Understanding pointwise convergence vs. uniform convergence example

Question

I'm trying to understand the difference between pointwise convergence and uniform convergence. I read this post and the last answer on it is the following:

$f_n\to f$ pointwise on $(a,b)$ if for each fixed $x\in(a,b)$, $|f_n(x)-f(x)|\to 0$ as $n\to\infty$. Notice this is a pointwise (local) criterion.

On the other hand, $f_n\to f$ uniformly on $(a,b)$ if $\sup_{a< > x<b}|f_n(x)-f(x)|\to 0$ as $n\to\infty$. This is a "global" criterion in that is requires the maximum of all the pointwise errors on $[a,b]$ to tend to zero.

As an example, $f_n(x)=x^n$, $0\le x\le 1$ converges pointwise to $f(x)=\begin{cases} 0, &0\le x<1,\\ 1, &x=1.\end{cases}$, because the first condition above holds, but the convergence is not uniform since the second condition does not hold.

I don't fully understand why the example isn't uniformly convergent, but here is my attempt:

$x^n\rightarrow 0$ when $x\in[0,1)$, $n\rightarrow\infty$

$x^n\rightarrow 1$ when $x=1$, $n\rightarrow\infty$,

so $x^n\rightarrow f(x)$ pointwise. This is pretty clear for me, but uniform convergence is a bit more troubling. Why doesn't uniform convergence hold in this example? Is it because we can never select such an $N$ that $|x^N-f(x)|<\epsilon$, for all $x\in[0,1]$? That is there is no upper bound for N, so that the condition holds for all $x\in[0,1]$?

Thank you for any help =)

Answer 1

The supremum $$ \sup_{0\le x\le1}|x^n-f(x)|=1 $$ for each $n\ge1$ since $$ \sup_{0\le x<1}|x^n-f(x)|=\sup_{0\le x<1}|x^n|=1 $$ for each $n\ge1$. Hence, $$ \sup_{0\le x\le1}|x^n-f(x)|\not\to0\quad\text{as}\quad n\to\infty $$ and the sequence $x^n$ isn't uniformly convergent. Actually, $$ \sup_{0\le x\le1}|x^n-f(x)|\to1\quad\text{as}\quad n\to\infty. $$

We can select $N$ for each $x\in[0,1]$ such that $$ |x^n-f(x)|<\varepsilon $$ for $n>N$. But we cannot select $N$ such that $$ \sup_{0\le x\le1}|x^n-f(x)|<\varepsilon $$ for $n>N$.