Let $T\_1$ and $T\_2$ be two bounded linear operators in a complex banach space $X$.



> 
> If $T\_1T\_2 = T\_2T\_1$, I want to know how to show that 
>  $$
> r(T\_1+T\_2) \leq r(T\_1) + r(T\_2),
> $$
>  where $r(A)$ denotes the spectral radius of $A$, and it is given by
>  $$
> r(A) = \inf\_{n>0} \|A^n\|^{1/n} = \lim\_{n\rightarrow \infty} \|A^n\|^{1/n}.
> $$
> 
> 
> 



If $T_1T_2 = T_2T_1$, why $r(T_1 + T_2) \leq r(T_1) + r(T_2)$?
For simplicity, assume that $\|T\| \leqslant 1$ and $\|S\| \leqslant 1$. Write
$$ (T + S)^n = \sum\_{k = 0}^n \tbinom{n}{k} T^k S^{n - k} . $$
Fix $\varepsilon > 0$, and let $m$ be large enough, so that $\|T^j\| \le (r(T) + \varepsilon)^j$ and $\|S^j\| \le (r(S) + \varepsilon)^j$ for $j \ge m$. Then, if $n > 2 m$, we have
$$ \begin{aligned} \|(T + S)^n\| & \leqslant m \tbinom{n}{m} (\|T^{n - m}\| + \|S^{n - m}\|) + \sum\_{k = m}^{n - m} \tbinom{n}{k} (r(T) + \varepsilon)^k (r(S) + \varepsilon)^{n - k} \\
&\leqslant m \tbinom{n}{m} ((r(T) + \varepsilon)^{n - m} + (r(S) + \varepsilon)^{n - m}) + (r(T) + r(S) + 2 \varepsilon)^n . \end{aligned} $$
Since $(m \tbinom{n}{m})^{1/n} \to 1$ as $n \to \infty$, we easily find that
$$
r(T + S) \leqslant \max\{r(T) + \varepsilon, r(S) + \varepsilon, r(T) + r(S) + 2 \varepsilon\} = r(T) + r(S) + 2 \varepsilon .
$$
This gives the desired inequality.


