The Spectrum of an Element in a Commutative Banach Algebra

The Spectrum of an Element in a Commutative Banach Algebra

Recall that if $X$ is an algebra over $\mathbb{C}$ and $x \in X$ then the spectrum of $x$ in $X$ is defined as follows. If $X$ is an algebra with unit then:

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

And if $X$ is an algebra without unit then:

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

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

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

Theorem 1: Let $X$ be a commutative Banach algebra and let $x \in X$, $\lambda \in \mathbb{C} \setminus \{ 0 \}$. Then $\lambda \in \mathrm{Sp}(X, x)$ if and only if $\lambda = f(x)$ for some multiplicative linear functional $f$ on $X$. If in addition $X$ is a commutative Banach algebra with unit then we also have that $0 \in \mathrm{Sp}(X, x)$ if and only if $0 = f(x)$ for some multiplicative linear functional $f$ on $X$
  • Proof: Let $x \in X$, $\lambda \in \mathbb{C} \setminus \{ 0 \}$, and let $\lambda \in \mathrm{Sp}(X, x)$. Then $\frac{1}{\lambda} x \in \mathrm{q-Sing}(X)$. Let:
(3)
\begin{align} \quad J = \left \{ \frac{1}{\lambda} x y - y : y \in X \right \} \end{align}
  • We now claim that $J$ is a proper modular two-sided ideal of $X$.
  • 1. Showing that $J$ is a two-sided ideal of $X$: Let $z \in X$ and let $\frac{1}{\lambda}xy - y \in J$. Then by the commutativity of $X$ we have that
(4)
\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:
(5)
\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 $XJ \subseteq J$ and $JX \subseteq J$, so $J$ is a two-sided ideal of $X$.
  • 2. Showing that $J$ is modular: Consider the point $\frac{1}{\lambda}x$. We have by commutativity that:
(6)
\begin{align} \quad X - \frac{1}{\lambda}xX = \left \{ y - \frac{1}{\lambda} xy : y \in X \right \} = \left \{ \frac{1}{\lambda} xy - y : y \in X \right \} = J \end{align}
(7)
\begin{align} \quad X - X \frac{1}{\lambda}x = \left \{ y - y \frac{1}{\lambda}x : y \in X \right \} = \left \{ \frac{1}{\lambda} xy - y : y \in X \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 $\frac{1}{\lambda} x \in X \setminus J$. To see why, assume instead that $\frac{1}{\lambda}x \in J$. Then there exists a $y \in X$ such that:
(8)
\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 commutativity of $X$ we also have that $y \circ \frac{1}{\lambda} x = 0$. Thus $\frac{1}{\lambda} x$ is quasi-invertible - which is a contradiction since $\frac{1}{\lambda}x \in \mathrm{Sing}(X)$. So indeed, $\frac{1}{\lambda}x \not \in J$ and so $J \neq X$.
  • Thus $J$ is a proper modular ideal of $X$ with modular unit. By one of the propositions on the Basic Theorems Regarding Ideals in a Linear Space 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 $X$ with modular unit $\frac{1}{\lambda}x$. Since $X$ is a commutative Banach algebra we have by the theorem on the Maximal Modular Two-Sided Ideals in a Commutative Banach Algebra page that $M$ is the kernel of some multiplicative linear functional $f$ on $X$, 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 $X$ is a Banach algebra we have by the theorem on the Multiplicative Functionals on Modular Units of Ker(f) in a Banach Algebra page that $f \left ( \frac{1}{\lambda} x \right ) = 1$ or equivalently, $\lambda = f(x)$ for some multiplicative linear functional $f$ on $X$.
  • $\Leftarrow$ Suppose that there exists a multiplicative linear functional $f$ on $X$ for which $\lambda = f(x)$. Then for each $y \in X$ we have that:
(9)
\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:
(10)
\begin{align} \quad J = \left \{ \left ( \frac{1}{\lambda} x \right ) y - y : y \in X \right \} \end{align}
  • Again we claim that $J$ is a proper modular two-sided ideal of $X$.
  • 1. Showing that $J$ is a two-sided ideal of $X$: $XJ \subseteq J$ and $JX \subseteq J$ follows similarly from earlier.
  • 2. Showing that $J$ is modular: $X - \frac{1}{\lambda}xX \subseteq J$ and $X - X \frac{1}{\lambda}x \subseteq J$ follows similarly from earlier, and so $\frac{1}{\lambda}x$ is a two-sided modular unit for $X$, 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 $X$ 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 a Linear Space 1 page we have that since $\frac{1}{\lambda}x$ is a modular unit for the proper ideal $M$ of $X$ 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 $\frac{1}{\lambda}x \in \mathrm{q-Inv}(X)$. Then there exists a $z \in X$ such that:
(11)
\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 $\frac{1}{\lambda}x \in \mathrm{q-Sing}(X)$ and so $\lambda \in \mathrm{Sp}(X, x)$. $\blacksquare$
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