Let $T$ be a bounded linear operator on some complex Banach space. We define its ascent by $\alpha(T) = \min \{ n \ge 0 \, / \, N(T^n) = N(T^{n+1}) \}$ and its descent by $\delta(T) = \min \{ n \ge 0 \, / \, R(T^n) = R(T^{n+1}) \}$ where $N(T)$ and $R(T)$ denote respectively the kernel and the range of $T$.
It's well known that if both $\alpha(T)$ and $\delta(T)$ are finite, then they are equal.
Suppose now we know that $\alpha(T)$ is finite, and $N(T^{\alpha(T)})$ is finite-dimensional. What can we say about $\delta(T)$ ? Is it finite ?
Thank you for your answers.