Uniform Convergence

1. Oct 5, 2005

philosophking

I haven't done uniform convergence since last year when I took analysis, and now I have this problem for topology (we're studying metric spaces right now) and I can't remember how to disprove uniform convergence:

f_n: [0,1] -> R , f_n(x)=x^n

Show the sequence f_n(x) converges for all x in [0,1] but that the sequence does not converge uniformily.

I think I have a pretty good concept of uniform convergence, that is, a sequence of functions is uniformily convergent on some domain if the closeness of the function values (in the range) does not depend on the closeness of the values (in thedomain). To disprove uniform convergence, should I show that there exists some epsilon such that |f_n(x) - f_n(y)| >/= epsilon? But I don't know how I would do that.

Thanks for the help!

2. Oct 5, 2005

fourier jr

um... what happens @ x=1?

3. Oct 6, 2005

philosophking

You get the sequence 1,1,1,1,... which converges? I guess I don't know where you're going with that.

4. Oct 6, 2005

Galileo

If the sequence converges uniformly, the limit function is continuous. What's the limit function in this case?

5. Oct 6, 2005

philosophking

I'm sorry, I'm not sure what you mean by "limit function". Although I think I now know where fourier was going.

The definition of uniform convergence does not depend on x. That means I can take any two x in the domain and their images must necessarily converge to less than epsilon. But say in this example I take x=0 and y=1. Then I can find a counterexample epsilon (.5) such that 1^n - 0^n is not less than or equal to 1/2 as n approaches infinity. I think I can now say that hence, the sequence is not uniform convergence.

EDIT: Also, just as a side-question, if the domain were (0,1) instead of [0,1], this would imply that f_n is uniformly convergent on the domain, right?

6. Oct 6, 2005

fourier jr

yes, that's right :)

7. Oct 6, 2005

fourier jr

oh yeah, & the "limit function" would be f(x) where:

$$f(x) = \lim_{n\rightarrow\infty} f_{n}(x)$$

8. Feb 12, 2011

carnegie

The open interval at 1 creates a problem, even though it does not include 1. Every N you choose will only create uniform convergence for a certain closed interval [0,b] where b<1. Once you get to the right of this b the distance to zero will be greater than epsilon.