Timeline for How do I apply Brouwer fixed-point theorem in this claim?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 17, 2021 at 7:17 | vote | accept | Guy Fsone | ||
| Feb 18, 2021 at 11:39 | |||||
| Feb 16, 2021 at 18:55 | comment | added | Martin Väth | Oh, stupid me! Thank you! I was already so confused that the positive definitness of the derivative (which I had used in my proof) and the boundedness of the functional from below (which you had used) seemed unrelated to me at a first glance... Of course, the former must imply the latter (and actual convexity of the functional). | |
| Feb 15, 2021 at 22:59 | comment | added | Guy Fsone | If $v\geq 0$ then $G(v)\geq 0$ this is clear. If $v\leq 0$ then $\zeta(t)t\leq 0$ for all $v\leq t\leq 0$ which implies $\int_v^0 \zeta(t)td t\leq 0$ that is $G(v)=\int_0^v \zeta(t)tdt\geq 0$ | |
| Feb 15, 2021 at 20:26 | comment | added | Martin Väth | I haven't calculated carefully, but I think that if you have $G(v)=G(|v|)$ the the derivative for e.g. strictly negative $v$ has the wrong sign. My reason to assume this without calculation is that in case $\zeta(0)\ne0$ the derivative at $0$ does not exist while I think that this should be the case for the "correct" definition of $G$. | |
| Feb 14, 2021 at 21:46 | history | edited | Guy Fsone | CC BY-SA 4.0 |
Complete solution with an alternative solution
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| Feb 14, 2021 at 18:13 | comment | added | Guy Fsone | for the negative $v \leq 0$ you have to reverse the integral \int_v^0 you will get $G\geq0$ | |
| Feb 14, 2021 at 13:17 | comment | added | Martin Väth | Why is $G$ non-negative? $v$ can be negative. | |
| Feb 13, 2021 at 10:10 | history | edited | Guy Fsone | CC BY-SA 4.0 |
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| Feb 12, 2021 at 23:54 | history | edited | Guy Fsone | CC BY-SA 4.0 |
added 54 characters in body
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| Feb 12, 2021 at 22:35 | history | answered | Guy Fsone | CC BY-SA 4.0 |