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Jun 24, 2020 at 18:01 comment added Nick S To complete @Nik Weaver answer, I think that it is easy to show that there exists some $\epsilon_0$ such that $l$ is the same for all $0 < \epsilon <\epsilon_0$ (i.e. $l$ does not increase as $\epsilon$ decreases) if and only if $f$ is periodic.
Aug 26, 2018 at 15:54 vote accept nanshan
Aug 26, 2018 at 15:51 answer added Nik Weaver timeline score: 7
Aug 26, 2018 at 15:23 history asked nanshan CC BY-SA 4.0