Prove that the sequence of functions $f_n(x)=x^ng(x), n \in \mathbb{N}$, converge uniformly on $[0, 1]$.

Question

Let $g: [0,1]\longrightarrow \mathbb{R}$ a continuous function with $g(1)=0$. Prove that the sequence of functions $f_n(x)=x^ng(x), n \in \mathbb{N}$, converge uniformly on $[0, 1]$.

I know that the function $f_n\quad$converges to the function $f(x)= 0$ on $[0,1]$. I tried to make a bipartition of the domain like $[0,\delta]$ and $[\delta,1]$. Any suggestions would be great!

Answer 1

Let $c>0$, there exists $d>0$ such that $|1-x|<d$ implies that $|f(x)|<c$ since $f$ is continuous at $1$. We may suppose $d<1/2$.

The function $f$ is continuous and defined on the compact $[a,b]$ there exists $M$ such that $|f(x)|<M$. There exists an integer $N$ such that $(1-d)^NM<c$.

Let $n>N$ if $x\leq 1-d, |x^nf(x)|\leq (1-d)^nM<c$,

If $x>1-d$, $1-x<d$ and $|f(x)|<c$ this implies that $|x^nf(x)|\leq |f(x)|<c$.

Answer 2

Let $\epsilon >0$ and choose $\delta \in (0,1)$ such that $|g(x)| <\epsilon$ for $1-\delta <x \leq 1$; choose $N$ such that $n >N$ implies $x^{n} <\frac {\epsilon} M$ for $0 \leq x \leq 1-\delta$ where $M=\sup \{|g(x)|: 0 \leq x \leq 1\}$. Can you check that $|x^{n}g(x)|<\epsilon$ for all $x$ if $n >N$?

[Note: In choosing $N$ I have used the fact that $(1-\delta)^{n} \to 0$ as $n \to \infty$].