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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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Jellyfish
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1 Answers1

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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.

Ian
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  • @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). – Ian Jun 27 '15 at 17:32
  • @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 $x<\delta,y=x/2$. – Ian Jun 27 '15 at 17:35
  • @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. – Ian Jun 27 '15 at 17:45
  • Sorry I am still editing my comment. – Jellyfish Jun 27 '15 at 19:44
  • @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 $\delta$). – Ian Jun 27 '15 at 19:46
  • for the video linked, if I choose $x = \delta / 2$ and $y = \delta/4$ just like you've suggested, we'll get $\mid x - y\mid = \mid \delta /2 - \delta / 4\mid = \mid \delta / 4\mid < \delta$. And, $\mid f(x) - f(y)\mid = \mid 1/(\delta/2) - 1/(\delta / 4)\mid = \mid 2/\delta - 4/\delta\mid = 2 /\delta$. And, it is only going to work if we choose $0<\delta<1$. But, if we choose $\delta > 1$, $\mid f(x) - f(y) \mid$ won't be $\geq1$. Two questions: 1. Does $\delta$ have to be in the same domain as for $x$ and $y$? 2. If we change the domain to $[0, \infty)$, would the same choice of x,y work? – Jellyfish Jun 27 '15 at 19:50
  • @Jellyfish Technically, as I put in my answer, the domain restrictions only apply to $x,y$; $\delta$ should technically be allowed to be anything in $(0,\infty)$. But if you can handle all $\delta \in (0,1)$ then you can handle all $\delta \in (0,\infty)$ by simply choosing the same $x,y$ that you used for $\delta=1/2$ when you deal with $\delta \geq 1$. Again this is intuitively because continuity is local. – Ian Jun 27 '15 at 19:53
  • Ok, thanks. I should've read your comment before make mine. Thanks again. – Jellyfish Jun 27 '15 at 19:54