Operators and power regularity

Question

Let $E$ be an infinite-dimensional complex Hilbert space.

An operator $A\in \mathcal{L}(E)$ is called power regular if for all $x\in E$, $r(x,A):=\lim_{n\rightarrow \infty} \|A^nx\|^{1/n}$ exits.

Assume that for all $x,y\in E$, $r(x,A),r(y,A)$ and $r(x+y,A)$ exists.

I want to show that $$r(x+y,A)\leq \max\{r(x,A),r(y,A)\} .$$

Answer 1

We have

\begin{align} \|A^n(x+y)\|^{1/n} &= \|A^nx+A^ny\|^{1/n} \\ &\le \Big(\|A^nx\|+\|A^ny\|\Big)^{1/n} \\ &\le \Big(\max\{\|A^nx\|, \|A^ny\}\| + \max\{\|A^nx\|, \|A^ny\}\|\Big)^{1/n}\\ &= \Big(2\max\{\|A^nx\|, \|A^ny\}\|\Big)^{1/n}\\ &= 2^{1/n }\max\left\{\|A^nx\|^{1/n}, \|A^ny\|^{1/n}\right\} \end{align}

Letting $n\to\infty$ gives

\begin{align} r(x+y, A) &= \lim_{n\to\infty}\|A^n(x+y)\|^{1/n} \\ &\le \lim_{n\to\infty} 2^{1/n }\max\left\{\|A^nx\|^{1/n}, \|A^ny\|^{1/n}\right\} \\ &= \max\left\{\lim_{n\to\infty}\|A^nx\|^{1/n}, \lim_{n\to\infty}\|A^ny\|^{1/n}\right\}\\ &= \max\{r(x, A), r(y, A)\} \end{align}