Question:
Let $f_n(x)=xn(1-x^2)^n$ on domain $[0,1]$. Verify that the pointwise limit function is the zero function through domain.
Prove that $\displaystyle\int f_n=n/(2n+2)$ on $[0,1]$ and deduce that
$\displaystyle\lim_{n\to\infty}\int f_n\neq\int\displaystyle\lim_{n\to\infty}f_n$.
Is the convergence of the sequence uniform?
What is $\displaystyle\lim_{n\to\infty}sup{|f_n(x)-f(x)|:0\le x\le1}$?
It is pretty clear that $f_n(x)\to0$ as $n\to\infty$ for $0\le x\le1$. Using integration by substitution we can get $\int f_n$.
$\displaystyle\lim_{n\to\infty}\int f_n= \dfrac{1}{2}$ and $\displaystyle\int\displaystyle\lim_{n\to\infty}f_n=0$.
Until now it is pretty easy and clear for me. The supremum part is where I don't understand.
$$ sup{|f_n(x)-f(x)|:0\le x\le 1=\dfrac{n}{(\sqrt{1+2n})\cdot (1+\dfrac{1}{2n})^n\to\infty $$.
I don't know how they get to this but my answer is kinda different. Since $f(x)$ is 0 then
$sup{|f_n(x)-f(x)|}=0$
and thus it is uniformly convergent. I think I am wrong and right at same time. Can somebody explain this step? Also there this written in answer which I think is kinda useless but you might have better idea why it was with answer
$f'_n(\dfrac{1}{\sqrt{2n+1}})=0$.
Why does $\sup_x |f_n(x)-0| \to 0$? $f_n(x) \to 0$ for each $x$ does not imply that the supremum tends to $0$. You have to find the maximum of $f_n(x)$ by setting the derivative equal to $0$ which gives $x=\frac 1 {\sqrt {2n+1}}$. The maximum value turns out to be $\frac n {\sqrt {2n+1}} (1-\frac 1 {2n+1})^{n}$ and the limit of this is $\infty$.