Suppose you are given a sequence of functions $f\_n \rightarrow F$ with a certain notion of convergence. Suppose that in your setting where this implies that $f\_n^{'} \rightarrow F^{'}$ with the same notion of convergence.


We will say that ``the rate of convergence of the sequence of derivatives is at least as fast as the rate of convergence of the sequence of functions'' if the following holds:


there exists a constant $C>0$ (independent of $n$) such that 


$|\frac{f\_n^{'}-F^{'}}{f\_n-F}| \leq C$


In plain language, what conditions should your sequence of functions satisfy
in order to guarantee that the rate of convergence under taking the derivative gets no worse?


In particular, let us suppose that we are in the setting where our functions
are complex-analytic, and the convergence is uniform convergence. In this situation,
we have the ``Weirstrass Uniform Convergence Theorem'' that tells us that the derivatives
converge uniformly as well. But can we conclude anything about the rate of convergence?


Any type of reference where a problem of this type is considered would be much appreciated.



Reference request: Rate of convergence of sequence of functions
Consider the special case of holomorphic functions. We can assume that $F=0$. Denote by $D\_R$ the disk $\lbrace |z|\leq R\rbrace$. Denote by $Z(n, R)$ the number of zeros of $f\_n$ in the interior of $D\_R$. Then


$$2\pi Z(n,R)= \left|\int\_{\partial D\_R} \frac{f\_n'}{f\_n} dz\right| \leq \int\_{\partial D\_R} \left|\frac{f\_n'}{f\_n}\right| dz. $$


Let


$$f\_n(z) = \frac{1}{n!} \prod\_{k=1}^n\left( z-\frac{1}{k}\right). $$


Observe that


$$ \sup\_{z\in D\_R} |f\_n(z)|\leq \frac{(R+1)^n}{n!}, $$


so that $f\_n$ converges uniformly on compacts to $F\equiv 0$. On the other hand for a fixed $R>1$ we have


$$ Z(n, R)= n. $$


Now you see the problem.


