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 | |
|---|---|---|---|---|---|
| Feb 10, 2021 at 23:37 | history | edited | Martin Väth | CC BY-SA 4.0 |
Replaced boundedness hypothesis by a-priori bound.
|
| Feb 10, 2021 at 23:30 | history | edited | Martin Väth | CC BY-SA 4.0 |
Replaced boundedness hypothesis by a-priori bound.
|
| Feb 10, 2021 at 23:28 | comment | added | Martin Väth | I edited the reply to require an a-priori bound. This is usual technique needed to prove the existence of a solution in such cases. Without any such a-priori bound, I doubt that it is possible to apply Brouwer's fixed point theorem directly for the problem. | |
| Feb 10, 2021 at 23:25 | history | edited | Martin Väth | CC BY-SA 4.0 |
Replaced boundedness hypothesis by a-priori bound.
|
| Feb 10, 2021 at 22:44 | comment | added | Guy Fsone | Note that $\phi_k's\in L^\infty$ and thus $v_k\in L^\infty$ because $v_k\in \mathcal V_k$ that is why we have that $\zeta(v_k)v_k$ is bounded by assumption. | |
| Feb 10, 2021 at 22:42 | comment | added | Martin Väth | If $u\mapsto\zeta(u)u$ is (globally) bounded, then $F$ (and thus $G$) is globally bounded, that is, there is some $R$ such that $\lVert G(v)\rVert\le R$ for every $v$. As mentioned in the previous comment, instead of the boundedness, sublinear growth near $\infty$ is suffiicient. | |
| Feb 10, 2021 at 22:40 | comment | added | Guy Fsone | How to you get that the range of G is in a ball? I tough we have to find $G$ and $R>0$ such that $\|v\|\leq R\implies \|G(v)\|\leq R$? This would solve the problem. | |
| Feb 10, 2021 at 22:38 | comment | added | Martin Väth | I know, but I guess that some assumption was forgotten, because it was used in the question e.g. that $\zeta(v_k)v_k$ is bounded. Moreover, I am quite sure that the assertion is not provable from the finite-dimensional reduction if practically nothing more than the continuity of $u\mapsto z(u)u$ is assumed - local boundedness of the derivative is practically an empty hypothesis concerning Brouwer. It is sufficient that $\frac{\lVert F(u)\rVert}{\lVert u\rVert}\to0$ as $\rVert u\rVert\to\infty$, though. This is a bit weaker than global boundedness. | |
| Feb 10, 2021 at 22:25 | comment | added | Guy Fsone | The assumption says the derivative of $\zeta(t)t$ is locally bounded and not the function itself. But yes this assumption implies the local boundedness $\zeta(t)t$. | |
| Feb 10, 2021 at 22:22 | history | answered | Martin Väth | CC BY-SA 4.0 |