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Mar 30, 2020 at 0:47 comment added S. Maths Yes, this is obvious, because in locally compact metric spaces "locally Lipschitz" is equivalent to "globally Lipschitz on every compact subset". In your case the space is $[0,T]$ and the function is $F(u)$.
Mar 29, 2020 at 23:30 comment added Math I wanted to say that the same proof applies in the case where$F$ is globally lipschitz.
Mar 29, 2020 at 23:26 comment added Math Sorry, it should be "no matter if $F$ is locally Lipschitz or globally Lipschitz."
Mar 29, 2020 at 23:23 history edited Math CC BY-SA 4.0
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Mar 29, 2020 at 23:00 comment added S. Maths What do you mean by "no matter if $F$ is locally Lipschitz or locally Lipschitz."
Mar 29, 2020 at 20:17 vote accept Math
Mar 29, 2020 at 20:17
Mar 29, 2020 at 18:49 history answered Math CC BY-SA 4.0