The Spectral Radius of a Point in a Normed Algebra

# 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:
(1)
\begin{align} \quad \| a^k \|^{1/k} < r(a) + \epsilon \quad (*) \end{align}
• Now every $n \in \mathbb{N}$ with $n \geq k$ can be uniquely written in the form:
(2)
• 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:
(3)
\begin{align} \quad \lim_{n \to \infty} \frac{q(n)}{n} \leq \lim_{n \to \infty} \frac{k - 1}{n} = 0 \end{align}
• And so by the squeeze theorem, $\displaystyle{\lim_{n \to \infty} \frac{q(n)}{n} = 0}$. Therefore:
(4)
\begin{align} \quad \limsup_{n \to \infty} \frac{q(n)}{n} = 0 \end{align}
• Also, with the equality at $(**)$ we also have that:
(5)
\begin{align} \quad \lim_{n \to \infty} \frac{p(n)}{n} = \frac{1}{k} \end{align}
• Therefore:
(6)
\begin{align} \quad \limsup_{n \to \infty} \frac{p(n)}{n} = \frac{1}{k} \end{align}
• Hence for every $n \in \mathbb{N}$ with $n \geq k$ we have that:
(7)
\begin{align} \quad \limsup_{n \to \infty} \| a^n \|^{1/n} &= \lim_{n \to \infty} \| a^{p(n)k + q(n)} \|^{1/n} \\ &= \limsup_{n \to \infty} \| a^{p(n)k} a^{q(n)} \|^{1/n} \\ & \overset{(\dagger)} \leq \limsup_{n \to \infty} \| a^k \|^{p(n)/n} \| a \|^{q(n)/n} \\ & \leq \| a^k \|^{1/k} \\ & \overset{(*)} < r(a) + \epsilon \end{align}
• (Where the inequality at $(\dagger)$ comes from the fact that $\|ab\| \leq \|a\| \|b\|$ for all $a, b \in \mathfrak{A}$). Therefore:
(8)
\begin{align} \quad \limsup_{n \to \infty} \| a^n\|^{1/n} \leq r(a) \quad (*) \end{align}
• But we also have that:
(9)
\begin{align} \quad r(x) = \inf \{ \| a^n \|^{1/n} : n \in \mathbb{N} \} \leq \liminf_{n \to \infty} \| a^n \|^{1/n} \quad (**) \end{align}
• Thus combining $(*)$ and $(**)$ we see that $\displaystyle{\lim_{n \to \infty} \| a^n \|^{1/n}}$ exists and:
(10)
\begin{align} \quad r(a) = \lim_{n \to \infty} \| a^n \|^{1/n} \quad \blacksquare \end{align}

## 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)
\begin{align} \quad |a^n|^{1/n} = |a| \end{align}

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)
\begin{align} \quad \| f \|_{\infty} = \sup_{e \in E} |f(e)| \end{align}

Let $f \in B(E)$. Observe that for each $n \in \mathbb{N}$ we have that:

(13)
\begin{align} \quad \| f^n \|^{1/n} = \left ( \sup_{e \in E} |f(e)|^n \right )^{1/n} \end{align}

Therefore:

(14)
\begin{align} \quad r(f) = \lim_{n \to \infty} \left ( \sup_{e \in E} |f(e)|^n \right )^{1/n} \end{align}

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 $r(f) = M$.