Relationship between spectral rays commuting

Question

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

$$r_\sigma(T) = \lim_{n \to \infty} ||T^n||^{1/n}$$

Answer 1

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)} $

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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.