The Spectral Radius of a Point in a Normed Algebra
Definition: Let $\mathfrak{A}$ be a normed algebra and let $a \in \mathfrak{A}$. The Spectral Radius of $a$ is defined to be $r(a) = \inf \left \{ \| a^n \|^{1/n} : n \in \mathbb{N} \right \}$. |
The following proposition gives us an alternative way to compute the spectral radius of a point $a \in \mathfrak{A}$.
Proposition 1: Let $\mathfrak{A}$ be a normed algebra and let $a \in \mathfrak{A}$. Then $\displaystyle{r(a) = \lim_{n \to \infty} \| a^n \|^{1/n}}$. |
- Proof: Let $\epsilon > 0$. By the definition of infimum, for this given $\epsilon$ there exists a $k \in \mathbb{N}$ such that:
- Now every $n \in \mathbb{N}$ with $n \geq k$ can be uniquely written in the form:
- Where $p \geq 0$ and $0 \leq q \leq k - 1$ (this is just by the division algorithm). Now observe that since $q(n) \leq k - 1$ for every $n \in \mathbb{N}$, we have that:
- And so by the squeeze theorem, $\displaystyle{\lim_{n \to \infty} \frac{q(n)}{n} = 0}$. Therefore:
- Also, with the equality at $(**)$ we also have that:
- Therefore:
- Hence for every $n \in \mathbb{N}$ with $n \geq k$ we have that:
- (Where the inequality at $(\dagger)$ comes from the fact that $\|ab\| \leq \|a\| \|b\|$ for all $a, b \in \mathfrak{A}$). Therefore:
- But we also have that:
- Thus combining $(*)$ and $(**)$ we see that $\displaystyle{\lim_{n \to \infty} \| a^n \|^{1/n}}$ exists and:
Example 1
Consider the algebra $\mathbb{R}$. For each $a \in \mathbb{R}$ observe that for all $n \in \mathbb{N}$ we have that:
(11)Thus for each $a \in \mathbb{R}$ we have that $r(a) = |a|$.
Example 2
Let $\mathfrak{E}$ be a nonempty set and let $B(E)$ denote the set of all functions from $E$ to $\mathbb{R}$ that are bounded. Equip $B(E)$ with the operations of function addition, scalar multiplication, and pointwise function multiplication. It is easy to check that then $B(E)$ is an algebra. Furthermore, it is a normed algebra with norm:
(12)Let $f \in B(E)$. Observe that for each $n \in \mathbb{N}$ we have that:
(13)Therefore:
(14)For example, if $X = [a, b]$ and $f$ is continuous function on $[a, b]$ then there exists an $M \geq 0$ such that $\sup_{e \in E} |f(e)| = M$, and so $r(f) = M$.