Firstly, I am sorry, I am unable to put my question properly on the title dut to the character limit. The problem actually looks like-
Let, $g:[0,1]\to\Bbb{R}$ be a continuous function such that $g(1)=0$ and $\{f_n\}_n$ be a sequence of functions on $[0,1]$ defined by for each $n\in\Bbb{N}$ $f_n (x)=x^n g(x)\: \forall x\in [0,1]$. Prove that, $\{f_n\}_n$ converges uniformly on$[0,1]$.
Intuitively, the problem seems easy, but while writing it rigorously with the proper argument, I get stuck.
Can anybody give a proper argumentative solution to the problem?
Thank you for your assistance in advance.
Let $\varepsilon>0$. Then there exists $\delta>0$ such that for $1-\delta<x<1$ it is $|g(x)|<\varepsilon$, since $g$ is continuous at $1$ and $g(1)=0$. Then, for every $x\in(1-\delta,1]$ it is $|f_n(x)|=|x^ng(x)|\leq|g(x)|<\varepsilon$. Now, for $x\in [0,1-\delta]$ it is $|f_n(x)|\leq (1-\delta)^n|g(x)|\leq (1-\delta)^n\max_{t\in[0,1-\delta]}|g(t)|:=(1-\delta)^nM$. But $(1-\delta)^n\to 0$, so there exists $n_0$ such that for all $n\geq n_0$ it is $(1-\delta)^nM<\varepsilon$. Hence for all $x\in [0,1-\delta]$ it is $|f_n(x)|<\varepsilon$ for all $n\geq n_0$. But also for $x\in(1-\delta,1]$ and for all $n$ it is $|f_n(x)|<\varepsilon$, therefore for all $x\in[0,1]$ and for all $n\geq n_0$ it is $|f_n(x)|<\varepsilon$. This is precisely the definition of uniform convergence.