Show that f is uniformly continuous and that $f_n$ is equicontinuous

Question

$f_n: A \rightarrow \mathbb{R}$,$n \in \mathbb{N}$ is a sequence of functions defined on $A \subseteq \mathbb{R} $. Suppose that $(f_n)$ converges uniformly to $f: A \rightarrow \mathbb{R}$, and that each $f_n$ is uniformly continuous on $A$.

1.) Can you show that $f$ is uniformly continuous on A?

2.) Can you show that $(f_n)$ is equicontinuous?

1. We are given that$(f_n)$ converges uniformly to $f$. This means that for every $\epsilon>0$ there exists an $N\in \mathbb{N}$, such that $|f_n(x)-f(x)| <\epsilon$ whenever $n \ge N$ and $x \in A$. We have to show that $f$ is uniformly continuous on $A$, which means that for every $\epsilon >0$ there exists a $\delta>0$ such that $|x-y|<\delta$ implies |$f(x)-f(y)|<\epsilon$

2. We need to show that for every $\epsilon>0$ there exists a $\delta>0$ such that $|f_n(x)-f_n(y)|< \epsilon$ for all $n\in \mathbb{N}$ and $\|x-y|< \delta$ in $A$

Answer 1

For 1, let $\epsilon >0$. Then pick $n$ such that $|f_n(x) - f(x)| < \epsilon/3$ on $A$. By uniform continuity of $f_n$, there exists a $\delta$ such that $|x-y| < \delta \Longrightarrow |f_n(x)-f_n(y)| < \epsilon/3$. Now if $|x-y| < \delta$, $$ |f(x)-f(y)| = |f(x)-f_n(x)+f_n(x)-f_n(y)+f_n(y) -f(y)| \leq $$ $$|f(x)-f_n(x)| + |f_n(x)-f_n(y)| + |f_n(y) -f(y)| < \epsilon$$

Answer 2

For 2, let $\epsilon > 0$. Let $N$ be such that $n \geq N \Rightarrow |f_n(x) - f(x)| < \epsilon/3 $ on $A$. By (1), we know that $f$ is uniformly continuous so there exists a $\delta^*$ such that $|x-y|< \delta^* \Rightarrow |f(x)-f(y)| < \epsilon/3$.

For each $i < N$, there exists a $\delta_i$ such that $|x-y| < \delta_i \Rightarrow |f_i(x) - f_i(y)| < \epsilon$. Now let $\delta = \min\{\delta^*, \delta_1, ..., \delta_{N-1}\}$. I'll leave it to you that this $\delta$ will be good enough to show equicontinuity for all $(f_n)_{n \in \mathbb{N}}$