Proposition 1

Let A be a normed algebra. For every $x\in A$, the sequence $(\|x^n\|^{1/n})_{n\geqslant 1}$ is convergent and its limit $\varrho (x)$ is equal to inf$_{n\geqslant 1}\|x^n\|^{1/n}$. Moreover, for every norm $x\mapsto \|x\|_1$ defining the topology of A, one also has

$\varrho (x) =$ lim$_{n\rightarrow+\infty}\|x^n\|^{1/n}_1=$ inf$_{n\geqslant 1}\|x^n\|^{1/n}_1$.

If $x$ is nilpotent, one has $\|x^n\|^{1/n}_1= 0$ for every sufficiently large integer $n$ and every norm $x\mapsto \|x\|_1$ defining the topology of A.

Suppose now that $x$ is not nilpotent, and put $\alpha_n=$ $\|x^n\|$. One has $\alpha_n>0$ for every integer $n\geqslant 1$, and $\alpha_{n+m}\leqslant \alpha_n\alpha_m$ for all $n, m\in \mathbf{N}$ according to (1). Lemma 2, applied to the sequence $a_n=$ log($\alpha_n$), shows the existence of the limit $\varrho (x)$ and the formula $\varrho (x) =$ inf$_{n>0}\alpha_n^{1/n}$.

Let $x\mapsto \|x\|_1$ be a norm defining the topology of A. There exist real numbers $a >0$ and $b >0$ such that

for all $x\in A$ (EVT, II, p. 7, cor. 2). Consequently,

for all $n\geqslant 1$, whence, by passing to the limit, or by taking the greatest lower bound, the equality

$\varrho (x) =$ lim$_{n\rightarrow+\infty}\|x^n\|^{1/n}_1=$ inf$_{n>0}\|x^n\|^{1/n}_1$.