# Which entire functions satisfy $\,\lvert\,f(z)\rvert \leq \lvert z\rvert^k$?

Which entire holomorphic functions satisfy $\,\lvert\,f(z)\rvert \leq \lvert z\rvert^k$, for all $z\in\mathbb{C}$?

So I've shown that $\,\lvert\, f(z)\rvert \leq \lvert z\rvert ^k \implies f(z)$ is a polynomial of degree at most $k$. Therefore we may write

$f(z)=\sum_{0}^{k} c_nz^n$ where $c_n$ is Taylorian coefficients of $f$. So I see that this should certainly mean that $c_0=0$ since $\,\lvert\,f(0)\rvert=0$. But after that, I have no idea how to proceed. I guess that $c_k \leq 1$ but have no idea to prove it, not to mention how other coefficients would vary according to value of $c_k$.

Any helps are appreciated!

• possible duplicate of Entire function bounded by a polynomial is a polynomial – mrf Nov 28 '14 at 8:13
• Really not a duplicate @mrf (and other closers). OP has already proven that $f$ is a polynomial of degree $\le k$, they want more information about the coefficients. – Najib Idrissi Nov 28 '14 at 8:37

Put $z=\frac{1}{N}$, where $N \in \mathbb{N}$. Then we have:

$$\left|\sum_{n=0}^{k}\frac{c_{n}}{N^n}\right|\leq \frac{1}{N^{k}}$$

Multiply both sides by $N^k$:

$$\left|\sum_{n=0}^{k}c_{n}N^{k-n}\right|\leq 1$$

By triangle inequality:

$$|c_k|-\sum_{n=0}^{k-1}N^{k-n}|c_n|\leq \left|\sum_{n=1}^{k}c_{n}N^{k-n}\right|$$

And inequality:

$$|c_k|-\sum_{n=0}^{k-1}N^{k-n}|c_n|\leq 1$$

Can by satisfy if only if $c_0=c_1=\cdots=c_{k-1}=0$.

• Thanks for the solution, but could you explain a bit more why the last inequality implies that all coefficients must be equal? and I guess this means that $f(z)=0$ for all z in $\mathbb{C}$? – user160738 Nov 22 '14 at 13:46
• No, note that $c_0=c_1=\cdots=c_{k-1}=0$ and $c_k$ can be $\neq 0$. You get $c_0=c_1=\cdots=c_{k-1}=0$ when you calculate limits of both sides when $N \to \infty$. – agha Nov 22 '14 at 13:50
• Oh wow cool, so you are basically saying that since last inequality should be satisfied for all $N \in \mathbb{N}$ we must have all other coefficients to be 0? – user160738 Nov 22 '14 at 13:57
• Yes, so $f(z)=c_kz^{k}$, when $|c_k|\leq 1$. – agha Nov 22 '14 at 13:58
• Thanks a lot I got it now! – user160738 Nov 22 '14 at 14:00

We have that $$\left|\frac{f(z)}{z^k}\right|\le 1, \,\,\,\, \text{whenever \,\,z\in\mathbb C\smallsetminus\{0\}},$$ and hence the function $g(z)=z^{-k}f(z)$, is holomorphic in $\mathbb C\setminus\{0\}$ and has removable singularity at $z=0$. Therefore it can be extended holomorphically at $z=0$; in fact it extends to a bounded entire function. Thus $g(z)$ is constant, i.e., $g(z)=c$, for some constant $\,\lvert c\rvert\le 1$.

Therefore, $\,f(z)=cz^k$, with $\lvert c\rvert\le 1$.

Note. Let's assume instead that $\lvert\,f(z)\rvert\le \lvert z\rvert^a$, for $a>0$, which is not an integer, then $f\equiv 0$.

Say $a=n+b$, where $n=\lfloor a\rfloor$ and $b\in (0,1)$. Then $g(z)=z^{-n}f(z)$ has a removable singularity at $z=0$, and thus $g$ extends to an entire function, satisfying $\lvert g(z)\rvert\le \lvert z\rvert^b$, for all $z$. But, as $g$ is continuous at $z=0$, then $g(0)=0$, and hence $g(z)=zh(z)$, for some entire function $h$, and hence $\lvert h(z)\rvert\le \lvert z\rvert^{b-1}$. Thus $h$ is bounded, and therefore constant, which implies that $f(z)=cz^{n+1}$, for some $c$. But $\lvert cz^{n+1}\rvert\le \lvert z\rvert^{n+b}$, for all $z$, implies that $c=0$.

• Thanks for the solution very neat one :) – user160738 Nov 22 '14 at 14:02
• Sorry, but doesn't the first inequality is true if and only if $z \neq 0$? – user160738 Nov 22 '14 at 14:04
• I don't see how the Note follows. Could you explain that briefly? – Darrin Jan 6 '15 at 23:33
• See updated answer. – Yiorgos S. Smyrlis Jan 7 '15 at 0:02