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It's well known that if $ f : \bf (X,d) \to \bf (Y,e) $ is a uniformly continuous function then $f$ maps bounded set to bounded set.Does the converse hold ? More Precisely,

Suppose $f : X \to Y$ is a (continuous) bounded map (i.e. $f$ maps bounded set to bounded set).Does this implies that $f$ is uniformly continuous?

Arpit Kansal
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The classic function $f:(0,1) \to [0,1]$ $$f(x) = \sin(1/x)$$ is a counterexample.

Crostul
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There is a well known lemma saying that if a function $f$ defined on $\mathbb{R}$ such that $\int_0^{+\infty}f(x)dx$ converge and $f$ is uniformly continuous then $lim_{x\rightarrow +\infty} f(x)dx = 0$.

Consider the function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $f(x)=\sin(x^2)$. It satisfies your hypothesis and the integral of $f$ converges and its limit at $+\infty$ is not $0$ hence it's not uniformly continuous.

Hicham
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  • Dear @Hicham,Thank you! Perhaps easy way to show that $f$ is not unifomly continuous is to note that $f' \to \infty$ as $x \to \infty $.Could you please give me a reference for the lemma? – Arpit Kansal Dec 23 '15 at 11:24
  • Dear Arpit, you're welcome. The derivative of the function in my example vanishes an infinite number of times near infinity, so it can't goes to infinity. And even if the derivative goes to infinity it is not obvious for me how we can prove the non-uniform continuity. I'll try to fin a counter example to this later. It is in general difficult to prove unirorm continuity on unbounded domains. That why I like this example, because it gives easily the non uniform continuity based on a general lemma. – Hicham Dec 23 '15 at 14:22
  • For a reference check this link http://math.stackexchange.com/questions/92105/f-uniformly-continuous-and-int-a-infty-fx-dx-converges-imply-lim-x – Hicham Dec 23 '15 at 14:23
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    Thank you.Check this:http://math.stackexchange.com/questions/1571906/does-there-exist-a-f-bf-r-to-bf-r-which-is-differentiable-uniformly-contin – Arpit Kansal Dec 23 '15 at 14:25
  • Thanks Arpit for the link. – Hicham Dec 23 '15 at 14:49
  • Sorry, I am quite curious: I have never seen the fact that $\int_0^{\infty} \sin x^2 dx$ is convergent. Could you please give me alink for a proof of this fact? – Crostul Dec 23 '15 at 23:11
  • It is known as Fresnel integral, we know even the value of this integral $$\int_0^{+\infty}\sin(x^2)dx = \sqrt{\frac{\pi}{8}}$$. see the following link https://en.wikipedia.org/wiki/Fresnel_integral. – Hicham Dec 23 '15 at 23:18
  • For a proof of convergence, we can use a variable change and an integration by part to write $$\int_0^{x}\sin(x^2)=\int_0^{x}\frac{\sin u }{2\sqrt u}du =[\frac{1-\cos u}{2\sqrt u}]_0^x+\int_0^{x}\frac{1-\cos u}{4u^{3/2}}du$$ from which you can see that all converge when $x$ goes to infinity – Hicham Dec 23 '15 at 23:30