$(x_n)$ be sequence such that $x_n \to x$ in $B$. And $r(x_n)=0 \ \forall n$ . Then does it imply $r(x)=0$??

Question

let $B$ is commutative unital Banach algebra. $(x_n)$ be sequence such that $x_n \to x$ in $B$. And $r(x_n)=0 \ \forall n$ . Then does it imply $r(x)=0$?? where $r(x)$ denote spectral radius of $x$ . I think it is true. but little confuse how to prove it. i am thinking of $x_n^m \to x^m$ for all $m \in \mathbb{N}$ and then use $r(x)=\lim _{m\to \infty} ||x^m||^{\frac{1}{m}}$

Is this correct way or there is better way to do this. please suggest.

Thanks in advanced.

Answer 1

Because you are in a commutative Banach algebra, you have $$ \sigma(x)=\{f(x):\ f\in\mathcal C(B)\}, $$ where $\mathcal C(B)$ is the set of characters. For any $f\in\mathcal C(B)$, $$ f(x)=\lim f(x_n)=0, $$ since $f(x_n)=0$ for all $n$ (as $\sigma(x_n)=\{0\}$).