Timeline for Deduce that a function is zero on interval $[0,M]$
Current License: CC BY-SA 4.0
12 events
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| Apr 21, 2023 at 6:59 | history | edited | Grandes Jorasses | CC BY-SA 4.0 |
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| Oct 30, 2022 at 20:07 | vote | accept | Grandes Jorasses | ||
| Oct 30, 2022 at 19:48 | answer | added | Saúl RM | timeline score: 1 | |
| Oct 30, 2022 at 19:34 | comment | added | Grandes Jorasses | It is just based on the idea that the integral of a nonnegative function over a set of Lebesgue measure greater than zero cannot be zero. The contradiction comes from this fact. (The measure is also absolutely continuous wrt the Lebesgue measure and has positive density). | |
| Oct 30, 2022 at 19:11 | comment | added | Iosif Pinelis | Your sketch of proof is incomprehensible to me overall, as is almost any part of it. | |
| Oct 30, 2022 at 19:08 | comment | added | Grandes Jorasses | Sketch of proof: suppose $g$ Is not the zero function, is continuous and has finitely many fluctuations above and below zeros. Consider the biggest neighbour of $0 \in B$ such that $g > 0$ (could be also negative, doesn’t change reasoning). If $g$ is zero around 0, then take the neighbourhood starting from the point where $g$ is nonzero. Such a neighbourhood exists as $g$ is continuous and nonzero. Then $B$ will have the form $B=[0, y)$. Then $\int_{0}^{y} g(z) d \mu_y (z) > 0$ which is not possible. Thus $g$ must be zero up to $y$. Repeat this procedure for the finitely many oscillations. | |
| Oct 30, 2022 at 18:56 | history | edited | Grandes Jorasses | CC BY-SA 4.0 |
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| Oct 30, 2022 at 18:21 | comment | added | Grandes Jorasses | @SaúlRM I agree with you. My fault. When I requested bounded variation I was really trying to avoid cases like the topologist’s sine, but I forgot about the existence of the case that you mention. Indeed the function you propose is of bounded variation but has infinitely many fluctuations. My idea of proof works only if we exclude that case. I will share it later. How is it called the class of functions that cannot have infinitely many fluctuations? | |
| Oct 30, 2022 at 17:39 | comment | added | Iosif Pinelis | I think it could be helpful if you share your proof, even assuming that $g$ has only finitely many fluctuations from above to below zero. | |
| Oct 30, 2022 at 15:33 | comment | added | Saúl RM | A function of bounded variation can fluctuate infinitely many times from above to below $0$, e.g. $f:[0,1]\to\mathbb{R}$, $f(0)=0$ and $f(x)=x^2\sin(1/x)$ for $x>0$. In fact for that same function and fixed $\theta\in(0,1]$ it should not be difficult to find a measure $\mu_\theta$ as you say so that $\int g(z)d\mu_\theta(z)=0$ right? | |
| S Oct 30, 2022 at 15:21 | review | First questions | |||
| Oct 30, 2022 at 20:08 | |||||
| S Oct 30, 2022 at 15:21 | history | asked | Grandes Jorasses | CC BY-SA 4.0 |