Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Another Qual question here, For the function $$\sum_{n=0}^\infty z^{2^n}$$, Prove the following:

i) $f$ converges to a function analytic in the open unit disk $D$,

ii) $f(z) =z+f(z^2)$ and

iii) $f(z)$ can not be analytically continued past any point on the unit circle.

I can even see (ii) very easily, but I can not see how can I prove rigorously (i) and (iii). Help please.

share|improve this question
    
For i), note that this is a power series, and power series always define analytic functions within their disk of convergence, but I don't know what you are allowed to assume. –  Jonas Meyer Dec 24 '12 at 5:53
1  
A rough idea for iii): Using the fact that the set of numbers that are $2^n$th roots of $1$ for some $n$ is dense in the unit circle, can you show that $f$ is unbounded as you approach any point on the circle? –  Jonas Meyer Dec 24 '12 at 6:13
    
Sorry, I can not. Would you please! –  Deepak Dec 24 '12 at 6:16
    
Deepak: I do not mean immediately, within 3 minutes, can you show this. I meant to be suggesting an approach, which would take some thought to work through. I wish you well. –  Jonas Meyer Dec 24 '12 at 6:20
add comment

2 Answers 2

up vote 4 down vote accepted

We have $f(z) = \sum_n a_n z^n$, where $a_n = \begin{cases} 1 & \text{if}\ \exists k\ n = 2^k \\ 0 & \text{otherwise} \end{cases}$

i) The radius of convergence is given by $\frac{1}{R} = \limsup_n \sqrt[n]{a_n}= \lim_n \sqrt[2^n]{1} = 1$. Hence $R=1$, and $f$ is defined on $D$.

ii) $f(z) = \sum_{n=0}^\infty z^{2^n} = z+\sum_{n=1}^\infty z^{2^n} = z+\sum_{n=0}^\infty z^{2^{n+1}} = z+\sum_{n=0}^\infty (z^2)^{2^{n}} = z+f(z^2)$. It follows by induction that $f(z) = \sum_{k=0}^{n-1} z^{2^k} + f(z^{2^n})$ for any $n$.

iii) Note that $\lim_{r \uparrow 1}f(r) = \infty$ (for $r$ real). If $w^{2^n} = 1$ ii) gives $f(rw) = \sum_{k=0}^{n-1} (rw)^{2^k} + f(r^{2^n})$, and we have $\lim_{r \uparrow 1}|f(rw)| = \infty$. Let $\Omega_n = \{w| w^{2^n} = 1\}$, and $\Omega = \cup_n \Omega_n$. It is easy to see that $\Omega$ is dense in $\partial D$, and hence the set $\{z \in \partial D | \lim_{r \uparrow 1}|f(rz)| = \infty \}$ is dense in $\partial D$. Hence $f$ can not be continued in any neighborhood of any point in $\partial D$.

share|improve this answer
add comment

$f$'s radius of convergence is 1 by Hadamard's theorem. So in the open disk it gives an analytic function.

For III, if $f$ can be analytically continued at some point, then we should have a functional element around that point in the unit circle via analytical continuation. Continue this process, this would imply $f$ can be analytically continued in the whole circle except maybe 1 point. So we may assume one of $\pm 1, \pm i$ is in the region it converges. But this cannot hold, since the value are all $\infty$.

share|improve this answer
    
I see. now fixed. –  Bombyx mori Dec 24 '12 at 6:10
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.