# Reference request: Rate of convergence of sequence of functions

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.

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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.

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Thanks for pointing this out. I fixed the missing subscripts. –  Liviu Nicolaescu Feb 10 '12 at 10:14
Thanks very much for the answer that you provided. At the time of your post, I was very new to MO, and I didn't even know how to make comments so as to thank you at the time. :-) –  Euplio M. Apr 20 '12 at 4:32