Skip to main content
9 events
when toggle format what by license comment
Apr 17 at 21:24 vote accept MathLearner
Apr 17 at 19:21 answer added Iosif Pinelis timeline score: 1
Apr 17 at 17:57 comment added MathLearner @IosifPinelis It should be very clear now.
Apr 17 at 17:56 history edited MathLearner CC BY-SA 4.0
deleted 51 characters in body
Apr 17 at 17:27 comment added Pietro Majer Note that the first assumption is superfluous . Since $f$ is continuous , it is uniformly continuous on $[0,2\pi]\times[0,T]$, so $\epsilon\mapsto f(\cdot,\epsilon)$ is continuous wrto the uniform norm.
Apr 17 at 16:49 comment added Iosif Pinelis I think this should be further clarified. Of course, the matter here is the clear placement of quantifiers $\exists$ and $\forall$.
Apr 17 at 16:09 history edited MathLearner CC BY-SA 4.0
added 416 characters in body
Apr 17 at 15:31 comment added Iosif Pinelis Perhaps, it makes sense to highlight the difference between this question and the previous one.
Apr 17 at 15:26 history asked MathLearner CC BY-SA 4.0