Converge uniformly on close interval implies on open interval

Question

Suppose $f_n(x)$ converges uniformly to $f(x)$ on $(a,b)$. Does this imply that $f_n(x)$ converges uniformly to $f(x)$ on $[a,b]$?

I have a theorem in my book that states the argument is correct if $f_n(x)$ are all continuous on $[a,b]$. But is it true for other cases?

Answer 1

No, it is not true in general. Define $f_n\colon[0,1)\longrightarrow\mathbb R$ by$$f_n(x)=\begin{cases}1&\text{ if }x=0\\\frac xn&\text{ otherwise.}\end{cases}$$Then $(f_n)_{n\in\mathbb N}$ converges to the null function on $(0,1)$, but not on $[0,1)$.

Answer 2

Correct me if wrong:

$f_n(x) = 0$ for $x \in (a,b).$

$f_n(x)=1$ for $x=a$, or $x=b.$

Theorem:

$ f_n$, continuos, converges uniformly on [a,b] to f(x) implies $f(x)$ continuos on $[a,b]$

Hence choose a sequence $f_n(x)$ that is not continuous at $x=a$ or $x=b.$