Prove if $f_n(x)=n^2 x^n(1-x)^2$ converges pointwise and/or uniformly on $I=[0,a]$, where $a<1$

Question

I have the following problem and I am kind of stuck in the second part of it. So the problem says...

"Find the pointwise limit of the given sequence and determine whether or not the convergence is uniform, giving reasons for your answer.

$f_n(x)=n^2 x^n(1-x)^2$ on $I=[0,a]$, where $a<1$"

And my answer is:

If $x=0$ then $f_n(0)=0$.

If $0<x\le a$ and $a<1$, then $0<(1-x)^2\le a$ is equivalent to $0<(1-x)^2<1$, and so $x^n\to 0$ as $n\to\infty$. Hence $f_n(x)\to 0$.

Therefore, $f_n(x)\to f(x)=0$ pointwise.

Now here is where I do not know what to do - proving if it converges uniformly or not. I solved another problem which was the same but with $I=[0,1]$ by letting $x=1-1/n$, which gave me $f_n(x)=(1-1/n)^n\to 1$ as $n\to\infty$. I think this could be used here as well, but it would be weird, I do not see why my professor would give us 2 exercises that are so similar and give each of them the same weight in the assignment (more than other exercises). :/ How would you show if it is uniformly convergent?

Thanks a lot!

Answer 1

We have

$$\forall x\in[0,a],\qquad|f_n(x)|\le n^2 a^n\xrightarrow{n\to\infty}0$$ so the sequence $(f_n)$ is uniformly convergent to the zero function.

Answer 2

Consider that $f_n(x)$ is a positive and bounded function on the $I=[0,1]$ interval.
$f_n(x)$ has the only stationary point, zero of $f_n'(x)$: $$x_0=\frac{n}{n+2}\in I,$$ hence: $$ 0\leq f_n(x) \leq f_n\left(1-\frac{2}{n+2}\right) = \frac{4n^2}{(n+2)^2}\left(1-\frac{2}{n+2}\right)^n=4\left(1-\frac{2}{n+2}\right)^{n+2}\leq\frac{4}{e^2},$$ and $f_n(x)$ is increasing on the $I_n=\left[0,1-\frac{2}{n+2}\right]$ interval.

This gives that for any big enough $n$ the maximum of $f_n(x)$ over the $J_n=\left[1-\frac{\log n}{n}\right]$ interval is just:

$$ f_n\left(1-\frac{\log n}{n}\right)=\log^2(n)\cdot\left(1-\frac{\log n}{n}\right)^n\leq\frac{\log^2 n}{n},$$ so the uniform convergence claim follows by observing that $1-\frac{\log n}{n}\to 1$ and $\frac{\log^2 n}{n}\to 0$.