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Show that $f(x) = \frac{2}{x}$ is not uniformly continuous on $(0,1]$

WLOG Suppose, $0< \delta \leq 1.$ Let, $\epsilon = 1$ and $x = \frac{\delta}{2}, y = x + \frac{\delta}{3}, x,y \in (0,1]$

Skipping all the details: $\mid f(x) - f(y)\mid =...= \mid \frac{24}{5\delta}\mid \geq 1 = \epsilon$, since $0<\delta\leq 1$.

I have seen people restricting the $\delta$ like I did here, but I have never done this before and I thought the only condition applies to $\delta$ is $\delta >0$(though restricting the $\delta$ does make things easier). So, since this is one of my HW question, am I allowed to do it like this (just to make sure I don't make any stupid mistake)? Do you think my approach is correct? Please let me know. Thanks.

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4 Answers 4

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You should choose $\varepsilon$ before $\delta$. Also $|x-y|< \delta$ must hold. So, it is better to say:

Assume for a contradiction that $f$ is uniformly continuous on $(0,1]$ and let $\varepsilon =1$. Then one can find $\delta' > 0$ such that the uniform continuity definition hold.

Let $\delta = \min \{ \delta' , 1 \}$

Now let $$x= \frac{\delta}{2} \quad \text{ and } \quad y= \frac{\delta}{3}$$

so that $$|x-y| < \delta$$

Hence,

$$\left| f(x) - f(y)\right| = \left| \frac{4}{\delta} - \frac{6}{\delta} \right| \geq 2 > \varepsilon$$

Contradiction.

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  • $\begingroup$ Thank you! @quid also explained the same thing. $\endgroup$
    – Jellyfish
    Commented Jul 27, 2015 at 18:00
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Let us see what you do. You assume a function is uniformly continuous. You take $\epsilon = 1$. Then you have some $\delta$ such that for all $x,y$ with $|x-y| < \delta$ something should hold. But if this something holds for all $|x-y| < \delta$ then to say it holds for all $|x-y| < \delta'$ for some $\delta' < \delta$ is a weaker assumption.

Thus you can always assume a smaller delta and thus in particular you can assume $\delta$ is less than $1$ if it helps.

You could write:

Assume for a contradiction that $f$ is uniformly continuous, and chose $\epsilon = 1$. There exists a $\delta > 0 $ such that for all $x,y$ with $|x-y|< \delta $ one has $|f(x)-f(y)| < 1$. WLOG we can assume $\delta < 1$. Now let $x= \delta/2$ ...

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  • $\begingroup$ So, what do you think, I am ok if I do this? $\endgroup$
    – Jellyfish
    Commented Jul 27, 2015 at 17:51
  • $\begingroup$ More or less. You should phrase it a bit differently. I will include a specifc proposal now. $\endgroup$
    – quid
    Commented Jul 27, 2015 at 17:53
  • $\begingroup$ Ok, thank you so much. $\endgroup$
    – Jellyfish
    Commented Jul 27, 2015 at 17:54
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I am really unsure that this is right, I think it's wrong but that's what I tried :Consider $x_{n}=\frac{1}{n}$, $y_{n}=\frac{1}{2n}$. $x_{n}-y_{n}=\frac{1}{2n}$. $|f(x_{n})-f(y_{n})|=2n$. Let $\epsilon =1$. Let $\delta>0$. Pick $n$ such that $\frac{1}{n}<\delta $. $|x_{n}-y_{n}|<\frac{1}{n}<\delta$ and $|f(x_{n})-f(y_{n}|>1$

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Use this theorem : A real valued continuous function is uniformly continuous in an interval $(a,b)$ or $(a,b]$ or $[a,b)$ iff it can be extended continuously on $[a,b]$ .

Now $2\over x$ is continuous on $(0,1]$ but not continuous at the point $0$ as $1\over x$ does not take any value at $0$ since it cannot even be defined at $0$. So the continuous extension not being possible it is not uniformly continuous on $(0,1]$.

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