Question:
Find a sequence of differentiable functions $f_n$ on $\mathbb{R}$ that converge uniformly to a differentiable function $f$, such that $f'_n$ converges pointwise but not uniformly to $f'$.
Attempt:
I have tried a number of possibilities, such as $f_n=x^n$ or $f_n=\frac{x^n}{n}$ but I don't know what the right approach is to construct the function. I am initially thinking that it's easiest to construct such a sequence of functions on the interval $[0,1]$ so that in the limit of $n$, part of the function goes to $0$ and the other part goes to $1$. However, this would make the resulting $f$ non-differentiable.
Best Answer
I think that$$f_n(x) = \dfrac{\sin(nx)}{n^2x}$$ will work. We have $f_n = \frac{1}{n}\textrm{sinc}(nx)$ and $\textrm{sinc}$ is uniformly bounded by $1$, so $f_n$ converges uniformly to zero.
Further, $$f_n' = \dfrac{\cos nx}{nx} - \dfrac{\sin nx}{n^2 x^2}$$ which should converge pointwise to zero, but $f_n'(\pi/n) = -1/\pi$ so the convergence can't be uniform. (These derivatives are all continuous too, with $f_n'(0) = 0$; you can convince yourself by writing out the Taylor series, $\cos y / y \sim 1/y - y/2 + O(y^3);$ $\sin y/y^2 \sim 1/y - y/6 + O(y^3)$ so the singularities "cancel"; alternately, you can use L'Hopital's rule).
(Here's a sketch that the derivatives converge pointwise to $0$: knowing that $f_n'(0) = 0$ for all $n$, just need to show that $$f_n' = \dfrac{1}{nx}(\cos nx - \textrm{sinc }nx)$$ can be made small for $x\neq 0$. The difference of cosine and sinc is bounded by $2$, and $\dfrac{1}{nx}$ can be made arbitrarily small for fixed $x \neq 0$.)