If $f$ is a nonconstant entire function such that $|f(z)|\geq M|z|^n$ for $|z|\geq R$, then $f$ is a polynomial of degree atleast $n$.

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 ?

• 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 $|z|<R$ May 4 '20 at 4:16
• @BrianMoehring That's true , but no info has been given of $f(z)$ when $|z|<R$ May 4 '20 at 4:27
• You are told $f$ is entire. How many zeros can an entire function have on a bounded set? May 4 '20 at 4:29
• @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 May 4 '20 at 5:32
• You may want to look at my answer, now edited.
– zhw.
May 13 '20 at 18:26

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.)

• 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 $|z| \leq R_1$. May 9 '20 at 20:13

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 $$d\ge n$$.

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.

• So, why the downvote?
– zhw.
May 9 '20 at 18:56
• 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 May 9 '20 at 20:16
• @N.S. I edited by previous answer.
– zhw.
May 13 '20 at 18:26
• This is a lovely answer! +1 Sep 17 at 10:17