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Hello I am trying to understand why the sequence of functions $f_n=nx(1-x^2)^n$ does not converge uniformly to 0 on the interval $[0,1]$ but does on the interval $[a,1]$ where $a\in (0,1)$.

I know that it does not converge on $[0,1]$ as $\int_0^1 nx(1-x^2)^n=1/2$ which is not equal to $\int_0^1 0=0$.

However if I then consider $\mbox{lim}_{n\rightarrow\infty}\sup|f_n(x)|$ I am confused as to why this does not go to 0 on $[0,1]$ but does on $[a,1]$

(unless in the first case i can choose my $x$ to be something like $\frac{1}{n}$ but in the second case I can't-im a bit confused at this bit though)

Thanks for any help-(I hope my post is clear enough, sorry if it is not I am new here)

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Find the maximum of your functions in $[a,1]$. –  Mariano Suárez-Alvarez Oct 26 '11 at 19:25
    
You might try to use that $(1-\frac{1}{n})^n$ converges to $e^{-1}$ as $n\rightarrow \infty$. Therefore, if $x=\frac{1}{\sqrt{n}}$, $f(x)$ is close to $\sqrt{n} e^{-1}$. –  Thomas Andrews Oct 26 '11 at 19:31
    
Yeah Im confused how to find the max of my functions on [a,1]. –  hmmmm Oct 26 '11 at 19:38
    
Hint: For a differentiable function, $g$, the maximum value of $g(x)$ on an interval $[u,v]$ is either at a point where $g'(x)=0$ or it is at $x=u$ or $x=v$. Now, note for $a>0$ and $n$ large enough, $f_n'(x)$ is not zero for $x\in[a,1]$. –  Thomas Andrews Oct 26 '11 at 19:43
    
Ok cool so the max is at x=a and so $lim_{n\rightarrow\infty}|an(1-a^2)^n|\rightarrow 0$ so it is uniformly convergent. Thanks very much –  hmmmm Oct 26 '11 at 19:53
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up vote 1 down vote accepted

$\int_0^1 nx(1-x^2)^n=1/2$ looks incorrect and might be $\int_0^1 nx(1-x^2)^n = \frac{n}{2(n+1)}$ or $\int_0^1 (n+1)x(1-x^2)^{n} = \frac{1}{2}$ or $\int_0^1 nx(1-x^2)^{n-1} = \frac{1}{2}$ (probably the last of these looking at your comments). But that does not matter much here.

On your original problem, if you find the value of $x$ which gives the maximum of $f_n(x) = nx(1-x^2)^n$ and then substitute that back in to find $\max(f_n(x))$ you should see that this function has a peak close to $x=0$ which gets higher as you increase $n$. This is why there is not uniform convergence.

For large enough $n$, $f_n(x)$ can be as close as you want to $0$ for $x \in [a,1]$ for fixed $a$ with $0 \lt a \lt 1$, because the high values of $f_n(x)$ will all occur when $x \in (0,a)$.

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