How do I show $f_n(x)=x^n(1-x)$ uniformly converges to $0$ on $[0,1]$?

Question

Please read before posting a hint or answer as I know this question is probably elementary and there are standard hints and answers:

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Show that $f_n(x)=x^n(1-x)$ uniformly converges to $0$ on $[0,1]$.

We have to show that $\forall \epsilon \ \exists N \ \forall x \in [0,1]$ such that $n \ge N \implies |f_n(x)-0|<\epsilon$.

So, I must show $|x^n(1-x)| < \epsilon$ if $n \ge N$.

We see that $x^n(1-x)=x^n-x^{n+1}$ and since $x \in [0,1]$ then $|x^n(1-x)|=x^n(1-x)$.

So, we must show $x^n(1-x) < \epsilon$ if $n \ge N$.

I am having trouble beyond this and so I am trying to first go about it by showing pointwise convergence first, but I can't even do this. How do I even go about showing pointwise convergence?

Answer 1

Pointwise convergence: Say, let $x=1/3$, do you know what value is $\lim_{n\rightarrow\infty}f_{n}(1/3)$? If so, what if for any other $x\in[0,1]$, that is, for fixed $x\in[0,1]$, what is the value of $\lim_{n\rightarrow\infty}f_{n}(x)$?

Answer 2

Hint (pointwise convergence): How would you compute $\lim_{n \to \infty}(2/3)^n(1 - 2/3)$? What about $\lim_{n \to \infty}(1)^n(1 - 1)$? If you can find these limits, then you can surely find the pointwise limit of $\lim_{n \to \infty}x^n(1 - x)$

Hint (uniform convergence): In terms of $n$, what is the maximum value attained by the function $f_n(x) = x^n(1 - x)$ over $[0,1]$? What happens to this maximum as $n \to \infty$?

Answer 3

Recall Bernoulli's inequality: $( 1+x )^n \geq 1 + nx$ for $x \geq -1$.

Use this to show $x^n \rightarrow \infty $ for $x>1$ and hence $x^n \rightarrow 0$ for $x \in (0,1)$. Pointwise convergence is then obvious.

Once you have these pointwise bounds, you can then try to cook up a uniform bound using what we had above or otherwise.

Answer 4

Hint 1: When $x\ge1-\epsilon$, show that $x^n\color{#C00}{(1-x)\le\epsilon}$.

Hint 2: When $x\lt1-\epsilon$ and $n\ge\frac{\log(\epsilon)}{\log(1-\epsilon)}$, show that $\color{#C00}{x^n}(1-x)\color{#C00}{\le\epsilon}$.

Answer 5

By the AM-GM inequality, for any $n\geq 1$ and any $x\in[0,1]$ we have

$$\begin{eqnarray*} 0\leq x^n (1-x) = \frac{1}{n} x^n (n-nx) &=& \frac{1}{n}\text{GM}\left(x,x,x,\ldots,x,n-nx\right)^{n+1}\\&\color{red}{\leq}&\frac{1}{n}\text{AM}(x,x,x,\ldots,x,n-nx)^{n+1}=\frac{n^n}{(n+1)^{n+1}}\end{eqnarray*}$$ and $\color{red}{\leq}$ holds as an equality only if $x=n-nx$, i.e. at $x=\frac{n}{n+1}$. $$ 0\leq f_n(x)\leq \frac{1}{n+1} $$ then gives that $f_n(x)$ is uniformly convergent to zero on $[0,1]$.