Timeline for Convexity of an equivalent norm
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 14, 2023 at 3:16 | comment | added | PPB | I think for element $e_n$ belongs to the set $A_0(x)$, $(e_n)$ will not have any weakly convergent subsequence and hence $A_0(x)$ will not be compact. | |
| Sep 11, 2023 at 12:14 | comment | added | Jochen Wengenroth | I doubt very much that this is true because spheres in Banach or Hilbert spaces are usually not weakly closed. I would try to modify the standard example of the unit vectors $e_n$ in $\ell^2$ which belong to the sphere but converge weakly to $0$. | |
| Sep 11, 2023 at 3:59 | comment | added | PPB | @ Iosif Pinelis, I am sorry for the mistake. D is the convex combination of elements from $(B_{l_2} \cup B)$. | |
| Sep 11, 2023 at 3:57 | history | edited | PPB | CC BY-SA 4.0 |
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| Sep 10, 2023 at 14:29 | comment | added | Iosif Pinelis | What does $(B_{l_2} \cup B)$ mean? The same as $B_{l_2} \cup B$? If so, then $D$ is not convex and hence $\mu_D$ is not a norm. | |
| Sep 9, 2023 at 16:13 | history | edited | Michael Hardy | CC BY-SA 4.0 |
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| Sep 9, 2023 at 10:43 | comment | added | PPB | Thank you for your comment. Sorry that I missed one statement. No doubt reflexivity gives a weakly convergent subsequence. But I need to check whether $(x_n)$ has a weakly convergent subsequence converging to an element in $A_0(x)$ or not. | |
| Sep 9, 2023 at 10:40 | history | edited | PPB | CC BY-SA 4.0 |
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| Sep 9, 2023 at 10:32 | history | edited | PPB | CC BY-SA 4.0 |
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| Sep 9, 2023 at 9:31 | comment | added | Jochen Wengenroth | Since $\|\cdot\|$ and the usual norm of $\ell_2$ are equivalent, the sequence $(x_n)_n$ is bounded and hence has a weakly convergent subsequence by reflexivity of $\ell^2$. | |
| Sep 9, 2023 at 9:24 | history | asked | PPB | CC BY-SA 4.0 |