Timeline for Generalize upper semicontinuous regularization using Borel Hierachy
Current License: CC BY-SA 3.0
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| S Apr 24, 2017 at 12:56 | history | bounty ended | CommunityBot | ||
| S Apr 24, 2017 at 12:56 | history | notice removed | CommunityBot | ||
| Apr 16, 2017 at 14:07 | history | edited | Idonknow | CC BY-SA 3.0 |
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| Apr 16, 2017 at 13:27 | history | edited | Idonknow | CC BY-SA 3.0 |
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| Apr 16, 2017 at 13:14 | history | edited | Idonknow | CC BY-SA 3.0 |
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| Apr 16, 2017 at 12:49 | history | edited | Idonknow | CC BY-SA 3.0 |
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| Apr 16, 2017 at 12:14 | history | edited | Idonknow | CC BY-SA 3.0 |
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| Apr 16, 2017 at 11:29 | comment | added | Joel David Hamkins | Ah, I missed that detail. But can't you just take $f_n(x)=f(x)+\frac 1n$? This shifts the whole graph up by a constant, makes it strictly decreasing, and doesn't seem to change the upper semi-continuity class 2 part. | |
| Apr 16, 2017 at 11:21 | comment | added | Idonknow | If we assume that $f_n$ is upper semicontinuous, then the answer is negative, as stated in the link in the post. If $f_n = f$, then $(f_n)$ is not strictly decreasing to $f$. | |
| Apr 16, 2017 at 11:18 | comment | added | Joel David Hamkins | But now I wonder if that really is what you mean. Nothing seems to prevents me taking $f_n=f$. Do you mean that $f_n$ should be upper semicontinuous, or what? | |
| Apr 16, 2017 at 11:06 | history | edited | Idonknow | CC BY-SA 3.0 |
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| Apr 16, 2017 at 11:05 | comment | added | Joel David Hamkins | You haven't stated in the question that you want the $f_n$ also to be upper semi-continuous class 2. | |
| Apr 16, 2017 at 10:59 | comment | added | Liviu Nicolaescu | Can you provide some context? | |
| S Apr 16, 2017 at 10:53 | history | bounty started | Idonknow | ||
| S Apr 16, 2017 at 10:53 | history | notice added | Idonknow | Draw attention | |
| Apr 14, 2017 at 2:38 | history | edited | Idonknow |
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| Apr 14, 2017 at 0:32 | history | asked | Idonknow | CC BY-SA 3.0 |