Dual fixed point

Question

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 $\langle e,\nu\rangle\ne 0$?

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.

Answer 1

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.