6

Let $f$ be an holomorphic function on the unit ball with $f(0)=0$. Prove that $\sum_{n=1}^{\infty}f(z^n)$ is uniformly locally convergent in the unit ball.

My attemp:

It is suffice to prove that $\sum_{n=1}^{\infty}f(z^n)$ converges uniformly in any $\overline{B_0(r)}$ with $r<1$.

$f'$ is also an holomorphic function, defined on a compact set and therfore bounded. Let's say $|f'|\leq M$. Then, for every $z\in \overline{B_0(r)}$, $|f(z^n)-f(0)|\leq M|z^n-0|$, $|f(z^n)|\leq M|z^n| = M|z|^n \leq Mr^n$.

The series $\sum_{n=1}^{\infty}r^n$ converges and therefore by the M test our series converges uniformly. Is it correct?

Thanks.

catch22
  • 3,057

2 Answers2

3

Let's do brute force!

If $f(z)=\sum_{n\geq1}a_nz^n$, then as a formal series $$\sum_{k\geq1}f(z^k)=\sum_{n\geq1}\left(\sum_{d\mid n}a_d\right)z^n.$$ If $f$ converges in the unit disc, we know from the Cauchy-Hadamard formula that $\limsup_{n\to\infty}|a_n|^{1/n}\leq1$, and we have to show that then $$\limsup\left|\sum_{d|n}a_d\right|^{1/n}\leq1.$$ One should be able to prove that

If $(x_n)_{n\geq1}$ is a sequence of non-negative real numbers such that $\limsup_{n\to\infty}x_n^{1/n}\leq1$, then $\limsup_{n\to\infty}(x_1+\cdots+x_n)^{1/n}\leq 1$.

and using this our result follows.

Mariano Suárez-Álvarez
  • 135,076
  • The statement $\limsup_{n\to \infty} x_n^{1/n} \le 1$ means, in particular, that for each $\epsilon >0$ $\exists N(\epsilon)$ s. t. $\forall n > N(\epsilon)$ the inequality $x_n^{1/n} <1+\epsilon $ holds. Since a finite set of the terms of a sequence is of no importance for its upper limit, we may assume the last one holds for each natural $n$. Therefore, we have $$(x_1+\dots+x_n)^{1/n} \le (n(1+\epsilon)^n)^{1/n}.$$ Passing to the upper limit, we obtain $$\limsup_{n \to \infty}(x_1+\dots+x_n)^{1/n} \le1+\epsilon . $$ This is it as $\epsilon$ is an arbitrary positive number. – user64494 Jul 25 '13 at 10:16
  • Can you please explain why $\sum_{k\geq1}f(z^k)=\sum_{n\geq1}\left(\sum_{d\mid n}a_d\right)z^n$ ? – catch22 Jul 26 '13 at 08:35
  • By adding series, we have $$\sum_{k\geq1}f(z^k)=\sum_{n\geq1}a_nz^n+ \sum_{n\geq1}a_nz^{2n}+\sum_{n\geq1}a_nz^{3n}+\dots=$$ $$a_1z+(a_2+a_1)z^2+(a_3+a_1)z^3+(a_4+a_2)z^4+\dots=\sum_{n\geq1}\left(\sum_{d \mid n}a_d\right)z^n.$$ – user64494 Jul 27 '13 at 06:38
  • But why can we change the order of item? – Zeldovich Yakov Mar 07 '21 at 01:21
0

The answer is no because $M$ depends on $r$ as $M:=\max \{|f'(z)|: z \in \overline{B_0(r)} \}.$

user64494
  • 5,811