$f_n$ uniform converge to f in $[a,b)$ means it uniform converge in $[a,b]$?

Question

I have the given : $f_n$ uniform converge to $f$ in $[a,b)$ I am asked for the following:

1. if $f_n$ is Continuous for every $n\in\mathbb{N}$ is it true that $f_n$ uniform converge to $f$ in $[a,b]$?

2. if both $f$ and $f_n$ are Continuous is it true that $f_n$ uniform converge to $f$ in ?

I not sure I understand how to go from $[a,b)$ to $[a,b]$ thats is the main trick I think required here here

Answer 1

Suppose that $f_n\colon[a,b]\longrightarrow\Bbb R$ is defined as $f_n(x)=0$ and that you have$$\begin{array}{rccc}f\colon&[a,b]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<b\\1&\text{ otherwise.}\end{cases}\end{array}$$Then $(f_n)_{n\in\Bbb N}$ converges uniformly to $f$ on $[a,b)$, but not on $[a,b]$.

However, if $f$ is continuous then, yes, uniform convergence on $[a,b)$ implies uniform convergence on $[a,b]$. Take $\varepsilon>0$ and take $N\in\Bbb N$ such that, if $n\geqslant N$ and if $x\in[a,b)$, then $|f_n(x)-f(x)|<\frac\varepsilon2$. Then, by the continuity of all the functions involved, if $n\geqslant N$ and if $x\in[a,b]$, then $|f_n(x)-f(x)|\leqslant\frac\varepsilon2<\varepsilon$.

Answer 2

This is true. In fact, if $F: [a;b] \to \Bbb R$ is any continuous function, then $$\sup_{x \in [a;b]} |F(x)| = \sup_{x \in [a;b)} |F(x)|$$ Now, since uniform convergence is defined using the supremum, there is no distinction between $[a,b]$ and $[a,b)$. In other words $$\lim_n \sup_{x \in [a;b]} |f_n(x)-f(x)| = \lim_n \sup_{x \in [a;b)} |f_n(x)-f(x)|$$