Prove the limit function $f$ of the sequence of differentiable functions $f_n$ is differentiable If every function in the sequence of functions $(f_n)$
with domain $I:[0,1]$ is differentiable and $f_n \to f$ uniformly, then $f$ is differentiable.
I forget the method to prove differentiability a little bit, should I try to prove the continuity of $f$?
 A: That is not the statement of the theorem of differentiability of the limit of a sequence of functions, which is less simple than the theorem for integrability.
Namely, there are 3 hypotheses:


*

*Each $f_n$ is differentiable.

*The sequence of derivatives converges uniformly on every bounded closed interval contained in $I$ ($I$ is not necessarily closed, nor bounded) to a function $g$.

*There exists $x_0\in I$ such that the sequence $\bigl(f_n(x_0)\bigr)$ converges.


Then: 
a) $(f_n)$ converges  uniformly on every bounded closed interval contained in $I$ to a function $f$.
b) $f$ is differentiable and $f'=g$.
A: I would just like to spell out Bernard's point. 
For a. Consider $x \in [x_0,y] \subseteq [0,1]$. 
$$ |f_n(x) - f_m(x) | \le | (f_n - f_m )|_{x_0}^x | + |f_n(x_0) - f_m(x_0) \le ||f'_n - f'_m||_\infty |x_0-y| + \varepsilon. $$ 
Thus, $f_n$ is uniformly convergent $[x_0,y]$.  
For b. With out loss of generality, it sufices to prove for right derivative. Let $x<y \in [0,1)$. For $t \in [x,y] \subseteq [0,1)$, define
$$ h_n(t):= \int_x^t f_n' \, ds$$
If $f'_n \xrightarrow{u} g$ on $[x,y]$, then for all $t \in [x,y]$,
$$ \Big| h_n - \int_x^t g \, ds \Big| \le ||h_n - g||_{\infty} |x-y|$$
by Triangle inequality. Thus, $h_n \xrightarrow{u} \int_x^t g \, ds $ on $[x,y]$. But we also know $h_n \xrightarrow{u} f(t)-f(x)$ by FTC. By uniqueness of limits, $f(t) -f(x)= \int_x^t g \,ds.$ Hence, $f'(x)=g(x)$. 
