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Jun 2, 2015 at 19:06 comment added paul garrett @weather, but, still, despite the weak dual not being complete, it is quasi-complete, so all is not lost, etc.
Mar 4, 2015 at 11:30 history edited Peter Michor CC BY-SA 3.0
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Mar 4, 2015 at 11:23 history edited Peter Michor CC BY-SA 3.0
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Mar 4, 2015 at 10:23 comment added weather The above answer contains two errors. The two topologies do not have the same bounded sets in general and the finest locally convex topology which agrees on the unit ball with the weak topology is, by definition, the bounded weak topology which, in the case of a Hilbert space, is complete, something which the weak topology is not.
Mar 4, 2015 at 10:10 history answered Peter Michor CC BY-SA 3.0