Question: Let $f$ be entire such that $(1+|z|^k)^{-1}\frac{d^m f}{dz^m}(z)$ is bounded for some $k, m$. Prove $d^nf/dz^n(z)$ is $0$ for sufficiently large $n$. How large should $n$ be, in terms of $k$ and $m$?

Answer: I do actually have the solution for this question, but I am having a hard time understanding it, so I'm hoping someone can walk me through the details. First, we have:

$$\frac{d^mf}{dz^m}(z) \le c(1+|z|^k),$$

for some constant $c$. Now, we would like to show that $d^mf/dz^m(z)$ has a pole at infinity of order at most $k$. I'm getting stuck here, because in order to show that $d^mf/dz^m(z)$ has a pole at infinity, we need to show that $d^mf/dz^m(1/z)$ has a pole at $0$. In order to do that, we need to show that $|d^mf/dz^m(1/z)| \to \infty$ as $z \to 0$. I am getting really confused here, because I don't see how what we've been given allows us to show this.

I think that we have:

$$\frac{d^mf}{dz^m}(1/z) \le c\left(1+\frac{1}{|z|^k}\right),$$

but that doesn't tell me anything useful, because even though the RHS approaches infinity as $z \to 0$, that only tell us something useful if the inequality goes the other way, right? What am I doing wrong/misunderstanding here?

I thought maybe I would just proceed on through the proof pretending like the inequality was the other way, but got stuck again.The next line says that we have a pole of order at most $k$ at infinity. I'm not sure why that's true.

The rest of the proof is ok - I have worked through the proof that if $df^m/dz^m$ has a pole of order at most $k$ at infinity, it is a polynomial of degree at most $k$, and thus we let $n = m+k+1$ and we are done. But, I haven't taken complex in quite a while, and I just can't quite wrap my head around those other two points. A detailed explanation would be really great. Thanks!

  • 2
    $\begingroup$ Liouville theorem says a bounded entire function is constant, and also an entire function bounded by $|z|^a$ (for $|z|$ large enough) is a polynomial of degree $\lfloor a \rfloor$ $\endgroup$
    – reuns
    Aug 29, 2016 at 23:29
  • $\begingroup$ Good point user1952009 $\endgroup$
    – mick
    Aug 30, 2016 at 0:22

1 Answer 1


According to the question, we have $$\Big|(1+|z|^{k})^{-1}f^{(m)}\Big|\leq M,\ \text{for some}\ k\ \text{and}\ M.$$

Thus, for all $z\in\mathbb{C}$, we have $$|f^{(m)}(z)|\leq M|1+|z|^{k}|=M(1+|z|^{k}),\ \text{for some}\ k\ \text{and}\ M.$$

Set $g(z):=f^{(m)}(z),$ which is also an entire function.

Thus, for all $z_{0}\in\mathbb{C}$, we can always find an open set that $g$ is holomorphic in and contains the closure of a disc $D(z_{0}, R)$, for $R$ large enough, such that, by Cauchy's inequalities, we have for all $\ell\geq 0$, \begin{align*} |g^{(\ell)}(z_{0})|&=|f^{(m+\ell)}(z_{0})|\\ &\leq\dfrac{\ell!\sup_{z\in \partial D}|g(z)|}{R^{\ell}}\\ &\leq \dfrac{\ell!\sup_{z\in\partial D}M(1+|z|^{k})}{R^{\ell}}\\ &=(\ell! M)\Big(\dfrac{1}{R^{\ell}}+\dfrac{R^{k}}{R^{\ell}}\Big) \end{align*}

Taking $R\longrightarrow\infty$, as long as $\ell\geq k$, we have $$g^{(\ell)}(z_{0})=f^{(m+\ell)}(z_{0})\longrightarrow 0,\ \text{for all}\ z_{0}\in\mathbb{C}.$$

Thus, as long as $$n=m+\ell\geq m+k,$$ we have $$f^{(n)}(z)\equiv 0.$$

Interesting enough, this question is extremely similar to a question I posted here:Prove that a specific analytic function has $n^{th}$ derivative identically zero for sufficiently large $n$.

Except that in my question, it is $(1+|z|^{k})f^{(m})$ that is bounded, not the inverse of it. I attach this question here for people to have more interesting similar reference.


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