{F}$ of $\mathbb{C}$.
Assume that $A$ contains all the bounded functions in $\mathbb{F}^{X}$. Then
every element of $A$ is a sum of two units.

###### Proof.

Let $f\in A$. We define two functions $g$ and $h$ in $\mathbb{F}^{X}$ as
follows:

$g(x)=\begin{cases}f(x)+3&\text{if }|f(x)|\leq 2\\\ f(x)-1&\text{ if
}|f(x)|>2\end{cases}$

$h(x)=\begin{cases}\frac{1}{f(x)+3}&\text{if }|f(x)|\leq 2\\\
\frac{1}{f(x)-1}&\text{ if }|f(x)|>2.\end{cases}$

We have $g\in A$ by Remark 3.3, and $h\in A$ since $h$ is a bounded function.
Also $gh=I_{X}$, so $g$ is invertible. We have $f=(f-g)+g$, and $f-g$ is
invertible since $\frac{1}{f-g}$ is bounded. Hence $f$ is a sum of two units
in $A$. ∎

######  Corollary 3.5.

Let $A$ be a function algebra over a subfield $\mathbb{F}$ of $\mathbb{C}$.
Assume that if $f\in A,g\in\mathbb{F}^{X}$, and $|g|\leq|f|$, then $g\in A$.
Then $A$ contains all the bounded functions in $\mathbb{F}^{X}$, so every
element of $A$ is a sum of two units.

###### Proof.

Let $g$ be a bounded function in $\mathbb{F}^{X}$. Let $n$ be a positive
integer such that $|g(x)|\leq n$ for all $x\in X$. Thus $|g|\leq nI_{X}$,
implying that $g\in A$. By Proposition 3.4, every element of $A$ is a sum of
two units in $A$. ∎

######  Remark 3.6.

In the setting of the next Proposition 3.7, the ideal $L$ is the set generated
by the indicators of all finite subsets of $X$. Thus $L$ consists of all
functions $t\in A$ such that $t(x)=0$ for all $x\in X$, except finitely many
elements, that is, the set $t^{-1}(0)$ is cofinite in $X$.

######  Proposition 3.7.

Let $A\subseteq\mathbb{F}^{X}$ be a function algebra that contains the
indicators of all singletons of $X$, and let $L$ be the ideal of $A$ generated
by these indicators. Then

  1. (1)

$A$ is a GKŻ algebra $\Leftrightarrow A/L$ is a GKŻ algebra (Here $A/L$ is the
zero algebra when $L=A$).

  2. (2)

If there exists in $A$ an element $f$ such that $f-\lambda I_{X}$ is
invertible modulo $L$ for every scalar $\lambda\in\mathbb{F}$, then the only
linear functionals $\Lambda$ on $A$ such that $\Lambda(u)\neq 0$ for all
$u\in\operatorname{{\mathcal{U}}}(A)$ and $\Lambda(\mathbf{1})=1$, are the
projections $p_{x}$ for $x\in X$. Thus $A$ is a GKŻ algebra.

###### Proof.

  1. (1)

By Proposition 3.2, the only linear functionals $\Lambda$ on $A$ such that
$\Lambda(u)\neq 0$ for all $u\in\operatorname{{\mathcal{U}}}(A)$,
$\Lambda(\mathbf{1})=1$, and $\Lambda(L)\neq(0)$ are the projections $p_{x}$
for $x\in X$. Hence, $A$ is a GKŻ algebra $\Leftrightarrow A/L$ is a GKŻ
algebra.

  2. (2)

By assumption, the spectrum of $f+L$ in $A/L$ is empty, so $A/L$ is a GKŻ
algebra by Remark 2.23. By (1), $A$ is a GKŻ algebra.

∎

######  Remark 3.8.

Here is an alternative proof of Proposition 3.7 using Proposition 2.24.

By Proposition 3.7, it is enough to show that if $f\in A$ and $f$ is a unit
modulo $L$, then there exists $f_{0}\in\operatorname{{\mathcal{U}}}(A)$ such
that $f-f_{0}\in L$. By assumption, there exists $g\in A$ such that $fg=1+t$
for some $t\in L$. Define for $x\in X$:

$f_{0}(x)=\begin{cases}f(x)&\text{ if }t(x)=0\text{ (that is, if
}f(x)g(x)=1)\\\ 1&\text{ if }\text{otherwise}.\end{cases}$

Define $g_{0}$ similarly. Clearly, $f_{0}g_{0}=I_{X}$, and if $t(x)=0$ for
some $x\in X$, then $(f-f_{0})(x)=0$, so by Remark 3.6, $f-f_{0}\in L$, as
required. ∎

######  Corollary 3.9.

If the set $X$ is finite, then a subspace of $\mathbb{F}^{X}$ that contains
the indicators of all singletons of $X$ is equal to $\mathbb{F}^{X}$. Hence,
by Proposition 3.7, $\mathbb{F}^{n}$ is a GKŻ algebra for all positive
integers $n$.

Let $\mathbb{F}$ be a subfield of $\mathbb{C}$. A function
$f\in\mathbb{F}^{X}$ is called discrete if the set $f(X)$ is discrete in
$\mathbb{C}$, and the sets $f^{-1}(\lambda)$ are finite for all
$\lambda\in\mathbb{C}$.

######  Remark 3.10.

If $\mathbb{F}^{X}$ contains a discrete function, then $|X|=\aleph_{0}$. On
the other hand, if $X=\\{x_{n}\ (n\in\mathbb{N})\\}$, where the elements
$x_{n}$ are distinct, and $f\in\mathbb{F}^{X}$, then $f$ is discrete if and
only if $\lim_{n\to\infty}|f(x_{n})|=\infty$.

######  Theorem 3.11.

Let $A\subseteq\mathbb{F}^{X}$ be a unit generated function algebra over a
field $\mathbb{F}$ contained in $\mathbb{C}$. Assume that $A$ contains the
indicators of all subsets of $X$, and a discrete function $f$. Then the only
linear functionals $\Lambda$ on $A$ such that $\Lambda(u)\neq 0$ for all
$u\in\operatorname{{\mathcal{U}}}(A)$ and $\Lambda(\mathbf{1})=1$, are the
projections $p_{x}$ for $x\in X$. Thus $A$ is a GKŻ algebra.

###### Proof.

By Proposition 3.7 and in the same notation, it is enough to show that
$f-\lambda\mathbf{1}$ is invertible modulo $L$ for all $\lambda\in\mathbb{F}$.
Define the function $f_{\lambda}:X\to\mathbb{C}$ as follows:

$f_{\lambda}(x)=\begin{cases}1&\text{if }f(x)=\lambda I_{X}\\\ f(x)-\lambda
I_{X}&\text{if }f(x)\neq\lambda I_{X}.\end{cases}$

Since $f$ is discrete, $f(x)=\lambda I_{X}$ just for finitely many $x$’s.
Hence

$(f(x)-\lambda I_{X})-f_{\lambda}\in L.$

Since the function $f_{\lambda}$ is discrete, there exists $m=|\min
f_{\lambda}(X)|$, and $m>0$, implying that
$|\frac{1}{f_{\lambda}}(x)|\leq\frac{1}{m}$, so $\frac{1}{f_{\lambda}}\in A$,
and $f_{\lambda}$ is invertible in $A$. Since ${f(x)-\lambda I_{X}}\equiv
f_{\lambda}\pmod{L}$, we infer that ${f(x)-\lambda I_{X}}$ is invertible
modulo $L$ as required. ∎

######  Corollary 3.12.

Let $A\subseteq\mathbb{F}^{X}$ be a function algebra over a subfield
$\mathbb{F}$ of $\mathbb{C}$. Assume that $A$ contains all the bounded
functions in $\mathbb{F}^{X}$, and a discrete function $f$. Then every element
of $A$ is a sum of two units, the only linear functionals $\Lambda$ on $A$
such that $\Lambda(u)\neq 0$ for all $u\in\operatorname{{\mathcal{U}}}(A)$ and
$\Lambda(\mathbf{1})=1$, are projections. Thus $A$ is a GKŻ algebra.

###### Proof.

$A$ contains the indicators of all subsets of $X$ since the indicators are
bounded functions. By Proposition 3.4, every element of $A$ is a sum of two
units. We conclude the proof by Theorem 3.11. ∎

######  Corollary 3.13.

Let $A\subseteq\mathbb{F}^{X}$ be a function algebra over a subfield
$\mathbb{F}$ of $\mathbb{C}$. Assume that if $f\in A,g\in\mathbb{F}^{X}$, and
$|g|\leq|f|$, then $f\in A$, and that $A$ contains a discrete function. Then
every element of $A$ is a sum of two units, $A$ contains all the bounded
functions in $\mathbb{F}^{X}$, and the only linear functionals $\Lambda$ on
$A$ such that $\Lambda(u)\neq 0$ for all $u\in\operatorname{{\mathcal{U}}}(A)$
and $\Lambda(\mathbf{1})=1$, are the projections $p_{x}$ for $x\in X$. Thus
$A$ is a GKŻ algebra.

###### Proof.

By Corollary 3.5, $A$ contains all bounded functions. We conclude the proof by
Corollary 3.12. ∎

######  Corollary 3.14.

Let $\mathcal{H}$ be a set of functions in $[2,\infty)^{X}$ satisfying the
following conditions:

  1. (1)

For each $h_{1},h_{2}\in\mathcal{H}$ there exists $h\in\mathcal{H}$ such that
for all $x\in X$, $h_{1}(x)+h_{2}(x)\leq h(x)$.

  2. (2)

For each $h_{1},h_{2}\in\mathcal{H}$ there exists $t\in\mathcal{H}$ such that
for all $x\in X$ $h_{1}(x)h_{2}(x)\leq t(x)$.

Let $A=\\{f\in\mathbb{F}^{X}\mid\exists h\in\mathcal{H}\textrm{ such that
}\forall x\in X,\;|f(x)|\leq h(x)\\}$. We have:

  1. (a)

$A$ is a unital $\mathbb{F}$-subalgebra of $\mathbb{F}^{X}$ containing
$\mathcal{H}$.

  2. (b)

Every element of $A$ is a sum of two units.

  3. (c)

If $A$ contains a discrete function, then the only linear functionals
$\Lambda$ on $A$ such that $\Lambda(u)\neq 0$ for all
$u\in\operatorname{{\mathcal{U}}}(A)$ and $\Lambda(\mathbf{1})=1$, are
projections. Thus $A$ is a GKŻ algebra.

###### Proof.

(a) By item (1), we obtain inductively that for every $f\in A$ and every
integer $n\geq 1$, we have $|nf|\leq h$ for some $h\in\mathcal{H}$, so
$\lambda f\in A$ for all $\lambda\in\mathbb{F}$. It is now easy to complete
the proof of (a). We conclude the proof using Corollary 3.5. ∎

For an application of Corollary 3.14 to distribution algebras see §4.1.

##  4\. The GKŻ property for algebras of distributions

As usual, $\mathcal{D}(\mathbb{R})$ is the space of test functions (compact
supported complex-valued infinitely differentiable functions on $\mathbb{R}$),
and $\mathcal{D}^{\prime}(\mathbb{R})$ is the space of distributions on
$\mathbb{R}$. We study three subspaces of $\mathcal{D}^{\prime}(\mathbb{R})$
that are algebras with convolution as multiplication.

###  4.1. Algebra of periodic distributions

For background on periodic distributions and its Fourier series theory, we
refer the reader to [20, p. 527-529]. For $v\in{{\mathbb{R}}}$, the
translation operator
$\mathbf{S}_{v}:\mathcal{D}^{\prime}(\mathbb{R})\rightarrow\mathcal{D}^{\prime}(\mathbb{R})$,
is given by

$\langle{\mathbf{S}_{v}}t,\varphi\rangle=\langle
t,\varphi(\cdot+{v})\rangle\textrm{ for all
}\varphi\in{\mathcal{D}}({\mathbb{R}}),\textrm{ and for all
}t\in\mathcal{D}^{\prime}(\mathbb{R}).$

A distribution $t\in{\mathcal{D}}^{\prime}({\mathbb{R}})$ is called periodic
with period $v\in{\mathbb{R}}\setminus\\{\mathbf{0}\\}$ if
${\mathbf{S}_{v}}t=t.$ We define ${\mathcal{D}}^{\prime}_{v}(\mathbb{R})$ to
be the set of the periodic distributions with period $v$. As is well known,
${\mathcal{D}}^{\prime}_{v}(\mathbb{R})$ is a complex algebra with convolution
as multiplication. Moreover, using Fourier transforms, one obtains that
${\mathcal{D}}^{\prime}_{v}(\mathbb{R})$ is isomorphic as a complex algebra to
the algebra $\mathcal{S}^{\prime}(\mathbb{Z})$ of all complex-valued maps on
$\mathbb{Z}$ of at most polynomial growth, that is,

$\mathcal{S}^{\prime}(\mathbb{Z}):=\bigg{\\{}\mathbf{a}:\mathbb{Z}\rightarrow\mathbb{C}\;\Big{|}\;\begin{array}[]{ll}\exists\textrm{
an integer }k\geq 1\;\textrm{ such that}\\\ \forall
n\in\mathbb{Z},\;|\mathbf{a}(n)|\leq(2+|n|)^{k}\end{array}\bigg{\\}}.$

Since $\mathcal{S}^{\prime}(\mathbb{Z})$ is a function algebra satisfying the
assumptions of Theorem 3.14, by letting $\mathcal{H}$ to be the set of all
functions $h:\mathbb{Z}\to[2,\infty)$ of the form $h(n)=(2+|n|)^{k}$ for all
integers $n$, where $k\geq 1$ is an integer, we obtain

######  Proposition 4.1.

All the linear functionals on the complex function algebra
$\mathcal{S}^{\prime}(\mathbb{Z})$ not vanishing on units and preserving
unity, are projections. Hence the algebra
$\mathcal{D}^{\prime}_{v}(\mathbb{R})$ of periodic distributions with period
$v\in\mathbb{R}\setminus\\{0\\}$ satisfies the GKŻ property.

###  4.2. The algebra
$\mathbb{C}\delta_{0}+\mathcal{D}^{\prime}_{+}(\mathbb{R})$

Let $\mathcal{D}^{\prime}_{+}(\mathbb{R})$ denote the set of all distributions
$t\in\mathcal{D}^{\prime}(\mathbb{R})$ having their distributional support
${\textrm{supp}{\;}}t$ contained in the half line $(0,\infty)$. Then
$\mathcal{D}^{\prime}_{+}(\mathbb{R})$ is an algebra without unity, with
convolution as multiplication. Let $\delta_{0}$ denote the Dirac distribution
supported at $0\in\mathbb{R}$. Clearly,
$\mathcal{A}:=\mathbb{C}\delta_{0}+\mathcal{D}^{\prime}_{+}(\mathbb{R})$ is a
unital subalgebra of $D^{\prime}(\mathbb{R})$, and $\mathcal{A}$ is isomorphic
to the unitisation of $\mathcal{D}^{\prime}_{+}(\mathbb{R})$.

######  Proposition 4.2.

The unique linear functional on $\mathcal{A}$ not vanishing on units and
preserving unity, is the functional induced by the canonical homomorphism
$\mathcal{A}\to\mathcal{A}/\mathcal{D}^{\prime}_{+}(\mathbb{R})=\mathbb{C}$.
Hence the algebra $\mathbb{C}\delta_{0}+\mathcal{D}^{\prime}_{+}(\mathbb{R})$
satisfies the GKŻ property.

###### Proof.

Since $\mathcal{A}$ is a subalgebra of the topological algebra
$\mathcal{D}^{\prime}_{+}(\mathbb{R})$, by Proposition 2.26, it is enough to
prove that $\lim_{n\to\infty}a^{\ast n}=0$ for all $a\in\mathcal{A}$. This
follows from ${\textrm{supp}{\;}}(s^{\ast n})\subseteq
n\;\\!{\textrm{supp}{\;}}s$ for all positive integers $n$, and so, for any
test function $\varphi\in\mathcal{D}(\mathbb{R})$,
$({\textrm{supp}{\;}}(s^{\ast
n}))\cap({\textrm{supp}{\;}}\varphi)\neq\emptyset$ for only finitely many
$n\in\mathbb{N}$. ∎

###  4.3. The algebra $\mathcal{E}^{\prime}(\mathbb{R})$

Let $\mathcal{E}^{\prime}(\mathbb{R})$ denote the space of all distributions
$t\in\mathcal{D}^{\prime}(\mathbb{R})$ that have compact support. Thus
$\mathcal{E}^{\prime}(\mathbb{R})$ is an algebra with convolution as
multiplication.

The Dirac distribution with support equal to $\\{a\\}$, where
$a\in\mathbb{R}$, will be denoted by $\delta_{a}$. We let $G=\\{\delta_{a}\mid
a\in\mathbb{R}\\}$. Thus $G$ is a group isomorphic to $(\mathbb{R},+)$ by the
map $\mathbb{R}\to G\ (a\mapsto\delta_{a})$.

For the sake of completeness, in the next Proposition 4.3, we reproduce the
characterisation of units in $\mathcal{E}^{\prime}(\mathbb{R})$ and its proof
from [15]:

######  Proposition 4.3.

[16, Proposition 4.2]
$\operatorname{{\mathcal{U}}}(\mathcal{E}^{\prime}(\mathbb{R}))=\mathbb{C}^{\bullet}G$
(a direct product of two groups).

###### Proof.

Clearly,
$\mathbb{C}^{\bullet}G\subseteq\operatorname{{\mathcal{U}}}(\mathcal{E}^{\prime}(\mathbb{R}))$.
For the converse inclusion, suppose that $t$ is invertible in
$\mathcal{E}^{\prime}(\mathbb{R})$. Then there exists a distribution
$s\in\mathcal{E}^{\prime}(\mathbb{R})$ such that $t\ast s=\delta_{0}$. By the
theorem on supports [7, Theorem 4.3.3, p.107], we have

$\textrm{c.h.supp}(t\ast s)=\textrm{c.h.supp}(t)+\textrm{c.h.supp}(s),$

where, for a distribution $a\in\mathcal{E}^{\prime}(\mathbb{R})$, the notation
$\textrm{c.h.supp}(a)$ is used for the closed convex hull of
${\textrm{supp}{\;}}a$, that is, the intersection of all closed convex sets
containing ${\textrm{supp}{\;}}a$. So we obtain

$\\{0\\}=\textrm{c.h.supp}(\delta_{0})=\textrm{c.h.supp}(t\ast
s)=\textrm{c.h.supp}(t)+\textrm{c.h.supp}(s),$

from which it follows that $\textrm{c.h.supp}(t)=\\{a\\}$ and
$\textrm{c.h.supp}(s)=\\{-a\\}$ for some $a\in\mathbb{R}$. But then also
${\textrm{supp}{\;}}t=\\{a\\}$ and ${\textrm{supp}{\;}}s=\\{-a\\}$. As
distributions with support in a point $p$ are linear combinations of
${\scaleobj{0.9}{\delta_{p}}}$ and its derivatives
${\scaleobj{0.9}{\delta_{p}^{{\scaleobj{0.81}{(n)}}}}}$ [20, Theorem 24.6,
p.266], $t$ and $s$ have the form

$t={\scaleobj{0.93}{\sum_{n=0}^{N}}}\;\\!t_{n}\delta_{a}^{(n)},\textrm{ and
}s={\scaleobj{0.93}{\sum_{m=0}^{M}}}\;\\!s_{m}\delta_{-a}^{(m)},$

for some integers $N,M\geq 0$ and $t_{n},s_{m}\in\mathbb{C}$ ($0\leq n\leq N$,
$0\leq m\leq M$). Then $t\ast s=\delta_{0}$ implies that $N=M=0$ and
$t_{0}s_{0}=1$, since for each $a\in\mathbb{R}$, the elements
$\delta_{c}^{(n)}$, where $c\in\\{0,-a,a\\}$ and $n\geq 1$ is an integer, are
linearly independent over $\mathbb{C}$ (if $a=0$, then $c=0$). It follows that
$t_{0}\neq 0$. Thus we have that
$t=t_{0}\delta_{a}\in\\{c\delta_{p}:p\in\mathbb{R},\;0\neq c\in\mathbb{C}\\}.$
This completes the proof. ∎

######  Proposition 4.4.

$\mathcal{E}^{\prime}(\mathbb{R})$ is not a GKŻ algebra.

###### Proof.

Using Proposition 4.3, we see that the distributions in
$\operatorname{span}(\operatorname{{\mathcal{U}}}(\mathcal{E}^{\prime}(\mathbb{R})))$
have finite support, so
$\operatorname{span}(\operatorname{{\mathcal{U}}}(\mathcal{E}^{\prime}(\mathbb{R})))$
is properly contained in $\mathcal{E}^{\prime}(\mathbb{R})$. Since the
distributions $\delta_{a}\ (a\in\mathbb{R})$ are linearly independent over
$\mathbb{C}$, it follows that the algebra
$\operatorname{span}(\operatorname{{\mathcal{U}}}(\mathcal{E}^{\prime}(\mathbb{R})))$
is isomorphic to the group ring $\mathbb{C}[G]$. From Corollary 2.10 it
follows that $\mathcal{E}^{\prime}(\mathbb{R})$ is not a GKŻ algebra. ∎

We can prove directly that $\mathcal{E}^{\prime}(\mathbb{R})$ is not a GKŻ
algebra.

Let $\varphi$ be a function in $C^{\infty}(\mathbb{R})$, and let
$\Lambda=\Lambda_{\varphi}:\mathcal{E}^{\prime}(\mathbb{R})\to\mathbb{C}$ be
the functional $\Lambda(t)=\langle t,\varphi\rangle$ for all
$t\in\mathcal{E}^{\prime}(\mathbb{R})$. We have for all $a\in\mathbb{R}$:
$\Lambda(\delta_{a})=\varphi(a)$. For all $a,b\in\mathbb{R}$:
$\Lambda(\delta_{a})\cdot\Lambda(\delta_{b})=\Lambda(\delta_{a+b})=a+b$, and
$\Lambda(\delta_{a})\Lambda(\delta(b))=\varphi(ab)$. Hence we have:

  1. (1)

$\Lambda$ preserves units if and only if $\varphi(a)\neq 0$ for all
$a\in\mathbb{R}$.

  2. (2)

$\Lambda(\delta_{0})=1$ if and only if $\varphi(0)=1$.

  3. (3)

$\Lambda(uv)=\Lambda(u)(v)$ for all
$u,v\in\operatorname{{\mathcal{U}}}(\mathcal{E}^{\prime}(\mathbb{R}))$ if and
only if $\varphi(a+b)=\varphi(a)\varphi(b)$ for all $a,b\in\mathbb{R}$.

The only continuous functions $\varphi$ with range contained in $\mathbb{R}$
satisfying the conditions stated in (2) and (3) above are the functions
$c^{x}$, where $c\in(0,\infty)$. Thus the function $\varphi(x)=1+x^{2}$ for
$x\in\mathbb{R}$ does not satisfy this conditions, but it does satisfy the the
properties in (1) and (2). Hence the corresponding functional
$\Lambda_{\varphi}$ preserves units and the unity, but it is not
multiplicative. It follows that $\mathcal{E}^{\prime}(\mathbb{R})$ is not a
GKŻ algebra.

For every $\varphi$ the linear functional on
$\Lambda:\mathcal{E}^{\prime}(\mathbb{R})\to\mathbb{C}$ given by
$\Lambda(t)=\langle t,\varphi\rangle$ for all
$t\in\mathcal{E}^{\prime}(\mathbb{R})$, is continuous, when
$\mathcal{E}^{\prime}(\mathbb{R})$ is equipped with the weak dual/weak-$\ast$
topology $\sigma(\mathcal{E}^{\prime}(\mathbb{R}),\mathcal{E}(\mathbb{R}))$;
see e.g. [20].

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