Please help me prove: $v(a+b)\leq v(a)+v(b)$, and $v(ab)\leq v(a)v(b)$ where $v(x)=\inf{\{\vert x^n \vert}^{1/n}: n\in\mathbb{N}\}$ I'm reading a book functional analysis, and reading and have seen an example of somebody please help me if you can. The example that I've seen is the following: 
If $A$ is a normed algebra and $a,b\in A$ so that $ab=ba$, then $$v(a+b)\leq v(a)+v(b),$$ and $$v(ab)\leq v(a)v(b).$$
Remarks: Let $A$ is a normed algebra whith unit. For $x\in A$ number $$v(x)=\inf{\{\vert x^n \vert}^{1/n}: n\in\mathbb{N}\}$$ is spectral radius of element $x$.
Please help me. Thanks for your attention. 
 A: Denote $v_n(a)=|a^n|^{1/n}$ and $w_n(a)=\ln |a^n|$. It is easy to check that $w_{k+l}(a)\leq w_k(a)+w_l(a)$. By Fekete's lemma we get there exist the limit
$$
\lim\limits_{n\to\infty}\frac{w_n(a)}{n}=\inf\limits_{n\in\mathbb{N}}\frac{w_n(a)}{n}
$$
Since $\ln v_n(a)=w_n(a)/n$, then from previous equality we get
$$
v(a)
=\inf_{n\in\mathbb{N}} v_n(a)
=\lim\limits_{n\to\infty} v_n(a)
$$
Note that
$$
v_n(ab)
=|(ab)^n|^{1/n}
=|a^nb^n|^{1/n}
\leq(|a^n||b^n|)^{1/n}
=|a^n|^{1/n}|b^n|^{1/n}
=v_n(a)v_n(b)
$$
then we take limits as $n\to\infty$ and get
$$
v(ab)\leq v(a)v(b)\tag{1}
$$
We will need one more property of $v$ function. For any $\lambda\in\mathbb{C}$ we have
$$
v_n(\lambda a)
=|(\lambda a)^n|^{1/n}
=|\lambda ^n a^n|^{1/n}
=(|\lambda|^n |a^n|)^{1/n}
=|\lambda| |a^n|^{1/n}
=|\lambda|v_n(a)
$$
hence after taking the limits we see that
$$
v(\lambda a)=|\lambda| v(a)\tag{2}
$$
To prove $(2)$ consider arbitrary $s>v(a)$ and $t>v(b)$. Then define $c=s^{-1}a$ and $d=t^{-1}b$. From $(2)$ we know that $v(c)<1$ and $v(d)<1$. From definition of $v$ we derive that there exists $M\in\mathbb{N}$ such that $|c^n|^{1/n}<1$ and $|d^n|^{1/n}<1$ for  all $n>M$. Hence there exist $\alpha>0$ such that $|a^n|<\alpha$, $|b^n|<\alpha$ for all $n\in\mathbb{N}$. Now we have estimation
$$
|(a+b)^n|
=\left|\sum\limits_{k=0}^n{n \choose k}a^k b^{n-k}\right|
\leq\sum\limits_{k=0}^n{n \choose k}|a^k| |b^{n-k}|
=\sum\limits_{k=0}^n{n \choose k}s^k t^{n-k}|c^k| |c^{n-k}|\\
\leq \alpha^2\sum\limits_{k=0}^n{n \choose k}s^k t^{n-k}
=\alpha^2(s+t)^n
$$
Note that binomial formula is valid because $ab=ba$. Therefore
$$
v(a+b)
=\lim\limits_{n\to\infty}|(a+b)^n|^{1/n}
\leq\lim\limits_{n\to\infty}\alpha^{2/n}(s+t)
=s+t
$$
Since $s>v(a)$ and $t>v(b)$ are arbitrary, then
$$
v(a+b)\leq v(a)+v(b)\tag{3}
$$
