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


*

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

*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, $0<x<1, \lim_{x \to \infty} x^n(1-x) = 0$
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. 
 A: Your answer for the pointwise convergence is correct. In the other case, 
$\sup_{x\in[a,b]} f_n(x) = 0$ is not correct. Here it is a graph for $f_n$ between 0 and 1 for several n, indicating the supremum. We have that $\sup f_n=(1-\frac{1}{n+1})^n/(n+1)\rightarrow 1/e*0=0$ so the convergence is uniform.

A: 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 $n \to \infty$.
A: Your proof for pointwise convergence looks good. But I don't know how you get $$\sup_{x\in[a,b]} f_n(x) = 0$$ (because that is not true). You want to show that $f_n$ converges to $f$ uniformly on $[0,1]$. You can use the criteria you mentioned. A hint for that:
we obviously have $$f_n(x)-f(x)=f_n(x)=x^n(1-x),$$ this function is continuous and differentiable in $[0,1]$. To get $$\sup\limits_{x\in[0,1]}|f_n(x)-f(x)|$$ you can just use the derivative to check for maxima/minima. Once you have found these, you can calculate $$\lim\limits_{n\to\infty}\sup\limits_{x\in[0,1]}|f_n(x)-f(x)|$$ and if this is $0$, you have proven that it $f_n$ converges uniformly to $f$ on $[0,1]$.
A: Uniform convergence:
For any $\epsilon>0$ choose some $N>1/\epsilon$.
The maximum of $f_n$ occurs at $x=n/(n+1)$, and $f_n(n/(n+1))=(n/(n+1))^n/(n+1)$. So we have for any $x\in[0,1]$:
$$
\begin{aligned}
|f_n(x)|&\leq f_n(n/(n+1))=\left(\frac{n}{(n+1)}\right)^n\frac{1}{(n+1)}\\
&\leq \frac{n}{(n+1)}\frac{1}{(n+1)} \leq \frac{1}{n}
\end{aligned}
$$
So now $\forall n> N$, $\forall x\in[0,1]$, $|f_n(x)|<\epsilon$. Thus $f_n$ converges to 0 uniformly on $[0,1]$.
