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Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be a entire function. Suppose that there are $M$, $r>0$ and $n\in \mathbb{N}$ such that $\left|f(z)\right|<M\left|z\right|^n$ for all $z \in \mathbb{C}$ with $\left|z\right|\geq r$.

Show that $f$ polynomial of degree at most $n$.

Remark: I tried to follow the proof of Liuville's Theorem but I have complications with the condition $\left|z\right|\geq r$..

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  • $\begingroup$ Is it $\mid f(z)\mid <M\mid z\mid^n$? $\endgroup$ – Tsemo Aristide Feb 6 '16 at 21:15
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Write $f(z)=\sum_n a_nz^n$ and $g(z)= {{f(z)}\over{z^n}}$, $g(z)=a_0/z^n+...+a_{n-1}/z+\sum_{l\geq n}a_lz^{l-n}$. Since $\mid g(z)\mid <M$ for $\mid z\mid >r$, we deduce that $\mid \sum_{l\geq n}z^{l-n}\mid-\mid a_0/z^n+...a_{n-1}/z\mid<M$. This implies that the entire function $h(z)=\sum_{l\geq n}a_nz^{l-n}$ is bounded so it is a constant. Thus $a_l=0, l\geq n+1$.

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Well, first of all we know that, as $f$ is an entire function, we can write it as a power series around zero: $$f(z)=\sum\limits_{m\geq 0}a_mz^m,\:\:\:\forall z\in\mathbb{C},$$ where, by Taylor, $a_m=f^{(m)}(0)/m!$.

Now, if we consider the Cauchy's inequality, we have that $$|f^{(m)}(0)|\leq m!\frac{\sup\limits_{|z|=r}|f(z)|}{r^m}\leq m!\frac{Mr^n}{r^m}=Mm!r^{n-m},\:\:\:\forall m\in\mathbb{N},\: r>0.$$ Note that if $m > n$, then $0\leq f^{(m)}(0)|\to 0$ when $r\to\infty$. So, we know that $a_m=f^{(m)}/m!=0$ for all $m>n$ and $$f(z)=\sum\limits_{m=0}^na_mz^m,\:\: \forall z\in\mathbb{C}.$$ So we have proved that $f$ is a polynomial of degree at most $n$.

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