# f is a meromorphic function satisfying $\vert f(z)\vert\leq\vert z\vert^n$ then f is a rational function

I would like to see how the following question can be proved:

Let $$f\in\mathcal{M}(\mathbb{C})$$ satisfying $$\vert f(z)\vert\leq M\vert z\vert^n$$ for all $$z\in\mathbb{C}\setminus P(f)$$ with $$\vert z\vert>r$$ for some finite constants $$M,r$$ and some $$n\in\mathbb{N}$$. show that $$f$$ is a rational function

I tried with Cauchy integral formula and using that $$ord(f)=0$$ for all $$z\in\mathbb{C}\setminus P(f)$$ but it didnt get me anywhere

• You wrote “ for some finite constants $M$, $r$ and some $n\in\mathbb N$”, but where is that $M$? – José Carlos Santos Apr 23 at 15:06
• changed it, $\vert f(z)\vert\leq M\vert z\vert^n$ – Roni Ben Dom Apr 23 at 15:09

As $$\overline D(0, r)$$ is compact, $$f$$ has a finite number of poles in this disc. So by multiplying by the polynomial $$P=\prod (z-\alpha_i)^{m_i}$$ for $$\alpha_i$$ the poles and $$m_i$$ their multiplicity, you still have $$|Pf(z)|\leq |Q|$$ for a certain polynomial $$Q$$, and $$z$$ outside the disc. But $$|Pf|$$ is continuous function on this compact disc so we have $$C$$ such that $$|Pf|\leq C$$ on the disc. Then, $$|Pf|\leq C + |Q|$$ everywhere. Then, you can see that there is a polynomial $$R$$ such that $$|Pf|\leq|R|$$ everywhere. This is a classical result that an entire function bounded by a polynomial is a polynomial (see for example this thread), so we are done.
• Why is $\vert Pf\vert$ continuous on the disc? – Roni Ben Dom Apr 23 at 16:26
• $|Pf|$ is continuous at each point where $f$ is, so everywhere except at the poles in the disc. At the poles, continuity is assured locally because we are deleting the pole thanks to $P$, by the definition of the multiplicity of a pole – elidiot Apr 23 at 16:28