For a function $f:X\to Y$ in which $X$ and $Y$ are metric spaces i say that $f$ is locally uniformly continuous if $\forall x\in X, \exists I_x$ a nbd of $x$ s.t.$f|I_x$ (the restriction) is uniformly continuous. I was wandering for a continuous function that isn't locally uniformly continuous, but i searched in Mathstackexchange but i found only this, the problem is that this function has a problem only in one point i would see a continuous function s.t. no point of his domain has a nbd in which the function is uniformly continuous (we can call this a continuous strongly not locally uniformly continuous function). Obviously, by Cantor's theorem the domain cannot have point with compact nbds.
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Do you want $f$ to be continuous? – zhw. Apr 03 '18 at 15:52
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omg, i forgot the fundamental condition, surely i want $f$ to be contunuous, otherwise is sufficient the Dirichlet's function as cuonterexample, now i'll change the question, thanks. – G. Ottaviano Apr 03 '18 at 16:03
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@zhw Continuity and local continuity of $f$ are equivalent. – DanielWainfleet Apr 04 '18 at 15:31
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@DanielWainfleet I was commenting on the question as originally stated. – zhw. Apr 04 '18 at 15:55
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@zhw.... I thought that was the case. I let my comment stand anyway to see whether it was. – DanielWainfleet Apr 04 '18 at 16:41
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On $\mathbb R,$ let $f(x)= 0, x<0, f(x) = 1, x\ge 0.$ Let $q_1,q_2, \dots$ be the rationals. For $x\in \mathbb R,$ define
$$g(x) = \sum_{n=1}^{\infty}\frac{f(x-q_n)}{2^n}.$$
The sum converges uniformly on $\mathbb R,$ and the $n$th summand is continuous everywhere except at $q_n,$ where it jumps by $1/2^n.$ Thus $g$ is continuous at each irrational.
Let $X$ be the set of irrationals, $Y=\mathbb R.$ We then see $g|_X: X\to Y$ is continuous. Let $x_0\in X$ and let $r>0.$ Claim: $g$ is not uniformly continuous on $(x_0-r,x_0+r)\cap X.$ Proof: $(x_0-r,x_0+r)$ contains a rational $q_m.$ I think you can see what is going to happen: At $q_m,$ $g$ jumps by $1/2^m.$ That will remain true when we restrict $g$ to $X.$ Thus there will be irrationals $x_1,x_2,$ arbitrarily close to $x_0,$ such that $g(x_2)-g(x_1) > 1/2^m.$ It follows that $g|_X$ is not uniformly continuous on $(x_0-r,x_0+r)\cap X.$
zhw.
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Does there exist an example with $Y=\Bbb R$ where $X$ is complete, connected and locally connected? (E.g. Hilbert space) – DanielWainfleet Apr 04 '18 at 15:45
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@G.Ottaviano If you think my answer is a good one, please consider giving it a √. – zhw. Apr 12 '18 at 16:05
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@G.Ottaviano Do you see the green mark next to my answer at https://math.stackexchange.com/questions/1780336/how-to-prove-that-if-f-is-continuous-a-e-then-it-is-measurable/1780447#1780447? – zhw. Apr 14 '18 at 16:28
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