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if f is holomorphic, is Σf(z^k) holomorphic?

pantboio

Member
Nov 20, 2012
45
Let $D=D(0,1)$ be the unit disc centered at 0. Let $f$ be holomorphic in D, with $f(0)=0$. Show that

$g(z)=\sum_{k=1}^{\infty} f(z^k)$

defines an holomorphic function in $D$.

I've argued as follows: since $f$ is holomorphic, then $f$ is locally the sum of a power series

$f(z)=\sum_{n=0}^{\infty}a_n z^n$
and since $f(0)=a_0=0$ by assumption, we can write

$f(z)=\sum_{n=1}^{\infty}a_nz^n$


which holds for all $z$ in $D$.
Hence

$g(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}a_n z^{nk}$

How can i deal with this double index summation?
 
Last edited by a moderator:

chisigma

Well-known member
Feb 13, 2012
1,704
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

Let $D=D(0,1)$ be the unit disc centered at 0. Let $f$ be holomorphic in D, with $f(0)=0$. Show that

$g(z)=\sum_{k=1}^{\infty} f(z^k)$

defines an holomorphic function in $D$.

I've argued as follows: since $f$ is holomorphic, then $f$ is locally the sum of a power series

$f(z)=\sum_{n=0}^{\infty}a_n z^n$
and since $f(0)=a_0=0$ by assumption, we can write

$f(z)=\sum_{n=1}^{\infty}a_nz^n$


which holds for all $z$ in $D$.
Hence

$g(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}a_n z^{nk}$

How can i deal with this double index summation?
Welcome on MHB!...

Inverting Your last relation You obtain...

$\displaystyle g(z)= \sum_{n=1}^{\infty} a_{n}\ \sum_{k=1}^{\infty} (z^{n})^{k} = \sum_{n=1}^{\infty} a_{n}\ \frac{z^{n}}{1-z^{n}}$ (1)

Because $\displaystyle \lim_{ n \rightarrow \infty} 1-z^{n}=1$ for $|z|<1$, it exists an N for which forall n>N is...

$\displaystyle \frac{|z^{n}|}{|1-z^{n}|}< c\ |z^{n}|$ (1)

... where c> 1 is a constant, so that the series (1) converges for |z|<1 and g(z) is holomorphic...

Kind regards

$\chi$ $\sigma$
 

pantboio

Member
Nov 20, 2012
45
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

Welcome on MHB!...

Inverting Your last relation You obtain...

$\displaystyle g(z)= \sum_{n=1}^{\infty} a_{n}\ \sum_{k=1}^{\infty} (z^{n})^{k} = \sum_{n=1}^{\infty} a_{n}\ \frac{z^{n}}{1-z^{n}}$ (1)

Because $\displaystyle \lim_{ n \rightarrow \infty} 1-z^{n}=1$ for $|z|<1$, it exists an N for which forall n>N is...

$\displaystyle \frac{|z^{n}|}{|1-z^{n}|}< c\ |z^{n}|$ (1)

... where c> 1 is a constant, so that the series (1) converges for |z|<1 and g(z) is holomorphic...

Kind regards

$\chi$ $\sigma$
First of all, thank you for your help.
Secondly, i hope i've completely understood your argument. It is quite clear until you get the estimation

$|\frac{z^n}{1-z^n}|\leq c |z^n|$

where the RHS is the n-th term of a convergent series (geometric with $|z|<1$)

Then i think i can conclude the following

$|a_n||\frac{z^n}{1-z^n}|\leq c|a_n| |z^n|$

and RHS is the n-th term of a convergent series since

$f(z)=\sum_{n=1}^{\infty}a_n z^n$

is convergent, and absolutely convergent, in the unit disc by assumption.

A little last remark; i think the possibility to invert the order of summations is granted by some absolute convergence, but in this case which is the absolutely convergent series which allows me to reverse indexes?

Best regards
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

... a little last remark... i think the possibility to invert the order of summations is granted by some absolute convergence, but in this case which is the absolutely convergent series which allows me to reverse indexes?...
Effectively this 'little last remark' is very 'insidious'(Evilgrin)... searching on 'Monster Wolfram'...

General Mathematical Identities for Analytic Functions: Summation

... I found that the identity...

$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} a_{k,n} = \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} a_{k,n}$ (1)

... is subject to the restriction...

$\displaystyle a_{k,n} = \mathcal {O} \{(k^{2}+ n^{2})^{- r}\},\ r>1$ (2)

... for the absolute convergence of (1)...

A very interesting problem that requires a little time!(Nerd) ...

Kind regards

$\chi$ $\sigma$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

Let $D=D(0,1)$ be the unit disc centered at 0. Let $f$ be holomorphic in D, with $f(0)=0$. Show that

$g(z)=\sum_{k=1}^{\infty} f(z^k)$

defines an holomorphic function in $D$.

I've argued as follows: since $f$ is holomorphic, then $f$ is locally the sum of a power series

$f(z)=\sum_{n=0}^{\infty}a_n z^n$
and since $f(0)=a_0=0$ by assumption, we can write

$f(z)=\sum_{n=1}^{\infty}a_nz^n$


which holds for all $z$ in $D$.
Hence

$g(z)=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}a_n z^{nk}$

How can i deal with this double index summation?
Another way to prove this would be to use the proof of Schwarz's Lemma to say that the function $h(z) = \begin{cases}f(z)/z&(z\ne0) \\ f'(0)&(z=0)\end{cases}$ is holomorphic in $D$ and therefore bounded in any smaller disc $D_r = \{z:|z|<r\}$, where $|r|<1.$ Say $|h(z)|\leqslant M_r$ whenever $|z|\in D_r.$

Thus $|f(z)|\leqslant M_r|z|\ (z\in D_r).$ Therefore $$|g(z)| \leqslant \sum_{k=1}^{\infty}|f(z^k)| \leqslant \sum_{k=1}^{\infty}M_r|z|^k = \frac{M_r|z|}{1-|z|} \leqslant \frac{M_rr}{1-r}\ (z\in D_r).$$ Hence $g(z)$ is a uniform sum of holomorphic functions on $D_r$ and therefore holomorphic there. Since $r<1$ is arbitrary it follows that $g(z)$ is holomorphic on the whole of $D$.
 

pantboio

Member
Nov 20, 2012
45
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

Another way to prove this would be to use the proof of Schwarz's Lemma to say that the function $h(z) = \begin{cases}f(z)/z&(z\ne0) \\ f'(0)&(z=0)\end{cases}$ is holomorphic in $D$ and therefore bounded in any smaller disc $D_r = \{z:|z|<r\}$, where $|r|<1.$ Say $|h(z)|\leqslant M_r$ whenever $|z|\in D_r.$

Thus $|f(z)|\leqslant M_r|z|\ (z\in D_r).$ Therefore $$|g(z)| \leqslant \sum_{k=1}^{\infty}|f(z^k)| \leqslant \sum_{k=1}^{\infty}M_r|z|^k = \frac{M_r|z|}{1-|z|} \leqslant \frac{M_rr}{1-r}\ (z\in D_r).$$ Hence $g(z)$ is a uniform sum of holomorphic functions on $D_r$ and therefore holomorphic there. Since $r<1$ is arbitrary it follows that $g(z)$ is holomorphic on the whole of $D$.
Thans for the response. I have understood your answer but i can't see the role played by schwartz's lemma in it. I mean, do i actually need Schwartz' lemma to state that? is it equivalent if i say:

$f(z)=a_0+a_1z+a_2z^2+\ldots$

but

$f(0)=0=a_0$

hence

$f(z)=a_1z+a_2 z^2+\ldots$

Therefore

$\frac{f(z)}{z}=a_1+a_2 z+...$

is holomorphic in $D(0,1)$ hence is continuous on compact sets $\overline{D(0,r)}$ and so it is bounded and so on....