# Prove $f(x)=\frac{3x+2}{2x-1}$ is uniformly continuous on $D_r= (-\infty, 1/2 -r] \cup [1/2 +r, \infty)$

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.

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?

• 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! – user592838 Oct 11 '18 at 3:04
• @Ashley, you are welcome. It is customary to accept the answer that you found most useful. – Ennar Oct 11 '18 at 8:06
• sorry about that, just accepted it! – user592838 Oct 15 '18 at 18:54

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...

• 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 – user592838 Oct 10 '18 at 19:08