Timeline for Weak sequential continuity vs strong continuity
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Oct 26, 2020 at 22:26 | comment | added | Bill Johnson | It is easy to show that a Banach space is a Shur space (every weakly convergent sequence is norm convergent) if and only if the norm is weakly sequentially continuous. Hint: Show that in every non Shur space there is a sequence of unit vectors that converges weakly to zero. | |
| Oct 26, 2020 at 14:05 | review | Close votes | |||
| Nov 5, 2020 at 18:45 | |||||
| Oct 26, 2020 at 13:50 | comment | added | Motaka | Thank you for pointing out this. | |
| Oct 26, 2020 at 13:47 | history | edited | Motaka | CC BY-SA 4.0 |
added 10 characters in body
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| Oct 26, 2020 at 12:10 | comment | added | Jochen Glueck | I assume that $T$ is allowed to be non-linear again, as in your previous question? If yes, then I suggest to replace "an operator" in your first sentence either with "an (in general, non-linear) operator" or with "a mapping". | |
| Oct 26, 2020 at 11:10 | history | asked | Motaka | CC BY-SA 4.0 |