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I also need to prove that f(x) is not Uniformly continuous on its domain, $D= (-\infty, \frac{1}{2})\cup(\frac{1}{2},\infty)$. I have tried algebreically manipulating |f(x)-f(y)| and obtain $|f(x)-f(y)|=\frac{7|x-y|}{|2x-1||2y-1|}$, but I am not sure how to obtain a delta that doesn't depend on y for $x,y \in D_r$. I am also not sure why it is not uniformly continuous on domain D.

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For the first part, if $x\in D_r$, then either $x\geq 1/2 + r$ or $x\leq 1/2 - r$, that is, either $x-1/2 \geq r$ or $x-1/2\leq -r$. This is equivalent to saying $|x-1/2|\geq r$. Using what you got, we now have

$$|f(x)-f(y)| = \frac{7|x-y|}{|2x-1| |2y-1|} \leq \frac{7|x-y|}{4r^2}.$$ You can now choose $\delta$ that depends only on $\varepsilon$ and $r$.

For the second part, as you said in the comments: it is enough to show that there are sequences $(x_n)$ and $(y_n)$ such that $x_n-y_n\to 0$ but $f(x_n)-f(y_n)\not\to 0$. If you think about it, there is obviously an issue near $1/2$. So, why not choose sequences that converge to $1/2$ one from above and the other from below?

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  • $\begingroup$ This is very clarifying! You explained proving uniform continuity on Dr very clearly. Your last sentence is where I was stuck the most, and I got it instantly after constructing such sequences! Thank you! $\endgroup$ – user592838 Oct 11 '18 at 3:04
  • $\begingroup$ @Ashley, you are welcome. It is customary to accept the answer that you found most useful. $\endgroup$ – Ennar Oct 11 '18 at 8:06
  • $\begingroup$ sorry about that, just accepted it! $\endgroup$ – user592838 Oct 15 '18 at 18:54
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I don't know what tools you are allowed to use. But if you can use differentiability, then you can easily prove the first part.

First, if a function is differentiable, and the derivative is bounded, then it is also Lipschitz (this is rather obvious, follows from Lagrange theorem). Then, you can easily prove that a Lipschitz continuous function is uniformly continuous. Now, look at your function and its derivative in the given set, and conclude.

For the second part, argue by contradiction: assume such uniform $\delta$ exist for a given $\varepsilon$. Can you find two points that are closer than $\delta$ but further apart than $\varepsilon$? The answer is yes, provided you get close enough to 1/2...

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  • $\begingroup$ On my phone so formatting Is off. Unfortunately, I cannot use differentiation. I think I was able to show it’s UC on DR by showinging |2x-1| |2y-1| >= 4r^2. To show it’s not UC in D, I know I need to find sequences such that xn-yn —> 0 but fxn-fyn does not. However, I cannot think of such sequencss $\endgroup$ – user592838 Oct 10 '18 at 19:08

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