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Ok, I know the same question has already been asked here, and I am not looking for an answer even though my proof looks kind of the same. But, I need to know whether or not I am on the right track. Also, the choice of $y$ on the other proof doesn't many much sense to me. So, here it goes:

Show that $f(x) = x^2$ is not uniformly continuous on $[0,\infty)$

This is what I did:

Suppose, it is. Then, fix $\epsilon = 1 > 0.$ Let, $x < \delta \in [0, \infty)$ and $y = 2x \in [0,\infty)$. Then, according to the definition,

$$\forall \epsilon > 0, \quad\exists \delta > 0\quad\text{such that} \quad\forall x,y \in [0,\infty),\qquad\mid x-y\mid < \delta\quad \implies\quad\mid f(x) - f(y)\mid < \epsilon$$

If we replace $y = 2x$, then $$\mid x -2x\mid = \mid -x\mid = x < \delta.$$ So, that holds. Now, $$\mid f(x) - f(y)\mid = \mid x^2 - 4x^2\mid = \mid -3x^2\mid = 3x^2 > \epsilon = 1,$$ which is a contradiction depending on the choice of $x$.

Is it correct? Can I do that? Thanks.

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Jellyfish
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    No this proof is not correct. Did you study thoroughly the answer to the question on the other page? What did you fail to get about it? – Did Jun 26 '15 at 20:18
  • Absolute continuity is stronger than uniform continuity. – TomGrubb Jun 26 '15 at 20:18
  • @Jellyfish You can't choose $x$ as big as you want because you need to satisfy $|x|<\delta$ in order to be able to choose $y=2x$. Indeed, you could choose $y=x+\delta/2$. – Surb Jun 26 '15 at 20:18
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    @JackD'Aurizio ??? What's your definition of "absolutely continuous"? The term has a standard definition, and absolutely continuous functions certainly need not be sublinear. Same for uniformly continuous functions - I don't see how absolute continuity has anything to do with the question. Maybe you were thinking of $Lip_1$ functions? – David C. Ullrich Jun 26 '15 at 20:19
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    @JackD'Aurizio Do you think this can help an OP asking for a proof that $x\mapsto x^2$ is not uniformly continuous? – Did Jun 26 '15 at 20:19
  • @Did, I didn't understand why did he choose $y = x+ \delta /2$. And, there is no clarification on that. I mean I can't just randomly come up with such choices. – Jellyfish Jun 26 '15 at 20:25
  • They had to pick some $y$ such that $|y-x|<\delta$, obviously $y=x+\delta/2$ does the job. Note that any other choice of $y$ as $y=x+c$ with $|c|<\delta$ would work as well. Does this allow you to go to the next problem you have? – Did Jun 26 '15 at 20:27
  • @Did, Oh, I get it. $x, y$ should depend on delta in order for it to work, right? If that's true, can I choose $x$ to be, say $\delta /2$ and $y$ to be $\delta/3$ and hope for it to work? – Jellyfish Jun 26 '15 at 20:31
  • Ok, I had a little lapse there too. Absolutely continuous functions are in fact bounded, hence sublinear. Not that this has any relevance, and not that it really makes sense to say an absolutely continuous function is sublinear... – David C. Ullrich Jun 26 '15 at 20:32
  • You might - but uniform continuity requires that x can be anywhere. And do not forget that you try to disprove uniform continuity... – Did Jun 26 '15 at 20:33
  • @Did, $x$ can be anywhere within its domain, right? – Jellyfish Jun 26 '15 at 20:35
  • Yeah, and the domain is [0,oo). Note that f would be uniformly continuous on [0,42], say... – Did Jun 26 '15 at 20:37
  • @Jack No I am all right, thank you. And yes, it was better to make this answer disappear... – Did Jun 26 '15 at 20:41

2 Answers2

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Your proof is wrong but the idea of contradiction is good. The problem in setting $y=2x$, is that the condition $|x-y|<\delta$ forces you to choose $|x|<\delta$ and so you can't guarantee a contradiction with $3x^2 >\epsilon$ ($\delta$ might be extremely small).

Suppose by contradiction that for $\epsilon =1$ there exists $\delta >0$ such that $|f(x)-f(y)|<\epsilon$ for every $|x-y|<\delta$.

Note that $\delta$ here is given and you don't know what it is, it can be anything very small or very big (but usually very small).

You want to find $x,y>0$ such that $|x-y|<\delta$ and $|f(x)-f(y)|>\epsilon$, this would imply a contradiction.

Your idea to choose $y$ so that $x$ comes out when you evaluate $|f(x)-f(y)|$ (for getting the contradiction) is good. Since you need $|x-y|<\delta$ you have to choose $y\in ]x-\delta,x+\delta[$.

For example, we can take $y=x+\delta/2$ to get $$|f(x)-f(y)|=|x^2-(x+\delta/2)^2| =|x\delta +\delta^2/4|=x\delta + \delta^2/4\qquad \qquad \forall x\geq 0$$

Finally, choose $x$ big enough to get the contradiction. That is, we want $x\geq 0$ such that $$|f(x)-f(y)|=x\delta+\delta^2/4\geq \epsilon = 1 \implies x\geq \dfrac{1-\delta^2/4}{\delta}.$$ It follows that any $x$ such that $x\geq \dfrac{1-\delta^2/4}{\delta}$ will lead to a contradiction.

You can try to find such bound for $y=x+\alpha$ with $|\alpha|<\delta$. This is a good exercise to check if you really understood the proof.

Surb
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  • you can choose $y=x+\alpha$ for any $0<\alpha<\delta$ if you prefer, e.g. $y=x+\delta/123798$. The important thing is to be sure that $|x-y|<\delta$. – Surb Jun 26 '15 at 20:51
  • @Jellyfish I edited my answer. – Surb Jun 26 '15 at 21:16
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    Thank you so much for your detail answer. – Jellyfish Jun 26 '15 at 21:38
  • @Jellyfish you're welcome :). – Surb Jun 26 '15 at 22:02
  • Sorry, few more question. Are we always going to choose such $y$ like you did for whenever we prove a function that is not uniformly continuous. And, we are only choosing $y$, not $x$, right? And, why are we always choosing $y$? Is $x$ fixed in here? Sorry for asking so many questions. – Jellyfish Jun 26 '15 at 22:29
  • For any $|\alpha|<\delta$ It doesn't matter if you choose $y=x +\alpha$ or $x=y+\alpha$. Again what is important is to choose them so that $|x-y|<\delta$. This assumption is part of the definition of uniform continuity so yes it generalizes easily but this is not always the easiest way to proceed. – Surb Jun 26 '15 at 22:39
  • I am sorry, one more question. Even though now I understand the concept of a function that is not uniformly continuous, I am still failing to understand how did you get $x \geq \frac{1+\delta^2}{2\delta}$. I would really appreciate if you could elaborate on that. Thanks again. – Jellyfish Jun 29 '15 at 19:42
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    @Jellyfish I edited my answer. Is it ok now? (don't be sorry, it's a pleasure to answer someone who is really interested in understanding the answer :) ) – Surb Jun 29 '15 at 21:00
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    Sorry! one more thing. I think there is a computational mistake where you computed $\mid x^2 - (x+\delta/2)^2\mid$ I think it should be $\mid x^2 - (x^2 + 2x\delta/2 + \delta^2 / 4)\mid = \mid -x\delta-\delta^2/4\mid = x\delta + \delta^2/4$. But, then again, it's more about understanding the idea that counts. Thank you so much. – Jellyfish Jul 05 '15 at 00:43
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    @Jellyfish good job :), thanks for pointing out this miscalculation. I edited in consequence. Should be correct now :). Thank you for your feedback. – Surb Jul 05 '15 at 13:59
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No, your proof method is invalid. Uniform continuity means, effectively, that if I give you any value $\epsilon > 0$, no matter how small, there is some number $\delta$ such that the neighborhood of size $\delta$ around every point $x$ is mapped to a neighborhood of size $\epsilon$ or smaller around $f(x)$. Importantly, this $\delta$ has to be the same for all $x$. What you've shown, on the other hand, is that the neighborhoods of size $\delta$ around the specific point $x = \delta$ (in other words, the intervals $(0, 2 \delta)$) will be mapped to a neighborhood around $f(\delta)$ whose size is greater than 1 if we choose $\delta$ to be sufficiently large. These are not the same proposition.

The easiest way to actually prove that a function is not uniformly continuous is to show that $\delta$ can't exist. In other words, you need to show that for a given $\epsilon$ and $\delta$, you can always find an $x$ value such that the $\delta$-neighborhood around $x$ will be mapped to a neighborhood of $f(x)$ whose size is greater than $\epsilon$.

Michael Seifert
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