Uniform convergence and differentiation How does one show that given that $g\in C^1(a,b)$, the sequence of functions 
$$
g_n=n\left(g\left(x+{1\over n}\right)-g(x)\right)
$$ 
converges uniformly on all closed intervals in $(a,b)$? I assume the limit function is $g'(x)$.
 A: Ok, so one of the proof is the following. Let $I$ be a closed subinterval of $(a,b)$ so $g'\in C(I)$ and hence uniformly continuous on $I$. Let us put $n_0$ such that $I' = I\cup\{x+1/n:x\in I\}\subset (a,b)$. By Lagrange Mean Value Theorem for $n>n_0$ we have:
$$
g_n(x) = g'(\xi_n(x))
$$
where $\xi_{n}(x)\in[x,x+1/n]$.
Since $g'$ is uniformly continuous on a closed subinterval $I'\subset(a,b)$ it means that for any $\epsilon>0$ there is $\delta(\epsilon)>0$ such that if $|x-y|<\delta(\epsilon)$ where $x,y\in I'$ it holds that $|g'(x)-g'(y)|<\epsilon$. 
That means, that for any $\epsilon>0$ there is $N>\max\left\{n_0,\frac{1}{\delta(\epsilon)}+1\right\}$ such that for any $n>N$ it holds that
$$
\sup\limits_{x\in I}|g_n(x) - g'(x)|\leq\sup\limits_{x\in I'}|g'(\xi_n(x)) - g'(x)|<\epsilon
$$
which proves the uniform convergence $g_n\to g'$ on $I$.
A: Starting fresh.  
Letting $h_n=\frac{1}{n}$, then $g_n(x) = \frac{g(x+h_n) - g(x)}{h_n}$
For any $x$, by the mean value theorem, $g_n(x) = g'(y)$ for some $y\in [x,x+h_n]$.
Let $C=[u,v]$ be our closed interval. Note that $g_n$ is defined on $C$ only if
$v+h_n<b$.  Let $n_0$ be the least $n$ such that $v+h_n<b$, and let $C'=[u,v+h_{n_0}]$.
Since $C'$ is compact, and $g'$ is continuous, $g'$ must be uniformly continuous on $C'$.  Therefore, given $\epsilon>0$ there is a $\delta>0$ such that if $|x-y|<\delta$, $|g'(x)-g'(y)|<\epsilon$ for $x,y\in C'$.
Choose $N>\max(n_0,\frac{1}{\delta})$.  Then for every $x\in C$, and any $n>N$, there is a $y$ in $[x,x+h_n]\subset C'$ such that $g_n(x)=g'(y)$.  But then, since $x,y\in C'$:
$$|g_n(x)-g'(x)| = |g'(y)-g'(x)| < \epsilon$$
since $|y-x| < h_n < \frac{1}{N} < \delta$.
So the $g_n$ converge uniformly to $g'$.
So this essentially results from:


*

*The intermediate value theorem when derivatives are continuous

*The fact that a continuous function on a compact set is uniformly continuous on that set

*Currently, the proof uses that $h_n$ decreases and has limit 0, but you can prove it with general $h_n\rightarrow 0$, even allowing some $h_n=0$ if you define $g_n(x)=g'(x)$ for those $n$.

