I was wondering if I could get some clarification on a specific property regarding uniform convergence.
For example, consider $f_{n}(x)=x^{n}$ for $x \in [0,1]$ and $f_{n}(x)=x^{n}$ for $x \in [0,1)$. I understand that the first sequence of functions is not uniformly convergent and the second is. I have proved both cases.
However, I am attempting to do the same problem with the following Theorem and with some different bounds:
Suppose the sequence $\{ f_{n} \}$ of real-valued functions on the set $E$ converges P.W. to $f$ on $E$. For each $n \in \mathbb{N}$, set \begin{align*} M_{n}= \sup_{x \in E} |f_{n}(x)-f(x)| \end{align*} Then $\{ f_{n} \}$ converges uniformly to $f$ on $E \iff \lim_{n \to \infty}M_{n}=0$.
My problem arises when we try attempt to examine the uniform convergence of a set like $x \in (-1,1)$ versus $[-a,a]$ where $0<a<1$. I want to say that both are uniform, but supposedly for the $[-a,a]$ case it is uniformly convergent and the other is not. In particular, how come for $(-1,1)$ we get that $\lim_{n \to \infty} M_{n}=1$? While for the other we get that $\lim_{n \to \infty} M_{n}=0$?
The uniform convergence of $f_n(x)=x^n$ to $0$ on $[-a,a]$ for any $a\in (0,1)$ is not because $[-a,a]$ is compact. It is because $\|f_n\|=\sup \{|f_n(x)| a\geq |x|\}=a^n$ tends to $0$ as $n\to \infty.$
But $f_n(x)=x^n$ does not converge uniformly to $0$ on $[0,1)$ because for each $n$ and each $r>0$ we can find $x_n\in (0,1)$ with $f_n(x_n)>1-r.$ So $\|f_n\|\geq \sup \{1-r : r>0\}=1.$
Observe that extending the domain from $[0,1)$ to the compact set $[0,1]$ will not change this : If $f_n$ does not converge uniformly to $0$ on a non-empty subset $A$ of the common domain $B$ of all the $f_n,$ it can't converge uniformly to $0$ on $B.$ Because if $\|f_n|_A\|=\sup \{|f_n(x)|: x\in A\}$ doesn't converge to $0,$ then $\|f_n\|\geq \|f_n|_A\| ,$ and can't converge to $0$ either.
It may be helpful to look at the graphs of $f_n(x)=x^n$ for $x\in [0,1)$ for various values of $n,$ and see what happens to $z_n,$ where $f_n(z_n)=1/2.$