Timeline for An integral on the interval depending on the integrand
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| May 22, 2023 at 7:15 | comment | added | Fawen90 | Dear fedja, you have proved my desired claim for $c=1/e$. Do you think it also holds for $c=1$? I may totally create an alternative post if needed. Many thanks | |
| Apr 3, 2023 at 20:43 | comment | added | fedja | @Fawen90 1) We'll have equality if $\tau\le 1$. If $1$ is in between, then $\tau\wedge 1$ becomes the $\tau$ I'll use in my argument and then we only have an inequality. The case $\widetilde\tau>1$ is trivial. 2) Yep, it should be $\widetilde F$. | |
| Apr 3, 2023 at 20:30 | vote | accept | Fawen90 | ||
| Apr 3, 2023 at 20:30 | comment | added | Fawen90 | Thanks again for your help. I do appreciate your trick | |
| Apr 3, 2023 at 20:29 | comment | added | Fawen90 | Fantastic reasoning! Indeed you have proved a stronger result for all measurable and bounded functions $f:\mathbb R_+\to\mathbb R_+$. This is exactly what I need. Just one typo, $\widetilde F=\max(F,e^{-2})$ instead of $F=\max(F,e^{-2})$. Also, I think we should have equality $\int_0^\tau F=\int_0^{\widetilde\tau}\widetilde F$. Do you think so? | |
| Apr 3, 2023 at 16:14 | comment | added | fedja | @Fawen90 Yeah, ask as many questions as you want :-) | |
| Apr 3, 2023 at 14:01 | comment | added | Fawen90 | Thank you very much fedja for your answer. I believe that I need to read very carefully your reasoning, and please allow me to come back towards you if it's not clear to me | |
| Apr 3, 2023 at 12:24 | history | answered | fedja | CC BY-SA 4.0 |