Relationship between spectral rays commuting Show sing binomial Newton that if $S$ and $T$ are continuous operators on Banach space $B$ such that $ST = TS$ then
$$r_\sigma(T + S) \leq r_\sigma(T) + r_\sigma(S)$$
where $r_\sigma$ is spectral radius of operetors. Is an fact that for every $T$ are continuous operators on Banach space that 
$$r_\sigma(T) = \lim_{n \to \infty} ||T^n||^{1/n}$$
 A: I recently wrote up a solution to this very problem, which I've copied below.  Note that in this context, $\mathfrak A$ is the space of bounded operators on $B$.
I hope you find this helpful.
$
\newcommand{\f}{\mathfrak}
\DeclareMathOperator{\rad}{r}
\newcommand{\eps}{\varepsilon}
\DeclareMathOperator{\spec}{spec}
\newcommand{\lp}{\left(}
\newcommand{\rp}{\right)}
$

Let $A,B \in \f A$.  We which to show that $\rad(A + B) \leq \rad(A) + \rad(B)$.  Let $\eps > 0$ be arbitrary.  Noting that $\|A^n\|^{1/n} \to \rad(A)$, we may select a an $m$ such that for all $p \geq m$, we have 
$$
\|A^p\| \leq (\rad(A) + \eps)^p, \qquad 
\|B^p\| \leq (\rad(B) + \eps)^p
$$
Since $A$ and $B$ commute, the binomial theorem applies. So, we have for $n > 2m$
\begin{align*}
\|(A + B)^n\| & \leq 
\left \|
\sum_{p=0}^m \binom np A^pB^{n-p}
\right\| 
\\ & \quad +
\left \|
\sum_{p={m+1}}^{n-m-1} \binom np A^pB^{n-p}
\right\| + 
\left \|
\sum_{p=0}^m \binom np A^{n-p}B^{p}
\right\|
\\ & \leq 
\sum_{p=0}^m \binom np 
\lp 
    \|A\|^p(\rad(B) + \eps)^{n-p} + 
    (\rad(A) + \eps)^{n-p}\|B\|^p
\rp 
\\ & \quad + \sum_{p=m+1}^{n - m - 1} \binom np
(\rad(A)+\eps)^{p}(\rad(B) + \eps)^{n-p}
\\ & \leq 
    (\rad(B) + \eps)^n \overbrace{n^m}^{\binom np \leq n^p \leq n^m}
    \lp 
        \sum_{p=0}^m  \|A\|^p(\rad(B) + \eps)^{-p}
    \rp
\\ & \quad +
    (\rad(A) + \eps)^{n} n^m
    \lp
        \sum_{p=0}^m (\rad(A) + \eps)^{-p} \|B\|^p
    \rp
\\ & \quad + \sum_{p=0}^{n} \binom np
(\rad(A)+\eps)^{p}(\rad(B) + \eps)^{n-p}
%
%
\\ & = 
n^m (\rad(B) + \eps)^n c(A,B,m) + n^m (\rad(A) + \eps)^n c(B,A,m)
\\ & \quad + (\rad(A) + \rad(B) + 2\eps)^n
\end{align*}
Thus, we have 
\begin{align*}
\lim_{n \to \infty}\frac{\|(A+B)^n\|}{(\rad(A) + \rad(B) + 2 \eps)^n} &\leq 
\lim_{n \to \infty} n^m \lp \frac{\rad(B) + \eps}{\rad(A) + \rad(B) + 2 \eps} \rp^n c(A,B,m)
\\ & \quad +
\lim_{n \to \infty} n^m \lp \frac{\rad(A) + \eps}{\rad(A) + \rad(B) + 2 \eps} \rp^n c(B,A,m)
% \\ & \quad 
+ 1 = 1
\end{align*}
Thus, we may conclude that
$$
\rad(A + B) = \lim_{n \to \infty} \|(A+B)^n\|^{1/n} \leq 
\rad(A) + \rad(B) + 2\eps
$$
Since $\eps$ was arbitrary, conclude that $\rad(A + B) \leq \rad(A) + \rad(B)$ as desired.
