Is there a sequence of positive numbers $\epsilon_1,\epsilon_2,...$ with the following property?

Question

Is there a sequence of positive numbers $\epsilon_1,\epsilon_2,...$ with the following property? If $g_n\in C[0,1]$(continuous functions on $[0,1]$) and $g_n\rightarrow0$ pointwise, then $\epsilon_ng_n\rightarrow0$ uniformly.

Attempt) Since $g_n\in C[0,1]$, $|g_n|\leq M_n$ for each $n$. I guess since $g_n\rightarrow0$ pointwise, $M_n\leq M$ for some $M$. Then if we take $\epsilon_n=\frac{1}{n}$. $\sup\limits_{x\in [0,1]}|\epsilon_ng_n(x)-0|\leq\frac{M}{n}\rightarrow0$. So $\epsilon_ng_n\rightarrow0$ uniformly.

Answer 1

Given any sequence of positive numbers $\epsilon_n$, you can find $g_n$ such that $g_n(0)=0$ and $g_n(x) = 0$ on $[1/n, 1]$, but $\epsilon_n g_n(1/(2n)) =1$.

Answer 2

Suppose $\sup_{x\in [0,1]}|g_n(x)|=M_n$ then if there exists such an $\epsilon_n$ sequence.Then $\epsilon_n M_n \to 0$ for every such $M_n$.but we can find a sequence of continuous functions $g_n(x)$ such that $M_n=1/\epsilon_n$,then $\epsilon_n.1/\epsilon_n \to 0$ which is a contradiction. construction of $g_n$ is $g_n(0)=0$ and $g_n(x)=0$ if $x\in [1/n,1]$ and $g_n(1/2n)=1/\epsilon_n$,join these points continuously.