> [!definition] > > Let $\alg$ be a unital [[Banach Algebra|Banach algebra]] and $\spec{x}$ be its [[Spectrum|spectrum]], then > $ > \rho(x) = \sup\bracs{\abs{\lambda}: \lambda \in \spec{x}} > $ > is its **spectral radius**. Moreover, > $ > \rho(x) = \limv{n}\norm{x^n}^{1/n} > $ > *Proof*. By binomial shenanigans, > $ > \lambda^ne - x^n = (\lambda e - x)\sum_{j = 0}^{n - 1}\lambda^jx^{n - 1 - j} > $ > In which case, if $(\lambda^n e - x^n)$ is invertible, then so is $(\lambda e - x)$. If $\lambda \in \spec{x}$, then $\lambda^n \in \spec{x^n}$, so $\abs{\lambda}^n \le \norm{x^n}$. Hence $\rho(x) \le \liminf \norm{x^n}^{1/n}$. > > On the other hand, let $\phi \in \alg^*$, then $\phi \circ R$ is analytic for $\abs{\lambda} > \rho(x)$, and > $ > \phi \circ R(\lambda) = \sum_{k = 0}^\infty \lambda^{-k-1}\phi(x^k) > $ > So for any $\abs{\lambda} > \rho(x)$, $\abs{\lambda^{-n-1}\phi(x^n)} \le C_\phi$ for all $n$. Viewing each term as a linear map > $ > S_n: \alg^* \to \complex \quad \phi \mapsto \lambda^{-k-1}\phi(x^k) > $ > then by the [[Uniform Boundedness Principle]], there exists $C \in \real$ such that $\norm{\lambda^{-k}x^k} < C$ for all $k \in \nat$. So $\norm{x^n}^{1/n} \le C^{1/n}\abs{\lambda}$. Therefore $\limsup\norm{x^n}^{1/n} \le \rho(x)$. > [!theorem] > > Let $\alg$ be a unital [[C-Star Algebra|C-star algebra]] and $x \in \alg$ such that $x = x^*$, then $\rho(x) = \norm{x}$. > > *Proof*. $\norm{x^2} = \norm{x}^2$.