Could someone please use the following theorem to show how a pointwise convergent function is sometimes uniform convergent using the following theorem through the means of an example.
Example: $f_n:[0,1] \rightarrow \mathbb{R}$ defined by $f_n(x)=n^2x(1-x)^n$. This function is pointwise convergent to the zero function but is not uniformly convergent on $[0,1]$.
Thanks!
Hint. Take $f_n(x)=n^ax(1-x)^n$ and $E=[0,1]$ for $a\geq 0$. Then $f_n$ is pointwise convergent to $f=0$. Show that
$$M_n=\sup_{x\in E}|f_n(x)-f(x)|=f_n\left(\frac{1}{n+1}\right)=\frac{n^a}{n+1}\left(1+\frac{1}{n}\right)^{-n}$$ which goes to $0$ iff $a<1$ (why?). For such $a$, the sequence $(f_n)_n$ is uniformly convergent to $f$ in $[0,1]$.
P.S. Note that $f_n$ is non negative in $[0,1]$ and $f_n'(x)=n^a(1-x)^{n-1}(1-(n+1)x)$. Hence $f_n$ attains its maximum value at $1/(n+1)$.