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Jan 24, 2020 at 10:54 vote accept Safwane
Jan 13, 2020 at 20:25 review Reopen votes
Jan 14, 2020 at 12:25
Jan 13, 2020 at 19:52 history closed Alexandre Eremenko
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Jan 13, 2020 at 19:45 answer added Arno timeline score: 5
Jan 12, 2020 at 14:20 comment added Andrés E. Caicedo And MR1837868 De Marco, Giuseppe For every $\epsilon$ there continuously exists a $\delta$.Amer. Math. Monthly 108 (2001), no. 5, 443–444.
Jan 12, 2020 at 14:18 comment added Andrés E. Caicedo You may want to see also MR1745893 (2000m:54014) Enayat, Ali $\delta$ as a continuous function of $x$ and $\epsilon$. Amer. Math. Monthly 107 (2000), no. 2, 151–155.
Jan 12, 2020 at 14:15 review Close votes
Jan 13, 2020 at 19:55
Jan 12, 2020 at 10:27 comment added GH from MO Then trivially the answer to your question is no. For any positive valued function, there is a smaller positive valued non-continuous function. See my first remark.
Jan 12, 2020 at 10:25 comment added Safwane @GHfromMO: It can be considered as one of those reals which I assume depends on $x$
Jan 12, 2020 at 10:24 comment added GH from MO No, I meant that for a given $x$, infinitely many positive $\delta$'s work (if $\delta$ is ok than any smaller $\delta$ is also ok), so it is not clear what you mean by $\delta(x)$.
Jan 12, 2020 at 10:23 comment added GH from MO Note also that on any compact interval $I\subset\mathbb{R}$ you can choose $\delta=\delta(I)$ to be constant, since $f'(x)$ is positive and uniformly continuous on $I$.
Jan 12, 2020 at 10:22 comment added Safwane @GHfromMO: Did you mean that we must add some thing about the set of $x$.
Jan 12, 2020 at 10:15 comment added GH from MO You have not defined $\delta$ uniquely as a function of $x$. For example, if $f(x)=x$, then one can take $\delta=1$ for $x\in\mathbb{Q}$ and $\delta=2$ for $x\not\in\mathbb{Q}$, which is not continuous (obviously).
Jan 12, 2020 at 8:56 answer added Fedor Petrov timeline score: 9
Jan 12, 2020 at 8:31 history asked Safwane CC BY-SA 4.0