How to prove $r(T_1 + T_2) \leq r(T_1) + r(T_2)$ when $T_1T_2 = T_2T_1$ for bounded linear operators?

Question

Suppose $T_1$ and $T_2$ are bounded linear operators in a complex banach space and $r(A)$ is the spectral radius of $A$, satisfying $$ r(A) = \inf_{n>0} \|A^n\|^{1/n} = \lim_{n\rightarrow \infty} \|A^n\|^{1/n} = \sup_{x\in \sigma(A)} \|x\|, $$ where $\sigma(A)$ is the spectrum of $A$. If $T_1T_2 = T_2T_1$, we want to prove that $$ r(T_1+T_2) \leq r(T_1) + r(T_2) \tag{*}. $$

I was thinking maybe I can use $$ (T_1+T_2)^n = \sum_{k=1}^n \begin{pmatrix}n \\ k\end{pmatrix}T_1^kT_2^{n-k} $$ to get some useful results. But still I cannot figure out how to prove $(*)$.

Answer 1

$\exists n_0$, s.t. $\|T_1^{n_0}\| \leq (r(T_1)+\epsilon )^{n_0}$ and $\|T_2^{n_0}\| \leq (r(T_2)+\epsilon )^{n_0}$. $$ \begin{split} \|(T_1+T_2)^n\| &\leq \sum_{k=1}^n \begin{pmatrix}n \\ k \end{pmatrix} \|T_1^k T_2^{n-k}\|\\ &\leq \sum_{k=1}^n \begin{pmatrix}n \\ k \end{pmatrix} \|T_1^{n_0}\|^{\lfloor k/n_0 \rfloor} \|T_1\|^{r_k} \|T_2^{n_0}\|^{\lfloor (n-k)/n_0 \rfloor} \|T_2\|^{r_{n-k}} \\ &\leq \frac{\|T_1\|^{n_0}\|T_2\|^{n_0}}{\|T_1^{n_0}\|\|T_2^{n_0}\|}\sum_{k=1}^n \begin{pmatrix}n \\ k \end{pmatrix} \|T_1^{n_0}\|^{k/n_0}\|T_2^{n_0}\|^{(n-k)/n_0} \\ & = Const \cdot (\|T_1^{n_0}\|^{1/{n_0}} + \|T_2^{n_0}\|^{1/{n_0}})^{n} \end{split} $$ Let $n\rightarrow \infty$, we have $$ r(T_1+T_2) \leq r(T_1) + r(T_2) + 2\epsilon. $$

Answer 2

You need a small variation of your idea. First, we may assume without loss of generality that $\|T_1\|\leq1$, $\|T_2\|\leq1$. Now you take $n_0$ such that $\|T_1^{n}\| \leq (r(T_1)+\epsilon )^{n}$ and $\|T_2^n\|\leq(r(T_2)+\epsilon)^n$ for all $n\geq n_0$. Also, since $\|T_1^k\|^{1/k}\to r(T_1)$ and $\|T_2^k\|^{1/k}\to r(T_2)$, there exist $c_1,c_2>0$ with $$ \|T_1^k\|^{1/k}\leq c_1,\ \ \ \ \|T_2^k\|^{1/k}\leq c_2\ \ \ \text{ for all } k. $$

Now, for $n\geq 2n_0$, let $d_j=r(T_j)+\epsilon$ and $$R=1+\max\left\{ \left(\frac{c_1}{d_1}\right)^k+\left(\frac{c_2}{d_2}\right)^{n-k}:\ k=0,1,\ldots,n_0\right\}. $$ The key fact is that $R$ does not depend on $n$. So \begin{align} \|(T_1+T_2)^n\| &\leq \sum_{k=1}^n {n \choose k} \|T_1^k\|\,\| T_2^{n-k}\|\\ \ \\ &\leq \sum_{k=1}^{n_0-1} {n\choose k}c_1^k\,d_2^{n-k}+\sum_{k=n_0}^{n-n_0}{n\choose k}d_1^k\,d_2^{n-k}+\sum_{k=n-n_0+1}^n{n\choose k}d_1^k\,c_2^{n-k}\\ \ \\ &= \sum_{k=1}^{n_0-1} {n\choose k}d_1^k\,d_2^{n-k}\,\left(\frac{c_1}{d_1}\right)^k+\sum_{k=n_0}^{n-n_0}{n\choose k}d_1^k\,d_2^{n-k}+\sum_{k=n-n_0+1}^n{n\choose k}d_1^k\,d_2^{n-k}\,\left(\frac{c_2}{d_2}\right)^{n-k}\\ \ \\ &\leq R\,\sum_{k=0}^n{n\choose k}\,d_1^kd_2^{n-k} =R\,(d_1+d_2)^n. \end{align} Then $$ r(T_1+T_2)=\lim_{n\to\infty}\|(T_1+T_2)^n\|^{1/n}=(d_1+d_2)\,\lim_{n\to\infty}R^{1/n}=d_1+d_2=r(T_1)+r(T_2)+2\epsilon. $$ As $\epsilon$ was arbitrary, we get $$r(T_1+T_2)\leq r(T_1)+r(T_2).$$