uniform convergence of a functional sequence

Question

Is this sequence of functions $$f_n(x)=n^3x(1-x)^n$$ converges uniformly for $x\in[0,1]$. I need some help on this.

Answer 1

Pointwise $f_{n}(x_0)\rightarrow 0 $ in $[0,1]$. So in order to show that it converges uniformly you just need to show that $$\sup_{x\in[0,1]}|f_n(x)-0|\rightarrow0$$ Now take the $x-$derivative $$f'_{n}(x)=n^3(1-x)^n-n^4x(1-x)^{n-1}=n^3(1-x)^{n-1}(1-x-nx)$$ You can easily find that the maximum point is $$x_{max}=\frac{1}{n+1}\ ,$$ so $$\sup_{x\in[0,1]}|f_n(x)-0|=\max_{x\in[0,1]}|f_n(x)|=|f_n(x_{max})|=n^3\frac{1}{n+1}\left(\frac{n}{n+1}\right)^n\sim \frac{n^2}{e}$$ that is divergent.

Answer 2

Before asking for uniform convergence, we have to seek for pointwise convergence. If $x\in [0,1)$, then $n^3x(1-x)^n\to 0$ because $n^3a^n\to 0$ for $a\in (0,1)$.

Hence $f_n(x)\to 0$ for any $x\in [0,1]$. Since $f(1/n)=n^2(1-1/n)^n$ and $f(1/n)=(1-1/n)^n\to e^{-1}\gt 0$, $\sup_{x\in [0,1]}|f_n(x)|\to +\infty$ as $n$ goes to infinity.