Uniform convergence of the derivative

Question

Let g be a continuously differentiable function on $\Bbb R$. Let $f_n(x)= n(g(x+ 1/n)-g(x))$. Prove that $f_n \longrightarrow g'$ on $[-R,R]$ for each $R \gt0$.

Well, I see that the problem here is to find a uniform estimate of $\vert \frac{ g(x+ \frac{1}{n}) - g(x)}{n} - g'(x)\vert$. But how can I argue that this estimate exists?

Answer 1

$$ \frac{g(x+1/n)-g(x)}{1/n}=g'(x+\theta(x,n)),\quad\text{where } 0\le\theta(x,n)\le\frac1n, $$ by Lagrange's theorem. The function $g'(x)$ is continuous and hence uniformly continuous on any finite interval; hence $g'(x+\theta(x,n))\to g'(x)$ uniformly on any finite interval.

Answer 2

$f_n$ clearly converges pointwise to $g'$. We need to show the convergence is uniform on the compact set $[-R,R]$.

By the Mean Value Theorem, $f_n(x)=\frac{g(x+1/n)-g(x)}{1/n}=g'(c_n)\quad $ for some $x<c_n<x+1/n$.

Let $\epsilon >0$. As $g'$ is uniformly continuous on the compact set $[-R,R]$ there is a $\delta >0$ such that $\vert g'(x)-g'(y)\vert <\epsilon $ whenever $\vert x-y\vert <\delta$. Now choose $N$ so large that $1/N<\delta $.

Then, if $x\in [-R,R]$ and if $n>N$, we have $f_n(x)=g'(c_n);\quad x<c_n<x+1/n<x+1/N<x+\delta$

so $\vert f_n(x)-g'(x)\vert =\vert g'(c_n)-g'(x)\vert <\epsilon ,\quad $which is what we need.