1. ## Compactness

Consider the next set.

$\displaystyle \Lambda = \left\{ {f \in C\left( {\left[ {0,1} \right],\mathbb{R}} \right);\left| {f\left( x \right) - f\left( y \right)} \right| \leqslant \pi \left| {x - y} \right| \wedge \left\| f \right\|_\infty \leqslant 1} \right\}$

I have been ask to prove that $\displaystyle \Lambda in \left( {C\left( {\left[ {0,1} \right],\mathbb{R}} \right),\left\| . \right\|_\infty } \right)\$ is compact.

I really dont know how to do it. I have no idea where to start from. Please help.

Regards.

2. Use Arzela-Ascoli: The pointwise boundedness follows from the bound on the supremum norm, and the equicontinuity from the uniform Lipschitz condition.

3. The problem is that in my course the Arzela Th. havent been taught....

4. Here's an outline of an easy proof in this case:

1. Ennumerate the rationals in [0,1], use a diagonal argument to prove that there is a subsequence (of your function sequence) that converges at every rational.

2. Prove that this limit function (with the rationals as domain) satisfies the same Lipschitz condition.

3. Prove that there is an extension of this function to [0,1] and, finally, that it satisfies the conditions.

5. Originally Posted by Jose27
Here's an outline of an easy proof in this case:

1. Ennumerate the rationals in [0,1], use a diagonal argument to prove that there is a subsequence (of your function sequence) that converges at every rational.

2. Prove that this limit function (with the rationals as domain) satisfies the same Lipschitz condition.

3. Prove that there is an extension of this function to [0,1] and, finally, that it satisfies the conditions.
Ok, I`ll try that.

Just for the record, could you do the exercise using the Arzela Th ?

Thank you!
Felipe

6. Well, my first post gives you how to use the theorem, you would just have to check that the limit function is in your set, but that is easy to see (take pointwise limits).

7. Why the equicontinuity follows from the uniform Lipschitz condition.¿¿??

8. $\displaystyle |f(x)-f(y)| \leq \pi |x-y| <\varepsilon$ whenever $\displaystyle |x-y|< \delta = \varepsilon / \pi$

9. oK, Thanks.

But.... the Arzela theorema saids also that the set must be close.... How do I demostrate that?

10. Originally Posted by orbit
oK, Thanks.

But.... the Arzela theorema saids also that the set must be close.... How do I demostrate that?
You need only notice that the norm map $\displaystyle n$ ($\displaystyle n(x)=\|x\\_\infty$) is (of course) continuous and the map $\displaystyle \varphi_{x,y}:C[0,1]\to\mathbb{R}:f\mapsto |f(x)-f(y)|$ is continuous since the absolute value map and the evaluation functionals $\displaystyle e_z:C[0,1]\to\mathbb{R}:f\mapsto f(z)$ are continous (and differences, are continous etc.). But, we have that $\displaystyle \displaystyle \Lambda=\bigcap_{x,y\in[0,1]}\varphi_{x,y}^{-1}\left([0,\pi|x-y|]\right)\cap n^{-1}([0,1])$

11. Originally Posted by Jose27
Here's an outline of an easy proof in this case:

1. Ennumerate the rationals in [0,1], use a diagonal argument to prove that there is a subsequence (of your function sequence) that converges at every rational.

2. Prove that this limit function (with the rationals as domain) satisfies the same Lipschitz condition.

3. Prove that there is an extension of this function to [0,1] and, finally, that it satisfies the conditions.
Im sorry but I don´t get how to prove that there is a subsequence (of your function sequence) that converges at every rational.

12. Originally Posted by orbit
Im sorry but I don´t get how to prove that there is a subsequence (of your function sequence) that converges at every rational.
A possibly easier way than proving it full out might be to just prove it's totally bounded and complete. You get complete for free since it's a closed (as I showed) subspace of a complete space and the fact that it's totally bounded is not too bad.

13. Originally Posted by Drexel28
A possibly easier way than proving it full out might be to just prove it's totally bounded and complete. You get complete for free since it's a closed (as I showed) subspace of a complete space and the fact that it's totally bounded is not too bad.
Would you show me how to prove is T.B??

14. Originally Posted by orbit
Would you show me how to prove is T.B??
Will you try first?

15. Hi. Believe me, I tried to do this exercise. But I could not, that´s why I posted it. Im not lazy or anything like that.It´s just Im taking analysis for the first time and it´s been a very hard course for me. That´s all.
Regards