Timeline for Does $f(t) \le \int_0^t (t-s)^{-\frac{1}{2}} [f(s) + |f(s)|^{\beta}] \, \mathrm d s$ imply $f=0$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 27 at 22:54 | comment | added | zeraoulia rafik | @TerryTao , Can we utilize the Banach fixed-point theorem, to approach that problem ? | |
| Mar 27 at 18:07 | comment | added | Terry Tao | One can compare with the initial value problem $f'(t) = f(t)^\beta, f(0)=0$ (or equivalently $f(t) = \int_0^t f(s)^\beta\ ds$), which famously has a non-trivial solution $f(t) = C t^{\frac{1}{1-\beta}}$ where $C = (1-\beta)^{1/(1-\beta)}$ when $0 < \beta < 1$. | |
| Mar 27 at 8:18 | comment | added | Akira | I am sad that it is not true but thank you so much for your answer! | |
| Mar 27 at 8:07 | vote | accept | Akira | ||
| Mar 27 at 3:22 | history | edited | Terry Tao | CC BY-SA 4.0 |
added 71 characters in body
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| Mar 27 at 1:37 | history | answered | Terry Tao | CC BY-SA 4.0 |