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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 $c$.

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What do you have to show for uniform convergence? What function is a candidate for $f_n$ to converge against? –  Listing May 7 '11 at 23:04
Let $f=0$, then for which values of $c$ does $f_n$ converge uniformly to $f$ on $[0,1]$? –  Susie Q May 7 '11 at 23:21
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1 Answer

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 $$ \begin{align*} \sup_{0<x<1} |f_n(x)| &= n^c\cdot \frac{1}{\sqrt{1+2n}} \cdot \left(1-\frac{1}{{1+2n}}\right)^n\\ \lim_{n \rightarrow \infty} \sup_{0<x<1} |f_n(x)| &= \lim_{n \rightarrow \infty} \frac{n^c}{\sqrt{1+2n}} \cdot \left(1-\frac{1}{{1+2n}}\right)^n\\ &=\lim_{n \rightarrow \infty} \frac{n^c}{\sqrt{1+2n}} \cdot \left(\left(1+\frac{-1}{{1+2n}}\right)^{(-1-2n)}\right)^{\frac{n}{-1-2n}}\\ &=\lim_{n \rightarrow \infty} \frac{1}{\sqrt{2}n^{0.5-c}}\cdot e^{-0.5} \end{align*} $$

so we need $$c<0.5 $$ for uniform convergence.

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