Let $f_n$ is a sequence of differentiable function on some opens set $E \subset \mathbb{R}$. Suppose $f_n$ converges uniformly to $f$ and $f'_n$ converges uniformly to $g$ then $f'=g.$ What is the counter example to show that uniform convergence cannot be replaced by pointwise convergence.
More precisely I am looking for a sequence of differentiable functions $f_n$ which converges pointwise to a function $f$ and $f'_n$ converges pointwise to a function $g$ but $f' \neq g.$
P.S: Examples where $f$ is not differentiable would also suffice but examples where $f$ is differentiable will be great.
Simple and easy to verify examples are appreciated. Thanks in advance.
Let's take a set of continuous functions that converges pointwise at a function with jumps on some interval. An easy example is:
$g_n(x)=(1-x)^n, 0 < x <1, g_n(x)=(1+x)^n, -1 <x <0, g(0)=1$;
Then clearly $g_n$ is continuous on $(-1,1)$ and converges pointwise to $0$ except at $0$ where $g_n(0) \to 1$
Consider $f_n$ a primitive (antiderivative) of $g_n$ eg
$f_n(x)=\int_0^xg_n(t)dt, -1<x<1$
so expliciting the above one has that
$f_n(x)=\frac{1-(1-x)^{n+1}}{n+1}, 0 < x <1, f_n(x)=\frac{(1+x)^{n+1}-1}{n+1}, -1<x<0, f_n(0)=0$
But clearlly $f_n'=g_n$ by construction, while $f_n \to 0$ on $(-1,1)$