The Spectrum of an Element in a Normed or Banach Algebra

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

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

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

When $X$ is a Banach algebra, we obtain the following theorem.

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

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

**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 $1 - \frac{1}{\lambda} x \in \mathrm{Inv}(X)$ by the theorem on the Invertibility of 1 - x When r(x) < 1 in a Banach Algebra page, and thus $\lambda - x \in \mathrm{Inv}(X)$. So $\lambda \not \in \mathrm{Sp}(X)$. Thus if $\lambda \in \mathrm{Sp}(X)$ 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:

\begin{align} \quad r(x) = \max \{ |\lambda| : \lambda \in \mathrm{Sp}(X, x) \} \end{align}

- Observe that as a result, $\mathrm{Sp}(X, x)$ is a bounded subset of $\mathbb{C}$. Consider the map $F : \mathbb{C} \to X$ defined for all $z \in \mathbb{C}$ by $F(z) = z - x$. Note that $F(z) \in \mathrm{Inv}(X)$ if and only if $z \in \mathbb{C} \setminus \mathrm{Sp}(X, x)$. So:

\begin{align} \quad F^{-1}(\mathrm{Inv}(X)) = \mathbb{C} \setminus \mathrm{Sp}(X, x) \end{align}

- Clearly $F : \mathbb{C} \to X$ is a continuous mapping. We know that $\mathrm{Inv}(X)$ is an open set in $X$ from The Set of Invertible Elements Inv(X) is an Open Subset of X page. Thus by continuity of $F$ and the equality above, $\mathbb{C} \setminus \mathrm{Sp}(X, x)$ is open in $X$, i.e., $\mathrm{Sp}(X, x)$ is closed in $X$.

- Since closed and boundedness implies compactness in $\mathbb{C}$, we see that $\mathrm{Sp}(X, x)$ is compact in $\mathbb{C}$. $\blacksquare$