Theorem 1
If $\{f_n\}$ is a sequence of continuous functions on $E$,
and if $f_n$ converges uniformly to $f$ on $E$, then $f$ is continuous on $E$.
The inverse is not true.
That is, a sequence of continuous functions may converge to a continuous function,
although the convergence is not uniform.
Here is an example. Let $f_n=n^2x(1-x^2)^n$ ($0 \le x \le 1 $).
I want to show the inverse of theorem 1 is not true by using the theorem 2 below.
Theorem 2
Suppose $\lim_{n \to\infty}f_n(x)=f(x)$ ($x \in E$). Put $M_n=\sup_{x \in E} | f_n(x)-f(x)|$.
Then $f_n\rightarrow f$ uniformly on $E$ if $M_n \rightarrow 0$ as $n \rightarrow \infty$
In that example, $\lim_{n \to\infty}f_n(x)=0$ ($0 \le x \le 1 $).
Then, I think, $M_n \rightarrow 0$ as $n \rightarrow \infty$ which is not true!
I need your help. How can I solve this?