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If $A$ is a linear operator on a normed vector space $X$ and $A^k$ is compact for some positive integer $k$, is it possible that $A$ is unbounded?

If not, how to prove that $A$ is bounded?

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2 Answers 2

up vote 6 down vote accepted

Of course, if $k=1$ then $A$ is bounded. Otherwise, it doesn't need to be true. Let $H$ a separable Hilbert space with $\{e_n\}$ a Hilbert basis and define $$Ae_{kj+l}=\begin{cases} je_{kj+l+1}&\mbox{ if }0\leq l<k-1\\\ 0 &\mbox{ if }l=k-1. \end{cases}$$ We have $A^k=0$ so $A^k$ is compact but $A$ is not bounded.

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Thanks! It's a good example. –  henryforever14 Mar 27 '12 at 21:02
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Hmm, I don't quite agree that $A^k=0$. The closure of $A^k$ is $0$. (But maybe the OP meant to assume that the closure of $A^k$ is compact.) –  Hendrik Vogt Mar 28 '12 at 13:52

This is almost the same as Davide's answer:

Let $S:\ell_2\rightarrow\ell_2$ be any unbounded operator. Define $T:\ell_2\oplus_2\ell_2\rightarrow\ell_2\oplus_2\ell_2$ via $T(x,y) = \bigl(0,S(x)\bigr)$. Then $T^k=0$ for any $k>1$ and thus compact, but $T$ is unbounded.

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Thanks! It's a nice example. –  henryforever14 Mar 27 '12 at 21:02

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