Termwise differentiation of sequence of rational functions when the uniform limit is analytic

Question

Given a sequence $\{f_n(x)\}$ of rational functions which converges uniformly to the analytic function $\{g(x)\}$ on $[a, b]$ ($f_n(x)$ are defined on $[a, b]$ and hence are analytic), what can we say about the sequence of derivatives $\{f'_n(x)\}$ -- does it necessary converge to $\{g'(x)\}$ on $[a, b]$?

Answer 1

Derivatives can seriously misbehave. Let $h_n(x) = n^{-1}\sin (n^2x)$ on $[0,2\pi]$. By the Weierstrass theorem, there is a polynomial $p_n$ such that $\sup |p_n - h_n| <\frac1{2n}$ on $[0,2\pi ]$. Then $p_n\to 0$ uniformly. But $p_n$ oscillates $n^2$ times between $\pm \frac{1}{2n}$, which implies $$ \int_0^{2\pi}|p_n'| \ge n $$ So, we do not have either uniform convergence or convergence in the $L^1$ sense.

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That said, we do have convergence in the sense of distributions on the interval $(a,b)$. This is simply because uniform convergence implies distributional convergence, which passes to derivatives automatically. Note this has nothing to do with functions being rational/analytic. The downside is, it's a very weak mode of convergence. But it's something.