Timeline for Perhaps an application of Hardy's inequality
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jan 26, 2023 at 18:08 | comment | added | Willie Wong | If $y = 0$, then you don't have the lower boundary $y f(y)^2$ that shows up from the fundamental theorem of calculus (it equals $0$). In this case you get the estimate $\int_0^1 (f)^2 \leq 4 \int_0^1 x^{\alpha} (f')^2$, which is same as what you get by just taking the expression in my answer and taking the limit $y\to 0$. | |
| Jan 26, 2023 at 18:07 | history | edited | Willie Wong | CC BY-SA 4.0 |
added 1 character in body
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| 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 16:12 | comment | added | user253963 | I voted his as best answer because he posted it first and I can only select one. But yours I would upvote 10 times if it were possible. Thanks. | |
| Jan 26, 2023 at 16:05 | comment | added | user253963 | I understood. Thanks for the emotions. | |
| Jan 26, 2023 at 16:04 | comment | added | Willie Wong | @user253963: if $C = 6 \lambda^p$, then $C \lambda^{-p} = 6$, so the the answer above proves $\int_{\lambda^{-k}}^1 f^2 \leq C \lambda^{-p} \int_{\lambda^{-k}}^1 x^\alpha (f')^2$. | |
| Jan 26, 2023 at 16:02 | comment | added | Willie Wong | There is a 4 in $4 + 2 \lambda^{-(1-\alpha)k}$. | |
| Jan 26, 2023 at 15:57 | comment | added | user253963 | where does the number 4 appear? Please. | |
| Jan 26, 2023 at 15:55 | comment | added | user253963 | $C = 6\lambda^{-p}$. the sign of $p$ is negative. | |
| Jan 26, 2023 at 15:49 | history | answered | Willie Wong | CC BY-SA 4.0 |