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> Let $E$ be a Banach space, let $T:E\to E$ have norm $1$ and let $\nu\in E^\*\setminus\{0\}$ be such that $T^\*\nu=\nu$. Under which conditions there is $e\in E$ such that $Te=e$ and $\langle e,\nu\rangle\ne 0$?
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This is true for reflexive spaces. Indeed, WLOG $\|\nu\|=1$. Consider $D=\nu^{-1}(1)\cap \overline{B}\_{E}$, which is weakly compact and non-empty. Then $T\_{D}$ is weak-to-weak continuous self-map of $D$. Therefore, there is a fixed point due to Tychonoff fixed point theorem. **Thus, I am looking for conditions significantly weaker than reflexivity.**


REMARK: In fact, the argument works if there is a linear topology on $E$ such that $T$ and $\nu$ are continuous with respect to that topology and $\overline{B}\_{E}$ is compact in that topology.



Dual fixed point
**Part 1 of the answer.** In terms of $T$, the property you are looking for is characterized by the *mean ergodic theorem*:


**Theorem.** Let $E$ be a Banach space and let $T$ be a bounded linear operator on $E$ that is *power-bounded* in the sense that $\sup\_{n \in \mathbb{N}\_0}\|T^n\| < \infty$. Then the following assertions are equivalent:


(i) $T$ is *mean ergodic*, i.e. the sequence of *Cesàro means* $(\frac{1}{n}\sum\_{k=0}^{n-1} T^k)\_{n \in \mathbb{N}}$ converges strongly.


(ii) The fixed space of the dual operator $T^\*$ is separated by the fixed space of the operator $T$, i.e. for every $0 \not= \nu \in \ker(1 - T^\*)$ there exists $0 \not= e \in \ker(1-T)$ such that $\langle \nu, e \rangle \not= 0$.


(iii) The space $E$ is the topologically direct sum of the fixed space $\ker(1-T)$ and the closure of the range of $1-T$.


This is a classical theorem in operator theory. You can find it, for instance, in Theorem 8.20 of [T. Eisner, B. Farkas, M. Haase, R. Nagel: *Operator Theoretic Aspects of Ergodic Theory* (2015)] (actually, the theorem there contains several more equivalent assertions and is not only stated for power-bounded operators but for the slightly larger class of *Cesàro bounded* operators).


The result for reflexive spaces that is stated in the question is a special case of the above theorem since every power-bounded operator on a reflexive Banach space is mean ergodic [op. cit., Theorem 8.22].


**Part 2 of the answer.** The question on which spaces every power-bounded (or contractive) operator is mean-ergodic is a very subtle one.


* In the paper [V. P. Fonf, M. Lin, P. Wojtaszczyk: Ergodic Characterizations of Reflexivity of Banach Spaces (Journal of Functional Analysis, 2001)] it is shown that a Banach space with a Schauder basis is necessarily reflexive if every power-bounded operator on it is mean ergodic.
* In the paper [V. P. Fonf, M. Lin, P. Wojtaszczyk: A non-reflexive Banach space with all contractions mean ergodic (Israel Journal of Mathematics, 2010)] it is shown that there exist non-reflexive Banach spaces an which every operator $T$ of norm $\|T\| \le 1$ is mean ergodic.


