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

real-analysissequence-of-functionuniform-convergence

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

Best Answer

  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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