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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.

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    $\begingroup$ Every term in the last sum of yours simplifies to $1$. $\endgroup$
    – mrf
    Apr 19, 2016 at 12:41
  • $\begingroup$ Much more holds: this function cannot be continuously extended past of the unit disk $D$, just because for all $a \in \partial D$ the limit $$\lim_{z \to a} f(z)$$ does not exist. $\endgroup$
    – Crostul
    Apr 19, 2016 at 12:41
  • $\begingroup$ @Crostul This has been my confusion, so if I just consider $lim_{r \rightarrow 1} \sum_{n=0}^ \infty r^{2^n}e^{i \theta 2^n}$ for any angle it will diverge to infinity? In essense the entire unit circle is a pole? Why would Stein and Shakarchi put the reader through all this trouble of choosing a specific angle based on some $p$ and $k$ if any angle at all would do? Is divergence a lot easier to show in their points? $\endgroup$
    – Mike
    Apr 19, 2016 at 12:45
  • $\begingroup$ @mrf Thanks, you mean the second one of the two sums on the right-hand side? $\endgroup$
    – Mike
    Apr 19, 2016 at 12:47
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    $\begingroup$ I would consider $\lim_{r \to 1^-} f(r e^{ 2 i \pi k / 2^m})$ with $k,m$ integers, show that it diverges, hence there is a singularity in the neighborhood of any point of the unit circle (because the set $\{\frac{2 \pi k}{2^m} \ \mid \ (k,m) \in \mathbb{Z}^2 \ \}$ is clearly dense in $\mathbb{R}$). And since the analytic continuation relies on some path of intersecting disks where the function is holomorphic, there is no such path for leaving the unit disk out, hence no possible analytic continuation of $f(z)$ outside the unit disk. $\endgroup$
    – reuns
    Apr 19, 2016 at 13:18

3 Answers 3

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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 $D(0;1)$.

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Yes, your "I understand from the hint" paragraph is good (I especially liked "plowing ahead"). Working with the sum, I wouldn't right off the bat start taking limits. What we can say is that for $0\le r <1,$

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

Now the $n$th summand in first sum on the right is simply $r^{2^n}.$ (That's why those weird points on the boundary were chosen.) In the second sum, move the absolute values inside the sum. This shows $(1)$ is at least

$$\sum_{n=k+1}^\infty r^{2^n} - \sum_{n=0}^k r^{2^n} \ge \sum_{n=k+1}^\infty r^{2^n} - (k+1).$$

Thus we're done if we show the first sum on the right has limit $\infty.$ One way to do this is let $N\in \mathbb N, N>k+1 .$ Then

$$\sum_{n=k+1}^\infty r^{2^n} > \sum_{n=k+1}^N r^{2^n}.$$

This implies

$$\tag 2 \lim_{r\to 1^-} \sum_{n=k+1}^\infty r^{2^n} \ge N-k.$$ Since $N$ is arbitrarily large, the left side of $(2)$ is $\infty,$ and we're done.

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In the comments I made the claim that a holomorphic function in $\mathbb D$ cannot have radial limit $\infty$ at each boundary point. There was some interest in this, which is why I'm posting another answer (4 years later!). I'll here sketch a proof for the analogous result in the upper half plane $H$ to make things a little easier.

Thm: Suppose $f$ is holomorphic on $H.$ Then it is impossible to have

$$\lim_{y\to 0^+}|f(x+iy)|=\infty\, \text {for all } x\in \mathbb R.$$

Proof: Suppose otherwise. The first step is to use Baire. For $m=1,2,\dots,$ let$$E_m=\{x\in \mathbb R: |f(x+iy)|\ge 1,$$ $0<y\le 1/m\}.$ Then each $E_m$ is closed and $\mathbb R=\cup E_m.$ By Baire there exists $m$ such that $E_m$ contains a nontrivial interval. Thus there exists a rectangle $R=(a,b)\times (0,h)$ with $|f|\ge 1$ on $R.$

We then have $1/f$ holomorphic on $R$ with $|1/f|\le 1$ in $R.$ So $1/f\in H^\infty(R)!$ Furthermore, $1/f(x+iy) \to 0$ as $y\to 0^+$ for $x\in (a,b).$ It is well known that such a function is identically $0.$ So we conclude $1/f\equiv 0,$ contradiction.

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  • $\begingroup$ What I was missing in mt understanding of the argument is the classical representation theorem of functions in $H^\infty(B(0;1))$ as the Poisson kernel acting on $L_\infty(\mathbb{S}^1)$. One may need to use the Koebe-Riemman conformal map to complete the argument I suppose. Things are clear now, thanks! $\endgroup$
    – Mittens
    Oct 12, 2020 at 19:51

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