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

$$f_n(x) = (n+1)x^n(1-x)$$, 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.

## 1 Answer

1. For $$0 and $$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 $$|f_n(-1)| = 2(n+1) \to +\infty$$

• 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?
– ZKK
May 12, 2021 at 14:23
• 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 May 12, 2021 at 14:29
• 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) May 12, 2021 at 14:31