Endomorphism/K/Powers/Boundedness/Fact/Proof

Proof

(1) implies (2). Let ${}v\in V$. We can work with an arbitrary norm on ${}V$ and on the endomorphism space; for example, with the maximum norm. Because of

${}\Vert {\varphi ^{n}(v)}\Vert \leq \Vert {\varphi ^{n}}\Vert _{\rm {max}}\cdot \Vert {v}\Vert \,,$

${}\varphi ^{n}(v)$ is bounded. From (2) to (3) is clear. If (3) is fulfilled, and

${}v=a_{1}v_{1}+\cdots +a_{m}v_{m}\,$

is a linear combination, then

${}\varphi ^{n}{\left(a_{1}v_{1}+\cdots +a_{m}v_{m}\right)}=a_{1}\varphi ^{n}(v_{1})+\cdots +a_{m}\varphi ^{n}(v_{m})\,.$

The boundedness of the sequences ${}\varphi ^{n}(v_{j})$ implies the boundedness of this sum sequence. The equivalence between (4) and (5) is clear, because over ${}\mathbb {C}$ the Jordan normal form exists, and the eigenvalues and their multiplicities can be read off from the Jordan blocks. From (2) to (5). We may assume ${}{\mathbb {K} }=\mathbb {C}$. Let

${}J={\begin{pmatrix}\lambda &1&0&\cdots &0\\0&\lambda &1&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\\vdots &\cdots &0&\lambda &1\\0&\cdots &\cdots &0&\lambda \end{pmatrix}}\,$

be a Jordan block of the Jordan normal form. In case

${}\vert {\lambda }\vert >1\,,$

for a corresponding eigenvector ${}v$ we obtain

${}\Vert {\varphi ^{n}(v)}\Vert =\vert {\lambda }\vert ^{n}\Vert {v}\Vert \,,$

directly contradicting boundedness. So let

${}\vert {\lambda }\vert =1\,,$

and assume that the length of the Jordan block is at least two. Due to exercise, we have

${}{\begin{pmatrix}\lambda &1&0&\cdots &0\\0&\lambda &1&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\\vdots &\cdots &0&\lambda &1\\0&\cdots &\cdots &0&\lambda \end{pmatrix}}^{n}{\begin{pmatrix}1\\1\\0\\\vdots \\0\end{pmatrix}}={\begin{pmatrix}\lambda ^{n}+n\lambda ^{n-1}\\\lambda ^{n}\\0\\\vdots \\0\end{pmatrix}}\,.$

Here, the first component is

${}\lambda ^{n}+n\lambda ^{n-1}=\lambda ^{n-1}(\lambda +n)\,,$

which is not bounded contradicting the condition.

To conclude from (5) to (1), we may consider each Jordan block separately, because, according to exercise, stability is compatible with a direct sum decomposition. For the first type, the statement follows from fact. For the type ${}\lambda$ with ${}\lambda =1$, the statement is clear, because the norms of the powers equal ${}1$.

To fact