Uniformly continuous #3

**Also sprach Zarathustra** · January 3rd 2010, 10:11 AM

Prove that if f : (a,b)-->R is is uniformly continuous function in (a,b) so f is bounded function.

**Jose27** · January 3rd 2010, 01:28 PM

Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $f(x_n)<f(x_{n+1})$ 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?.

**Plato** · January 3rd 2010, 01:56 PM

Originally Posted by Jose27

Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $x_n<x_{n+1}$ 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.

**Also sprach Zarathustra** · January 3rd 2010, 01:58 PM

hmmm... I don't understand...
What is Cauchy subsequence ?
"Since f is uniformly cont. it sends Cauchy seq..." why is that?

**Abu-Khalil** · January 3rd 2010, 02:55 PM

Originally Posted by Jose27

Assume it's not bounded then pick a (unbounded) sequence $(x_n) \subset (a,b)$ such that $f(x_n)<f(x_{n+1})$ 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.

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 $i,j\geq K$ .

**Jose27** · January 3rd 2010, 02:56 PM

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.

**Also sprach Zarathustra** · January 3rd 2010, 03:29 PM

Dear Jose! Can you please write me why the existence of such K ==> boundedness of f?

**Jose27** · January 3rd 2010, 03:43 PM

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.

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