I'm trying to solve this problem, which comes from the book mentioned in the title.
Suppose $A$ is a Banach algebra, $m$ is an integer, $m\geq2$, $K<\infty$, and $$\|x\|^m \leq K\|x^m\| $$ for every $x\in A$. Show that there exist constants $K_n<\infty$, for $n=1,2,3,\ldots$, such that $$\|x\|^n\leq K_n\|x^n\|, \quad\forall x\in A. $$
I tried to work separating in three different cases. Suppose $x\neq 0$, for the case $x=0$ is always true.
$\underline{n=1}:$ This case is trivial, with $K_1=1$.
$\underline{2\leq n\leq m}:$ Write $m=n+p$, for $0\leq p < n$. Then, for all $x\in A$, we have $$\|x\|^n = \|x\|^{m-p} = \frac{\|x\|^m}{\|x\|^p} \leq K\frac{\|x^m\|}{\|x\|^p} = K\frac{\|x^{n+p}\|}{\|x\|^p} \leq K\frac{\|x^{n+p}\|}{\|x^p\|} = K\bigg\|x^n \frac{x^p}{\|x^p\|}\bigg\| \leq$$ $$\leq K\|x^n\|\bigg\|\frac{x^p}{\|x^p\|}\bigg\| = K\|x^n\|. $$ Therefore, we can make $K_n=K$ for all $2\leq n\leq m$.
$\underline{n>m}$: This is the case which is giving me trouble. I tried writing
1) $n=m+p$, for $0<p<n$
2) $n=qm+r$, for $1\leq q<n$ and $0\leq r < m$.
But none of this two approaches seems to work.
I appreciate a help to work this last case. Also, if there is some mistakes in my previous arguments, please point me out. Thank you a lot.
Well, $$||x||^{m^2}=(||x||^m)^m\le(K||x^m||)^m=K^m||x^m||^m\le K^mK||(x^m)^m||=K^{m+1}||x^{m^2}||.$$Now your argument for $n\le m$ also works for $n\le m^2$...