$g_n:I \rightarrow \mathbb R$ for $I\in [0,1]$, $g_n(a)=a^n(1-a)$ show $g_n$ converges pointwise. does it converge uniformly (proof)
My attempt: I am pretty confuse. I am try to understand the difference between pointwise and uniform convergence.
For pointwise suppose $a\in [0,1) \implies a^n \rightarrow 0 \implies a^n(1-a) \rightarrow 0$
if $a=1$ then $(1-a)=0 \implies a^n(1-a)=0$
does this look okay? how to prove it is uniformly convergent?
That looks good! For uniform convergence, let $\epsilon > 0$. If $a>1-\epsilon$, then $1-a < \epsilon$ so that $g_n(a) < \epsilon$. If $a \leq 1-\epsilon$, then $a^n \leq (1-\epsilon)^n$ which will be less than $\epsilon$ for large enough $n$, and thus $g_n(a) < \epsilon$. This gives uniform convergence to 0.
$g_n'(x) = nx^{n-1} - (n+1)x^n = x^{n-1}\left(n-(n+1)x\right) = 0 \iff x = 0, \dfrac{n}{n+1} \to \text{sup } g_n(x) = \left(\dfrac{n}{n+1}\right)^n - \left(\dfrac{n}{n+1}\right)^{n+1}= \left(\dfrac{n}{n+1}\right)^n\cdot \dfrac{1}{n+1}\to \dfrac{1}{e}\cdot 0 = 0\to \text{ convergence is uniform on [0,1]}$.
