# Real analysis

1. Nov 2, 2013

### Lee33

1. The problem statement, all variables and given/known data
Show that the sequence of functions $x,x^2, ...$ converges uniformly on $[0,a]$ for any $a\in(0,1)$, but not on $[0,1]$.

2. The attempt at a solution

Is this correct? Should I add more detail? Thanks for your help!

Let $\{f_n\} = \{x^n\}$, and suppose $f^n \to f$. We must show that for $\epsilon>0$, there exists an $N$ such that $d(f,f^n)<\epsilon$ whenever $n>N$ for all $x.$

For $a\in (0,1)$, it is clear to see that $x^n\to 0$ as $n$ approaches infinity. We must then show $|x^n|<\epsilon$ whenever $n$ is greater than some $N$.

On $[0,a]$, $x^n$ attains its max at $x=a$ so $x^n<a^n$. Then note that $a^n$ decreases with increasing $n$, so we choose $N$ such that $a^N<\epsilon$.

$\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.

Last edited: Nov 2, 2013
2. Nov 2, 2013

### LCKurtz

You need more for the case $x\in [0,1]$. You haven't shown that the convergence of $x^n$ to the function$$f(x) = \left\{\begin{array}{rl} 0,& 0\le x<1\\ 1, & x=1 \end{array}\right.$$is not uniform. After all, that is the function to which it converges for each $x$. Why isn't the convergence uniform?

3. Nov 2, 2013

### Lee33

LCKurtz - Can you elaborate?

For $[0,1]$, $\{f_n\}$ does not converge uniformly on $[0,1]$ because at $x=1$, $f^n = (1)^n =1\ne 0$ for all $n$.

Last edited: Nov 2, 2013
4. Nov 2, 2013

### LCKurtz

You know the definition of uniform convergence. Given $\epsilon > 0$ you can find $N$ such that if $n>N$, $\|f_n-f\| < \epsilon$. To show this is false you need to find an $\epsilon$ for which you can't find an $N$ that works. You already know the problem is near $x=1$. Think about that.

Last edited: Nov 2, 2013
5. Nov 2, 2013

### Lee33

Okay, thanks.

6. Nov 2, 2013

### LCKurtz

Please note I edited my post. It's late and the first one needed fixing.

7. Nov 3, 2013

### Lee33

Its okay, thanks for the help!