Proving $f_n(x)$ not uniformly convergent

Question

Let $f_n(x)$ be a sequence of functions that converge for $f(x)$ on $x\in[a,b]$ but it's not uniformly convergent on the same range of $x$.

Prove that $f(x)$ is not uniformly convergent on $x\in(a,b)$.

Here is my attempt : using contrapositive I assumed $f(x)$ is uniformly convergent on $x\in(a,b)$ then $\sup_{x\in(a,b)}|f_n(x)-f(x)|\to 0$.

I'm assuming that from this point I should conclude that $\sup_{x\in[a,b]}|f_n(x)-f(x)| \to 0$ which makes the statement true but I'm not sure what should I do next and is it possible that ($\limsup_{x\in(a,b)}|f_n(x)-f(x)|=\limsup_{x\in[a,b]}|f_n(x)-f(x)|$).

If it was the other way around it would be much easier I suppose.

Answer 1

Assume $f_n$ converges uniformly to $f$ on $(a,b)$ and pointwise to $f$ on $[a,b]$. Then, for all $\epsilon>0$, there exists

1. $N_1(\epsilon)$ st. for all $n>N_1(\epsilon)$ and $a<x<b$, $|f_n(x)-f(x)|<\epsilon$,
2. $N_2(\epsilon)$ st. for all $n>N_2(\epsilon)$, $|f_n(a)-f(a)|<\epsilon$,
3. $N_3(\epsilon)$ st. for all $n>N_3(\epsilon)$, $|f_n(b)-f(b)|<\epsilon$

Therefore, if $N(\epsilon)= \max\{N_1(\epsilon),N_2(\epsilon),N_3(\epsilon)\}$, we obtain: for all $\epsilon>0$, there exist $N(\epsilon)$ st. for all $n>N(\epsilon)$ and $x\in [a,b]$, $|f_n(x)-f(x)|<\epsilon$.

This is a contradiction, since $f_n$ is not uniformly convergent on $[a,b]$.

Answer 2

An elementary fact about supremums is that $\sup(A \cup B) = \max{\{\sup A, \sup B\}}$ for any subsets $A$ and $B$ of the extended real line.

Using this fact, we have that $$ \sup_{x\in[a,b]} \lvert f_n(x) - f(x) \rvert = \max{\{\lvert f_n(a) - f(a)\rvert, \lvert f_n(b) - f(b)\rvert, \sup_{x\in(a,b)} \lvert f_n(x) - f(x) \rvert\}}. $$

If $f_n$ is uniformly convergent on $(a, b)$, then the above expression is the maximum of three sequences, each of which tends to zero, so it must tend to zero as well. This proves the contrapositive that you were after.