Timeline for Perhaps an application of Hardy's inequality
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2023 at 16:58 | comment | added | Iosif Pinelis | @user253963 : What do you mean by "an inequality" and "that way"? You should strive for clarity, specificity, and precision in what you say -- otherwise, it is hardly possible to understand what you are saying. Give it a good deal of time to reread and rethink well what you want to ask before actually asking. Also, I don't see any problems caused by $\lambda^{-k}$. | |
| Jan 26, 2023 at 16:44 | comment | added | user253963 | Would you get an inequality that way, but instead of $\lambda^{-k}$ being 0? that is, the integrals would be from 0 to 1. Because I'm noticing that what's causing the problem is $\lambda^{-k}$ as one of the integration factors. | |
| Jan 26, 2023 at 16:26 | comment | added | Iosif Pinelis | @user253963 : I have added an explicit example showing that your desired bound is impossible. | |
| Jan 26, 2023 at 16:24 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 484 characters in body
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| Jan 26, 2023 at 16:11 | vote | accept | user253963 | ||
| Jan 26, 2023 at 16:05 | comment | added | user253963 | I understood. Thanks for the emotions. | |
| Jan 26, 2023 at 15:51 | comment | added | Willie Wong | Then just do as what Iosif said and take the limit as $\lambda \to \infty$. The RHS is bound by $\lambda^{n-p}$ times the $H^1_0$ norm of the function $f$ and so goes to zero, but the LHS doesn't. The kind of result you want is patently impossible. | |
| Jan 26, 2023 at 15:35 | comment | added | user253963 | The constant $C$ can depend on $\lambda$, but I don't want it to make the term $\lambda^{-p}$ disappear. For example, we can have $C= \lambda^{n}$, so $C\lambda^{-p}$ = $\lambda^{n-p}$, but that $n-p < 0$. | |
| Jan 26, 2023 at 15:29 | comment | added | user253963 | No, I want the inequality to be $\lambda^{-p}$. The constant $C$ cannot take this term. If $C=2\lambda^{-p}$, then in the inequality I would miss the term $\lambda^{-p}$. I don't want something like that to happen. | |
| Jan 26, 2023 at 15:16 | comment | added | user253963 | Why $C= 2\lambda^{p}$? | |
| Jan 26, 2023 at 14:27 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |