Let $[a,b] \subseteq \Bbb R$, let $f_n:[a,b] \to \Bbb R \space\space$ such that $\forall n>0 \space\space f_n$ is uniformly continuous on $[a,b]$. Let $f:[a,b] \to \Bbb R$, Suppose that $f_n \to f \space$ then $f\space$ is also uniformly continuous on $[a,b]$
This is a question that came to my mind when I was looking at a proof of this where $f_n \rightrightarrows f$.
It was raised to my awareness that the statement is not true in this case because its p.w convergence but Im not sure where the proof I state beneath fails, and why it doesnt fail for uniform convergence? any explanation would be really helpful.
Incorrect proof: let $\epsilon > 0\space$, $f_n$ is uniformly continuous so there exists $\delta_0 > 0$ such that for every $x,y \in [a,b]$ if $|x-y| < \delta_0$ then $|f_n(x) - f_n(y)| < \frac{\epsilon}{3}$
define $\delta = \delta_0$ let $x,y \in [a,b]$ such that $|x-y| < \delta$. since $f_n \to f$ then there exists $N_{0} \in \Bbb N \space\space , N_{0}>0$ such that for every $n \geq N_{0}$. $|f_n(x) - f(x)| < \frac{\epsilon}{3}$
And there exists $N_{1} \in \Bbb N \space\space , N_{1}>0$ such that for every $n \geq N_{1}$. $|f_n(y) - f(y)| < \frac{\epsilon}{3}$
let $n = \max\{N_{0}, N_{1} \} + 1$, So: $$|f(x) - f(y)| = |f(x) - f_n(x) + f_n(x) - f_n(y) + f_n(y) - f(y)| \\\ \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| < 3\frac{\epsilon}{3} = \epsilon$$
The main flaw is that the value of $n=\max(N_1,N_2) + 1$ is based on the quantities $x,y,\epsilon$. Therefore, as $x,y$ vary, so will this $n$. So, you cannot pick any "universal" $n$ at the start to use in your final triangle inequality, as varying $x,y$ will change the $n$, which will change the $\delta_0(n)$ you discuss at the beginning of your proof.