# Uniform Convergence Proof (new question)

1. Jul 31, 2009

### JG89

1. The problem statement, all variables and given/known data

Prove that the sequence $$f_n(x) = x^{\frac{2^n-1}{2^n}}$$ converges uniformly in the interval [0,1].

2. Relevant equations

3. The attempt at a solution

First notice that the f_n do converge to f(x) = x for all x in [0,1]. By definition, if for every positive epsilon the difference $$x^{\frac{2^n - 1}{2^n}} - x$$ (this difference is positive so I omitted the absolute value signs) can be made smaller than epsilon for ALL x in the interval at the same time, provided n is sufficiently large, then we say the f_n converge uniformly to f.

The maximum value for the function $$g(x) = x^{\frac{2^n - 1}{2^n}} - x$$ in the interval [0,1] occurs when $$x = (1 - \frac{1}{2^n})^{2^n}$$.

Let M denote the x value at which g attains its maximum in [0,1]. Then after some algebraic manipulation (it's going to be very messy to show, please assume the algebra is right for this step) we can write
$$g(M) = (1 - \frac{1}{2^n}})^{2^n - 1}(\frac{1}{2^n}) < \frac{1}{2^n}$$.

Now, since g(M) is the maximum of g, we can write $$x^{\frac{2^n - 1}{2^n}} - x < g(M) < \frac{1}{2^n}$$ where the inequality holds for all x in [0,1]. Since the right hand side goes to 0 for increasing n, and is greater than f_n(x) - f(x) for all x in [0,1] at the same time, it follows the sequence converges uniformly.

This is the first time I have proved anything about uniform continuity, so I am not sure if my method was correct. It may be overcomplicated, but does the proof hold?

2. Jul 31, 2009

### snipez90

I just checked your algebra for you, and it's all good. Nice work! By the way, once you show that |g(M)| < 1/(2^n), and that 1/(2^n) -> 0 as n -> 0, you're basically done. Also, finding the maximum of |f_n(x) - f(x)| is a good way of verifying uniform convergence.

3. Jul 31, 2009

### JG89

Thanks for checking it! :)