# Complex Analysis Proof Application of Liouville's Theorem

I am studying for my Complex Analysis and have been trying to answer this question for some time now. My professor gave a hint to use the power series expansion of $f(z)$ (to manipulate $f(z)$) and then to use or apply Liouville's Theorem. I have tried this, but am getting nowhere. Any help, suggestions, or tips would be much appreciated. Please note that this is for an elementary course on complex analysis so please do not give too advanced an explanation. Thank you. I have attached two pictures, one of the Cauchy estimates and one of the actual question.

• Can you give us the statement of the version of Liousville's Theorem you received in your class? – roo Apr 4 '17 at 7:42
• Yes, of course. – kemb Apr 4 '17 at 7:53
• If F is entire and if there is a constant M such that |F(z)|<M for all z, then F is identically constant. – kemb Apr 4 '17 at 7:54
• @kemb Which book is this from if you don't mind? – Ovi Feb 24 at 2:04

I would do Jennifer's proof, but if you want it explicitly using Liouville's theorem, use the same as her but exactly with $f^{(m)}(z)$, where $F$ of your Liouville's theorem statement is $f^m$.

By the Cauchy inequality with $z_0 = 0$ and $r\geq R_0$, $$|f^{(m)}(0)|\leq\frac{m!}{r^m}\max_{|z|=r}|f(z)| \leq \frac{n!}{r^n}A|z|^m = (\ast).$$ As $|z| = r$, $$(\ast) = n! A = M$$ and by Liouville's theorem $f^{(m)}$ is constant, which implies that $f$ is a polynomial of degree $m$ or less (because it's entire and you can write it like a power serie as Jennifer said).

• This makes sense. I am slightly confused however by (*)=n!A=M, could you explain this step? Thank you. – kemb Apr 4 '17 at 8:18
• Look under "$\max$" of the Cauchy inequality. – Abelois Apr 4 '17 at 8:19
• I'm not sure I follow. – kemb Apr 4 '17 at 10:04

You don't have to use Liouville's theorem.

You can write $f(z)=\sum_{n=0}^\infty a_nz^n$. With Cauchy inequality, $\forall n,|a_n|\leq\frac{\max_{|z-z_0|=r}|f(z)|}{r^n}$. So with the hypoyhesis of the question : $\forall n,\forall r \geq R_0,|a_n|\leq\frac{Ar^m}{r^n}={Ar^{m-n}}$.

When $n>m$ if you make tend $r\rightarrow \infty$, you have $|a_n|=0$. So $f(z)=\sum_{n=0}^m a_nz^n$ so $f$ is a polynomial of degree $m$ or less.

We will show this result by induction on $m$.

By Liouville's theorem, the statement is true for $m=0$, indeed in this case $f$ is bounded so $f$ is constant, so $f$ is a polynomial of degree at most $0$.

Suppose true for $m-1$. Consider $$g(z)=\left\{\begin{array}{ll} \frac{f(z)-f(0)}{z}, & z\neq 0\\ f'(0), & z=0\end{array}\right.$$ Then $g$ is entire, and for sufficiently large $|z|$, $$|g(z)|\le \left|\frac{f(z)}{z}\right|+\left|\frac{f(0)}{z}\right|\le|Az^{m-1}|+|f(0)|\le C|z|^{m-1}$$ for some constant $C$.

Hence by induction hypothesis, $g$ is a polynomial of degree at most $m-1$.
So $f$ is a polynomial of degree at most $m$.

• Hi, but my Professor said that I should use the theorem – kemb Apr 4 '17 at 8:03
• Is there any way to use the theorem? – kemb Apr 4 '17 at 8:03
• Well you prove Liouville's theorem with Cauchy estimates, so you can proabably prove this result with Liouvilles theorem but it will be longer, more curbersome – Jennifer Apr 4 '17 at 8:08
• Is there any way you could show me, I'm really trying to understand how to use the theorem to do the proof. – kemb Apr 4 '17 at 8:12
• Ok I will add it. – Jennifer Apr 4 '17 at 8:15