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I am currently listening to some functional analysis lecture and solved the following exercise:

Let E be a Banach space, $T,S:E\rightarrow E$ bounded linear Operators, such that $\exists k \in \Bbb N:S^k=T $.

If: $\sup_{n \in \Bbb N} \Vert T^n \Vert \in \Bbb R $ and $T $ is mean ergodic

$ \implies \sup_{n \in \Bbb N} \Vert S^n \Vert \in \Bbb R $ and $S $ is mean ergodic

PS: T is defined to be mean ergodic if and only if $\forall x \in E: \lim_{N\to \infty} \frac 1N\sum_{i=0}^{N-1}(T^i(x)) \in E$ (converges in the Norm of E)

My question is: Does 'the other direction' hold too? Is this an equivalence?

It seems very likely to me, but i could not proof the mean ergocity. For reflexive spaces its true since $T$ is power bounded, but in the general case i have no idea.

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  • $\begingroup$ It would be useful to to add the definition of a mean ergodic transformation. $\endgroup$
    – User
    Commented Jun 28 at 20:52
  • $\begingroup$ The statement is wrong, I found a counterexample: $\endgroup$
    – Simon Colt
    Commented Jun 29 at 6:29
  • $\begingroup$ Let M be the multiplikation Operator on c (convergent sequences of complex Numbers) defined by $M(x) = a \cdot x$ while $a = (\frac {1-n}n)_{n \in \Bbb N}$. If interested one can check by himself that M suffices the required conditions, while $ M^2 $ is not mean ergodic. $\endgroup$
    – Simon Colt
    Commented Jun 29 at 6:36

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