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$f: (0,+\infty) \to (0,+\infty)$ $f(x) = 1/x$, prove that f is not uniformly continuous.

Firstly, I negated the definition of uniform convergence obtaining:

$\exists \epsilon > 0 $ s.t. $\forall \delta > 0 $ with $|x-y| < \delta$ & $x,y \in (0, + \infty)$ and $|f(x) - f(y)| = \left|\dfrac{x-y}{xy}\right| \geq \epsilon$

so I choose $\epsilon = 1$ and $ x = \delta/2 \in (0,+\infty)$ and $y = \delta /4 \in(0,+\infty)$ so $|x-y| = \delta/2 < \delta$ and $|f(x) - f(y)| = |2/\delta|$ How do I show that this is greater than or equal to epsilon?

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3 Answers 3

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The condition you want is $|2/\delta|\ge \epsilon = 1$, which holds true as long as $|\delta|\le 2$. So you're set for small $\delta$. Can you see a way to manage the case when $\delta>2$ accordingly?

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    $\begingroup$ I'll give it a go - but is there something specifically wrong with mine? $\endgroup$
    – Warz
    Jan 20, 2014 at 19:40
  • $\begingroup$ @Warz See my edit. $\endgroup$ Jan 20, 2014 at 19:48
  • $\begingroup$ Could we not just set $\epsilon = 1/\delta $ then we would still get $ |2/\delta| > 1/\delta = \epsilon$? $\endgroup$
    – Warz
    Jan 20, 2014 at 19:53
  • $\begingroup$ No, you don't get to set $\epsilon$ in terms of $\delta$. $\endgroup$ Jan 20, 2014 at 19:55
  • $\begingroup$ hmm, I don't really see what we could do then - I always thought of delta as small so didn't really consider the other case. $\endgroup$
    – Warz
    Jan 20, 2014 at 19:56
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Does the function take Cauchy sequences to Cauchy sequences? What does it take the sequence of numbers $x_n = n^{-1}$ to ?

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Here a full answer (that i writte too to practice) but take into account that I am just a student so I hope it is correct.

1 - First let recall the definition of a non uniformly continuous function.
It exists at least one $\epsilon_0>0$ such that for every $\delta>0$ that we can choose it will always exists at least $x$ and $y$ that verifies $|x-y|<\delta$ but $|f(x)-f(y)|>\epsilon_0$.
More formally: $\exists \epsilon_0>0 \; , \forall \delta>0 \; : \; \exists |x-y|< \delta \Rightarrow |f(x)-f(y)| \geq \epsilon_0$

2 - Now let pay attention to the following inequalities:
(1): for any $\delta>0$ given it exists $N=Max(1; \left \lceil 1/ \delta \right \rceil)$ s.t. $1/N < \delta$ . Moreover all $n \geq N$ verifies too this inequality (by assumption we are in $(0; \infty)$ ).
(2): $\forall n \in \mathbb{N} $ we have $|\frac{1}{1/n}-\frac{1}{n+1/n}|=|n-\frac{n}{n^2+1}|=|n(1-\frac{1}{n^2+1})|=|\frac{n^3}{n^2+1}| \geq 1/2$

3 - Now we can writte:
$\exists \epsilon_0 = \frac{1}{4}>0$ such that for any $\delta>0$ it will always exists ,with $n \geq N$ as define in (1), at least two points $x_n=n$ and $y_n=n+1/n$ that despite that verifying $|x_n-y_n|=|1/n| < \delta$ (by (1)) $ \Rightarrow|f(x_n)-f(y_n)|=|\frac{n^3}{n^2+1}|>1/4$.
More formally: $\exists \epsilon_0 = \frac{1}{4}>0 \; , \forall \delta>0 \; : \; \exists \; x_n = n, \; y_n=n+\frac{1}{n}$ with $n \geq max(1; \left \lceil 1/ \delta \right \rceil)$
By (1): $|x_n-y_n|< \delta$
By (2): $\Rightarrow |f(x_n)-f(y_n)|=|\frac{n^3}{n^2+1}| \geq \epsilon_0 = \frac{1}{4}$
Q.E.D.

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