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Dec 13, 2021 at 12:22 comment added pietro siorpaes ah ah! You are totally right, I got confused between the intersection of $V$ with the unit ball in $L^p$, and the unit ball in $L^0$, which are of course very different things
Dec 12, 2021 at 14:41 comment added Jochen Wengenroth That is not a valid argument. Locally compact means that some neighbourhood of $0$ is compact. But the unit ball of the $L^p$-norm on $V$ need not be a $L^0$-neighbourhood. (Your argument would implay that every compact operator between Banach spaces had finite dimensional range.)
Dec 11, 2021 at 16:25 history answered pietro siorpaes CC BY-SA 4.0