# Prove $1/x$ is not uniformly continuous

$$f: (0,+\infty) \to (0,+\infty)$$ $$f(x) = 1/x$$, 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?

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?

• I'll give it a go - but is there something specifically wrong with mine?
– Warz
Jan 20, 2014 at 19:40
• @Warz See my edit. Jan 20, 2014 at 19:48
• Could we not just set $\epsilon = 1/\delta$ then we would still get $|2/\delta| > 1/\delta = \epsilon$?
– Warz
Jan 20, 2014 at 19:53
• No, you don't get to set $\epsilon$ in terms of $\delta$. Jan 20, 2014 at 19:55
• 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.
– Warz
Jan 20, 2014 at 19:56

Does the function take Cauchy sequences to Cauchy sequences? What does it take the sequence of numbers $x_n = n^{-1}$ to ?

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.