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Given a finite rank (bounded) operator $F:H\to H $ when H is a separable Hilbert space, is it true that the image of the closed unit is compact? I know it is pre-compact, so the question is whether it's closed.

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  • $\begingroup$ Hilbert Spaces are reflexive, hence you can use this: math.stackexchange.com/questions/270862/… $\endgroup$ – Theo Bendit Mar 24 '18 at 8:30
  • $\begingroup$ Looks great!. However, in elementary notions, What is the meaning of "weakly compact"? (I.e. without using terminology from weak topology theory) $\endgroup$ – Or Kedar Mar 24 '18 at 8:42
  • $\begingroup$ It's hard to describe in elementary terms. In the context of Hilbert Spaces, weakly compact sets have the property that, given a sequence $(x_n)$ in the set, there exists a subsequence $(x_{n_k})$ such that $x_{n_k}$ converges weakly within the set. That is, for some $x$ in the set, we have $\langle x_{n_k}, u \rangle \to \langle x, u \rangle$ for all $u$ in the space. Closed, bounded convex subsets of a Hilbert space have this property, for example. $\endgroup$ – Theo Bendit Mar 24 '18 at 16:47

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