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Jul 24, 2021 at 16:38 vote accept Nate River
Jul 24, 2021 at 13:20 history edited Alexandre Eremenko CC BY-SA 4.0
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Jul 24, 2021 at 13:14 history undeleted Alexandre Eremenko
Jul 24, 2021 at 0:57 history deleted Alexandre Eremenko via Vote
Jul 24, 2021 at 0:56 history edited Alexandre Eremenko CC BY-SA 4.0
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Jul 23, 2021 at 23:34 comment added Nate River Oh what I meant was that if $g’$ was allowed here, we could emulate that construction with $g$ replaced by $g’$ and the proof would hold verbatim. But because we have only $g$ here, and the “regularity gap” is between $C^1$ + Lipschitz and $C^2$, the same proof won’t work.
Jul 23, 2021 at 23:31 comment added Leo Moos @NateRiver I'm not sure I follow - the functional in the linked answer doesn't seem to depend on the derivative.
Jul 23, 2021 at 23:14 comment added Nate River @Leo Moos - this would apply if there was a $g’$ in the argument for f, but the case where it depends only on $t, g$ is much more subtle.
Jul 23, 2021 at 20:42 comment added Leo Moos Could you clarify what you intend for $f$ - there seems to be a typo? Also, the last claim seems very strong, could you provide some details? It seems a bit counterintuitive given that a $C^1$ can be approximated by $C^2$ functions. (Personally I thought that the answer given to a related question ought to apply here too.)
Jul 23, 2021 at 13:26 history edited Alexandre Eremenko CC BY-SA 4.0
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Jul 23, 2021 at 13:18 history answered Alexandre Eremenko CC BY-SA 4.0