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I am not asking for any particular question today rather I need some help to understand the concept of a function that is not uniformly continuous. Ok, so If I understand this correct, to prove a function that is not uniformly continuous, we have to find some $\epsilon>0$ and some $x,y$ within their domain such that these violate the definition of uniform continuity(if I want to get a contradiction), right? But, my problem is, I am always getting stuck when finding such $x$ and $y$. For example, you can see I asked this question here yesterday, and my choices of $x$ and $y$ weren't the correct ones, and the idea I've got from the answers and the comments is that choosing $x$ or $y$ values with $\delta$ in it might do the trick. But, that's not always the case because if you watch this youtube video of a function that is not uniformly continuous, you can see the instructor didn't choose the $y$ value with $\delta$ in it. So, my question is, is there any rule of thumb when working with functions that are not uniformly continuous? Also, if I choose such $x$ and $y$ values with $\delta$, does it mean I am guaranteed to get a answer that can give me a contradiction?

I am new to real analysis, so I need all the help I can get to understand this subject. Also, I don't know if this question fits here, but then again I don't know where to ask such questions. Thanks so much.

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There's more going on than you've said here. It's actually just one more thing, but it's a very important thing.

Uniform continuity says that

$$(\forall \varepsilon > 0)(\exists \delta > 0)(\forall x,y \in D) \: |x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon.$$

When you go to negate this, you get $$(\exists \varepsilon >0)(\forall \delta > 0)(\exists x,y \in D) \: |x-y|<\delta \text{ and } |f(x)-f(y)|\geq \varepsilon.$$

The key thing that you've missed is that there is a universal quantifier over $\delta$. This means that you need to pick a universal $\varepsilon$, such that for any given $\delta$, you can choose $x,y$ (probably depending on $\delta$) such $x,y$ are $\delta$-close but $f(x),f(y)$ are not $\varepsilon$-close.

In thinking about this it may help to realize that if an existential quantifier follows a universal quantifier, then the existentially quantified variables can be treated as functions of the preceding universally quantified variables*. So $x,y$ in the statement above are functions of $\delta$ (maybe trivial ones that don't actually depend on $\delta$, but probably not). On the other hand $\varepsilon$ is not a function of $\delta$, since it comes before that quantifier.

For example, let's prove that $f(x)=x^2$ is not uniformly continuous on $[0,\infty)$. Pick $\varepsilon=1$ and let $\delta>0$. Now let's find two points $x,y$ with $|x-y|<\delta$ but $|x^2-y^2|\geq1$. To do that, notice that $|x^2-y^2|=|x+y||x-y|$. So to have $|x^2-y^2| \geq 1$, it is necessary and sufficient to have $|x+y| \geq \frac{1}{|x-y|}$. Putting the pieces together, you can choose $x=\frac{1}{\delta}$ and $y=x+\delta/2$. Checking our work:

$$\left | \left ( \frac{1}{\delta} + \frac{\delta}{2} \right )^2 - \frac{1}{\delta^2} \right | = \left | 1 + \frac{\delta^2}{4} \right | \geq 1$$

and $|x-y|=\delta/2<\delta$.

* There are some set-theoretic issues with this statement in general, but they do not matter in this context.

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  • $\begingroup$ @Jellyfish That's the thing, it doesn't work like that--you have to introduce the variables left-to-right (except if $f$ has an actual discontinuity). $\endgroup$ – Ian Jun 27 '15 at 17:32
  • $\begingroup$ @Jellyfish In the video you linked the style is a little bit different, but at the end of the day $x$ and $y$ really do depend on $\delta$, albeit not through an explicit formula. It would have been a little bit clearer if they had said $x=\frac{\delta}{2}$ and $y=\frac{\delta}{4}$, instead of $x<\delta,y=x/2$. $\endgroup$ – Ian Jun 27 '15 at 17:35
  • $\begingroup$ @Jellyfish I would recommend working with the negation of the statement as I wrote it, rather than trying to assume the assertion and derive a contradiction. I admit, knowing when to use which of these two approaches is not easy, but here I think the easier answer is the one I've given. $\endgroup$ – Ian Jun 27 '15 at 17:45
  • $\begingroup$ Sorry I am still editing my comment. $\endgroup$ – Jellyfish Jun 27 '15 at 19:44
  • $\begingroup$ @Jellyfish That's OK; you can add an extra case to deal with that. Basically, if $x=\delta/2$ and $y=\delta/4$, then $\frac{1}{y}-\frac{1}{x}=\frac{2}{\delta}>1$ provided $\delta<2$. So to handle arbitrary $\delta$, just choose $x=\min \{ \delta,1 \}$ and $y=x/2$. This sort of problem usually happens, since at the end of the day continuity only really cares about small $\varepsilon$ (while discontinuity only really cares about small $\delta$). $\endgroup$ – Ian Jun 27 '15 at 19:46

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