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

Consider the function $f(x)=x^2$. Fix a point $P \in \mathbb{R}, P>0$ and let $\epsilon >0$. In order to guarantee that $|f(x)-f(P)|<\epsilon$, we must have (for x>0) $|x^2-P^2|<\epsilon$, i.e., $|x-P|<\frac{\epsilon}{x+P}$.

From here, I start not to understand: Since x will range over a neighborhood of P, we see that the required $\delta$ in the definition of continuity cannot be larger than $\epsilon/2P$. In fact the choice $|x-P|<\delta=\epsilon/(2P+1)$ will do the job.

Can anyone tell me where does $\epsilon/2P$ from? And why $|x-P|<\delta=\epsilon/(2P+1)$ will do the job?

enter image description here

Mariana
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    You are asked to prove that $x^2$ is NOT uniformly continuous on $[0,\infty)$. Then, given any $0<\delta=1/n$ there exists a number $\varepsilon =2$, such that for $x=n$ and $y=n+1/n$ we have $|x-y|=\delta =\frac1n$ and $|x^2-y^2|=2+\frac1{n^2}>2=\varepsilon$ – Mark Viola Jun 16 '21 at 00:04
  • Can you tell me why $|x-P|<\delta=\epsilon/(2P+1)$ will do the job? – Mariana Jun 16 '21 at 00:06
  • Will do what job? – Joe Jun 16 '21 at 02:18
  • @Joe to show $f(x)=x^2$ is not uniformly continuous. – Mariana Jun 16 '21 at 02:25
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    This text seems horribly written and confusing. – Ted Shifrin Jun 16 '21 at 03:56
  • @TedShifrin I agree! It really confuses me!!! – Mariana Jun 16 '21 at 03:58
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    Yes, I was confused why you were saying that some particular $\delta$ "will do the job", because to me, showing that some function is *not* uniformly continuous means showing that $\exists \epsilon>0,\forall\delta>0,\exists x,y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|>\epsilon$ – Joe Jun 16 '21 at 11:15

2 Answers2

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If $x - P = \frac{\epsilon}{2P}$, then $x^2 - P^2 = \epsilon + \frac{\epsilon^2}{4P^2} > \epsilon$. So the neighborhood $(x-\delta, x+\delta)$ should not contain $P+\frac{\epsilon}{2P}$.

angryavian
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  • Thanks. Can you tell me why $|x-P|<\delta=\epsilon/(2P+1)$ will do the job? – Mariana Jun 15 '21 at 23:52
  • @Mariana I had assumed you were quoting part of someone else's proof. Without further context, I don't know what "job" the choice of $\delta = \epsilon / (2P+1)$ will achieve, since the question is asking you to prove that $f$ is not uniformly continuous. Mark Viola's comment sketches a proof for this. – angryavian Jun 16 '21 at 02:48
  • Yes, this is from my analysis textbook. I add the picture. – Mariana Jun 16 '21 at 03:42
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I think the statement is just poorly written. Intuitively, the $\epsilon$/$\delta$ definition of continuity means that everytime you pick an $\epsilon$, you get a $\delta$ such that every point really close of $P$ (at most $\delta$ : $|x - P| < \delta$ ) has its image really close to $f(P)$ (meaning $|f(x)-f(P)|< \epsilon$), the continuity of $f$ forces the images of nearby $x$ to be really close to the image of $P$.

Back to your statement, it just says that for a given $\epsilon$, if your $x$ and $P$ are farther away than $\epsilon/2P$ then the distance between $f(x)$ and $f(P)$ will be greater than $\epsilon$: The points farther than $\epsilon/2P$ are not close enough to one anothers to make the images at most $\epsilon$ close.

You can check the maths, like @angryavian did. Pick $\epsilon > 0$,if we have $$ |x-P|> \epsilon/2P $$ It means that $$ |x^2-P^2| = |(x-P)(x+P)| = |x-P|(x+P)$$ Then $$|x^2-P^2| = |x-P|(x+P)> \epsilon(x+P)/2P$$ it is quite clear that if $x>P$, $|x^2-P^2| > \epsilon$ the images are further away than we want them (namely at most $\epsilon$). It means $\epsilon/2P$ is too big of a $\delta$ for $\epsilon$.

$|x-P|<\delta=\epsilon/(2P+1)$ does the job because for small $\epsilon$ : $$ |x-P|<\delta=\epsilon/(2P+1) $$ means $$|x^2-P^2|<\epsilon(x+P)/(2P+1)<\epsilon\frac{2P+\epsilon/(2P+1)}{2P+1}$$ We obtain for $\epsilon$ small enough $$|x^2-P^2|<\epsilon$$ meaning the "job" is done

Uniform continuity means there is a $\delta$ that does not depend on $P$. If $f$ is uniformly continuous, the required proximity between $x$ and $P$ can depend only on $\epsilon$ no matter where $P$ is. Intuitively, you can see why it is not true for $x^2$ on $\mathbb{R}$

RiezFrechetKolmogorov
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  • How to get $\epsilon(x+P)/(2P+1)<\epsilon\frac{2P+\epsilon/(2P+1)}{2P+1}$? – Mariana Jun 16 '21 at 13:27
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    This is because $x$ is at most $P + \epsilon/(2P+1)$ because of the way we picked $\delta$. To be more specific $x+P < P + \frac{\epsilon}{2P+1}$ + P – RiezFrechetKolmogorov Jun 16 '21 at 17:46
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    Now I understand. But the last question: how to get $|x^2-P^2|<\epsilon$ in the last step? I only know now we build that $|x^2-P^2|<\epsilon\frac{2P+\epsilon/(2P+1)}{2P+1}$ – Mariana Jun 16 '21 at 19:40
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    The final trick is that $\epsilon$ can be really small and if it is small enough $\epsilon/(2P+1)$ is smaller than $1$. Then $2P+\epsilon/(2P+1)<2P+1$ so that $\frac{2P+\epsilon/(2P+1)}{2P+1}<1$ meaning the right side part becomes is $\epsilon\frac{2P+\epsilon/(2P+1)}{2P+1} < \epsilon \times 1$. – RiezFrechetKolmogorov Jun 16 '21 at 21:23