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Let $f_n : [0,1] \rightarrow \mathbb{R}$, and $\alpha \in \mathbb{R}$ $$ f_n(x)=n^{\alpha}x(1-x)^n$$

To show the point wise convergence, I divided the situation into three cases:

  • $x=0$ then $f_n(x)=0$, thus $f=0$ for $x=0$
  • $x=1$ then $f_n(x)=0$, thus $f=0$ for $x=1$

Now where I get stuck is showing that $f_n(x)$ converges to the null function when $0

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    A more interesting question is to ask whether or not it converges uniformly. When $\alpha \geq 1$ this is not the case since $f_n(1/n)$ is asymptotic to $n^{\alpha-1}/e$. When $\alpha<1$ we can use calculus to maximize $f_n$ and see... – 2017-02-22

1 Answers 1

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For $0

Let $a_n:=n^{\alpha}q^n$. Then $a_n^{1/n} \to q$ for $n \to \infty$. Now let $r \in (q,1)$. There is $N \in \mathbb N$ such that

$a_n^{1/n} N$, hence

$0N$. This gives: $a_n \to 0$ for $n \to \infty$.

Consequence: $f_n(x) \to 0$ for $n \to \infty$