## New conterexample on Ritt operators and \mathcal{R}-boundedness

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Loris Arnold
(Université de Franche-Comté)

Let $T$ be a bounded operator on a Banach space $X$. We say that $T$ is a ($\mathcal{R}$-)Ritt operator if it satisfies that the two sequence $(T^n)_{n \geq 0}$ and $(nT^n(T - I))_{n\geq 0}$ are ($\mathcal{R}$-)bounded. It is possible to construct a Ritt operator which is not $\mathcal{R}$-Ritt, but until now, without knowing whether $(T^n)_{n \geq 0}$ or $(nT^n(T - I))_{n\geq 0}$ (or both) is not $\mathcal{R}$-bounded. We will give some preliminaries about the topic and we will construct a Ritt operator such that the sequence $(T^n)_{n \geq 0}$ is not $\mathcal{R}$-bounded.