# $f_n\to f,$ uniformly on $S$ if and only if $f$ is continuous on $S$ and $\left| f_{k+n}(x)-f(x) \right|<\epsilon$

Let $\{f_n\}$ be a sequence of continuous functions, defined on a compact set $S$ and assume that $\{f_n\}$ converges pointwise on $S$ to a limit function $f.$ Prove that $f_n\to f,$ uniformly on $S$, if and only if, the following two conditions hold:

1. The limit function is continuous on $S;$

2. For every $\epsilon>0,$ $\exists\,m>0$ and $\delta>0,$ such that $n>m$ and $\left| f_k(x)-f(x) \right|<\delta$ implies $\left| f_{k+n}(x)-f(x) \right|<\epsilon,\;\forall\,x\in S$ and $\forall\,k\in \Bbb{N}.$

MY WORK SO FAR

Let $\epsilon>0.$ Assume that $f_n\to f,$ uniformly on $S$. Then, $\exists\,N=N(\epsilon)$ such that $,\forall\,n\geq N,\forall\,x\in S$ \begin{align} \left| f_{n}(x)-f(x) \right|<\frac{\epsilon}{3}\end{align} In particular, for $n=N,$ \begin{align} \left| f_{N}(x)-f(x) \right|<\frac{\epsilon}{3},\quad\forall\,x\in S\end{align} And for $x=x_0,$ \begin{align} \left| f_{N}(x_0)-f(x_0) \right|<\frac{\epsilon}{3}\end{align} Since $\{f_n\}$ is continuous at $x_0,$ then there exists $\delta>0,$ such that $\forall x\in S$ with $|x-x_0|<\delta$, we have \begin{align} \left| f_{n}(x)-f_{n}(x_0) \right|<\frac{\epsilon}{3},\forall\,n\in \Bbb{N}\end{align} In particular, for $n=N,$ \begin{align} \left| f_{N}(x)-f_{N}(x_0) \right|<\frac{\epsilon}{3},\text{whenever}\;\;|x-x_0|<\delta\end{align} Hence, for $|x-x_0|<\delta,$ \begin{align} \left| f(x)-f(x_0) \right|&=\left| f(x)-f_{N}(x)+f_{N}(x)-f_{N}(x_0)+f_{N}(x_0) -f(x_0)\right|\\&\leq\left| f(x)-f_{N}(x)\right|+\left|f_{N}(x)-f_{N}(x_0)\right|+\left|f_{N}(x_0) -f(x_0)\right|\\&<\left| f(x)-f_{N}(x)\right|+\left|f_{N}(x)-f_{N}(x_0)\right|+\left|f_{N}(x_0) -f(x_0)\right|\\&<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}=\epsilon\end{align} And we're done!

Please, any hint or help in proving 2? Thanks for your time and efforts!

• I am a bit confused about 2. As you note, there exists $N \in \mathbb N$ such that $|f_n(x)-f(x)|<\varepsilon$ for every $x \in S$ whenever $n \geq N$. Just take $m=N$. Then $n>m$ implies $k+n>m$ for every $k \in \mathbb N$... – Matt A Pelto Sep 19 '18 at 2:35