Samy_A
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There must be a typo here, since ##\Psi(\phi)## is antilinear in ##\phi## and ##\phi(\psi)## is linear in ##\phi##.strangerep said:Aargh, sorry! I just realized I made a typo in my post #28. The proposition should have said:
Proposition:
For every weak-* continuous antilinear functional ##\Psi## on ##H^\times##, there is a vector ##\psi\in H## such that $$\Psi(\phi) ~=~ \phi(\psi) ~,~~~~~ \mbox{for all}~ \phi\in H^\times ~.$$I.e., the difference is that ##\phi## can be any element of ##H^\times##, not merely ##H##.
It could be as follows:
Proposition:
For every weak-* continuous antilinear functional ##\Psi## on ##H^\times##, there is a vector ##\psi\in H## such that $$\Psi(\phi) ~=~ \psi(\phi) ~,~~~~~ \mbox{for all}~ \phi\in H^\times ~.$$
Well, not by much.strangerep said:I guess that means your proof must change significantly?
I already established (assuming the proof is correct) that there is a vector ##\psi\in H## such that ##\Psi(\phi) ~=~ \psi(\phi) ~~~~ \mbox{for all}~ \phi\in H~.##
But ##H## is dense in ##H^\times## with the weak-* topology, so since the weak-* continuous functionals ##\Psi, \psi## are equal on a dense subset of ##H^\times##, they must be equal on all of ##H^\times##