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$f_n(x) = (n+1)x^n(1-x)$, on which interval does $f_n$ converge uniformly?

Consider $x\in(-1,1]$, it is easy to see that $f_n$ converges to $f(x) = 0$ as $n\to\infty$, but stuck in constructing a proof followed strictly from the definition of uniform convergence.

Also, I wonder if (-1,1] the only set where $f_n$ converges uniformly.

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  1. For $0<r<1$ and $x\in[-r,r]$, you have : $$|f_n(x)| \leq 2(n+1) r^n \overset{n\to\infty}{\longrightarrow}0 $$ so $(f_n)$ converges uniformly on $[-r,r]$

  2. $(f_n$) does not converge uniformly on any neighborhood of $-1$ and $1$, since : $$f_n\left(1-\frac1n\right) = \frac{n+1}n \left( 1 - \frac 1n\right)^n \to e^{-1}\neq 0$$ and $$|f_n(-1)| = 2(n+1) \to +\infty$$

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  • $\begingroup$ really impressed by how brilliantly you show that $f_n$ does not converge uniformly around x = 1. Just a little bit of concern that although picking the number $1-{1\over n}$ will work for all the finite cases, is it legal to do it when $n$ goes to infinity? $\endgroup$
    – ZKK
    May 12, 2021 at 14:23
  • $\begingroup$ To find $1-1/n$, I computed the derivative of $f_n$ to find its maximum on $[0,1]$. Since this calculation is not part of the formal proof, I didn't include it $\endgroup$
    – SolubleFish
    May 12, 2021 at 14:29
  • $\begingroup$ If you consider an interval $I\subset (-1,1]$ which contains $1$, then (for $n$ big enough) you have $1-1/n \in I$ and therefore $\sup_I |f_n| > e^{-1}/2$ (for $n$ big enough) $\endgroup$
    – SolubleFish
    May 12, 2021 at 14:31

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