The Spectrum of an Element in a Commutative Banach Algebra over C

# The Spectrum of an Element in a Commutative Banach Algebra over C

Recall that if $\mathfrak{A}$ is an algebra over $\mathbb{C}$ and $x \in \mathfrak{A}$ then the spectrum of $x$ in $\mathfrak{A}$ is defined as follows.

If $\mathfrak{A}$ is an algebra with unit then:

(1)\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, x) = \{ \lambda \in \mathbb{C} : \lambda - x \in \mathrm{Sing}(\mathfrak{A}) \} \end{align}

And if $\mathfrak{A}$ is an algebra without unit then:

(2)\begin{align} \quad \mathrm{Sp}(\mathfrak{A}, x) = \{ 0 \} \cup \left \{ \lambda \in \mathbb{C} \setminus \{ 0 \} : \frac{1}{\lambda}x \in \mathrm{q-Sing}(\mathfrak{A}) \right \} \end{align}

When $\mathfrak{A}$ is a normed or Banach algebra over $\mathbb{C}$ then on The Spectrum of an Element in a Normed or Banach Algebra over C page we proved that $\mathrm{Sp}(\mathfrak{A}, x)$ is a nonempty compact subset of $\mathbb{C}$.

When the additional assumption that $\mathfrak{A}$ is a commutative Banach algebra is made, then we can completely determine $\mathrm{Sp}(\mathfrak{A}, x)$.

Theorem 1: Let $\mathfrak{A}$ be a commutative Banach algebra over $\mathbb{C}$ and let $x \in \mathfrak{A}$, $\lambda \in \mathbb{C} \setminus \{ 0 \}$. Then $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$ if and only if $\lambda = f(x)$ for some multiplicative linear functional $f$ on $\mathfrak{A}$. If $\mathfrak{A}$ is a commutative Banach algebra over $\mathbb{C}$ with unit then we also have that $0 \in \mathrm{Sp}(\mathfrak{A}, x)$ if and only if $0 = f(x)$ for some multiplicative linear functional $f$ on $\mathfrak{A}$ |

**Proof:**Let $x \in \mathfrak{A}$, $\lambda \in \mathbb{C} \setminus \{ 0 \}$, and let $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$. Then $\frac{1}{\lambda} x \in \mathrm{q-Sing}(\mathfrak{A})$. Let:

\begin{align} \quad J = \left \{ \frac{1}{\lambda} x y - y : y \in \mathfrak{A} \right \} \end{align}

- We now claim that $J$ is a proper modular two-sided ideal of $\mathfrak{A}$.

**1. Showing that $J$ is a two-sided ideal of $\mathfrak{A}$:**Let $z \in \mathfrak{A}$ and let $\frac{1}{\lambda}xy - y \in J$. Then by the commutativity in $\mathfrak{A}$ we have that

\begin{align} \quad z \left [ \frac{1}{\lambda}xy - y \right ] = \frac{1}{\lambda}x\underbrace{[yz]}_{\in X} - [yz] \in J \end{align}

- And also:

\begin{align} \quad \left [ \frac{1}{\lambda}xy - y \right ] z = \frac{1}{\lambda}x\underbrace{[yz]}_{\in X} - [yz] \in J \end{align}

- The above equalities show that $\mathfrak{A}J \subseteq J$ and $J\mathfrak{A} \subseteq J$, so $J$ is a two-sided ideal of $\mathfrak{A}$.

**2. Showing that $J$ is modular:**Consider the point $\frac{1}{\lambda}x$. We have by commutativity that:

\begin{align} \quad (1 - \frac{1}{\lambda}x)\mathfrak{A} = \left \{ y - \frac{1}{\lambda} xy : y \in \mathfrak{A} \right \} = \left \{ \frac{1}{\lambda} xy - y : y \in \mathfrak{A} \right \} = J \end{align}

(7)
\begin{align} \quad \mathfrak{A}(1 - \frac{1}{\lambda}x) = \left \{ y - y \frac{1}{\lambda}x : y \in \mathfrak{A} \right \} = \left \{ \frac{1}{\lambda} xy - y : y \in \mathfrak{A} \right \} =J \end{align}

- Therefore $\frac{1}{\lambda}x$ is a two-sided modular unit for $J$, so $J$ is modular.

**3. Showing that $J$ is proper:**We claim that $\frac{1}{\lambda} x \in \mathfrak{A} \setminus J$. To see why, assume instead that $\frac{1}{\lambda}x \in J$. Then there exists a $y \in \mathfrak{A}$ such that:

\begin{align} \quad \frac{1}{\lambda} x = \frac{1}{\lambda}xy - y \\ \quad \frac{1}{\lambda} x + y - \left [ \frac{1}{\lambda} x \right ] y = 0 \\ \quad \frac{1}{\lambda} x \circ y = 0 \end{align}

- By the commutativity in $\mathfrak{A}$ we also have that $y \circ \frac{1}{\lambda} x = 0$. Thus $\frac{1}{\lambda} x$ is quasi-invertible - which is a contradiction since $\frac{1}{\lambda}x \in \mathrm{Sing}(\mathfrak{A})$. So indeed, $\frac{1}{\lambda}x \not \in J$ and so $J \neq \mathfrak{A}$.

- Thus $J$ is a proper modular ideal of $\mathfrak{A}$ with modular unit. By one of the propositions on the Basic Theorems Regarding Ideals in an Algebra 1 page we have that $J$ is contained in some maximal modular proper ideal, call it $M$. Furthermore, by a different proposition on that page we have that since $\frac{1}{\lambda}x$ is a modular unit for $J$ and $J \subseteq M$ we have that $\frac{1}{\lambda}x$ is a modular unit for $M$.

- So $M$ is a maximal modular two-sided ideal of $\mathfrak{A}$ with modular unit $\frac{1}{\lambda}x$. Since $\mathfrak{A}$ is a commutative Banach algebra we have by the theorem on the Characterization of Maximal Modular Two-Sided Ideals in a Commutative Banach Algebra over C page that $M$ is the kernel of some multiplicative linear functional $f$ on $\mathfrak{A}$, that is, $M = \ker (f)$.

- Since $\frac{1}{\lambda} x$ is a modular unit for the ideal $M = ker(f)$ of some multiplicative linear functional $f$ and since $\mathfrak{A}$ is a Banach algebra we have by the resut on the For Algebras A - ker(f) is a Two-Sided Ideal of A for all Multiplicative Linear Functionals f page that $f \left ( \frac{1}{\lambda} x \right ) = 1$ or equivalently, $\lambda = f(x)$ for some multiplicative linear functional $f$ on $\mathfrak{A}$.

- $\Leftarrow$ Suppose that there exists a multiplicative linear functional $f$ on $\mathfrak{A}$ for which $\lambda = f(x)$. Then for each $y \in \mathfrak{A}$ we have that:

\begin{align} \quad f \left ( \left ( \frac{1}{\lambda}x \right ) y - y \right ) = \frac{1}{\lambda} f(x)f(y) - f(y) = \frac{1}{\lambda} \lambda f(y) - f(y) = 0 \end{align}

- Let:

\begin{align} \quad J = \left \{ \left ( \frac{1}{\lambda} x \right ) y - y : y \in \mathfrak{A} \right \} \end{align}

- Again we claim that $J$ is a proper modular two-sided ideal of $\mathfrak{A}$.

**1. Showing that $J$ is a two-sided ideal of $\mathfrak{A}$:**$\mathfrak{A}J \subseteq J$ and $J \mathfrak{A} \subseteq J$ follows similarly from earlier.

**2. Showing that $J$ is modular:**$(1 - \frac{1}{\lambda}x) \mathfrak{A} \subseteq J$ and $\mathfrak{A}(1 - \frac{1}{\lambda}x) \subseteq J$ follows similarly from earlier, and so $\frac{1}{\lambda}x$ is a two-sided modular unit for $\mathfrak{A}$, i.e., $J$ is modular.

**3. Showing that $J$ is proper:**Again, it can be shown that $\frac{1}{\lambda}x \not \in J$.

- So again let $M$ be a maximal modular two-sided ideal of $\mathfrak{A}$ containing $J$ so that $\frac{1}{\lambda}x$ is a modular unit for $M$.

- Now by one of the propositions on the Basic Theorems Regarding Ideals in an Algebra 1 page we have that since $\frac{1}{\lambda}x$ is a modular unit for the proper ideal $M$ of $\mathfrak{A}$ we have that $\frac{1}{\lambda}x \not \in M$. We claim that this implies that $\frac{1}{\lambda}x$ is quasi-singular.

- Suppose instead that $\frac{1}{\lambda}x \in \mathrm{q-Inv}( \mathfrak{A})$. Then there exists a $z \in \mathfrak{A}$ such that:

\begin{align} \quad 0 &= \frac{1}{\lambda}x \circ z \\ &= \frac{1}{\lambda}x + z - \frac{1}{\lambda}xz \end{align}

- Rearranging the above equation gives us that $\frac{1}{\lambda}x = \frac{1}{\lambda}xz - z \in M$, a contradiction. Thus $\frac{1}{\lambda}x \in \mathrm{q-Sing}( \mathfrak{A})$ and so $\lambda \in \mathrm{Sp}( \mathfrak{A}, x)$. $\blacksquare$