Timeline for "Essential values" of a function at a point?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Oct 23 at 10:05 | comment | added | David Gao | @LSpice Yeah, I only put $x = 0$ as the OP’s using intervals around $0$. This is equally valid for any point. | |
| Oct 23 at 8:02 | comment | added | LSpice | @NateRiver, re, oh, right, sorry. | |
| Oct 23 at 8:01 | comment | added | Nate River | That would not be independent of $x_0$ I think? In the case where $x_0$ is of interest, then OP’s $f_k$ should be the restriction to the intervals $[x_0 - 1/k, x_0 + 1/k]$. | |
| Oct 23 at 7:59 | comment | added | LSpice | @NateRiver, re, that was my first thought, but, if $\{\phi(f) : \phi\rvert_{C_0(\mathbb R)} = \delta_{x_0}\}$ is independent of $x_0$, then you get the same set if you take all characters $\phi$, omitting the restriction condition, don't you? | |
| Oct 23 at 7:56 | comment | added | Nate River | @LSpice I reckon it is not privileged, any other point admits the same analysis. | |
| Oct 23 at 7:54 | comment | added | LSpice | @DavidGao, re, why does $x = 0$ get privileged in this description? | |
| Oct 23 at 7:53 | history | edited | Nate River |
edited tags
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| Oct 23 at 7:48 | answer | added | Nate River | timeline score: 4 | |
| Oct 22 at 20:30 | history | edited | Sébastien Loisel | CC BY-SA 4.0 |
plural is better
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| Oct 22 at 20:05 | comment | added | Sébastien Loisel | @DavidGao that sounds about right. I'm trying to finish a paper and this is turning up. | |
| Oct 22 at 19:41 | comment | added | David Gao | There is a way of interpreting what $E$ is. Indeed, any character on $L^\infty(\mathbb{R})$ restricts to either the zero functional, or evaluation at some point $x \in \mathbb{R}$, on $C_0(\mathbb{R})$. $E$ is exactly $\{\phi(f)\}$ where $\phi$ ranges over all characters on $L^\infty(\mathbb{R})$ which restrict to evaluation at $x = 0$ on $C_0(\mathbb{R})$. In a sense, this is the “essential” value of $f$ at $0$. I don’t think it has a name or any use though. (Also note that this depends crucially on the topology of $\mathbb{R}$, not just the measure space structure.) | |
| Oct 22 at 19:39 | comment | added | Sébastien Loisel | @DavidGao thanks for that, yes I am aware. | |
| Oct 22 at 19:39 | comment | added | Sébastien Loisel | @BillJohnson oh I think you meant, the spectrum of the operator that is pointwise multiplication by the function, yes I get it. | |
| Oct 22 at 19:37 | comment | added | David Gao | It should be noted that $E$ may not be a singleton. For example, if $f = 1_{[0, 1]}$, then $E$ is $\{0, 1\}$. | |
| Oct 22 at 19:25 | comment | added | Sébastien Loisel | I'm not sure what the Gelfand transform of a function is, could you remind me? | |
| Oct 22 at 19:21 | comment | added | Bill Johnson | The essential range of a bounded measurable function is the range of the Gelfand transform of the function. So it is natural to think of the essential range as a substitute for the range. | |
| Oct 22 at 19:09 | history | edited | LSpice | CC BY-SA 4.0 |
`\operatorname`
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| Oct 22 at 19:02 | history | edited | Sébastien Loisel | CC BY-SA 4.0 |
\mathbb{R}
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| Oct 22 at 19:01 | comment | added | Sébastien Loisel | Thanks for that, yes I think I was thinking of essentially bounded functions. | |
| Oct 22 at 18:59 | comment | added | Aleksei Kulikov | To apply Cantor's theorem you need to assume that $E_k$ is compact (i.e. that $f$ is bounded) unless you want to allow for infinite values. Other than that, it does seem like we can assign to every $x$ its own closed set of potential values. | |
| Oct 22 at 18:54 | history | asked | Sébastien Loisel | CC BY-SA 4.0 |