real analysis - Pointwise convergence and uniform convergence of $f_n(x) = x^n(1-x)$

Ok, I am new to this pointwise and uniform convergence so don't mind if I make mistakes here.

Let: $f_n(x) = x^n(1-x), x \in [0,1]$

$f(x) = 0, x \in [0,1].$

1. Prove that $f_n$ converges to $f$ pointwisely on $[0,1]$




2. Prove that $f_n$ converges to $f$ uniformly on $[0,1].$

Attempt: For # 1: I've divided this into three cases.

Case 1. When, $x = 0, \lim_{n \to \infty} x^n(1-x) = 0$

Case 2. When, $x = 1, \lim_{n \to \infty} x^n(1-x) = 0$

Case 3. When, $0

I know in all cases the limit goes to 0, but I just wanted to make sure I don't miss anything.

For # 2: $f_n$ is uniformly converges to $f$ if $\sup_{\{a\leq x\leq b\}}\mid f_n(x) - f(x)\mid \to 0?$ Now what that'd be in this case? $\sup_{x\in[a,b]} f_n(x) = 0 = \sup_{x\in[a,b]} f \Rightarrow \mid 0 - 0\mid \to 0$? Is this what uniformly converges to $f$ mean here? If not, any explanation would be really appreciated. Thanks.

Answer

For #1 your approach is correct, and since $[0,1] = \{0\} \cup ( 0, 1 ) \cup \{ 1\}$ you have had all the possible cases.

In the second case, you do not have $\sup_{x \in [0,1]} f_n(x) = 0$, since for example $f_1\left(\frac{1}{2}\right) = \frac{1}{4}$.
However, you are right that $f_n$ converges uniform to 0 if $\sup_{x \in [0,1] } |f(x) | \to 0$, you can omit the $- f(x)$ since that is $0$. Note that $f_n(x) \ge 0$ for all $x \in [0,1]$, so we can omit the absolute value and just concentrate on the top of $f_n$.

And to find the top of $f_n$, we can differentiate $x^{n}- x^{n+1}$. This yields $nx^{n-1} - (n+1)x^n= x^{n-1}(n - (n+1) x)$. Take this to be $0$, which gives $x=0$ or $x=\frac{n}{n+1}$, note that should also consider $x=1$ since we may have that $f_n$ is strictly increasing on $[0,1]$. However, we have $f_n(0) =f_n(1)=0$, so the supremum of $f_n$ is $f_n \left( \frac{ n }{n+1} \right)= \frac{ n^n}{(n+1)^{n+1}}$, which goes to zero as $n \to \infty$.