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

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- Jan 3rd 2010, 10:11 AMAlso sprach ZarathustraUniformly 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 PMJose27
Assume it's not bounded then pick a (unbounded) sequence $\displaystyle (x_n) \subset (a,b)$ such that $\displaystyle f(x_n)<f(x_{n+1})$ for all $\displaystyle n$ then since this sequence is bounded we get a Cauchy subsequence $\displaystyle (x_{n_k})$ by Bolzano-Weierstrass. Since $\displaystyle f$ is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about $\displaystyle f(x_{n_k})$ by construction?.

- Jan 3rd 2010, 01:56 PMPlato
- Jan 3rd 2010, 01:58 PMAlso 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 PMAbu-Khalil
I believe that now it is easier to understand it.

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

Let $\displaystyle \epsilon>0$. Since $\displaystyle f$ is unif cont, there exists $\displaystyle \delta>0$ such as $\displaystyle |x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon.$ Since $\displaystyle \left(x_{n_k}\right)$ is Cauchy, there existe $\displaystyle K$ such as $\displaystyle |x_{n_i}-x_{n_j}|<\delta$ for all $\displaystyle i,j\geq K$. - Jan 3rd 2010, 02:56 PMJose27
- Jan 3rd 2010, 03:29 PMAlso sprach Zarathustra
Dear Jose! Can you please write me why the existence of such K ==> boundedness of f?

- Jan 3rd 2010, 03:43 PMJose27
So you know a Cauchy sequence is bounded, and

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