Use continuous limit theorem to prove if $f$ is uniformly differentiable, then $f'$ is uniformly continuous This is my first approaching to answer my own question in math stack exchange. Before that, I would define several terms according to Steven Abbott's Understand Analysis, which, by the way, the origin of this question.

*

*Derivative

Let $g:\mathbb{A}\rightarrow\mathbb{R}$ be a function defined on an interval $\mathbb{A}$. Given $c\in\mathbb{A}$, the derivative of $g$ at $c$ is defined by $g'(c)=\lim_{x\to c}\frac{g(x)-g(c)}{x-c}$.


*Continuous limit theorem

Let $(f_n)$ be a sequence of functions defined on $\mathbb{A}\subseteq\mathbb{R}$ that converges uniformly on $\mathbb{A}$ to a function $f$. If each $f_N$ is continuous at $c\in\mathbb{A}$, then $f$, is continuous at $c$.


*Uniformly differentiable

Given a differentiable function $f:\mathbb{A}\rightarrow\mathbb{R}$, let's say that $f$ is uniformly differentiable on $\mathbb{A}$ if, given $\epsilon>0\exists\delta>0$ such that if $0<|x-y|<\delta$, then $|\frac{f(x)-f(y)}{x-y}-f'(y)|<\epsilon$.
 A: Since $\mathbb{A}$ is an interval, it can be

*

*open interval

*closed interval

*semi-open-closed interval

Regardless, $\exists p,q\in\mathbb{R}$ such that $\mathbb{A}\subseteq[p,q]$ and I will assume closed interval since I deem it as the most complicated one.
$g_n$ will be defined as follow:
\begin{cases}
    \frac{f(x+\frac{q-x}{n})-f(x)}{\frac{q-x}{n}},& \text{if } x\in[p,q)\\
    f'(q),              & \text{if }x=q
\end{cases}
First I will prove all $(g_n)$ are continuous.
Case 1: $x\in[p,q)$
$g_n$ is a composition of various functions, including $f,\frac{q-x}{n}$. These functions are continuous at $x$. Therefore, the composition is continuous.
Case 2: $x=q$
Since $f'(q)$ exists, this means $\lim_{x\to q}\frac{f(x)-f(q)}{x-q}$ exists. By the sequential criterion for functional limits theorem, all $(y_n)\to q\land\forall n\in\mathbb{N},y_n\neq q$ will result in $\lim_{n\to\infty}\frac{f(y_n)-f(q)}{y_n-q}=f'(q)$, which exists.
Consider any sequence $(z_n)$ that approaches $q$ from the left and never touches $q$, $z_n+\frac{q-z_n}{n}$ will not touch $q$ either. Hence, this particular $z_n$ satisfies the condition of the above $(y_n)$. As a result, $\lim_{n\to\infty}\frac{f(z_n+\frac{q-z_n}{m})-f(q)}{\frac{q-z_n}{m}}=f'(q)=g_m(q)$.
From both cases, $\forall n\in\mathbb{N},g_n(x)$ is continuous at $A$. The last step is to prove the $(g_n(x)$ converges pointwise to $f'(x)$.
The problem assumed uniform differentiability. So for a given $\epsilon$, there is a $\delta$ response. Simply picking a $n_0\in\mathbb{N}$ such that $\forall n\geq n_0\frac{q-p}{n}<\delta$. By applying the definition of uniform differentiability will show that this applies to all $x$.
The process is completed. $(g_n)$ converges uniformly on $A$ on $f'$ and each of them is continuous. Henceforth, by continuous limit theorem, I declare $f'$ is continuous.
Please correct me if I am wrong.
