Assume $f_n(x) = n^cx(1-x^2)^n$
For which values of $c$ does ${f_n(x)}$ converge uniformly on $[0,1]$?
This is only part of the problem, I have already proven that ${f_n}$ converges pointwise on $[0,1]$ for all $c$.
Assume $f_n(x) = n^cx(1-x^2)^n$
For which values of $c$ does ${f_n(x)}$ converge uniformly on $[0,1]$?
This is only part of the problem, I have already proven that ${f_n}$ converges pointwise on $[0,1]$ for all $c$.
thats a nice exercise, you just find the maximum of f(x) in the interval, and make it go to zero. the maximum is obtained by differentiating the function:
$ f_n (x)=n^c x(1−x^2)^n $ f'_n (x)= n^c((1-x^2)^n-2nx^2(1-x^2)^{n-1}) = 0 \rightarrow x=\frac{1}{\sqrt{1+2n}} )
since $f(x)=0$ at $x=0$ and $x=1$, the maximum of $ |f_n (x)| $ is at that point.
so $ \begin{align*} \sup_{0
so we need $c<0.5 $ for uniform convergence.