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I have a question in my assignment :

If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$ for $|z|\geq R$ for some $n\in\mathbb N$ and some $M$ and $R$ in $(0,\infty)$ show that $f$ is a polynomial of degree atleast $n$.

Now , I defined a function $\ g(z) = \frac {1}{f(z)}\ $ such that $\ |g(z)| \le \frac{1}{M{|z|}^n}.$

Now , by using the cauchy inequality $$|g^{(n)}(z)| \le \frac{n!}{R^n |z|^nM}.$$

Considering that $ g(z) $ is an analytic function , it has a radius of convergence $ \infty $

$ \implies\ g^{(n)}(z) = 0.$

But if we go by this approach , then $ g^{(n)}(z) = 0 \ $ for any n . Also how can we be so sure that $ f(z) \neq 0 $ for any z ?

Is my reasoning correct or is there some other way to solve it ?

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    $\begingroup$ You need to first note $f$ has finitely many zeros, and then divide them out of $f$. Otherwise, the $g$ you've defined might not be analytic on $|z|<R$ $\endgroup$ Commented May 4, 2020 at 4:16
  • $\begingroup$ @BrianMoehring That's true , but no info has been given of $f(z)$ when $|z|<R$ $\endgroup$
    – Starboy
    Commented May 4, 2020 at 4:27
  • $\begingroup$ You are told $f$ is entire. How many zeros can an entire function have on a bounded set? $\endgroup$ Commented May 4, 2020 at 4:29
  • $\begingroup$ @BrianMoehring From what i have studied ,since f is analytic , it's also continuous. Hence , it is bounded on which means g(z) doesn't have any zeroes in the complex plane $\endgroup$
    – Starboy
    Commented May 4, 2020 at 5:32
  • $\begingroup$ You may want to look at my answer, now edited. $\endgroup$
    – zhw.
    Commented May 13, 2020 at 18:26

3 Answers 3

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As @Brian points out, $f$ has only finitely many zeros. Of course, $f(z)\neq 0$ if $|z|\geq R$. Since the set $B_R=\{z\mid |z|\leq R\}$ is compact, $f$ can only have finitely many zeros in $B_R$(use the identity theorem). Let $a_1,\ldots,a_k$ be the zeros of $f$ counted according to multiplicty. Let $$p(z)=(z-a_1)\cdots(z-a_k)=z^k+b_{k-1}z^{k-1}+\cdots+b_0.$$ For $|z|\geq R,$ we have $$|p(z)|\leq |z|^k\Bigl(1+\frac{|b_{k-1}|}{|z|}+\cdots+\frac{|b_{0}|}{|z|^k}\Bigl)\leq C|z|^k,$$ where $C=1+\frac{|b_{k-1}|}{R}+\cdots+\frac{|b_{0}|}{R^k}.$ Thus we have $$\frac{|z|^n|p(z)|}{|f(z)|}\leq \frac{|p(z)|}{M}\leq \frac{C|z|^k}{M},$$ for $|z|\geq R$.

Suppose that $n=k$. Then, by Liouville, we see that $\frac{p(z)}{f(z)}$ is a constant function and hence $f$ is a polynomial of degree $k=n$.

Suppose now that $n\lt k$. Then it is easy to see that $\frac{p(z)}{f(z)}$ is a polynomial of degree $\leq k-n$ (use the Cauchy's integral formula for derivatives. Click here for a proof.) But $\frac{p(z)}{f(z)}$ is a nowhere vanishing entire function. So $\frac{p(z)}{f(z)}$ is a constant and hence $f$ is a polynomial of degree $k\gt n$.

Finally, assume $n\gt k$. Then, by Liouville's theorem, $\frac{z^{n-k}p(z)}{f(z)}$ is a constant. So $f(z)=cz^{n-k}p(z)$ for some constant $c$ and degree of $f$ is $n$. But $f$ and $p$ share the same zeros with same multiplicities. So degree of $f$ is equal to degree of $p$, i.e., $n=k$, a contradiction. (One can also use the Rouche's theorem to conclude. See @N. S.'s comment below.)

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    $\begingroup$ The last step follows immediatelly from Rouche's Theorem: If $n > k$, then $f(z)+az^n$ is a polynomial of degree $n$ for some $0 <a <M$. Therefore, there exists some $R_1>R$ such that $f(z)+az^n$ has exactly $n$ roots inside $|z|<R_1$. But, $|f(z)|> |az^n|$ on $|z|=R_1$ and hence $ f(z)+az^n$ has at most $\deg(f)=k$ roots inside $|z| \leq R_1$. $\endgroup$
    – N. S.
    Commented May 9, 2020 at 20:13
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Let $g(z) = f(1/z), z\in \mathbb C \setminus \{0\}.$ Then $|g(z)|\ge M/|z|^n$ for $0<|z|<1/R.$ Now $g$ blows up at $0,$ so has a nonremovable singularity there. Thus $g$ has an essential singularity at $0,$ or a pole at $0.$ If the former, then Casorati-Weierstrass tells us $g(\{0<|z|<1/R\})$ is dense in $\mathbb C.$ But $ g(\{0<|z|<1/R\})\subset \{|z|>MR^n\},$ contradiction.

Therefore $g$ has a pole at $0.$ Thus there is a polynomial $p$ and an entire $h$ such that $g(z) = p(1/z)+h(z)$ in $\mathbb C\setminus \{0\}.$ Flipping back, we see $f(z)=p(z) + h(1/z)$ on $\mathbb C\setminus \{0\}.$ It follows that as $|z|\to \infty,$ $f(z)-p(z) \to h(0).$ Because an entire function with finite limit $L$ at $\infty$ equals $L$ everywhere (Liouville), we arrive at $f(z) - p(z) = h(0)$ everywhere. The growth rate of $f$ then implies $p$ has degree at least $n$ as desired.

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  • $\begingroup$ So, why the downvote? $\endgroup$
    – zhw.
    Commented May 9, 2020 at 18:56
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    $\begingroup$ I didn't downvote, but you probably need more details to show that $f(1/z)$ doesn't have an essential singularity at $0$. Pickard does that but looks like an overkill $\endgroup$
    – N. S.
    Commented May 9, 2020 at 20:16
  • $\begingroup$ @N.S. I edited by previous answer. $\endgroup$
    – zhw.
    Commented May 13, 2020 at 18:26
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Definitions

Consider the $\{z_k\}$ where $f(z_k)=0$. Since they all must be in $|z|\le R$, if there were infinitely many, there would be a limit point and then, by the Identity Theorem, $f$ would be identically $0$. At each $z_k$, there is an $d_k\in\mathbb{N}$, so that $f(z)=(z-z_k)^{d_k}g_k(z)$, where $g_k(z_k)\ne0$ and $g_k$ is entire. Therefore, $$ g(z)=\frac{f(z)}{\prod\limits_{k=1}^m(z-z_k)^{d_k}}\tag2 $$ is entire yet does not vanish. Since $|g(z)|\gt0$, we must have $|g(z)|\ge L$ on $|z|\le R$ (since $|g|$ is a continuous function and $|z|\le R$ is a compact set, $|g|$ attains its infimum on $|z|\le R$).

On $|z|\gt R$, $$ \begin{align} \prod_{k=1}^m|z-z_k|^{d_k} &\le\prod_{k=1}^m(|z|+|z_k|)^{d_k}\\ &\le\left[\prod_{k=1}^m\left(1+\frac{|z_k|}R\right)^{d_k}\right]|z|^d\\[6pt] &=C|z|^d\tag3 \end{align} $$ where $d=\sum\limits_{k=1}^md_k$.

Note that since $|z_k|\le R$, we have $C\le2^d$.


Show that $\boldsymbol{d\ge n}$

Inequalities $(1)$ and $(3)$ say that $$ |g(z)|\ge\frac MC|z|^{n-d}\tag4 $$ for $|z|\gt R$.

Let $h(z)=\frac1{g(z)}$, then $$ |h(z)|\le\left\{\begin{array}{} \frac1L&\text{for }|z|\le R\\ \frac CM|z|^{d-n}&\text{for }|z|\gt R \end{array}\right.\tag5 $$ Suppose $d\lt n$, then $h(z)$ is bounded and entire. Thus, by Liouville's Theorem, $h$, and therefore $g$, would be constant. This implies that $$ \begin{align} \frac{|f(z)|}{|z|^n} &=\frac{|g(0)|}{|z|^{n-d}}\prod_{k=1}^m\left|\frac{z-z_k}z\right|^{d_k}\\ &\hspace{-6pt}\overset{|z|\to\infty}\to0\tag6 \end{align} $$ which contradicts $(1)$. Therefore, $d\ge n$.


Show that $\boldsymbol{h}$ and $\boldsymbol{g}$ are Constant

For $|z|\gt R$, $(5)$ says that $|h(z)|\le\frac CM|z|^{d-n}$. Thus, for $r\gt R$, Cauchy's Integral Formula says $$ \begin{align} \left|h^{(k)}(0)\right| &=\frac{k!}{2\pi}\left|\int_{|z|=r}\frac{h(z)}{z^{k+1}}\mathrm{d}z\,\right|\\ &\le\frac{Ck!}Mr^{d-n-k}\tag7 \end{align} $$ So if $k\gt d-n$, we have $h^{(k)}(0)=0$. That is, $h$ is a polynomial of degree at most $d-n$. However, if $h$ has degree greater than $0$, it would have a root, which would be a pole for $g(z)$, and therefore, $g$ would not be entire. So $h$ and $g$ must be constant.


Conclusion

Since $g$ is a constant, $$ f(z)=g(0)\prod\limits_{k=1}^m(z-z_k)^{d_k}\tag8 $$ Therefore, $f$ is a polynomial of degree $d\ge n$.

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