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

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In summary: Cauchy's estimate, isn't it?Yes, it is essentially the same as Cauchy's estimate. I used the Schwarz's Lemma to justify the inequality $|f(z)| \leqslant M_r|z|$ for $|z| \in D_r$, which is then used to show that the series $\sum_{k=1}^{\infty} f(z^k)$ converges uniformly and therefore defines a holomorphic function. However, as you have pointed out, Cauchy's estimate can also be used to show that $f(z)/z$ is bounded in $D_r$, and therefore the series converges uniformly.
  • #1
pantboio
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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?
 
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  • #2
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

pantboio said:
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

$\chi$ $\sigma$
 
  • #3
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

chisigma said:
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:
  • #4
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

pantboio said:
... 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

$\chi$ $\sigma$
 
  • #5
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

pantboio said:
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 $D$.
 
  • #6
Re: if f is holomorphic, is $\sum f(z^k)$ holomorphic?

Opalg said:
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...
 

FAQ: If f is holomorphic, is Σf(z^k) holomorphic?

Is the sum of holomorphic functions also holomorphic?

Yes, if each individual function in the sum is holomorphic, then the sum is also holomorphic. This is because the sum of two holomorphic functions is still a holomorphic function.

Can a non-holomorphic function be written as the sum of holomorphic functions?

No, a non-holomorphic function cannot be written as the sum of holomorphic functions. This is because a non-holomorphic function would have singularities or poles, which cannot be canceled out by adding other holomorphic functions.

Does the order of the terms in the sum affect the holomorphicity of the overall function?

No, the order of the terms in the sum does not affect the holomorphicity of the overall function. As long as each individual term is holomorphic, the sum will also be holomorphic regardless of the order of the terms.

Can a power series with a non-holomorphic function as its terms be considered holomorphic?

No, a power series with a non-holomorphic function as its terms cannot be considered holomorphic. This is because a power series is a representation of a holomorphic function, and if any of its terms are not holomorphic, then the power series is not holomorphic.

Is the sum of infinitely many holomorphic functions always holomorphic?

Not necessarily. If the sum is a divergent series, then it may not be holomorphic. However, if the sum is a convergent series, then it will be holomorphic as long as each individual term is holomorphic.

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