$f_n(x) = x(1-x)^n$ converges uniformly to $0$ on $[0, 1]$?

Question

I am trying to show that the sequence of functions $f_n(x) = x(1-x)^n$ converges uniformly to $0$ on $[0, 1]$.

Well at $0$ and $1$, $f_n(x) = 0$ for all $n$.

So let $x \in (0, 1)$.

$f_n(x)$ converges uniformly to the zero function if there exists a sequence of values $\alpha_n$ that converges to $0$ and $|f_n(x) - f(x)| < \alpha$ for all $x$ and all $n$.

Well let $1 - x < k < 1$.

Then we have that $(1-x)^n < (1-k)^n$ and $(1-k)^n \to 0$ as $n \to \infty$.

And as $x \in (0, 1) \implies x(1-x)^n < (1-k)^n$

Hence $f_n(x)$ converges uniformly to the zero function.

Is that correct?

Answer 1

Hint: Notice that:

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

hence maximum at $x=\frac{1}{n+1}$ (Boundary terms evaluate to $0$).