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In Pedersen's Analysis Now, for example, you learn that a bounded operator on Hilbert space $T: \mathcal{H} \to \mathcal{H}$ is compact if and only if the image $T(B)$ of the unit ball is compact. It is pretty easy to see therefore that any bounded operator which carries the unit sphere to a compact set is compact. Does the converse hold, that is, if $T: \mathcal{H} \to \mathcal{H}$ is compact, is $T(S)$ compact, where $S$ is the unit sphere in $\mathcal{H}$?

With apologies if this is extremely obvious.

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up vote 8 down vote accepted

It may fail to be closed. If T is injective and compact on an infinite dimensional Hilbert space, then $0$ is in the closure of $T(S)$, and not in it.

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Even if it's rather simple, one still needs to put it so nicely! – Theo Buehler May 30 '11 at 5:55

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