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If we have a uniform continuous function $f: X \to Y$, then

  • $f$ takes a cauchy sequence in $X$ to a cauchy sequence in $Y$.

Now is this statement true: $f$ is uniformly continuous iff, given $\epsilon > 0$, there is an $N > 0$, such that for every $x,y \in I$,where $I$ is an interval ($x \neq y$) we have $$ \Biggl| \frac{f(x)-f(y)}{x-y} \Biggr| > N \ \Longrightarrow |f(x)-f(y)|< \epsilon.$$

If yes, how to prove it?

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Can you add the source of the problem, please? – Mariano Suárez-Alvarez Oct 29 '10 at 19:18
@Mariano: One of Junior's asked me this! – anonymous Oct 29 '10 at 19:29
up vote 3 down vote accepted

Well, the standard definition of uniform continuous on an interval $I$ is given by

$$\forall \epsilon > 0, \exists \delta > 0 \text{ such that } \forall x, y \in I, |x-y| < \delta \Rightarrow |f(x)-f(y)| < \epsilon$$

What you want to prove is that this is equivalent to

$$\forall \epsilon > 0, \exists N > 0 \text{ such that } \forall x, y \in I, \frac{|f(x)-f(y)|}{|x-y|} > N \Rightarrow |f(x)-f(y)| < \epsilon$$

Notice that

$$\frac{|f(x)-f(y)|}{|x-y|} > N \iff |x-y| < \frac{|f(x) - f(y)|}{N}$$

Proof (of your question):

$\Rightarrow$ : take $N$ such that $\delta=\frac{ \sup( f(x)) - \inf(f(y))}{N}$. The supremum and infimum are well-defined because of uniform continuity.

$\Leftarrow$ : take $\delta := \frac{\epsilon}{N}$

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Uniform continuity implies Cauchy continuity; not the converse. To see this, in a complete space (like in the reals), Cauchy continuity and continuity are equivalent, but uniform continuity is stronger than continuity.

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