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Mar 15, 2019 at 14:40 comment added Willie Wong After re-reading your answer four times, I finally figure out what I misunderstood: I kept thinking of the constant function 1/2 as being just like a characteristic function, but of course it is not in the set K since the OP is not allowing changing the coefficient.
Mar 14, 2019 at 15:29 comment added Mateusz Wasilewski @WillieWong No, I really wanted to say weak*, to correct the erroneous claim from the OP that this set is compact in the weak*-topology. The second part of the answer was devoted to the weak topology; it was not really necessary, but I thought that an explicit counterexample in this setting would be illuminating. Maybe I will edit the answer later to make it clearer.
Mar 14, 2019 at 14:55 comment added Willie Wong In the first sentence, do you mean to say that the set is not closed in the weak (and not weak-*) topology?
Mar 14, 2019 at 12:56 history answered Mateusz Wasilewski CC BY-SA 4.0