# Uniformly continuous #3

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• Jan 3rd 2010, 10:11 AM
Also sprach Zarathustra
Uniformly continuous #3
Prove that if f : (a,b)-->R is is uniformly continuous function in (a,b) so f is bounded function.
• Jan 3rd 2010, 01:28 PM
Jose27
Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $f(x_n) for all $n$ then since this sequence is bounded we get a Cauchy subsequence $(x_{n_k})$ by Bolzano-Weierstrass. Since $f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $f(x_{n_k})$ by construction?.
• Jan 3rd 2010, 01:56 PM
Plato
Quote:

Originally Posted by Jose27
Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $x_n for all $n$ then since this sequence is bounded we get a Cauchy subsequence $(x_{n_k})$ by Bolzano-Weierstrass. Since $f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $f(x_{n_k})$ by construction?.

Jose27, I think you want too 'clean up' that proof.
I know what you mean but you did not do it.
• Jan 3rd 2010, 01:58 PM
Also sprach Zarathustra
hmmm... I don't understand...
What is Cauchy subsequence ?
"Since f is uniformly cont. it sends Cauchy seq..." why is that?
• Jan 3rd 2010, 02:55 PM
Abu-Khalil
Quote:

Originally Posted by Jose27
Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $f(x_n) for all $n$. Since $\left(x_n\right)$ is bounded we get a Cauchy subsequence $(x_{n_k})$ by Bolzano-Weierstrass. Since $f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $f(x_{n_k})$ by construction?.

I believe that now it is easier to understand it.

Quote:

Originally Posted by Also sprach Zarathustra
hmmm... I don't understand...
What is Cauchy subsequence ?
"Since f is uniformly cont. it sends Cauchy seq..." why is that?

If $f$ is unif. cont and $\left(x_{n_k}\right)$ is Cauchy, then $f\left(x_{n_k}\right)$ is Cauchy too.

Let $\epsilon>0$. Since $f$ is unif cont, there exists $\delta>0$ such as $|x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon.$ Since $\left(x_{n_k}\right)$ is Cauchy, there existe $K$ such as $|x_{n_i}-x_{n_j}|<\delta$ for all $i,j\geq K$.
• Jan 3rd 2010, 02:56 PM
Jose27
Quote:

Originally Posted by Plato
Jose27, I think you want too 'clean up' that proof.
I know what you mean but you did not do it.

Yeah, already did. Sorry, it made no sense as it was.
• Jan 3rd 2010, 03:29 PM
Also sprach Zarathustra
Dear Jose! Can you please write me why the existence of such K ==> boundedness of f?
• Jan 3rd 2010, 03:43 PM
Jose27
Quote:

Originally Posted by Also sprach Zarathustra
Dear Jose! Can you please write me why the existence of such K ==> boundedness of f?

So you know a Cauchy sequence is bounded, and Abu-Khalil proved that if $f$ is a unif. cont. function and $x_n$ is a Cauchy seq. then $f(x_n)$ is also a Cauchy seq. but by how we constructed them $f(x_n) \rightarrow \infty$ (or $-\infty$ in which case invert all inequalities in my first post) clearly contradicting the fact that $f$ is unif. cont. so we conclude that our initial hypothesis is false ie. $f$ is bounded.