Let for $n = 0,1,2,...$ , $f_n : [0,1] \rightarrow \mathbb{R}$ defined by $f_n (x) = x^n$.
1) Is the convergence of {$f_n$}$_{n=0} ^\infty$ to $f$ locally uniformly on the interval $[0.1]$?
2) And on the interval $[0.1)$?
The definition of locally uniformly convergence is:
The sequence {$f_n$} converges locally uniformly to $f$ if for all $x \in D$ there is a neighborhood $U$ of $x$ in $\mathbb{R}^p$ such that $f_n | U \cap D$ is uniformly convergent, with $f : D \rightarrow \mathbb{C}$.
Where should I start with solving the first question? It is difficult to imagine what is meant by that neighborhood $U$.
If $f_n$ is to converge locally uniformly to some function $f:I\to\Bbb R$, then $f$ must be the pointwise limit of the $f_n$, that is $f(x)=\lim_{n\to\infty}f_n(x)$. This means $f(1)=1$ and $f([0,1))=\{0\}$, so $f$ is not continuous. However, if $f_n$ converges locally uniformly to $f$, then $f$ has to be continuous. Hence the $f_n$ do not converge locally uniformly on $I$.
On $J=[0,1)$, the situation is different. Here, the limit function is $f\equiv0$. Can you show that for each $d<1$, $f_n$ converge uniformly to $f$ on $[0,d)$, and conclude that $f_n$ converge locally uniformly on $J$ ?