Consider the sequence of function $f_{n}(x)=n^{2}x\bigl(1-x^{2}\bigr)^{n}$. Is this sequence of function convergent uniformly in $[0,1]$ ?
We find that $f$ converges point wise to the function $f(x)=0$.
$\mathbf {Method 1:}$ We find that, $$\lim_{n\to \infty}\int_{0}^{1}f_{n}(x)dx=\infty$$ but $$\int_{0}^{1}lim_{n\to \infty}f_{n}(x)dx=\int_{0}^{1}f(x)dx=0.$$ So $\bigl\{f_{n}(x)\bigr\}_{n}$ does not converge uniformly to $f$ in $[0,1]$.
$\mathbf{Method 2:}$ We consider $\bigl|f_{n}(x)-f(x)\bigr|=g(x)(say)$. Then $g(x)=n^{2}x\bigl(1-x^{2}\bigr)^{n}$. We evaluate $g'(x)$ , $g''(x)$. For extreme point we put $g'(x)=0$ & we get the values $x=\pm \frac{1}{\sqrt 3}.$
Now , $g''(\frac{1}{\sqrt 3})<0$. So, $g$ attains maximum value at $\frac{1}{\sqrt 3}$ & the maximum value is $$g(\frac{1}{\sqrt 3})=\frac{n^{2}}{\sqrt 3}\bigl(\frac{2}{3}\bigr)^{n}=M_{n}.$$ As $M_{n}\to 0$ as $n\to \infty$ so $\bigl\{f_{n}(x)\bigr\}_{n}$ converge uniformly to $f(x)$.
My question is " Why these two methods contradicts each other? " Where my mistake?
The maximum for $x(1-x^2)^n$ is attained in the same point where the function $\log x+n\log(1-x^2)$ takes its maximum, i.e. in a zero of: $$\frac{1}{x}-\frac{2nx}{1-x^2}$$ i.e. in $x=\frac{1}{\sqrt{2n+1}}$, so: $$ \sup_{x\in[0,1]} f_n(x) = \max_{x\in[0,1]} n^2 x(1-x^2)^n = \frac{n^2}{\sqrt{2n+1}}\left(1+\frac{1}{2n}\right)^{-n}\geq \frac{n^2}{\sqrt{(2n+1)e}}. $$