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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.

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1 Answer

Let $T$ be the right shift on $\ell^2$, i. e. $T(x) = (0, x_1, x_2, \ldots)$. Then $T$ is injective, so $\alpha(T) = 0$ and $\dim N(T^0) = 0$. On the other hand, $R(T^n) = \{x \in \ell^2\mid x_k = 0, k \le n\}$, so $\delta(T) = \infty$.

AB, martini.

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Yes, you're right. I forgot to say that $\alpha(T) > 0$ ... We can also suppose if necessary that $0$ is isolated in the spectrum of $T$. –  Ahriman Oct 10 '11 at 8:51
    
Modifying my example a little, I think we can have $\alpha(T) > 0$. Again over $\ell^2$, let $Tx = (x_2, 0, 0, x_3, x_4, \ldots)$, then for $n \ge 2$: $T^nx = (0, \ldots, 0, x_3, x_4, \ldots)$, so $N(T) = \langle e_1\rangle$, $N(T^n) = \langle e_1, e_2\rangle$ for $n \ge 2$. So $\alpha(T) = 2$, $\dim N(T^2) = 2$, $\delta(T) = \infty$. –  martini Oct 10 '11 at 9:29
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