Timeline for Does $\sum_{m=0}^{\infty} \left|c_m g(x)^{2m+1}\right|$ converge absolutely to an integrable function?
Current License: CC BY-SA 4.0
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| S Feb 8, 2022 at 14:03 | history | bounty ended | CommunityBot | ||
| S Feb 8, 2022 at 14:03 | history | notice removed | CommunityBot | ||
| Jan 31, 2022 at 13:04 | history | edited | UNOwen |
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| S Jan 31, 2022 at 13:01 | history | bounty started | UNOwen | ||
| S Jan 31, 2022 at 13:01 | history | notice added | UNOwen | Authoritative reference needed | |
| Jan 29, 2022 at 4:50 | history | edited | UNOwen |
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| Jan 29, 2022 at 1:56 | history | edited | UNOwen | CC BY-SA 4.0 |
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| Jan 28, 2022 at 16:50 | comment | added | UNOwen | I can say that $f$ is ergodic, if that helps. I think boundedness would be fine in certain situations (i.e. specific values of $\kappa$ and $\beta$). The integro-differential equation $\kappa \ddot{f}+\dot{f}=\beta\int_{0}^t J_1(f_t-f_s)e^{s-t}\mathop{ds}$ describes $f$ for real constants (but I'm interested in analysing the integral for a more general $f$). | |
| Jan 28, 2022 at 16:42 | comment | added | Iosif Pinelis | The boundedness of $f$ will certainly be enough. What can you say about your $f$? | |
| Jan 28, 2022 at 16:37 | comment | added | UNOwen | @IosifPinelis What further assumptions on $f$ would be required to make the answer yes? | |
| Jan 28, 2022 at 16:35 | comment | added | Iosif Pinelis | The answer is no without further assumptions on $f$. | |
| Jan 28, 2022 at 16:14 | history | asked | UNOwen | CC BY-SA 4.0 |