Show that $f(z)=\sum_{n=0}^{\infty}z^{2^n}$ can't be analytically continued past the unit disk.

I'm reading the problems of Stein and Shakarchi's Complex Analysis, Chapter 2 Problem 1 asks to show that $$f(z)=\sum_{n=0}^{\infty}z^{2^n}$$ cannot be analytically continued past the unit disk. (Hint: Suppose $\theta =\frac{2\pi p}{2^k}$ for $p,k$ positive integers, let $z=re^{i\theta}$ and show $\mid f(z)\mid\rightarrow\infty$ as $r \rightarrow1$).

I understand from the hint they want me to "pepper" the unit circle with points where the power expansion explodes so that it is dense with poles. I do not understand why they choose such particular points, but I assume that in retrospect it will show that those are the ones that I can show divergence for the easiest and are dense in the unit circle, so plowing ahead:

$$\lim_{r \rightarrow 1}\left| \sum_{n=0}^\infty r^{2^n}e^{ i2\pi p 2^{n-k}}\right| = \left| \sum_{n=0}^k e^{ \frac{i2\pi p}{2^{k-n}}} + \sum_{n=k+1}^\infty e^{ i2\pi p2^{n-k}} \right| $$

Where do I go from here? Is there some oversimplification of these sinusoids that I'm not seeing? Furthermore, once I manage to show this explodes, if I show that these numbers are dense on the unit circle I'm done, right?

Any insight is much appreciated.

This may be already answered by the OP himself. Here is a short answer any way.

First, recall that for any power series $f(z)=\sum_{n\geq0}a_nz^n$ with radius of convergence $1$, a point $p\in\mathbb{S}^1$ is called regular (for $f$) if there is an open ball $D(p;r)$, $r>0$, and an analytic function $g$ on $D(p;r)$ such that $f(z)=g(z)$ on $D(p;r)\cap D(0;1)$. Any point in $\mathbb{S}^1$ which is not regular for $f$ is said to be singular.

• It is easy to see that the set of regular points of $f$ is open and so the set of singular points of $f$ is closed.
• Furthermore, it is well known that the set of singular points is non empty (This can proven by contradiction using the properties of the radius of convergence).

The following result will be useful:

Lemma: If $f(z)=\sum_na_kz^k$ is a power series with radius of convergence $1$ and $a_k\geq0$, then $$\lim_{r\rightarrow1-}f(r)=\sum_ka_k$$

Proof: Since $a_n\geq0$, for each $0<r<1$ and $n\in\mathbb{N}$ $$f_n(r)=\sum^n_{k=0}a_kr^k\leq f(r)=\sum^\infty_{k=0}a_kr^k\leq \sum^\infty_{k=0}a_k $$ Clearly $f$ is monotone nondecreasing over $(0,1)$ and so, $\lim_{r\rightarrow1-}f(r)=\sup_{0<r<1}f(r)$ exists (as an extended real number). Putting things together, we obtain that

$$ \lim_{r\rightarrow1-}f_n(r)=\sum^n_{k=0}a_n\leq \lim_{r\rightarrow1-}f(r)\leq\sum_{k\geq0}a_k $$ Letting $n\rightarrow\infty$ gives $\sum_ka_k=\lim_{r\rightarrow1-}f(r)$.

For the OP, $f(z)=\sum_{n\geq0}z^{2^n}$ can be expressed as the power series $f(z)=\sum_ka_kz^k$ where $a_k=1$ if $k=2^n$ form some $n\in\mathbb{Z}_+$, and $0$ otherwise. Clearly it has radius of convergence $1$.

Thus, $\lim_{r\rightarrow1-}f(r)=\infty$ by the Lemma above.

On the other hand, for any $z\in B(0;1)$, $f(z^2)=f(z)-z$ from where we obtain (by induction) that $$ f(z^{2^n})=f(z)-\sum^{n-1}_{j=0}z^{2^j}, \qquad n\in\mathbb{N}$$

Along the line $rz_{k, m}=re^{2\pi i k2^{-m}}$, $0<r<1$, we have

$$ f(r^{2^m})=f\big(re^{2\pi ik2^{-m}}\big) - \sum^{m-1}_{j=0}r^{2^j}e^{2\pi ik 2^{j-m}} $$ The term $p_{m,k}=\sum^{m-1}_{j=0}r^{2^j}e^{2\pi ik 2^{j-m}}$ is bounded by $m+1$. By the Lemma above, $\lim_{r\rightarrow1-}f(r^{2^m})=\infty$, we conclude that $\lim_{r\rightarrow1-}|f(rz_{m,k})|=\infty$.

As a consequence, all points in $\mathcal{D}=\{z_{k, m}:k\in\mathbb{Z},m\in\mathbb{Z}\}$ are singular points of $f$. Since $\mathcal{D}$ is dense in $\mathbb{S}^1$, it follows that all points in $\mathbb{S}^1$ are singular for $f$; hence, $f$ cannot be analytically extended to any open domain $\Omega$ that properly contains $D(0;1)$.