5
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Dec 24 '12 at 5:53
  • 1
    $\begingroup$ 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? $\endgroup$ Dec 24 '12 at 6:13
  • $\begingroup$ Sorry, I can not. Would you please! $\endgroup$
    – Deepak
    Dec 24 '12 at 6:16
  • $\begingroup$ 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. $\endgroup$ Dec 24 '12 at 6:20
9
$\begingroup$

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 $\partial D$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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