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I'm trying to understand a step in the proof that the Neumann series converges:

Let $X$ be a Banach space and $T\in L(X)$ (the space of bounded, i.e. continuous linear maps $X\to X$). It is known that $L(X)$ is then also a Banach space with norm $\|\cdot\|$. Furthermore, $T$ satisfies

$\lim\operatorname{sup}_{m\to \infty} \|T^m\|^{\frac{1}{m}} < 1$.

Now, the author chooses a $\theta<1$ and an index $m$ such that $\|T^n\|\le \theta^n$ for all $n\ge m$. But how is this possible? We only know that a subsequence of $\|T^k\|^{\frac{1}{k}}$ converges to a value $<$ 1. Why must $\|T^n\|$ be smaller or equal than $\theta^n$ for all $n\ge m$ then?

Thank you!

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It's the $\limsup$. If the value of the limsup is $\alpha$ and if $\alpha<\theta (<1)$, then there is an $m$ as claimed (if infinitely many terms of the sequence exceed $\theta$, there would be a subsequence converging to a number exceeding $\alpha$; but this can't happen as $\alpha$ is the $\limsup$). – David Mitra Feb 7 '13 at 12:56
@DavidMitra: But why does the limit exist? There are sequences that have subsequences with a limit but do not have a limit themselves... – Sh4pe Feb 7 '13 at 12:59
?? Nothing was said in your post about a limit existing. – David Mitra Feb 7 '13 at 13:03
@DavidMitra: My comment before was a misunderstanding. Now I get what you wrote - I guess my understanding of the limit superior is not that good :/ Thank you! – Sh4pe Feb 7 '13 at 13:07
up vote 2 down vote accepted

Let $\alpha=\limsup\limits_{n\rightarrow\infty} \Vert T^n\Vert$.

You know quite a bit more than just that there is a subsequence of $(\Vert T^n\Vert)$ that converges to $\alpha$. From the definition of $\limsup$, you can deduce that $\alpha$ is the largest number that is a subsequential limit point of the sequence $(\Vert T^n\Vert)$.

So, if $\theta>\alpha$, it would follow that $\bigl|\,\{ n : \Vert T^n\Vert>\theta\}\,\bigr|<\infty$ (that is only finitely many terms of $(\Vert T^n\Vert)$ exceed $\theta$). Otherwise, you could find a subsequence that converges to a (possibly infinite) value $c\ge\theta>\alpha$.

Thus, if $\theta>\alpha$, there is an $m$ such that for all $n\ge m$, we have $\Vert T^n\Vert\le\theta$ (of course, this then implies that $\Vert T^n\Vert^{1/n}\le\theta^{1/n}$ for all $n\ge m$).

Note we need $\theta>\alpha$, here. The statement is not true for arbitrary $\theta$. The author chose $\alpha<\theta<1$.

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