An alternate proof of Liouville's theorem

Suppose that $$|f(z)|\leq A+B|z|^M$$ and that $$f$$ is entire. Show that for all coefficients $$c_j$$ with $$M in its power series extansion are $$0$$.

Attampt:

$$|f(z)|=\left|\sum_{k=0}^\infty c_kz^k\right| \leq A+B|z|^M \Rightarrow \\ \left|\sum_{k=0}^\infty c_kz^{k-M}\right|\leq{A\over |z|^M}+B\underset{z \to \infty}{\rightarrow}B$$ $$g(z):=\sum_{k=0}^\infty c_kz^{k-M}$$ is continuous and thus $$|g|$$ is bounded by some $$Q$$.

How can I finish the exercise?

• Do you already have this theorem for $M = 0$ (bounded entire function is constant)? – mihaild Apr 26 at 14:57
• This does not work since in general $g$ will have a pole at $z=0$. – Hans Engler Apr 26 at 15:01
• But we can first replace $f$ with $f(z) - c_0 - c_1 z - \ldots c_M z^M$, then $g$ will have no poles (and such replacement doesn't affect boundary or $c_j = 0$ for $j > M$). – mihaild Apr 26 at 15:11

It should be $$c_j=0$$ for $$j>M$$. By the way, it would be useful to express $$c_j$$ by the Cauchy Integral Formula:$$c_j=\frac{f^{(j)}(0)}{j!}=\frac{1}{j!}\frac{1}{2\pi i}\int_{\partial B_R}\frac{f(w)}{w^{j+1}}dw$$ Because then a bound for $$c_j$$ is easy to obtain: $$|c_j|\le\frac{2\pi R}{j!2\pi}\frac{A+BR^M}{R^{j+1}}\to 0 \text{ as }R\to\infty,\text{ for all }j>M$$
• I know Cauchy Integral Formula for $j=0$ can you please write the formula for the $j$'s derivative? Thanks – J. Doe Apr 26 at 15:23