Timeline for If $b\in C^1(E, \mathbb{R})$ and $b'$ is compact, then $b$ is weakly continuous — a reference request
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2023 at 21:09 | vote | accept | Alexandru Pirvuceanu | ||
| Jun 14, 2023 at 21:09 | comment | added | Alexandru Pirvuceanu | @IosifPinelis sorry, I have just read your post, I had a very long day. I accepted your answer, it fully solved my problem. I didn't expect this to be so straightforward, but I am really grateful for your answer, now I understand how this is proved. Thank you very much for your help! | |
| Jun 12, 2023 at 22:31 | answer | added | Iosif Pinelis | timeline score: 3 | |
| Jun 12, 2023 at 22:27 | comment | added | paul garrett | Ah! Thanks! That certainly was a possibility, but it was hard to guess for sure. :) Sometimes "prime" has some sort of duality sense... :) | |
| Jun 12, 2023 at 22:26 | comment | added | Alexandru Pirvuceanu | @paulgarrett $b'$ associates with each $x\in E$ the Frechet derivative of $b$ at $x$, which we denote by $b'(x)$. I have just added this in my edit, is the notation clear now? | |
| Jun 12, 2023 at 22:24 | history | edited | Alexandru Pirvuceanu | CC BY-SA 4.0 |
added 78 characters in body
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| Jun 12, 2023 at 22:19 | comment | added | paul garrett | Maybe there's a typo... since even after the edit you've not explained how $b'$ is related to $b$... ? | |
| Jun 12, 2023 at 21:30 | comment | added | Alexandru Pirvuceanu | @IosifPinelis Thank you, now I realise that this notion may be a bit vague. Does my edit clarify this? | |
| Jun 12, 2023 at 21:29 | history | edited | Alexandru Pirvuceanu | CC BY-SA 4.0 |
added 174 characters in body
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| Jun 12, 2023 at 14:16 | comment | added | Iosif Pinelis | What do you mean by "$b'$ is compact"? | |
| Jun 12, 2023 at 13:50 | history | asked | Alexandru Pirvuceanu | CC BY-SA 4.0 |