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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.

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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
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
up vote 6 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 $\partial D$.

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$f$'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 $\infty$.

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I see. now fixed. – Bombyx mori Dec 24 '12 at 6:10

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