Suppose that $f$ is holomorphic on all of $\mathbb{C}$ and that $$\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}f(z)$$

exists, uniformly on compact sets, and that this limit is not identically zero. Then the limit function $F$ must be a very particular kind of entire function. Can you say what kind ?


Functions of an order that are at most $1$, satisfy the following condition:

$$\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}f(z) = \Psi$$

In order to prove our Conjecture, one must construct an entire function that is at most of order of 1, and show that the $\lim_{n \rightarrow \infty}f(z)$ exists, and converages to some constant $\Psi$ the construction of our entire function can be seen in $\text{Lemma (1.1)}$

$\text{Lemma (1.1)}$

For arbitrary positive numbers $p$ and $\sigma$ one can construct an entire function of order $p$ and type $\sigma$ using in $(1.2)$



$\text{Lemma (1.2)}$

To attack our Conjecture, we substitute $(1.2)$ into $(1.1)$ the we have the following in $(1.3)$


$$\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}=\Psi$$


One performs the substitution to make clear one is trying to show the condition $\lim_{n \rightarrow \infty}(\frac{\partial}{\partial z})^{n}f(z) = \Psi$ is true

Since $(\frac{\partial }{\partial z})^{n}=(\frac{d}{dz})^{n}$, one can perform in $(1.4)$


$$\lim_{n \rightarrow \infty}(\frac{d}{d z})^{n}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}=\Psi.$$


At this stage of the proof, one wonders why didn't one just take $(\frac{\partial }{\partial z})^{n}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}$, one could that relies on a certain criteria being met which may not work for certain cases such as this one

$\text{Lemma 3}$

Changing $f(z)$ into a product representation, our recent developments now are as follows, in $(1.5)$


$$\lim_{n \rightarrow \infty}(\frac{d}{d z})^{n}\text{exp}\sum_{n=1}^{\infty}(ep\sigma/n)^{n/p}z^{n}=\Psi.$$

$$\lim_{n \rightarrow \infty}(\frac{d}{d z})^{n}\prod_{n=1}^{\infty} \exp (ep\sigma/n)^{n/p}z^{n}= \Psi.$$


The conservation, from sum to product was conducted by the following formula:

$$\exp \sum s_n = \prod e^{s_n}.$$

From $\text{Lemma 3}$ it seems like one can exploit the Logarithmic derivative of the product $\prod_{n=1}^{\infty} \exp (ep\sigma/n)^{n/p}z^{n}$, which at every point where $\prod_{n=1}^{\infty} \exp (ep\sigma/n)^{n/p}z^{n}$ is non-zero one would have $\lim_{n \rightarrow \infty} \sum_{n=1}^{\infty}\frac{a'n(z)}{1+a_{n}(z)}$. So far is my appoarch correct I feel what I have so far is not rigors, perhaps someone could help tighten my reasoning ?

  • 2
    $\begingroup$ I'm not sure I parsed the question correctly. You're assuming that the functions $F_n(z) = f^{(n)}(z)$ converge uniformly on compact sets to some function $F$. It's entire since each $f^{(n)}$ is, but I don't see why it should be constant (you're taking $\Psi$ to be the limit of the sequence as a function of $z$, right, not at an arbitrary fixed $z$?); take $f(z) = e^z$, for example. $\endgroup$
    – anomaly
    Commented Oct 12, 2017 at 20:20
  • $\begingroup$ Well, that's assumption posed by the question, but also one has to show that the limit is not zero, so one would have to pick a specific function of a certain order. $\endgroup$
    – Zophikel
    Commented Oct 12, 2017 at 20:27
  • $\begingroup$ It converges to $Ce^z$ for some $C$. This is easy to see. We can prove it with Taylor series, or simply observe. $\frac{d}{dz} F_n = F_{n+1}$ and $\lim_{n\to\infty} F_{n}'(z) = \lim_{n\to\infty}F_{n+1} (z) = F(z)$. So therefore $F$ is fixed by the derivative operator and must be $Ce^z$. $\endgroup$
    – user335907
    Commented Oct 12, 2017 at 20:28
  • $\begingroup$ Ahhhh, ok I didn't see it at first, was my initial appoarch wrong ? $\endgroup$
    – Zophikel
    Commented Oct 12, 2017 at 20:29
  • 1
    $\begingroup$ It's not that it's wrong, per se. It's that it seems a little convoluted and hard to follow (some of the indexes are wrong). Sometimes the answer is simpler than you first think, that's the trouble with math. Sometimes it's not about cracking a nut with a sledgehammer it's about finding a clever way to twist it open. $\endgroup$
    – user335907
    Commented Oct 12, 2017 at 20:32

1 Answer 1


Let $f_n = f^{(n)}$, and assume that $f_n$ converges locally uniformly to some $F$ (which is necessarily entire as well).

Then $f_n'$ converges locally uniformly (since $f_n' = f_{n+1}$) to the same function $F$. By general theory for uniform convergence, the limit of $f_n'$ is actually the derivative of $f$, i.e. $F = F'$.

  • $\begingroup$ Ahh, beat me to it. $\endgroup$
    – user335907
    Commented Oct 12, 2017 at 20:29
  • $\begingroup$ Can you clarify what you meant by general theory of uniform convergence ? $\endgroup$
    – Zophikel
    Commented Oct 12, 2017 at 20:37
  • 1
    $\begingroup$ If $f_n$ converges pointwise to $f$ (in fact convergence at a single point suffices) and $f_n'$ converges uniformly to $g$, then $f$ is differentiable and $f'=g$. This is true both in the real and complex setting. For holomorphic functions, it's even true in general that (locally) uniform convergence of $f_n$ implies locally uniform convergence of $f_n'$. $\endgroup$
    – mrf
    Commented Oct 12, 2017 at 20:41
  • $\begingroup$ Ahhh ok I did consider an appoarch like this but having the topological space $U$ map to a metric space, would his be an appropriate route to go ? $\endgroup$
    – Zophikel
    Commented Oct 12, 2017 at 22:29

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