Suppose $f_n :[a,b] \rightarrow \mathbb{R}$ are differentiable functions (need not be $C^1$) with $f_n \rightarrow 0$, $f_n ' \rightarrow g$ pointwise. Can we say that $g=0$ in some sense? (Say, a.e.)
In particular, is it possible for $g$ to equal $1$ everywhere?
cf) Interchanging pointwise limit and derivative of a sequence of C1 functions This question deals with the $C^1$ case.
Let $f(x) = \{x\} - \lfloor x \rfloor$ be the fractional part of $x$. We have $f'(x) = 1$ if $x \not\in \mathbb{Z}$ but $f'(x)$ undefined if $x \in \mathbb{Z}$. Set $f_n(x) = f(\sqrt{p_n} x)/\sqrt{p_n},$ where $p_n$ is the $n$th prime. We have $f_n \to 0$ pointwise. For the derivatives, for every $x$ there will be maximally one $n$ for which $f_n'(x)$ is not defined. After that $n$ we have $f_n'(x) = 1$ so $f_n' \to 1$ pointwise.
However, we want $f_n'$ to be defined everywhere, so let's try to fix that. Let $\varphi \in C_c^\infty(\mathbb{R})$ be a mollifier and take $(\epsilon_n)$ converging to $0$. Then set $g_n(x) = (f*\varphi_{\epsilon_n})(\sqrt{p_n} x)/\sqrt{p_n},$ where $\varphi_\epsilon(x) = \epsilon^{-1}\varphi(x/\epsilon)$. Then $g_n$ will be smooth and differ from $f_n$ only close to the jumps. Wouldn't this sequence satisfy $g_n \to 0$ pointwise and $g_n' \to 1$ pointwise?
