Timeline for Functional equation $\int_z^{2z} [f(x)-f(z)] dx = 0$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2019 at 22:09 | vote | accept | TomH | ||
| Jan 4, 2019 at 15:42 | answer | added | Robert Israel | timeline score: 5 | |
| Jan 4, 2019 at 15:39 | history | edited | TomH | CC BY-SA 4.0 |
Changed the interval to $[0,1]$
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| Jan 4, 2019 at 15:38 | comment | added | TomH | Again, good points! In my application $a=0$ and $0<b<\infty$. But it would be enough to have it for $[0,1]$ interval. I'll update the text once more. | |
| Jan 4, 2019 at 15:20 | comment | added | David E Speyer | With the update, the statement is no longer true: If $a<b<2a$ then the condition is vacuously true, but $f$ needn't be constant. And I suspect it is also false if $[a,b] \cap [2a,2b]$ is small but nonempty. The most natural formulation of the problem seems to be for functions on $[0, \infty)$, but is this what is relevant to your application? | |
| Jan 4, 2019 at 15:15 | review | Close votes | |||
| Jan 10, 2019 at 3:05 | |||||
| Jan 4, 2019 at 15:00 | comment | added | TomH | Good point, thanks! I updated the question slightly. | |
| Jan 4, 2019 at 15:00 | history | edited | TomH | CC BY-SA 4.0 |
Added quantifier for 2z.
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| Jan 4, 2019 at 13:53 | comment | added | user101142 | Do you want to truncate your domain of integration to $[z,2z] \cap [a,b]$? | |
| Jan 4, 2019 at 13:30 | history | asked | TomH | CC BY-SA 4.0 |