Prove that $x^n \rightarrow 0$ uniformly on $[0,b]$ with $0 < b < 1$, but not uniformly on $[0,1)$

Question

I have been working on this for a bit now and I am not sure what to do. Here is the question (also in the title), and then the work I have so far.

Question

Prove that $x^n \rightarrow 0$ uniformly on $[0,b]$ with $0 < b < 1$, but not uniformly on $[0,1)$

My attempt

Proof Let $\epsilon>0.$ To show $f_n(x)=x^n \rightarrow0$ uniformly on $[0,b]$ with $0 \lt b \lt 1$, we need to show $\forall \epsilon \gt 0$, $\exists N\in \Bbb N$ s.t.

$\vert f_n(x)-f(x) \vert \lt \epsilon$ when $n\geq N$ and $x\in [0,b]$.

Note that:

$\vert f_n(x)-f(x) \vert = \vert x^n-0 \vert = \vert x^n \vert \leq \vert b^n \vert $ (since $x\in[0,b])$.

I have also found that in order for $b^n \lt \epsilon$, we have that

$b^n \lt \epsilon \implies n \ln b \lt \ln \epsilon \implies n \lt \frac{\ln \epsilon}{\ln b}$

but that is it. I am also not sure how to prove that $f_n(x)$ does not converge uniformly on $[0,1)$.

Thank you!

Answer 1

To prove that convergence is not uniform on $[0,1)$, note that

$$\sup_{x \in [0,1)}|f_n(x) - 0| \geqslant f_n(1 - 1/n) = (1- 1/n)^n \underset{n \to \infty}\longrightarrow e^{-1} \neq 0$$