Gelfand theory for real Banach algebras

Abstract

We are concerned with the development of the more general real case of the classical theorem of Gelfand on representation of a complex commutative unital Banach algebra as an algebra of continuous functions defined on a compact Hausdorff space. To that end, we use only intrinsic methods which do not depend on the complexification of the algebra, and obtain two representation theorems for commutative unital real Banach algebras as algebras of continuous real (respectively, complex) functions on the compact space of real-valued (respectively, complex-valued) $\mathbb R$-algebra homomorphisms.

1 Introduction

In the classical literature, complex Banach algebras have received far greater attention than real Banach algebras perhaps due to the possibility of using the power of analytic function theory via the Gelfand transformation. Scalar multiplication in a complex algebra ${\mathcal {A}}$ is a map ${\mathbb {C}}\times {\mathcal {A}}\rightarrow {\mathcal {A}}$. By considering the restriction of this map to $\mathbb R\times {\mathcal {A}}$, ${\mathcal {A}}$ can be viewed as a real algebra. Hence every complex Banach algebra is also a real Banach algebra. Thus it is natural to ask what can be said about this larger class. This article intends to contribute to the theory of real Banach algebras by providing new proofs of two known results on the representation (via the Gelfand transform) of a real commutative unital Banach algebra ${\mathcal {A}}$ as a subalgebra of ${\mathcal {C}}_{{\mathbb {C}}} ({\mathcal {K}})$ (without any additional condition) or as a subalgebra of $\mathcal C_{{\mathbb {R}}} ({\mathcal {K}})$ for some compact Hausdorff space ${\mathcal {K}}$, under an easily checked condition involving the spectral radius. Our work can be seen as a supplement to the monograph on real function algebras by Kulkarni and Limaye [10], to which we refer for background.

In the study of Banach spaces the notion of a bounded linear functional is important. For Banach algebras, and in particular for the algebra of continuous functions on a compact Hausdorff space, the important idea is that of multiplicative linear functional. Since we deal both with real-valued and complex-valued multiplicative ${\mathbb {R}}$-linear functionals on a real Banach algebra ${\mathcal {A}}$, for clarity’s sake it will be convenient to set the terminology. By a real homomorphism of a real algebra ${\mathcal {A}}$ we mean a real-valued multiplicative $\mathbb R$-linear functional, i.e., a continuous nonzero $\mathbb R$-linear map $\phi :{\mathcal {A}}\rightarrow {\mathbb {R}}$ such that $\phi (ab)=\phi (a)\phi (b)$ for all $a,b\in {\mathcal {A}}$. In what follows we shall use $\Phi _{{\mathcal {A}}}^{{\mathbb {R}}}$ to denote the set of all nonzero real homomorphisms of ${\mathcal {A}}$. In turn, we put $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ to denote the set of all nonzero multiplicative ${\mathbb {R}}$-linear functionals from ${\mathcal {A}}$ into ${\mathbb {C}}$, or, in other words, $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ is the set of all nonzero ${\mathbb {R}}$-linear algebra homomorphisms from ${\mathcal {A}}$ into $\mathbb C$ (with ${\mathbb {C}}$ regarded as a two-dimensional real algebra).

In a commutative complex Banach algebra ${\mathcal {A}}$ with unit e of norm 1, the spectrum $\sigma (a)$ of an element $a\in {\mathcal {A}}$, defined as

$$\begin{aligned} \sigma (a):=\{\lambda \in {\mathbb {C}}: a-\lambda e \text{ is } \text{ not } \text{ invertible } \text{ in } {\mathcal {A}}\}, \end{aligned}$$

is a nonempty compact subset of the set of complex numbers ${\mathbb {C}}$. This fact, called the fundamental theorem of Banach algebras, is the basis for the classical Gelfand theory. An element $\lambda$ in $\sigma (a)$ gives rise to a multiplicative ${\mathbb {C}}$-linear functional $\phi :{\mathcal {A}}\rightarrow \mathbb C$: if $a-\lambda e$ is not invertible, the ideal $(a-\lambda e){\mathcal {A}}$ is contained in a maximal ideal and thus there is a multiplicative ${\mathbb {C}}$-linear functional $\phi :\mathcal A\rightarrow {\mathbb {C}}$ with $\phi (a)=\lambda$. Conversely, if $\phi :{\mathcal {A}}\rightarrow {\mathbb {C}}$ is a multiplicative $\mathbb C$-linear functional on ${\mathcal {A}}$ and $a\in \mathcal A$, then $a-\phi (a)e$ is not invertible. The complex Banach algebra ${\mathcal {A}}$ has thus a representation

$$\begin{aligned} \Lambda :{\mathcal {A}}\rightarrow {{\mathcal {C}}}_{\mathbb {C}}({\mathcal K}), \qquad a\hookrightarrow {\widehat{a}}, \end{aligned}$$

into the algebra of continuous complex-valued functions on the compact Hausdorff space ${\mathcal {K}}$ of nonzero multiplicative ${\mathbb {C}}$-linear functionals $\phi :{\mathcal {A}}\rightarrow \mathbb C$, with

$$\begin{aligned} \Vert {{{\widehat{a}}}}\Vert _{\infty , {{\mathcal {K}}}}=\sup _{\phi \in {{\mathcal {K}}}}|{\widehat{a}}(\phi )|=\sup _{\phi \in {\mathcal K}}|\phi (a)|= r(a),\qquad a\in {\mathcal {A}}, \end{aligned}$$

where

$$\begin{aligned} r(a):=\sup \{|\lambda |:\lambda \in \sigma (a)\}, \end{aligned}$$

is the spectral radius of a and $\Vert \cdot \Vert _{\infty , {\mathcal {K}}}$ denotes the standard norm on ${\mathcal {C}}_{\mathbb C}({\mathcal {K}})$. The operator $\Lambda$, called the Gelfand mapping or Gelfand representation of ${\mathcal {A}}$, is a ${\mathbb {C}}$-algebra homomorphism which in general need not be injective, surjective, or norm-preserving. It is, however, norm-decreasing since the Beurling–Gelfand formula guarantees that $\lim _{n\rightarrow \infty }\Vert a^{n}\Vert ^{1/n}$ exists for all $a\in {\mathcal {A}}$ and the value of this limit is exactly r(a). All these are well-known facts that the reader will find, for example, in [7, Ch. 2].

For real Banach algebras there are two corresponding results. The first one involves only real homomorphisms of ${\mathcal {A}}$ since, as we will see later on, the set ${\mathcal {K}}$ in Theorem 1.1 coincides with the set $\Phi _{\mathcal A}^{{\mathbb {R}}}$.

Theorem 1.1

Suppose ${\mathcal {A}}$ is a commutative real Banach algebra with unit e of norm 1. There is a representation $a\hookrightarrow {{\widehat{a}}}$ of ${\mathcal {A}}$ into ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}}),$ the algebra of continuous real-valued functions on a compact Hausdorff space ${\mathcal {K}},$ satisfying

$$\begin{aligned} \Vert {{{\widehat{a}}}}\Vert _{\infty , {{\mathcal {K}}}}= \sup _{x\in {\mathcal {K}}} |{{\widehat{a}}} (x)|=r(a), \end{aligned}$$

if and only if the spectral radius r satisfies the condition

$$\begin{aligned} r(a^2)\le r(a^2+b^2),\quad a,b\in {\mathcal {A}}. \end{aligned}$$

(1.1)

Recall that if ${\mathcal {A}}$ is a normed real algebra, the spectral radius r(a) of a is defined by

$$\begin{aligned} r(a)=\inf \{\Vert a^{n}\Vert ^{1/n} :n=1,2,\dots \}. \end{aligned}$$

Note that $r(a)\le \Vert a\Vert$ and that if $ab=ba$ for a, b in ${\mathcal {A}}$ then $r(ab)\le r(a)r(b)$ since $(ab)^n=a^n b^n$ for all $n\in {\mathbb {N}}$. We also have (see e.g. [10, p. 92–93]) that

$$\begin{aligned} r(a)=\lim \limits _{n\rightarrow \infty }\Vert a^{n}\Vert ^{1/n}, \end{aligned}$$

and if $ab=ba$ then

$$\begin{aligned} r(a+b)\le r(a)+r(b). \end{aligned}$$

Since, trivially $r(t a)=|t|r(a)$ for all $t\in {\mathbb {R}}$ and all $a\in {\mathcal {A}}$, the spectral radius defines a semi-norm on a commutative real Banach algebra ${\mathcal {A}}$.

The other result is more general:

Theorem 1.2

Suppose ${\mathcal {A}}$ is a commutative real Banach algebra with unit e of norm 1. There is a representation $a\hookrightarrow {{\widehat{a}}}$ of ${\mathcal {A}}$ into ${\mathcal {C}}_{{\mathbb {C}}}({\mathcal {K}}),$ the algebra of continuous complex-valued continuous functions on a compact Hausdorff space ${\mathcal {K}},$ satisfying

$$\begin{aligned} \Vert {{{\widehat{a}}}}\Vert _{\infty , {\mathcal {K}}}= r(a),\quad a\in {\mathcal {A}}. \end{aligned}$$

Here, ${\mathcal {K}}$ is the set $\Phi ^{{\mathbb {C}}}_{{\mathcal {A}}}$ of ${\mathbb {R}}$-linear algebra homomorphisms from ${\mathcal {A}}$ into ${\mathbb {C}}$ (regarded as a real algebra).

Theorems 1.1 and 1.2 have been previously obtained in the literature by various authors using complex function theory to some extent (see e.g. [1, 2, 9, 10]). In this note we attain those results in Sects. 2 and 3 respectively, by means of real Banach space theory in the former case and real harmonic function theory in the latter. The novelty in our approach is that we totally circumvent complex function theory. Finally, in Sect. 4 we exhibit connections between certain inequalities involving the algebra norm with isomorphic and isometric representations of the algebra as a subalgebra of the space of real-valued continuous functions on some compact Hausdorff space, and show that the spectral radius condition (1.1) is not sufficient for the Gelfand transformation to be an isomorphism. This solves a question that was left open in [2].

2 The state space approach to Theorem 1.1

A real-algebra homomorphisms from a commutative real Banach algebra ${\mathcal {A}}$ into ${\mathcal {C}}({\mathcal {K}})$ need not map into ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$, the algebra of continuous real-valued functions on $\mathcal K$, which might be of interest when dealing with real Banach algebras. Think, for example, of the complex numbers as an algebra over the reals. Some additional requirements on ${\mathcal {A}}$ are needed in order to guarantee the existence of a representation $a\hookrightarrow {\widehat{a}}$ of ${\mathcal {A}}$ into $\mathcal C({\mathcal {K}})$ that gives only real-valued functions in such a way that for all $a\in {\mathcal {A}}$,

$$\begin{aligned} \sup _{\phi \in {\mathcal {K}}} |\phi (a)|=r(a). \end{aligned}$$

An obvious necessary condition for ${\mathcal {A}}$ is the spectral radius condition (1.1) since the same condition holds for $\mathcal C_{{\mathbb {R}}}({\mathcal {K}})$. In this section we aim to show, using only real Banach space theory, that if inequality (1.1) is satisfied then there is a nonempty compact Hausdorff space ${\mathcal {K}}$ (actually, the set of ${\mathbb {R}}$-algebra homomorphisms from ${\mathcal {A}}$ into the reals) so that the map $a\hookrightarrow {{\widehat{a}}}$ is an ${\mathbb {R}}$-algebra homomorphism from ${\mathcal {A}}$ into ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$ satisfying

$$\begin{aligned} r(a)=\Vert {{\widehat{a}}}\Vert _{\infty , {\mathcal {K}}}=\sup _{\phi \in {\mathcal {K}}}|\phi (a)|. \end{aligned}$$

To that end, let ${\mathcal {A}}_r$ be the vector space ${\mathcal {A}}$ equipped with the spectral radius semi-norm r, and consider the state space of ${\mathcal {A}}$ with respect to r, given by

$$\begin{aligned} {\mathcal {S}}_r=\{\varphi \in {\mathcal {A}}^{*}_r: \Vert \varphi \Vert _r=\varphi (e)=1\}. \end{aligned}$$

Here ${\mathcal {A}}^*_r$ denotes the subspace of the dual space ${\mathcal {A}}^{*}$ of ${\mathcal {A}}$ consisting of all linear maps $\varphi :{\mathcal {A}}\rightarrow {\mathbb {R}}$ such that for some constant $C\ge 1$,

$$\begin{aligned} |\varphi (a)|\le C r(a), \quad a\in {\mathcal {A}}, \end{aligned}$$

and

$$\begin{aligned} \Vert \varphi \Vert _r:=\sup _{r(a)\le 1} |\varphi (a)|. \end{aligned}$$

Let us show that ${\mathcal {S}}_r$ is non-empty. Indeed, the spectral radius r is a sublinear functional on the linear subspace $E=\{\lambda e:\lambda \in {\mathbb {R}}\}$ of the normed space $({\mathcal {A}},\Vert \cdot \Vert )$. The map

$$\begin{aligned} \varphi :E\rightarrow {\mathbb {R}},\quad \lambda e\mapsto \lambda \end{aligned}$$

is linear and satisfies $\varphi (\lambda e)=r(\lambda e)=\lambda =\Vert \lambda e\Vert$. By the Hahn–Banach theorem there exists $\widetilde{\varphi }:{\mathcal {A}}\rightarrow {\mathbb {R}}$ linear such that $\widetilde{\varphi }|_{E}=\varphi$ and $\widetilde{\varphi }(a)\le r(a)$ for all $a\in {\mathcal {A}}$. In particular, we infer that $\widetilde{\varphi }\in \mathcal A_{r}^{*}$ and $\widetilde{\varphi }(e)=1=\Vert e\Vert$, hence $\widetilde{\varphi }\in {\mathcal {S}}_{r}$.

We use ${\mathcal {A}}_{+}$ to denote the set of squares in $\mathcal A$, that is,

$$\begin{aligned}{\mathcal {A}}_{+}=\{a^{2}:a\in {\mathcal {A}}\}.\end{aligned}$$

If ${\mathcal {A}}={\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$, then ${\mathcal {A}}_{+}$ is simply the positive cone of the space, that is, $\{f\in {\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}}):f\ge 0\}$. Recall that a nonempty subset of a real Banach space is said to be a cone if it is closed under addition and multiplication by nonnegative scalars. The closure of a cone is also a cone. For instance, the set

$$\begin{aligned} {\mathcal {C}}=\bigcup _{t> 0}t(e-{\mathcal {U}}), \end{aligned}$$

where ${\mathcal {U}}$ denotes the unit open ball of ${\mathcal {A}}$ is a cone. This is so because by the Square Root Lemma from Banach algebra theory (see e.g. [4, Proposition 4.2.6 (i)]), every element of the form $e-x$ with $\Vert x\Vert < 1$ is a square. Thus ${\mathcal {C}}$ is contained in ${\mathcal {A}}_{+}$.

By analogy we shall now consider the set

$$\begin{aligned} {\mathcal {C}}_{r}:=\bigcup _{t>0} t(e-{\mathcal {U}}_r), \end{aligned}$$

where ${\mathcal {U}}_r$ is the r-open unit ball of ${\mathcal {A}}$, i.e., ${\mathcal {U}}_r=\{a\in {\mathcal {A}}:r(a)<1\}$.

Lemma 2.1

Let ${\mathcal {A}}$ be a commutative real Banach algebra with unit.

• (i) If $r(a)<1$ then $e-a\in {\mathcal {A}}_{+}$. In particular,  ${\mathcal {C}}_{r}\subseteq {\mathcal {A}}_{+}$.

• (ii) Let $b\in {\mathcal {A}}$ be such that $e-b\not \in {\mathcal {C}}_{r}$. Then there exists $\varphi \in {\mathcal {S}}_{r}$ with $\varphi (b)\ge 1$.

Proof

(i) This is a consequence of [6, Lemma 1].

(ii) Clearly, ${\mathcal {C}}_{r}$ is r-open and convex in ${\mathcal {A}}_{r}$. Using the general form of the Hahn–Banach separation theorem, we can separate $e-b$ and ${\mathcal {C}}_{r}$ by an r-continuous linear functional $\psi \in \mathcal A_{r}^{*}$ with $\Vert \psi \Vert _r=1$. That is,

$$\begin{aligned} \psi (e-b)> \psi (t(e-u))=t\psi (e-u) \end{aligned}$$

for all positive numbers t and all u in ${\mathcal {U}}_{r}$. We deduce that $\psi (e-b)\ge 0$ and that $\psi (e-u)\le 0$ for all $u\in {\mathcal {U}}_r$. Since $\psi$ maps ${\mathcal {U}}_r$ onto the interval $(-1,1)$ we must have $\psi (e)=-1$. Let $\varphi =-\psi$. Then $\varphi (b)\ge 1$ and so $\varphi \in {\mathcal {S}}_r$. $\square$

The following lemma provides a characterization of the elements in ${\mathcal {S}}_{r}$.

Lemma 2.2

Suppose ${\mathcal {A}}$ satisfies (1.1).

• (i) Then $\varphi (x)\ge 0$ whenever $\varphi \in {\mathcal {S}}_{r}$ and $x\in {\mathcal {A}}_+$.

• (ii) Suppose $\psi \in {\mathcal {A}}^*_r$ is positive on ${\mathcal {A}}_+$. Then $\Vert \psi \Vert _r=\psi (e)$.

Consequently,  given $\psi \in {\mathcal {A}}_{r}^*$ with $\Vert \psi \Vert _r=1,$ we have $\psi \in {\mathcal {S}}_{r}$ if and only if $\psi \ge 0$ on ${\mathcal {A}}_{+}$.

Proof

(i) Take $x\in {\mathcal {A}}_{+}$ with $\Vert x\Vert < 1$. By the Square Root Lemma for Banach algebras, $e-x$ is a square too, so that by (1.1),

$$\begin{aligned} r(e-x)\le r( (e-x)+x)=r(e)=1. \end{aligned}$$

Since $\varphi \in {\mathcal {S}}_{r}$ we have

$$\begin{aligned} 1-\varphi (x)=\varphi (e)-\varphi (x) \le |\varphi (e-x)|\le 1, \end{aligned}$$

which yields $\varphi (x)\ge 0$.

(ii) Let $r(a)<1$. By Lemma 2.1(i), $e-a\in {\mathcal {A}}_{+}$ and so $\psi (e-a)\ge 0$, showing that $\psi (e)\ge \Vert \psi \Vert _r$. Conversely, $\Vert \psi \Vert _{r}\ge \psi (e)$ since $r(e)=1$. $\square$

Our next lemma will pave the way to the proof of Theorem 1.1.

Lemma 2.3

Suppose that ${\mathcal {A}}$ satisfies (1.1). Then the set $\partial _{\text {e}}{\mathcal {S}}_{r}$ of extreme points of ${\mathcal {S}}_r$ consists of multiplicative real-valued ${\mathbb {R}}$-functionals on ${\mathcal {A}}$. The same holds for its $\hbox {weak}^{*}$-closure ${\mathcal {K}}$ in ${\mathcal {A}}_r^{*}.$

Proof

Since ${\mathcal {S}}_{r}$ is convex and compact in the $\hbox {weak}^{*}$-topology of ${\mathcal {A}}_{r}^{*}$, the Krein–Milman theorem guarantees that $\partial _{\text {e}}{\mathcal {S}}_{r}$ is nonempty. Suppose that $\varphi$ lies in $\partial _{\text {e}}{\mathcal {S}}_{r}$. We claim that $\varphi (xy)=\varphi (x)\varphi (y)$ for all $x,y\in {\mathcal {A}}$. Notice that we can write every $x\in {\mathcal {A}}$ as

$$\begin{aligned} x=\frac{1}{2}(e+x)-\frac{1}{2}(e-x), \end{aligned}$$

and by the Square Root Lemma, when $\Vert x\Vert < 1$ both $\frac{1}{2}(e+x)$ and $\frac{1}{2}(e-x)$ belong to $\mathcal A_{+}$. Therefore, it suffices to show that $\varphi (a^2b)=\varphi (a^2)\varphi (b)$ for all $a,b \in {\mathcal {A}}$.

Given $a^2$ with $\Vert a^2\Vert < 1$, for all $b\in {\mathcal {A}}$ we have

$$\begin{aligned} \varphi (b)=\varphi (a^2b)+ \varphi ((e-a^2)b)=\psi _1(b)+\psi _2(b), \end{aligned}$$

where $\psi _{1}(b)= \varphi (a^2b)$ and $\psi _{2}(b)=\varphi ((e-a^2)b)$. By Lemma 2.2(i) the functionals $\psi _1$ and $\psi _2$ are positive on squares, therefore by Lemma 2.2(ii) they attain their norm on e, i.e.,

$$\begin{aligned} \Vert \psi _1\Vert _r=\psi _1 (e) \quad \text{ and }\quad \Vert \psi _2\Vert _r =\psi _2(e). \end{aligned}$$

If both norms are nonzero, since $\psi _1(e)+\psi _2(e)=\varphi (e)=1$, we can write $\varphi$ as a convex combination of elements in ${\mathcal {S}}_r$ in the following way

$$\begin{aligned} \varphi =\psi _1(e)\frac{1}{\psi _1(e)}\psi _1+\psi _2(e)\frac{1}{\psi _2(e)}\psi _2. \end{aligned}$$

The fact that $\varphi$ is an extreme point of ${\mathcal {S}}_{r}$ yields $\varphi =(1/\psi _1(e))\psi _1$ so that $\varphi (a^2)\varphi (b)=\varphi (a^2b)$ for all b, whenever $\Vert a^2\Vert < 1$, and hence for all $a^2$ and b. If $\Vert \psi _1\Vert =0$ then $\varphi (a^2b)=0$ for all b, in particular for $b=e$. It follows that $\varphi (a^2b)=0=\varphi (a^2)\varphi (b)$ for all b. The case when $\Vert \psi _2\Vert =0$ is similar. This result extends to the $w^*$-closure of $\partial _{\text {e}}{\mathcal {S}}_{r}$. $\square$

Completion of the proof of Theorem 1.1

Let ${\mathcal {K}}$ be the $\hbox {weak}^{*}$-closure of the set $\partial _{\text {e}}{\mathcal {S}}_{r}$ in ${\mathcal {A}}_r^{*}.$ We regard the elements $a\in {\mathcal {A}}$ as real-valued functions ${{\widehat{a}}}$ on ${\mathcal {K}}$, where ${{\widehat{a}}} (\varphi )=\varphi (a)$ for $\varphi \in {\mathcal {K}}$. The multiplicative nature of the functionals in ${\mathcal {K}}$ ensures that the linear map

$$\begin{aligned} \Lambda :{\mathcal {A}}\rightarrow {\mathcal {C}}_{{\mathbb {R}}}(\mathcal K), \quad a\hookrightarrow {\widehat{a}},\end{aligned}$$

(2.1)

is an $\mathbb R$-algebra homomorphism. In turn, the definition of ${\mathcal {S}}_r$ readily yields that

$$\begin{aligned} \Vert \Lambda (a)\Vert _{\infty , {\mathcal {K}}}=\Vert \widehat{a}\Vert _{\infty ,{\mathcal {K}}}\le r(a)\quad \text { for all } a\in {\mathcal {A}}, \end{aligned}$$

so $\Lambda$ is bounded.

For the reverse inequality, suppose $r(a^2)>1$. Then $e-a^2$ is not a square because, if it were, using (1.1) we would get

$$\begin{aligned} r(a^2)\le r(a^2+(e-a^2))=1. \end{aligned}$$

In particular, by Lemma 2.1(i), $e-a^2$ does not belong to ${\mathcal {C}}_{r}$ and so by Lemma 2.1(ii), there exists $\varphi \in \mathcal S_{r}$ with $\varphi (a^2)\ge 1$. We infer that $r(a^2)\le \varphi (a^2)$, and since $\Vert \varphi \Vert _{r}=1$, we conclude that $r(a^2)= \varphi (a^2)$.

The set of those $\varphi$ in ${\mathcal {S}}_r$ where $\varphi (a^2)=r(a^2)$ is $\hbox {weak}^{*}$-closed in $\mathcal A_{r}^{*}$, nonempty, and convex and thus by the Krein-Milman theorem it contains an extreme point, which is easily checked to be also an extreme point of ${\mathcal {S}}_r$. Therefore $r(a^2)\le \Vert {{\widehat{a}}}^2\Vert _{\infty ,{\mathcal {K}}}$. Since $r(a^2)=r(a)^2$ and $\Vert \widehat{a^2}\Vert _{\infty ,{\mathcal {K}}}=\Vert \widehat{a}\Vert _{\infty ,{\mathcal {K}}}^2$ the proof is finished. $\square$

In this context, one might ask whether every ${\mathbb {R}}$-algebra homomorphism from ${\mathcal {A}}$ to ${\mathbb {C}}$ is real-valued if condition (1.1) is satisfied. Proposition 2.5 shows that this is the case. The proof relies on the following important auxiliary result.

Lemma 2.4

Let ${\mathcal {A}}$ be a commutative real Banach algebra with unit. If $\phi \in \Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ then $|\phi (a)|\le r(a)$ for all $a\in \mathcal {A}$.

Proof

Let $a\in \mathcal {A}$. For any real numbers s and t we have

$$\begin{aligned} (a-se)^2+t^2e=(s^2+t^2)(1+(s^2+t^2)^{-1}(a-2as)), \end{aligned}$$

and if $(s^2+t^2)^{1/2}>3\Vert a\Vert$,

$$\begin{aligned} (s^2+t^2)^{-1}\Vert a-2as\Vert <1. \end{aligned}$$

It follows that $(a-se)^2+t^2e$ is invertible if $\sqrt{s^2+t^2}>3\Vert a\Vert$. Suppose $\phi \in \Phi _{\mathcal A}^{{\mathbb {C}}}$ and let $a\in \mathcal {A}$. Put $\phi (a)=s+it$. Then $\phi ((a-se)^2+t^2e)=0$ so that $(a-se)^2+t^2e$ is not invertible and thus

$$\begin{aligned}|\phi (a)|=\sqrt{s^2+t^2}\le 3\Vert a\Vert .\end{aligned}$$

Then, for $n\in {\mathbb {N}}$ we have

$$\begin{aligned}|\phi (a)|=|\phi (a^n)|^{1/n}\le (3\Vert a^n\Vert )^{1/n},\end{aligned}$$

and taking $n\rightarrow \infty$ yields $|\phi (a)|\le r(a)$ as we wished. $\square$

Proposition 2.5

Let ${\mathcal {A}}$ be a commutative real Banach algebra satisfying condition (1.1). Then every ${\mathbb {R}}$-algebra homomorphism in $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ is real-valued.

Proof

Suppose that for some $b\in {\mathcal {A}}$ we have $\phi (b)=s+it$, where $t\ne 0$. Set $c=(b-se)^2+t^2e$ and let $\Lambda$ be the homomorphism from ${\mathcal {A}}$ into ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$ defined in (2.1). By Lemma 2.2(i), $\Lambda ((b-se)^{2})\ge 0$ so that $\Lambda (c)\ge t^2$ is strictly positive on ${\mathcal {K}}$ and thus has an inverse f in ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$.

Now the image of ${\mathcal {A}}$ in $\mathcal C_{{\mathbb {R}}}({\mathcal {K}})$ under $\Lambda$ is dense so that there is a sequence $\{a_n\}$ in ${\mathcal {A}}$ such that $\Lambda (a_n)\rightarrow f$ and thus $\Lambda (a_nc-e)\rightarrow 0$ in ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$. It follows that $r(a_nc-e)\rightarrow 0$. But $\phi (a_nc)=\phi (a_n)\phi (c)=0$, a contradiction. $\square$

Our discussion below leads naturally to the following question: If ${\mathcal {A}}$ verifies condition (1.1), we just showed that all ${\mathbb {R}}$-algebra homomorphisms on ${\mathcal {A}}$ are real-valued, but are there any real-valued, ${\mathbb {R}}$-algebra homomorphisms of ${\mathcal {A}}$ outside the set ${\mathcal {K}}=\overline{\partial _{\text {e}}(\mathcal S_{r}))}^{w^{*}}$? The last proposition of this section answers this question in the negative.

Proposition 2.6

Let ${\mathcal {A}}$ be a commutative real Banach algebra with unit satisfying condition (1.1). Then the set ${\mathcal {K}}$ in Theorem 1.1 coincides with the set $\Phi _{{\mathcal {A}}}^{{\mathbb {R}}}$ of all real-valued,  ${\mathbb {R}}$-algebra homomorphisms of ${\mathcal {A}}$.

Proof

The inclusion ${\mathcal {K}}\subseteq \Phi _{{\mathcal {A}}}^{{\mathbb {R}}}$ was proved in Lemma 2.3. To see the other inclusion, let us first notice that every $\phi \in \Phi _{{\mathcal {A}}}^{{\mathbb {R}}}$ belongs to ${\mathcal {S}}_{r}$. This is easy since, by Lemma 2.4, $|\phi (a)|\le r(a)$ for all $a\in \mathcal {A}$ and, by Lemma 2.2(i), $\phi (e)=\Vert e\Vert _{r}=1$ since $\phi$ is positive on ${\mathcal {A}}_{+}$. Now to show that, actually, $\Phi _{{\mathcal {A}}}^{{\mathbb {R}}}\subseteq \partial _{\text {e}}{\mathcal {S}}_{r}$, assume that $\phi \in \Phi _{\mathcal {A}}^{{\mathbb {R}}}\subseteq {\mathcal {S}}_{r}$ is written in the form

$$\begin{aligned} \phi = t \phi _{1} + (1-t)\phi _{2}, \end{aligned}$$

for some $0<t<1$ and some $\phi _{1}, \phi _{2}$ in ${\mathcal {S}}_{r}$. The mapping $(a,b)\rightarrow \phi _{1}(ab)$ defines a real semi-inner product on $\mathcal {A}$, and so by the Cauchy–Schwarz inequality,

$$\begin{aligned} \phi _{1}^2(ab)\le \phi _{1}(a^{2})\phi _{1}(b^{2})\qquad a, b\in \mathcal {A}. \end{aligned}$$

Putting $b=e$ yields $\phi _1(a)=0$ whenever $\phi _1(a^{2})=0$. Since the functionals in ${\mathcal {S}}_{r}$ are positive on $\mathcal A_+$, we have $\phi _1(a^2)=0$ if $\phi (a^2)=0$, and thus since $\phi$ is multiplicative, $\phi _1(a^2)=0$ if $\phi (a)=0$. Combining with the above conclusion yields that $\phi _1(a)=0$ if $\phi (a)=0$. In particular, since $\phi (a-\phi (a)e)=0$ for all $a\in \mathcal {A}$, it follows that $\phi _{1}(a)= \phi (a)$ for all a in $\mathcal {A}$. Thus $\phi$ is an extreme point of ${\mathcal {S}}_{r}$. $\square$

3 The spectrum approach to Theorem 1.2

Consider now $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$, the set of nonzero real-algebra homomorphism from ${\mathcal {A}}$ into $\mathbb C$ (regarded as a two-dimensional real algebra). We can look at the elements of ${\mathcal {A}}$ as functions ${{\widehat{a}}}:\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}\rightarrow {\mathbb {C}}$ given by

$$\begin{aligned} \phi \hookrightarrow {{\widehat{a}}}(\phi )=\phi (a). \end{aligned}$$

The set $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ can then be endowed with the weakest topology making all these functions continuous. Equipped with this topology, called the Gelfand topology, $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ is a compact Hausdorff space known as the Gelfand space (also the carrier space) of ${\mathcal {A}}$. We will refer to the map

$$\begin{aligned} \Lambda :{\mathcal {A}} \hookrightarrow {\mathcal {C}}_{\mathbb C}\left( \Phi _{{\mathcal {A}}}^{{\mathbb {C}}}\right) , \quad a \hookrightarrow {{\widehat{a}}}\end{aligned}$$

as to the Gelfand mapping or Gelfand representation of the real algebra ${\mathcal {A}}$.

We look now at conditions implying that $\Phi _{\mathcal A}^{{\mathbb {C}}}$ is nonempty.

Let $\phi \in \Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$. Given $a\in {\mathcal {A}}$ put $\phi (a)=s+it$. Notice that $\overline{\phi }$, where $\overline{\phi }(a)=s-it$, is also in $\Phi _{\mathcal A}^{\mathbb C}$. Then $\phi ( (a-s^2e)+t^2e)=0$, so that the element $(a-se)^2+t^2e$ is not invertible. Conversely, if $b=(a-se)^2+t^2e$ is not invertible, then the ideal $b{\mathcal {A}}$ is contained in some maximal ideal and thus $\phi ( (a-se)^2+t^2e)=0$ for some $\phi \in \Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$, or equivalently, $\phi (a)=s+it$ or $\phi (a)=s-it$. The existence of nonzero ${\mathbb {R}}$-algebra homomorphisms from ${\mathcal {A}}$ into ${\mathbb {C}}$ is therefore equivalent to the set

$$\begin{aligned} \sigma _{{\mathbb {C}}}(a)=\{ (s,t)\in {\mathbb {R}}^2\, :\, (a-se)^2+t^2e \text{ is } \text{ not } \text{ invertible } \in {\mathcal {A}}\} \end{aligned}$$

being nonempty. This set, called the complex spectrum of an element a in a real Banach algebra ${\mathcal {A}}$, was introduced by Kaplansky in [8]. Notice that the complex spectrum of $a\in {\mathcal {A}}$ is closed in ${\mathbb {R}}^2$ and is symmetric about the s-axis.

The complex spectrum of an element in a real Banach algebra ${\mathcal {A}}$ is not empty, as we shall next show. This result was originally obtained by Oliver in [11], but his method does not give the connection with the spectral radius. This connection, and the relation of $\sigma _{{\mathbb {C}}}(a)$ to the Gelfand space $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ are provided by the following Proposition (cf. [1, 9, 10]). Unlike other proofs of these facts in the literature, we use only harmonic function theory, i.e., we do not use the fact that on suitable domains a real-valued harmonic function is the real part of an analytic function.

Proposition 3.1

Let ${\mathcal {A}}$ be a commutative real Banach algebra with unit. Given $a\in {\mathcal {A}}$ we have : 

• (a) Its complex spectrum $\sigma _{{\mathbb {C}}}(a)$ is nonempty.

• (b) An element $(s,t)\in {\mathbb {R}}^{2}$ belongs to $\sigma _{{\mathbb {C}}}(a)$ if and only if there are $\phi , \overline{\phi }$ in $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ with $\phi (a)=s+it$ and $\overline{\phi } (a)=s-it$. Thus $\sigma _{{\mathbb {C}}} (a)={{\widehat{a}}}\left( \Phi _{{\mathcal {A}}}^{{\mathbb {C}}}\right) .$

Proof

(a) For a in ${\mathcal {A}}$ and $\varphi$ in ${\mathcal {A}}^*$, the function

$$\begin{aligned} u(s,t)=\varphi [(a-se)((a-se)^2+t^2e)^{-1}] \end{aligned}$$

is defined on the complement of the complex spectrum of a. Now,

$$\begin{aligned} ((a-s)^2+t^2)^{-1}=(a^2-2as+s^2+t^2)^{-1}=\sum (-1)^n(s^2+t^2)^{-n-1}(a^2-2as)^n, \end{aligned}$$

converging for $s^2+t^2>3\Vert a\Vert$. Thus the function u(s, t) is infinitely often differentiable with respect to s and t for $s^2+t^2>3\Vert a\Vert .$

A straightforward computation involving only the definitions of first and second order partial derivatives shows that u satisfies Laplace equation, hence it is harmonic. As we saw in the proof of Lemma 2.4, the complex spectrum of a is contained in a disc with centre at the origin and radius 3||a||. Also, it is clear that u(s, t) approaches 0 as (s, t) tends to infinity.

Now, suppose $\sigma _{{\mathbb {C}}}(a)$ is empty for some $a\in {\mathcal {A}}$. Then u is harmonic on the plane and vanishes at infinity. By the maximum principle for harmonic functions, u is identically 0. But we can find (s, t) in the complex spectrum of a such that $(a-se)((a-se)^2+t^2e)^{-1}\ne 0$ and then, by the Hahn–Banach theorem, pick $\varphi \in {\mathcal {A}}^{*}$ for which $u(s,t)\ne 0$, a contradiction.

(b) Suppose $(s,t)\in \sigma _{{\mathbb {C}}}(a)$. Then $b=(a-se)^2+t^2e$ is not invertible and so there is a maximal ideal containing the ideal $b{\mathcal {A}}$. It follows that there is $\phi \in \Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ with $\phi (b)=0$. Then, either $\phi (a)=s+it$ or $\phi (a)=s-it$. But $\overline{\phi }$, where $\overline{\phi }(a)=\overline{\phi (a)}$, belongs to $\Phi _{{\mathcal {A}}}^{{\mathbb {C}}}$ if $\phi$ does.

Conversely, if $\lambda =s+it$ and $\phi (a)=\lambda$ then $\phi (a-\lambda e)=0$. Thus $(a-se)^2+t^2e=(a-\lambda e)(a-\overline{\lambda }e)$ is not invertible.

Part (c) was proved inside the proof of Proposition 2.5. $\square$

We are now in a position to tackle the proof of Theorem 1.2.

Proof of Theorem 1.2

Let $a\in {\mathcal {A}}$ and let $\varphi \in {\mathcal {A}}^*$. The Kelvin transform $F:{\mathbb {C}}\setminus \{(0,0)\}\rightarrow {\mathbb {C}}\setminus \{(0,0)\}$ given by

$$\begin{aligned} (s,t)\hookrightarrow \left( \frac{s}{s^2+t^2},\frac{t}{s^2+t^2}\right) \end{aligned}$$

maps the punctured plane onto the punctured plane, tends to infinity at the origin, and to 0 at infinity. Choose a positive number R such that $\sigma _{{\mathbb {C}}}(a)$ is contained in the open disc ${\mathbb {D}}(R)$ with centre at the origin and radius R. The function $v=u\circ F$ defined to be 0 at the origin is harmonic in ${\mathbb {D}}(R^{-1})$ and thus has a Fourier expansion

$$\begin{aligned} v(s,t)= \sum _{n=0}^{\infty } r^n(a_n \cos n\theta +b_n\sin n \theta ), \quad \text{ where } s=r\cos \theta \text{ and } t=r \sin \theta , \end{aligned}$$

converging for $0\le r<1/R$ and all $\theta \in {\mathbb {R}}$. For $\theta =0$ we get

$$\begin{aligned} v(s,0)=\sum _{n=0}^{\infty } a_ns^n, \end{aligned}$$

converging for $0\le s<1/R$. Now

$$\begin{aligned} v(s,0)=(u\circ F)(s,0)=u(1/s,0)=\varphi \left( \left( a-\frac{1}{s}e \right) ^{-1}\right) =-s\sum _{n=0}^{\infty } s^n\varphi (a^n), \end{aligned}$$

converging at least for $1/s\ge \Vert a\Vert$.

Comparing terms in the two series we find that for each $0<s<R^{-1}$ the latter one converges, in particular its terms are bounded. Since this is true for all $\varphi$, by the uniform boundedness principle there is a constant $K=K_s$ such that

$$\begin{aligned} s^n\Vert a^n\Vert \le K_s \quad \text{ for } \text{ all } n. \end{aligned}$$

Taking nth roots and letting n tend to infinity we conclude that $s r(a)\le 1$ for all $0<s<R^{-1}$. This shows that $r(a)\le R$. Since $\sigma _{{\mathbb {C}}}(a)$ is contained in ${\mathbb {D}}(R)$, taking the infimum over all such R yields

$$\begin{aligned} r(a)\le \inf \{R>0:\sigma _{{\mathbb {C}}}(a)\subseteq \mathbb D(R)\} =\sup _{\phi \in \Phi _{{\mathcal {A}}}^{{\mathbb {C}}}} |\phi (a)|= \Vert {\widehat{a}}\Vert _{\infty , \Phi _{\mathcal A}^{{\mathbb {C}}}}. \end{aligned}$$

The reverse inequality follows from Lemma 2.4. $\square$

4 Inequalities leading (or not) to isomorphic representations

A necessary and sufficient condition for a commutative real Banach algebra ${\mathcal {A}}$ to be isomorphic to a $\mathcal C_{{\mathbb {R}}}({\mathcal {K}})$-space is the fulfilment of the following inequality involving the norm of squares of the algebra:

$$\begin{aligned} \Vert a^2\Vert \le k\Vert a^2+b^2\Vert , \quad a,b\in {\mathcal {A}}, \end{aligned}$$

(4.1)

for some $k\ge 1$ (see [1, Theorem 3.6]).

What about conditions for isometric representations? An obvious necessary condition for ${\mathcal {A}}$ is the fulfilment of inequality (4.1) with $k=1$, i.e.,

$$\begin{aligned} \Vert a^2\Vert \le \Vert a^2+b^2\Vert , \quad \text{ for } \text{ all } a,b\in {\mathcal {A}}. \end{aligned}$$

(4.2)

However, the example of the real Banach algebra ${\mathcal {A}}={\mathcal {C}}_{{\mathbb {R}}}([0,1])$ equipped with the norm

$$\begin{aligned}\Vert f\Vert =\Vert f_+\Vert _{\infty ,I}+\Vert f_-\Vert _{\infty ,I},\quad f\in {\mathcal {A}},\end{aligned}$$

where $f_{+}=\max \{f,0\}$, $f_{-}=\max \{-f,0\}$, and $I=[0,1]$, shows that (4.2) is not sufficient for isometric representation. Indeed, under this norm $\mathcal C_{{\mathbb {R}}}([0,1])$ is a commutative real Banach algebra satisfying inequality (4.2) but it is only 2-isomorphic to $({\mathcal {C}}_{{\mathbb {R}}}([0,1]), \Vert \cdot \Vert _{\infty , [0,1]})$, i.e., the Gelfand transform

$$\begin{aligned} \Lambda :({\mathcal {C}}_{{\mathbb {R}}}([0,1]), \Vert \cdot \Vert )\rightarrow ({\mathcal {C}}_{{\mathbb {R}}}([0,1]), \Vert \cdot \Vert _{\infty , [0,1]}),\end{aligned}$$

which in this case is just the identity map, satisfies

$$\begin{aligned} \frac{1}{2}\Vert f\Vert \le \Vert \Lambda (f)\Vert _{\infty , I}\le \Vert f\Vert ,\qquad f\in {\mathcal {C}}_{{\mathbb {R}}}([0,1]), \end{aligned}$$

so that $\Vert \Lambda \Vert =1$ and $\Vert \Lambda ^{-1}\Vert = 2$.

Notice that in this example the norm of squares is preserved under the Gelfand representation, i.e.,

$$\begin{aligned}\Vert f^2\Vert =\Vert {{\widehat{f}}}^2\Vert _{\infty ,I},\quad f\in {\mathcal {A}}.\end{aligned}$$

This is, in general, the case if ${\mathcal {A}}$ satisfies (4.2) as we next show.

Proposition 4.1

Let ${\mathcal {A}}$ be a commutative real Banach algebra with unit. The following are equivalent : 

• (i) ${\mathcal {A}}$ satisfies (4.2).

• (ii) The Gelfand transform $\Lambda :{\mathcal {A}}\rightarrow {\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$ defined in (2.1) is an isomorphism which preserves the norm of squares.

Proof

(ii) $\Rightarrow$ (i) Suppose first that $a\hookrightarrow {{\widehat{a}}}$ is an isomorphism from ${\mathcal {A}}$ onto ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$ preserving the norm of squares, i.e., $\Vert \widehat{a^2}\Vert _{\infty , {\mathcal {K}}}=\Vert a^2\Vert$ for $a\in {\mathcal {A}}$. Then, by the hypotheses, the sum of two squares in ${\mathcal {A}}$ is a square in ${\mathcal {A}}$ and

$$\begin{aligned} \Vert a^2\Vert =\Vert \widehat{a^2}\Vert _{\infty , {\mathcal {K}}}\le \Vert \widehat{a^2}+\widehat{b^2}\Vert _{\infty , {\mathcal {K}}}= \Vert \widehat{a^2+b^2}\Vert _{\infty , {\mathcal {K}}}=\Vert a^2+b^2\Vert , \end{aligned}$$

as wished.

The proof of (i) $\Rightarrow$ (ii) follows essentially the methods of Sect. 2. Consider the state space of ${\mathcal {A}}$,

$$\begin{aligned} {\mathcal {S}}=\{\varphi \in {\mathcal {A}}^*: \Vert \varphi \Vert =\varphi (e)=1\}. \end{aligned}$$

The proofs that the elements of ${\mathcal {S}}$ are positive on ${\mathcal {A}}_{+}$ and that functionals which are positive on squares attain their norm at e under the assumption of condition (4.2) follow the steps of the proofs of the corresponding parts in Lemma 2.2 under the assumption of (1.1). From this it then follows just as in the proof of Lemma 2.3 that the extreme points of ${\mathcal {S}}$ are multiplicative on ${\mathcal {A}}$.

Thus, the set of extreme points of ${\mathcal {S}}$ consists of multiplicative real-valued functionals on ${\mathcal {A}}$, and the same thing then also holds for its $\hbox {weak}^{*}$-closure ${\mathcal {K}}$ in ${\mathcal {S}}$.

We now look at the elements $a\in {\mathcal {A}}$ as functions ${{\widehat{a}}}$ on ${\mathcal {K}}$, where ${{\widehat{a}}} (\phi )=\phi (a)$ for $\phi \in {\mathcal {K}}$. It follows from the definition of ${\mathcal {S}}$ that

$$\begin{aligned} \Vert {{\widehat{a}}}\Vert _{\infty ,{\mathcal {K}}}\le \Vert a\Vert \quad \text{ for } \text{ all } a\in {\mathcal {A}}. \end{aligned}$$

(4.3)

Continuing as in the completion of the proof of Theorem 1.1, we find that if $\Vert a^2\Vert >1$ then $\varphi (a^2)>1$ for some $\varphi$ in ${\mathcal {S}}$. We conclude that

$$\begin{aligned} \Vert a^2\Vert =\Vert {{\widehat{a}}}^2\Vert _{\infty , {\mathcal {K}}},\quad a\in {\mathcal {A}}, \end{aligned}$$

(4.4)

so that the Gelfand mapping preserves the norms of squares. To finish the proof we need only show that the algebra norm and the sup-norm are equivalent.

For $a\in {\mathcal {A}}$ we have,

$$\begin{aligned} \Vert a\Vert&= \frac{1}{4}\Vert (e+a)^2-(e-a)^2\Vert \\&\le \frac{1}{4}\Vert (e+a)^2\Vert +\frac{1}{4}\Vert (e-a)^2\Vert \\&\le \frac{1}{2}\Vert (e+a)^2+(e-a)^2\Vert \\&=\frac{1}{2}\Vert 2+2a^2\Vert \\&=\Vert e+a^2\Vert \\&\le 1+\Vert a^2\Vert , \end{aligned}$$

where to obtain the second inequality we used condition (4.2) twice. It follows that $\Vert a\Vert ^2\le 4$ if $\Vert a^2\Vert =1$, so that $\Vert a\Vert ^2\le 4\Vert a^2\Vert$ for all $a\in \mathcal A$. Combining with (4.4) gives

$$\begin{aligned} \Vert a\Vert ^{2}\le 4\Vert a^{2}\Vert =4\Vert {{\widehat{a}}}^2\Vert _{\infty , {\mathcal {K}}}=4\Vert {{\widehat{a}}}\Vert ^2_{\infty , {\mathcal {K}}}, \end{aligned}$$

which, together with (4.3) yields the equivalence of norms

$$\begin{aligned} \Vert {{\widehat{a}}}\Vert _{\infty , {\mathcal {K}}}\le \Vert a\Vert \le 2\Vert {{\widehat{a}}}\Vert _{\infty , {\mathcal {K}}},\quad a\in {\mathcal {A}}, \end{aligned}$$

and so the Gelfand representation is an isomorphism as claimed. $\square$

Remark 4.2

Notice that the standard norm on ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$ satisfies

$$\begin{aligned} \Vert f^2\Vert _{\infty ,{\mathcal {K}}}=\Vert f\Vert _{\infty , {\mathcal {K}}}^2,\quad f\in {\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}}), \end{aligned}$$

thus a necessary condition for isometric representation is

$$\begin{aligned} \Vert a\Vert ^2\le \Vert a^2+b^2\Vert \quad \text{ for } \text{ all } a,b\in {\mathcal {A}}. \end{aligned}$$

This condition was studied in [5] by Arens, who proved by means of the complexification of the algebra and using the theory of complex Banach algebras that it was sufficient for isometric representation. Arens’ result can also be obtained using the real algebra methods in this paper (see [2, Theorem 5.1] and [3, Theorem 1.1]).

We close this section and this note with an example of a commutative real Banach algebra which is not isomorphic to a $\mathcal C_{{\mathbb {R}}}({\mathcal {K}})$-space despite the fact that it verifies the spectral radius condition (1.1), thus solving a problem that naturally arose from our discussions in [2].

Proposition 4.3

There exists a commutative real Banach algebra ${\mathcal {A}}$ with unit that satisfies the spectral radius condition (1.1),  hence is representable as an algebra ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$ of real-valued continuous functions on a compact Hausdorff space ${\mathcal {K}}$, but is not isomorphic to ${\mathcal {C}}_{{\mathbb {R}}}({\mathcal {K}})$.

Proof

Let ${\mathcal {A}}$ be the real Banach algebra ${\mathcal {C}}_{{\mathbb {R}}}^{(1)}([0,1])$ of all continuously differentiable, real-valued functions on the unit interval $I=[0,1]$, equipped with the norm

Let us compute the spectral radius of a function f in $\mathcal A$. On one hand, for $n\in {\mathbb {N}}$,

On the other hand,

so that $r(f)=\Vert f\Vert _{\infty ,I}$.

Now, the homomorphisms of ${\mathcal {A}}$ are the evaluations at the points of [0, 1]. To see this, suppose $\phi$ is a homomorphism on ${\mathcal {A}}$, and that for all x in the interval [0, 1] there is g in the nullspace of $\phi$ for which $g(x)\ne 0$. Since $g^2$ is in the nullspace of $\phi$ we have for each x in [0, 1] an element in ${\mathcal {A}}$ which is nonnegative on [0, 1] and strictly positive on x. Since the unit interval is compact, a sum, say f, of finitely many of those functions is strictly positive on [0, 1].

But then f is invertible in the algebra and thus cannot be in the nullspace of $\phi$. We conclude that all the elements in the nullspace of $\phi$ vanish at some point of [0, 1] and thus $\phi$ is the evaluation at that point.

Clearly, the norm on ${\mathcal {A}}$ is not equivalent to the sup-norm, so that ${\mathcal {A}}$ is not isomorphic to some ${\mathcal {C}}_{\mathbb R}({\mathcal {K}})$. However, $r(f^2)\le r(f^2+g^2)$ for all f and g in ${\mathcal {A}}$. $\square$