Compact Operator on banach espace Hi I just dont know if this proposition is true, I think it is but I dont know how to start:
Let $X$ be a Banach space of infinite dimension, if $T \in B(x)$ and there is an $N$ such that $T^N=I$ then T is not compact.
I clearly know that $T^N$ is not compact but does that implie that $T$ isnt?
thx for your help
 A: Compact operators form an ideal and hence are stable under composition. To see this, apply $T$ to any bounded set and hence the image will be relatively compact. Rinse and repeat. Alternatively, if $T$ is compact, then for any sequence $x_n$, $\ \ Tx_n$ has a Cauchy subsequence, say $Tx_{n_k}$, and since $Tx_{n_k}$ is a sequence, $T(Tx_{n_k})$ has another Cauchy subseqence, so repeating this shows that $T^N$ has a Cauchy subsequence. 
A: Compact eoperators map bounded sets to precompact ones. A precompact set is strictly within it's compact completion, and a compact set in a metric space is also bounded (eg, cover your set with all possible balls of radius R and then take a finite subcover). 
Therefore for a bounded set $S$, $TS$ is bounded, and then so is $TTS, TTTS$, etc. Finally $T$ maps the bounded set $T^{k-1}S$ to a precompact one.
A: It does. The compacts are an ideal, so if $T$ is compact, then so is $T^k$ for any $k\in\mathbb N$. So, if any power of $T$ is not compact, then $T$ cannot be compact.
