Uniform bound on derivatives and uniform convergence If $f_n:[0,1]\to \mathbb{R}$ are differentiable, $|f_n'(x)|\leq C$ for all $n\in\mathbb{N}$ and $x\in [0,1]$, $f_n\to f$ uniformly, $f_n'(x)\to g(x)$ pointwise and $f$ is differentiable, can we conclude that $f'=g$? Equivalently, can we conclude that 
$$
\lim_{n\to \infty}\lim_{h\to 0}\frac{f_n(x+h)-f(x)}{h}=\lim_{h\to 0}\lim_{n\to \infty}\frac{f_n(x+h)-f(x)}{h}
$$
given the assumptions above? My guess is no, but I am unsure of a counter-example.

No. For a counterexample on the whole $\Bbb R$ (but the difference is unessential), consider the $C^\infty$ bump $$\Phi(x)=\begin{cases}\exp\frac1{x^2-1}&\text{if }-1<x<1\\ 0&\text{if }x\le-1\lor x\ge 1\end{cases}$$ and $g_n(x)=\Phi(nx)$, $f_n(x)=\int_{-\infty}^x g_n(y)\,dy$. Then: $$\begin{align} \lvert g_n(x)\rvert&\le \max_{x\in[-1,1]}\lvert\Phi(x)\rvert\\ g_n(x)&\to \begin{cases}\Phi(0)&\text{if }x=0\\ 0&\text{if }x\ne 0\end{cases}&\text{ pointwise}\\ f_n(x)&\to 0&\text{ uniformly}\end{align}$$
because $$\left\lvert\int_{-\infty}^x g_n(y)\,dy\right\rvert\le \int_{-\infty}^x\lvert \Phi(ny)\rvert\,dy= \frac1n\int_{-\infty}^{nx}\Phi(y)\,dy\le \frac1n\int_{-\infty}^\infty\Phi(y)\,dy$$ It is clear that the limit of $g_n$ isn't the derivative of anything because it hasn't got the IVP.
