- #1
AxiomOfChoice
- 533
- 1
Is it easy to show that [itex]T: X \to Y[/itex] is a compact linear operator -- i.e., that the closure of the image under [itex]T[/itex] of every bounded set in [itex]X[/itex] is compact in [itex]Y[/itex] -- if and only if the image of the closed unit ball [itex]\overline B = \{x\in X: \|x\|\leq 1\}[/itex] has compact closure in [itex]Y[/itex]? One direction is (of course) easy, since [itex]\overline B[/itex] is bounded, but I'm having trouble with the other one. Can someone help please? Thanks!