The Spectral Radius of a Point in a Normed Algebra

The Spectral Radius of a Point in a Normed Algebra

Definition: Let $X$ be a normed algebra and let $x \in X$. The Spectral Radius of $x$ is defined to be $r(x) = \inf \left \{ \| x^n \|^{1/n} : n \in \mathbb{N} \right \}$.

The following proposition gives us an alternative way to compute the spectral radius of a point $x \in X$.

Proposition 1: Let $X$ be a normed algebra and let $x \in X$. Then $\displaystyle{r(x) = \lim_{n \to \infty} \| x^n \|^{1/n}}$.
  • 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 \| x^k \|^{1/k} < r(x) + \epsilon \quad (*) \end{align}
  • Now every $n \in \mathbb{N}$ with $n \geq k$ can be uniquely written in the form:
(2)
\begin{align} \quad n &= p(n)k + q(n) \\ \quad 1 &= \frac{p(n)k}{n} + \frac{q(n)}{n}\quad (**) \end{align}
  • 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} \| x^n \|^{1/n} &= \lim_{n \to \infty} \| a^{p(n)k + q(n)} \|^{1/n} \\ &= \limsup_{n \to \infty} \| x^{p(n)k} x^{q(n)} \|^{1/n} \\ & \overset{(\dagger)} \leq \limsup_{n \to \infty} \| x^k \|^{p(n)/n} \| x \|^{q(n)/n} \\ & \leq \| x^k \|^{1/k} \\ & \overset{(*)} < r(x) + \epsilon \end{align}
  • (Where the inequality at $(\dagger)$ comes from the fact that $\|ab\| \leq \|a\| \|b\|$ for all $a, b \in X$). Therefore:
(8)
\begin{align} \quad \limsup_{n \to \infty} \| x^n\|^{1/n} \leq r(x) \quad (*) \end{align}
  • But we also have that:
(9)
\begin{align} \quad r(x) = \inf \{ \| x^n \|^{1/n} : n \in \mathbb{N} \} \leq \liminf_{n \to \infty} \| x^n \|^{1/n} \quad (**) \end{align}
  • Thus combining $(*)$ and $(**)$ we see that $\displaystyle{\lim_{n \to \infty} \| x^n \|^{1/n}}$ exists and:
(10)
\begin{align} \quad r(x) = \lim_{n \to \infty} \| x^n \|^{1/n} \quad \blacksquare \end{align}
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