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Prove that $f : A\subset \mathbb{R} \to \mathbb{R}$ is uniformly continuous in $A$ $ \iff$ for all

the sequences $(x_n), (y_n)\in A$ such that $$\lim_{n\to +\infty} ( x_n - y_n )= 0$$

then

$$\lim_{n\to +\infty} ( f(x_n) - f(y_n)) = 0.$$

$$Proof$$

Let's start with the first to demonstrate the implication:

$"{\Rightarrow}"$

in this case the hp are:

$ f $ uniformly continuous and $ (x_n), (y_n) $ sequences of numbers from the set $ A $ such that: $$\lim_{n\to+\infty}(x_n - y_n) = 0.$$

and we must show that: $$ \lim_{n\to +\infty} ( f(x_n) - f(y_n)) = 0.$$ Well, for the uniform continuity we have $$\forall\varepsilon> 0, \exists\delta> 0 \, \: \, \ \forall x, y \in A \ | x-y | <\delta \Rightarrow | f (x)-f (y) | <\varepsilon,$$

and by definition of limit: $$ \forall \varepsilon> 0, \exists \nu> 0 \, \, \, \: \, \, \, \ | (x_n-y_n) - 0 | = | x_n-y_n | <\varepsilon, \quad n> \nu $$

but then the fact that, thanks to the uniform continuity we have, $$ | x-y | <\delta \Rightarrow | f (x)-f (y) | <\varepsilon, $$

we have that $$ | f (x_n) - f (y_n) | <\varepsilon, \forall n> \nu $$

which is equivalent to saying

$$ \lim_ {n \to + \infty} (f (x_n) - f (y_n)) = 0. $$ $ "{\Leftarrow}" $

in this case hp are: $$ (1) \lim_{n\to +\infty} ( f(x_n) - f(y_n)) = 0 $$ $$ (2) \lim_{n\to+\infty}(x_n - y_n) = 0.$$ and we must show that $f$ is uniformly continuous: but if (1) and (2) are verified, $f$ is necessarily uniformly continuous . QED

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1 Answer 1

up vote 0 down vote accepted

Your forward implication is a bit off. In particular, you want $\delta$, not $\epsilon$, in the fourth displayed equation of your proof.

You also need to explicitly show that for any $\epsilon>0$, there is a $N$ (your $\nu$) that "works". You also need to explicitly fix your $\delta$.

You fix a value of $\epsilon$ first, and then state (and prove) that the appropriate $\delta$ and $N$ exist. Then you verify that the choices "work".

So, you could argue as follows:

Let $\epsilon>0$. Then, since $f$ is uniformly continuous on $A$, there is a $\delta>0$ (which we are now specifying) such that $$\tag{1}|x-y|<\delta \Rightarrow |f(x)-f(y)|<\epsilon \quad\text{ for all }x,y \text{ in } A .$$

Now, choose $N$ (as we may) so that $$\tag{2}|x_n-y_n|<\delta\quad\text{ for all }n>N.$$

We have found our $N$. Now we need to show that it actually works:

Then, if $n>N$, we have, by (2), that $|x_n-y_n|<\delta$. Then, from (1), we have $|f(x_n)-f(y_n)|<\epsilon$.

So we have shown that for any $\epsilon>0$, there is an $N$ so that $n>N$ implies $|f(x_n)-f(y_n)|<\epsilon$. Thus $(f(x_n)-f(y_n))\rightarrow 0$, as desired.


For the reverse implication, it seems that you are assuming your result...

Some hints:

Prove the contrapositive statement instead.

To do this, you could do the following:

Suppose that $f$ is not uniformly continuous on $A$. Then there would be an $\epsilon>0$ such that for any positive integer $N$, you can find points $x_n$ and $y_n$ with $|x_n-y_n|<{1\over N}$ but $|f(x_n) -f(y_n)|>\epsilon$.

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