In 《Analysis》of Tao Remark 14.7.2, He says theorem 14.7.1 is still true when the functions $f'_n$ are not assumed to be continuous:
14.7.1: Let $(f_n)$ be a sequence of differentiable functions defined on the closed interval$[,]$, and assume $(f'_n)$ converges uniformly to function $g$. If there exists a point $_0\in[a,b]$ where $f_n(_0)$ is convergent, then $(f_n)$ converges uniformly to a function $f$ and $f'=g$
And in Exercise 14.7.2 he says you could first prove that: with large enough m,n: $$ d_\infty(f'_n,f'_m)\le\varepsilon \Rightarrow |(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))|\le\varepsilon|x-x_0|\text{ }...(1)$$ How to prove it? Tao says without assuming $f'_n$ is continuous, you cannot use the fundamental theorem of calculus, but you can use mean-value theorem. but I don't think mean-value theorem can work there, here is my process to prove $(1)$:
from mean-value theorem we get:
$$f_n(x)-f_n(x_0)=f'_n(x_1)(x-x_0),\text{ for some }x_1\in(x,x_0)\text{ }...(2)$$
$$f_m(x)-f_m(x_0)=f'_m(x_2)(x-x_0),\text{ for some }x_2\in(x,x_0)\text{ }...(3)$$
If $|f'_n(x_1)-f'_m(x_2)|\le\varepsilon\text{ }...(4)$ is true, we can get $(1)$ from $(2)-(3)$.
If $x_1=x_2$, then for large enough m, n, we indeed have $|f'_n(x_1)-f'_m(x_1)|\le\varepsilon$ because function sequence $(f'_n)$ is uniform convergent. But if $x_1\neq x_2$, the uniform convergence of $(f'_n)$ can't conclude $|f'_n(x_1)-f'_m(x_2)|\le\varepsilon$ for large enough m,n.
So for all the above, my question is how to prove $(1)$ ?
Or can you directly prove 14.7.1 without assuming $f'_n$ to be continuous?
I got the answer myself, see $f_n(x)-f_m(x)$ as an whole.
Mean value theorem in this uniform convergence proof