The Spectrum of an Element in a Normed or Banach Algebra over C

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

So far we have only discussed properties of the spectrum of a point $x$ is an algebra $\mathfrak{A}$ over $\mathbb{C}$ with unit. When $\mathfrak{A}$ is a normed algebra (not necessarily with a unit), more can be said.

 Theorem 1: Let $\mathfrak{A}$ be an normed algebra over $\mathbb{C}$ and let $x \in \mathfrak{A}$. Then there exists a point $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$ such that $|\lambda| \geq r(x)$.

When $\mathfrak{A}$ is a Banach algebra, we obtain the following theorem.

 Theorem 2: Let $\mathfrak{A}$ be a Banach algebra over $\mathbb{C}$ and let $x \in \mathfrak{A}$. Then $\mathrm{Sp}(\mathfrak{A}, x)$ is a nonempty compact subset of $\mathbb{C}$ and moreover, $r(x) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, x) \}$.

Recall that a subset of $\mathbb{C}$ is compact if and only if it is closed and bounded.

In the proof below, we assume that $\mathfrak{A}$ has a unit. The Theorem is true regardless of this assumption.

• Proof: Suppose that $\lambda \in \mathbb{C}$ is such that $|\lambda| > r(x) > 0$ so that $|\lambda^{-1}| < r(x)^{-1}$. Then $r(\lambda^{-1}x) < 1$.
• Therefore $1 - \frac{1}{\lambda} x \in \mathrm{Inv}(\mathfrak{A})$ by the theorem on the Invertibility of 1 - a When r(a) < 1 in a Banach Algebra with Unit page, and thus $\lambda - x \in \mathrm{Inv}(\mathfrak{A})$. So $\lambda \not \in \mathrm{Sp}(\mathfrak{A}, x)$. Thus if $\lambda \in \mathrm{Sp}(\mathfrak{A}, x)$ then $\lambda \leq r(x)$.
• But by Theorem 1 there exists a $\lambda \in \mathbb{C}$ such that $|\lambda| \geq r(x)$. Therefore, there exists a $\lambda \in \mathbb{C}$ such that $r(x) = |\lambda|$ and:
(1)
\begin{align} \quad r(x) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(\mathfrak{A}, x) \} \end{align}
• Observe that as a result, $\mathrm{Sp}(\mathfrak{A}, x)$ is a bounded subset of $\mathbb{C}$. Consider the map $F : \mathbb{C} \to \mathfrak{A}$ defined for all $z \in \mathbb{C}$ by $F(z) = z - x$. Note that $F(z) \in \mathrm{Inv}(\mathfrak{A})$ if and only if $z \in \mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x)$. So:
(2)
\begin{align} \quad F^{-1}(\mathrm{Inv}(\mathfrak{A})) = \mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x) \end{align}
• Clearly $F : \mathbb{C} \to \mathfrak{A}$ is a continuous mapping. We know that $\mathrm{Inv}(\mathfrak{A})$ is an open set in $X$ from Inv(A) is an Open Subset of A in a Banach Algebra with Unit page. Thus by continuity of $F$ and the equality above, $\mathbb{C} \setminus \mathrm{Sp}(\mathfrak{A}, x)$ is open in $\mathfrak{A}$, i.e., $\mathrm{Sp}(\mathfrak{A}, x)$ is closed in $\mathfrak{A}$.
• Since closed and boundedness implies compactness in $\mathbb{C}$, we see that $\mathrm{Sp}(\mathfrak{A}, x)$ is compact in $\mathbb{C}$. $\blacksquare$