Let $g$ be a continuously differentiable function on $\mathbb{R}$. Let $f_n(x)= n(g(x+\frac{1}{n})-g(x))$, then $f_n \to g'$ uniformly on $[-R,R]$ for each $R>0$
(This question has been asked before by others but I want to discuss my approach and would like to learn why my approach is not working.)
Here's my approach: Clearly $f_n \to g'$ pointwise and since $g$ is continuously differentiable on $\mathbb{R} \implies g'$ is continuous on $[-R,R]$, where $R>0$ also it further implies that $g'$ is uniformly continuous on $[-R,R]$. So for $\epsilon>0$ there exists $\delta > 0$ such that for $x,y \in [-R,R]$ with $|x-y|<\delta \implies |g'(x) - g'(y)|<\epsilon.$ Now for any $x\in \mathbb {R}$ since $g$ is differntiable on $\mathbb{R}$, we can use MVT. So by MVT there exists $t_n \in (x,x+\frac{1}{n})$ such that $g'(t_n) = \frac{g(x+\frac{1}{n}) -g(x)}{\frac{1}{n}}$. Also by Archimedean property $\exists N \in \mathbb{N}$ such that $\frac{1}{N}< \delta$. So for $n \geq N$ and for $x\in [-R,R]$ we have $|x - t_n|< \frac{1}{n}\leq \frac{1}{N}<\delta$.
Hence $|g'(t_n) - g'(x)| < \epsilon$ i.e $|\frac{g(x+\frac{1}{n}) -g(x)}{\frac{1}{n}}- g'(x)| < \epsilon$, So $|f_n(x) -g'(x)|<\epsilon$ for all $x \in [-R,R]$ and for all $n \geq N$.
I think the problem lies with this argument is that when $x= R$ the $R + \frac{1}{N}$ and $t_N$ doesn't lie in $[-R,R]$.
Also, I just noticed that if $g'$ were uniformly continuous on $\mathbb{R}$, my argument might have worked.
So how to modify the argument to get the required result? Thanks in advance.