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I am studying for my exam and came across this question:

Suppose $f$ is entire and $r<R$. Prove that the family $\mathcal{F} = \{f_n = f(nz) | n \in \mathbb{N}\}$ for $z \in \mathbb{C}$ is normal on the annulus $r< |z|<R$ iff $f$ is constant.

This is my attempt:

$\Leftarrow$ If $f$ is constant, then $f(z) = z_0, \, \forall z \in \mathbb{C}$. Then since $f_n = f$ for all $n \in \mathbb{N}$. Thus $\mathcal{F} = \{f\}$, and therefore every sequence of $\mathcal{F}$ is the sequence of function $f$, and therefore wil converge uniformly on every compact subset, and thus is a normal family. Or I could also use Montels theorem and see that since $f_n(z) = f(z) = z_0$ for all $n \in \mathbb{N}$ and all $z \in \mathbb{C}$, we know that $\mathcal{F}$ is uniformly bounded by $z_0$ and therefore it is a normal family.

$\Rightarrow$ We know that $f$ is entire, therefore we only need to prove that it is bounded (and then can use Louiville's theorem to conclude that it is constant). Let $D_k(0)$ be a disk, then since it is compact we know that there exists a $B_k$ such that $|f_n(z)| \leq B_k$ for all $z \in \mathbb{C}, n \in \mathbb{N}$. I want to use the following theorem:

Corollary 4.6 Stein and Shakarchi: Suppose that $\Omega$ is a region with compact closure $\overline{\Omega}$. If $f$ is holomorphic on $\Omega$ and continuous on $\overline{\Omega}$ then $$sup_{z\in\Omega}|f(z)| \leq sup_{z \in \delta \Omega} |f(z)|.$$

But I am not sure how to since I don't know if on $D_R(0)$ it holds that $\mathcal{F}$ is uniformly bounded on $D_R$ since I am only given it is normal on $r<|z|<R$.

Can someone tell me if my attempts are correct and how I should go further in my second attempt?

fieke_2000
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  • If $\mathcal{F}$ is normal, you can find a sequence in $\mathcal{F}$ that converges uniformly on the circle $|z|=\rho$, where $\rho\in(r,R)$. Now use this and Maximum Modulus Principle (Corollary 4.6) to deduce that $f$ is bounded. – Sangchul Lee Dec 31 '21 at 09:55
  • Essentially a duplicate of https://math.stackexchange.com/q/422958/42969. That question only asks if $f$ is a polynomial, but the answer shows that $f$ is constant. – Martin R Dec 31 '21 at 10:02

1 Answers1

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First part is correct. Montel's Theorem is not required.

If $(f_n)%$ is normal then the sequence is uniformly bounded on any compact subset of the annulus. If $r<r_1<R_1<R$ then there exists $M$ such that $|f_n(z)| \leq M$ for $r_1 \leq |z|\leq R_1$. For any $z$ such that there the interval $(\frac {|z|} {R_1}, \frac {|z|} {r_1})$ has length $ >1$ pick an integer $n$ in it. It follows that $|f(z)|=|f(n \frac {z} n) |\leq M$ since $\frac z n \in [r_1,R_1]$. Conclude that $f$ is bounded, hence constant by Liouville's Theorem.

Kavi Rama Murthy
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  • From this we know that any $z$ such that $(\frac{|z|}{R_1}, \frac{|z|}{r_1})$ has length >1, we know that $|f(z)| \leq M$. So for such $z$ we know that $f$ is bounded, but what about the other $z$'s if that makes sense? For every $z$ for which $r<|z|<R$, will there always be a $r_k,R_k$ such that such an interval with length >1 exists? – fieke_2000 Dec 31 '21 at 10:21
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    @fieke_2000 Any continuous function is bounded on ${z: |z| \leq T}$ for any $T>0$. To apply Liouville's Theorem you only have to check boundedness for large values of $|z|$. – Kavi Rama Murthy Dec 31 '21 at 11:23
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    I have proved boundedness on ${z: |z|>\frac 1 {\frac 1 {r_1}-\frac 1 {R_1}}}$. @fieke_2000 – Kavi Rama Murthy Dec 31 '21 at 12:31
  • Oh right, thank you! – fieke_2000 Jan 26 '22 at 14:04