Uniform convergence and differentiable functions proof If $f_n \to f$ is pointwise on $[a,b]$ and each $f_n$ is differentiable, and $(f'_n)$ converges uniformly on $[a,b]$ to some function $g$, I want to prove that $f'=g$.
So if we let $\epsilon > 0 $ and fix $c \in [a,b]$, then to show $f'(c)$ exists and equals $g(c)$ we note that
$$\lim_{x \to c} \frac{f(x)-f(c)}{x-c} = f'(c)$$
So we must find some $\delta > 0 $ such that $\left| \frac{f(x)-f(c)}{x-c} - g(c) \right| < \epsilon$ whenever $0 < | x-c | < \delta$.
The next step is where I am having difficulty in understanding this proof.
Using the triangle inequality:
$$\left| \frac{f(x)-f(c)}{x-c} - g(c)\right| \leq \left| \frac{f(x)-f(c)}{x-c} - \frac{f_n(x)-f_n(c)}{x-c}\right| + \left|\frac{f_n(x) - f_n(c)}{x-c} - f'_n(c)\right| + |f'_n(c) - g(c)|$$
so I just want to make 100% sure on the intuition for adding/substracting $\frac{f_n(x)-f_n(c)}{x-c}$ and $f_n'(c)$
Firstly we are given $f'_n \to g$ so it is apparent we can make the third term arbitrarily small $|f'_n(c) - g(c)|$
Secondly for the middle term, is it because we assume $f_n(c)$ is differentiable? Or is it because $f'_n(c)$ is differentiable? My reasoning suggests the former, but the textbook says the latter, which I am suspecting is a typo.
Finally, and most importantly, how do we handle the first term to show it can be made arbitrarily small?
 A: As to the first term:
If $x \neq c$, let $$\varphi_n(x)=\frac {f_n(x)-f_n(c)}{x-c}$$ and $$\varphi(x)=\frac {f(x)-f(c)}{x-c}$$ It is true (see the algebraic limit properties of sequences) that $$\varphi_n(x) \rightarrow \varphi(x)$$ pointwise, but here this is not enough.
We need to prove the uniform convergence of $\{\varphi_n\}$.
Now, applying the MVT to the function $\,f_m-f_n\,$, one has $$\varphi_m(x)-\varphi_n(x)=f'_m(\alpha)-f'_n(\alpha)$$ with $\alpha \in [a,b]\,$ and depending on $x$ but the uniformly convergence of $\{f'_n\}$ assures the existence of $N_1$ such that $$|\varphi_m(x)-\varphi_n(x)|< \frac \varepsilon 3$$ for all $\,m,n \ge N_1$ and every $x \in [a,b]$.
Finally, from (see the absolute property of sequences) $$|\varphi_m(x)-\varphi_n(x)| \rightarrow |\varphi(x)-\varphi_n(x)|$$ it follows that (see the order property of sequences)  $$|\varphi(x)-\varphi_n(x)| \le \frac \varepsilon 3$$ for all $\,n \ge N_1$ and every $x \in [a,b]$.
A: Firstly, we can fix $x$ which makes the second term $|\frac{f_n(x) - f_n(c)}{x-c}-f_n^{'}(c)|$ sufficiently small, then
 $|\frac{f(x)-f(c)}{x-c} - \frac{f_n(x) - f_n(c)}{x-c}| = |\frac{f(x) - f_n(x)}{x-c} + \frac{f_n(c) - f(c)}{x-c}| \leq |\frac{f(x) - f_n(x)}{x-c}|+|\frac{f_n(c) - f(c)}{x-c}|$
By pointwise convergence of $f_n$ at $x$ and pointwise convergence at $c$, exist $N_1$ and $N_2$ such that for all $n>N_1$, we have $|f(x) - f_n(x)| < \frac{\epsilon}{2}$ and for all $n>N_2$, we have $|f(c) - f_n(c)| < \frac{\epsilon}{2}$. Then for all $n>max\{N_1,N_2\}$, we have $|\frac{f(x)-f(c)}{x-c} - \frac{f_n(x) - f_n(c)}{x-c}|<\epsilon$.
