How to prove uniform convergence for sequences $f_n=x^n(1−x)$ on $[0,1]$

Question

I have already shown that $f_n(0)=0$, $f_n(1)=0$, and for $0<x<1$, $f(x)\to0$. Therefore letting $f(x)=0$ for $0\le x\le 1$, $f_n\to f$ pointwise.

We were given the hint to use the fact that $(1+\frac {1}{n})^n\to e$

For other problems in class, we have started the problems like this:

Let $\epsilon>0$. Choose N such that ____$<\epsilon$. For $n<N$ and $0\le x\le 1$

$|f_n(x)-f(x)|$ so for this problem, I would have $|x^n(1-x)-0|$.

But I cannot figure out how to get from here to using the given hint to showing that $|f_n(x)-f(x)|<\epsilon$ in order to show that it converges uniformly on $[0,1]$.

Answer 1

Hint. Show that the non-negative function $f_n$ attains its maximum value in $[0,1]$ at $x_n=1-\frac{1}{n+1}$. Then $$\max_{x\in [0,1]} |f_n(x)-f(x)|=f_n(x_n).$$ Can you take it from here?

Answer 2

You can derive the function $f_n$:

$f_n'(x) = nx^{n-1} - (n+1)x^n = (n - x(n+1))x^{n-1}$

which is null for $x = \frac{n}{n+1}$. You check easily that it is the maximum of $f_n$. Then in that point:

$f_n\left(\frac{n}{n+1}\right) = \left(\frac{n}{n+1}\right)^n\left(1 - \frac{n}{n+1}\right) = \frac{1}{\left(1+\frac{1}{n}\right)^n}\frac{1}{n+1} \to e^{-1} \times 0 = 0$.

The maximum of the function goes to $0$, so the convergence is uniform.