Corollary of Liouville's Theorem

I need someone to verify my proof. I had Liouville's Theorem presented to me as

If $f$ is entire and bounded, then $f$ is constant.

I am then asked to prove

If $f$ is entire and there exists $k>0,R>0$ and $n\in \mathbb{N}$ such that $|f(z)|\leq k|z|^n$ for $|z|>R$. Then $f$ is a polynomial of at degree at most $n$.

My proof goes as follows:

From the assumptions of the theorem $\left|\frac{f(z)}{z^n}\right|\leq k$. So by Liouville's Theorem $\frac{f(z)}{z^n} = c$ for some constant $c\in\mathbb{C}$. Hence $f(z)=cz^n$.

This feels too simple and my main concern is the $|z|>R$ part, what if $|z|\leq R$, is there anything to say in this case?

• No, this proof doesn't work. $\frac{f(z)}{z^n}$ can have a singularity at 0 so it not entire (which is essential for Liouville's theorem, think of $\frac{1}{z}$) – bitesizebo Apr 21 '18 at 11:41
• I believe you are correct, this is the exact type of answer I was hoping for – pureundergrad Apr 21 '18 at 11:44
• You could replace $f$ by a related function $g$ for which $g(z)/z^n$ is bounded and entire. – Angina Seng Apr 21 '18 at 11:50

On the other hand, if $\displaystyle\sum_{n=0}^\infty a_nz^n$ is the Taylor series of $f$ at the origin, the, by Cauchy's inequalities, $$(\forall m\in\mathbb{N})(\forall r>0):|a_m|\leqslant\frac{\sup_{|z|=r}|f(z)|}{r^m}\leqslant kr^{n-m}.$$Therefore, if $m>n$,$$|a_m|\leqslant\lim_{r\to\infty}kr^{n-m}=0$$and so$$(\forall z\in\mathbb{C}):f(z)=a_0+a_1z+\cdots+a_nz^n.$$
To do this with Liouville's theorem, you can show that there is an $R'>0$ so that $$\left\lvert \sum_{k=0}^n a_k z^k \right\rvert \leq \lvert z \rvert^n$$ for $\lvert z \rvert > R'$, where $A=A(a_0,\dotsc,a_n)$. Now define $$g(z) = f(z) - \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} z^k.$$ Then $g$ is entire and has $g(0)=\dotsc=g^{(n)}(0)=0$, so $g(z) = z^{n+1} h(z)$ for some entire function $h$, and $g(z)/z^n$ is entire. Then by the triangle inequality, $$\left\lvert \frac{g(z)}{z^n} \right\rvert \leq A+k$$ for $z>R'$. $g(z)/z^n$ is entire, so it is analytic on $z \leq R'$, and hence also bounded there, so $g(z)/z^n$ is bounded everywhere and hence constant by Liouville.