Timeline for Deduce that a function is zero on interval $[0,M]$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| S Nov 1, 2022 at 8:08 | history | suggested | CommunityBot | CC BY-SA 4.0 |
space between g(z) and dz
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| Nov 1, 2022 at 6:23 | review | Suggested edits | |||
| S Nov 1, 2022 at 8:08 | |||||
| Oct 31, 2022 at 7:51 | comment | added | Grandes Jorasses | Thank you a lot! I will try to work on my own a bit more before asking. The density in $L^1$ might be a good path to try indeed! | |
| Oct 31, 2022 at 1:05 | comment | added | Saúl RM | Whether the $f_\theta$ from your example generate a dense subspace of $L^1$, or equivalently, if the functions $g_\theta(z)=\frac{1_{(0,\theta)}}{\sqrt{\theta-z}}$ generate a dense subspace of $L^1$, looks like it could be true but I'm not sure how to prove it. Maybe it's worth asking as a separate question | |
| Oct 31, 2022 at 0:54 | comment | added | Saúl RM | Btw even if $g$ is integrable respect to Lebesgue measure, it may not be integrable respect to $\mu_\theta$. If $g$ is bounded that's not a problem though. If you want check that if $\int g\cdot f_\theta dz=0$ for all $\theta$, then $g=0$, it would be enough to check that the subspace of $L^1([0,M])$ generated by the functions $f_\theta$ is dense in $L^1$. I don't know if that is true for your previous example though, to prove things like that the only tool I know is the Stone Weierstrass theorem and it doesn't seem to work here | |
| Oct 30, 2022 at 23:24 | comment | added | Grandes Jorasses | I mean, also the case of fixing $g$ and working with the measures is interesting to me! My original problem was the other way around, but it is nice also to think a bit more generally and see the setting from other perspectives. But yeah, if you happen to have some good idea, please just let me know, because at the moment I don't see a way to formalize my intuition (which might be wrong). | |
| Oct 30, 2022 at 23:13 | comment | added | Saúl RM | I see. For some reason I thought we were fixing $g$ and imposing that the integrals respect to any family of $\mu_\theta$ should be $0$. I would have to think about the case when you fix one family $(\mu_\theta)_\theta$ | |
| Oct 30, 2022 at 22:45 | comment | added | Grandes Jorasses | but we should still be able to find $ \theta^{\prime} \: \: : \: \: 0 \neq \int_{0}^{\theta^{\prime}} g(z) f_{\theta^{\prime}}(z) dz$. $\theta^{\prime}$ is contained in the interval where the function $g$ fluctuates infinitely. This is because $g$ must be independent of $\theta$. | |
| Oct 30, 2022 at 22:44 | comment | added | Grandes Jorasses | However, I think that if the family of measures is fixed as this one: $f_{\theta} (z) = \frac{\mathbb{1}_{[0,\theta)}(z)}{2\sqrt{\theta^2 - \theta z}}$ then isn't possible to find $g \neq 0$ independent of $\theta$ s.t. $0 = \int_{0}^{\theta} g(z) d \mu_{\theta} (z) = \int_{0}^{\theta} g(z) \underbrace{f_{\theta}(z)}_{> 0} dz \: \: \: \: \: \: \: \forall \: \theta \in [0,M]$. The reason is that one may try to construct a $g$ that fluctuates infinitely many times as the measures that you proposed, | |
| Oct 30, 2022 at 20:07 | comment | added | Grandes Jorasses | Thank you a lot! | |
| Oct 30, 2022 at 20:07 | vote | accept | Grandes Jorasses | ||
| Oct 30, 2022 at 19:53 | history | edited | Saúl RM | CC BY-SA 4.0 |
added 57 characters in body
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| Oct 30, 2022 at 19:48 | history | answered | Saúl RM | CC BY-SA 4.0 |