bounds for derivatives pass to the limit? Let $(f_n)_{n\in\mathbb{N}}$ a sequence of $C^1$ functions on $\mathbb{R}$ pointwise convergent to the $C^1$ function $f$.
I know that $f_n'(x)\geq a_n\cdot c(x)$ for all $x\in\mathbb{R},n\in\mathbb{N}$, with $c(x)>0$ and $a_n\rightarrow a>0$ as $n\to\infty$.
Is there a simple way to say that $f'(x)>0$?
By simple I mean that I would prefer not to die proving that $f_n'$ converges to $f'$...
 A: Your solution is  correct, but  the whole things is unnecessarily complicated because the assumption on derivative is stated in a cumbersome way: $f_n'(x)\ge a_n c(x)$... where we have to  talk about convergence of $a_n$ and continuity of $c$. All that really matters is that   

For every point $p$ there is a neighborhood $U$ and $\epsilon>0$ such that   $f_n'\ge  \epsilon $  in $U$ for all sufficiently large $n$.

Now the proof is transparent: since $f_n(x)-\epsilon x$ is nondecreasing in $U$ (for all sufficiently large $n$), the pointwise limit $f(x)-\epsilon x$ is also nondecreasing in $U$. Hence $f'(x)\ge \epsilon$ in $U$.
(You can check that the quoted condition is satisfied in your situation.)
A: I try to answer by myself. Assume $c$ is a continuous function.
Let $x>y$ and set $\underline{c}_{x,y}:=\min_{\xi\in[x,y]}c(\xi)>0$.
By the mean value theorem for all $n\in\mathbb{N}$ there is a $\xi_{n,x,y}\in[x,y]$ s.t.
$$f_n(y)=f_n(x)+f_n'(\xi_{n,x,y})(y-x) \geq f_n(x)+a_n\,c(\xi_{n,x,y})\,(y-x) \geq f_n(x)+a_n\,\underline{c}_{x,y}\,(y-x)$$
Passing to the limit $n\to\infty$
$$f(y) \geq f(x)+a\,\underline{c}_{x,y}\,(y-x)$$
Hence by the mean value theorem there exists $\xi_{x,y}\in\,]x,y[\,$ s.t.
$$f'(\xi_{x,y})\geq a\,\underline{c}_{x,y}$$
Now letting $y\to x+$ one obtains, by continuity of $f'$ and $c$, that
$$f'(x)\geq a\,c(x)>0$$
Is this reasonement correct?
