# The Image of the Closed Unit Ball Under Finite Rank Operators

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.

• Hilbert Spaces are reflexive, hence you can use this: math.stackexchange.com/questions/270862/… – Theo Bendit Mar 24 '18 at 8:30
• Looks great!. However, in elementary notions, What is the meaning of "weakly compact"? (I.e. without using terminology from weak topology theory) – Or Kedar Mar 24 '18 at 8:42
• 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. – Theo Bendit Mar 24 '18 at 16:47