Skip to main content
3 events
when toggle format what by license comment
Aug 2, 2021 at 14:31 comment added Bazin Is there not a "Schwartz Kernel Theorem" in that context saying that all linear bounded operators must have a (distribution) kernel?
Aug 2, 2021 at 6:05 comment added Jochen Glueck Well, but Douglas' result doesn't say that $C$ needs to be an integral operator, too.
Aug 1, 2021 at 16:49 history answered Bazin CC BY-SA 4.0