Determine if $f_n(x) = n^2x(1-x^2)^n$ converges uniformly on $E=[0,1]$. Determine if $f_n(x) = n^2x(1-x^2)^n$ converges uniformly on $E=[0,1]$. We can easily find the pointwise limit to be $f(x) = 0$ for all $x \in E$. $M_n = \sup\limits_{x\in E} |f_n(x)-f(x)| = \sup\limits_{x\in E} |f_n(x)| = \sup\limits_{x\in E} f_n(x)$ since $f_n$ is always positive on $E$. I am having trouble finding the supremum of the function.
 A: Notice that $f_n(\frac{1}{n}) = n(1-\frac{1}{n^2})^n \geq n(1-\frac{1}{n})^n$, which asymptotically looks like $\frac{n}{e}$, and thus tends to $\infty$.
Therefore there exists a sequence $x_n \to 0$ such that $f_n(x_n) \to \infty$, so the convergence is not uniform (here $x_n = \frac{1}{n})$.
A: Well, do it Calculus I style: $$f_n'(x) = n^2((1-x^2)^n - 2nx^2(1-x^2)^{n-1}).$$ We have $$f_n'(x) = 0 \iff (1-x^2)^{n-1}(1-x^2-2nx^2) = 0$$ So our critical points are $x = 0$, $1$ and the solutions to $$1-x^2-2nx^2 = 0,$$ that is: $$1-x^2(1+2n)=0 \implies x = \frac{1}{\sqrt{1+2n}},$$ since $0 \leq x \leq 1$. It is easy to check that this is indeed a maximum. So $$\|f_n\|_{\infty} = \frac{n^2}{\sqrt{1+2n}}\left(1-\frac{1}{1+2n}\right)^n = \frac{2^nn^{n+2}}{(1+2n)^{n+\frac{1}{2}}} \to +\infty,$$ but $\frac{1}{\sqrt{1+2n}}\to 0$, so the convergence is not uniform.
A: $$f'_n(x)=n^2(1-x^2)^n-2n^3x^2(1-x^2)^{n-1}=n^2(1-x^2)^{n-1}\left(1-(1+2n)x^2\right)$$
The supremum of $f_n$ Is reached at $x=\sqrt{\frac{1}{1+2n}}$. So
$$M_n=n^2\sqrt{\frac{1}{2n+1}}\left(\frac{2n}{2n+1}\right)^n$$
$M_n\to \infty$ and therefore the convergence is not uniform
