1. ## Uniform convergence

Hi!

Problem: Show that $f_{n}(x)=nx^{n}(1-x)$ converges "pointwise", donīt know if this is the correct english word for it. And show that the convergence is not uniform on $[0,1]$ .

Solution:

Itīs easy to find that $f_{n}(x) \to f(x)=0 \; \; , \forall x \in [0,1]$

How do I show that the convergence is not uniform?

We have that $\lim_{n\to \infty} f_{n}(x) = f(x)$ is continous on $[0,1]$ , so we have to proceed right?

By letting $d(x)=|f_{n}(x)-f(x)|$ , we have to show that $d(x) \to 0$ as $n\to \infty$

I tried taking the derivative but didnīt come up with anything good.

Thx!

2. I believe all the functions $f_n$ are positive and attain their maximum at $x=1/2$ (using the derivative). Hence the supremum norm of each $f_n$ is $n(1/2)^{n+1}$ and this implies uniform convergence!

Perhaps I have commited a mistake with elemental calculus, but the way of attacking is clearly this, to observe what's going on with the supremum of the $f_n$.

3. Originally Posted by Enrique2
I believe all the functions $f_n$ are positive and attain their maximum at $x=1/2$ (using the derivative). Hence the supremum norm of each $f_n$ is $n(1/2)^{n+1}$ and this implies uniform convergence!

Perhaps I have commited a mistake with elemental calculus, but the way of attacking is clearly this, to observe what's going on with the supremum of the $f_n$.
Actually taking the derivative shows that the max of $f_n$ occurs when $x=\frac{n}{n+1}$. So $f_n\left(\frac{n}{n+1}\right)=n\left(1-\frac{1}{n+1}\right)^n\frac{1}{n}=\left(1-\frac{1}{n+1}\right)^n\to e^{-1}$ (not sure about the final limit being $e^{-1}$ but i know it's not 0).
Thus $\sup|f-f_n|=e^{-1}\neq{0}$. So no uniform convergence for you.

4. I was supposed to show that the convergence was not uniform, so that is good =)

I donīt know where I got it wrong, I got that $x=\frac{-n}{n+1}$

thx