Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am looking for an example of a function $f$ that is 1) continuous on the closed unit disk, 2) analytic in the interior and 3) cannot be extended analytically to any larger set. A concrete example would be the best but just a proof that some exist would also be nice. (In fact I am not sure they do.)

I know of examples of analytic functions that cannot be extended from the unit disk. Take a lacuanary power series for example with radius of convergence 1. But I am not sure if any of them define a continuous function on the closed unit disk.

share|cite|improve this question
Take a continuous function on the unit circle which does not have derivatives at a dense set, and use Cauchy's formula to extend it to an analytic function in the interior. – Mariano Suárez-Alvarez Jan 5 '10 at 18:52
That does give an analytic function in the interior. But what about its behavior on the boundary? You don't know what happens. – Anweshi Jan 5 '10 at 19:10
That probably works but there is one small point I wonder about: if I take an arbitrary continuous function $f$ on the unit circle there is not always an analytic function $g$ on the unit disk that has $f$ as its boundary value. (There exists such a function only if the Dirichlet problem for $f$ happens to have an analytic solution as I see it.) Thus when I extend by the Cauchy formula and then go back to the unit circle I am not sure I end up with the same function as I started with nor that the function suddenly has not acquired derivatives everywhere. – Johan Jan 5 '10 at 19:14
Write up the complex Fourier series for the function on the boundary and throw away all the Fourier coefficients with negative index. What you're left with extends to the analytic function proposed in the answer, but that is useless. – Harald Hanche-Olsen Jan 5 '10 at 19:31

5 Answers 5

up vote 11 down vote accepted

Let $f(z) = \sum z^n/n^2$, which is continuous and bounded on the closed unit disc but not analytic near $1$. Then consider

$$\sum f(z^n)/n^2.$$

This should have a singularity at every root of unity; and should be analytic in the interior because it is uniformly convergent.

share|cite|improve this answer
$f(1) = \zeta(2)$. Why are you saying $f$ is not analytic near $1$? – Anweshi Jan 5 '10 at 19:20
Because its derivative is not bounded near $1$. – David Speyer Jan 5 '10 at 19:23
Your clarification(both here and in my deleted answer) had been most helpful. Thanks. – Anweshi Jan 5 '10 at 19:27
How do you know that the singularities in the second sum don't cancel each other out? – Harald Hanche-Olsen Jan 5 '10 at 19:37
I also ask about Hanche-Olsen's objection. – Anweshi Jan 6 '10 at 12:41

One can invoke Carathéodory's theorem.

If $U$ is a simply connected open subset of the complex plane with a Jordan curve as boundary then the Riemann map $f : U \to \mathbb D$ extends continuosly to the boundary and the extension is a homeomorphism $\partial U \to S^1$ at the boundary.

To obtain the sougth function, it suffices to consider a simply connected open set $U\subset \mathbb C$ having a nowhere analytic Jordan curve as boundary and take the inverse of the Riemann map of $U$.

share|cite|improve this answer
To be specific about an example: pick any non-real 'c' from the interior of the main cardioid of the Mandelbrot set. The Julia set for such a value of c is a Jordan curve. You can pick either the the interior or exterior components of the Fatou set (in the Riemann sphere) and get a uniformizing function that satisfies your requirements. The extension is even Hölder continuous. – Jacques Carette Mar 27 '10 at 14:27

I found some neat stuff in Remmert's Classical topics in complex function theory.

Fabry's gap theorem gives a way to construct many examples including some already mentioned. Stated for the unit disk, it says:

If $m_1,m_2,\ldots$ is a sequence of positive integers such that $\displaystyle{\lim_{n\to\infty}}\frac{m_n}{n}=\infty$ and if $\displaystyle{f(z)=\sum_{n=1}^\infty a_nz^{m_n}}$ has radius of convergence 1, then the unit disk is the domain of holomorphy of $f$.

For example, if $p_n$ is the $n^{th}$ prime, then $$f(z)=\sum_{n=1}^\infty \frac{z^{p_n}}{n^2}$$ converges uniformly on the closed disk and is therefore continuous. It is not analytically extendable to any larger set because it satisfies the hypotheses of Fabry's theorem.

An interesting result that yields many such functions in a nonconstructive way is a theorem of Fatou-Hurwitz-Pólya:

If $\displaystyle{f(z)=\sum_{n=0}^\infty a_n z^n}$ has radius of convergence 1, then the set of functions $$f_\epsilon(z)=\sum_{n=0}^\infty \epsilon_na_nz^n$$ for $\epsilon_n\in\{\pm1\}$ whose domain of holomorphy is the unit disk has cardinality $2^{\aleph_0}$.

Hausdorff showed further that if $\displaystyle{\lim_{n\to\infty} |a_n|^{1/n}}$ exists (and equals 1) then the set of such functions whose domain of holomorphy is not the unit disk is at most countable. This applies in particular to the function $\displaystyle{f(z)=\sum_{n=1}^\infty \frac{z^n}{n^2}}$, which therefore yields examples by changing the signs of the coefficients in all but countably many ways.

One more, this time an explicit example from Remmert: The series $$f(z)=1+2z+\sum_{n=1}^\infty\frac{z^{2^n}}{2^{n^2}}$$ is one-to-one and has real derivatives of all orders on the closed disk, and has the open disk as domain of holomorphy.

Reference: Remmert's Classical topics in complex function theory, pages 252-258. (Fatou-Hurwitz-Pólya is stated on a page without preview.)

share|cite|improve this answer

Here is a very concrete example:

$ g(z) = \sum_{n=0}^{\infty}\frac{z^{2^n + 1}}{2^n + 1}. $

The power series converges uniformly to a continuous function on the closed unit disk. Differentiating we obtain $g'(z) = f(z)$ with

$ f(z) = \sum_{n=0}^{\infty}z^{2^n}. $

This is the standard example of a function with a natural boundary. Clearly $f(x) \rightarrow +\infty$ as $x \rightarrow 1^{-}$ on the real axis. The functional equation

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

shows that $f(x) \rightarrow + \infty$ as $x \rightarrow (-1)^{+}$ on the real axis, then $|f(z)| \rightarrow \infty$ as $z$ tends radially to ${\pm}i$, and so on, so that $|f(z)|$ tends to $\infty$ as $z$ tends radially to any root of unity of order $2^m$. Hence $f(z)$ has a dense set of singularities on the unit circle, and so does $g(z)$, thus $g(z)$ has the unit circle as natural boundary.

share|cite|improve this answer
Why isn't the following used: $g(z)$ has the unit circle as natural boundary, because its derivative $f(z)$ blows up at all roots of unity and has the unit circle as the natural boundary. – Anweshi Jan 5 '10 at 21:20
The advantage of this example is the ease of showing rigorously that the function blows up at the roots of unity of form $\exp(2{\pi}ik/2^m)$, due to the functional equation. It is not obvious that this happens at other roots of unity; you would have to establish an estimate to exclude that there is not some weird kind of cancellation happening. And in fact, weird cancellation can happen! The function $h(z) = (1 - z)^{-1}(1 - z^2)^{-1}(1 - z^3)^{-1}{\cdots}$ blows up (radial approach) very fast at every root of unity. At most other points of the unit circle, it actually tends to zero. – engelbrekt Jan 6 '10 at 3:51
@engelbrekt,you mean $h(z)$ is without unit circle as natural boundary? – XL _at_China Jul 30 '14 at 0:07

I suggest this function: $$f(z)=\sum_{n=1}^\infty \frac{z^{n!}}{n^2}.$$ It converges uniformly on the closed unit disk, and the derivatives blow up as you approach any root of unity radially.

share|cite|improve this answer
I might add that the trick with n! in the exponent is a classic one. I did not invent it! (Not sure who did.) – Harald Hanche-Olsen Jan 5 '10 at 19:50
You could have replaced $n!$ by $n^n$ for instance, and then called it your own! – Anweshi Jan 5 '10 at 19:56
No. The point of $n!$ is that if $z$ is a root of unity then $z^{n!}=1$ for all sufficiently large $n$. This makes the blowup argument trivial. – Harald Hanche-Olsen Jan 5 '10 at 20:03
If this is a classic one, where else did you see it? – Anweshi Jan 6 '10 at 12:42
I suppose, you can use $n! + 1$ instead in the exponent. Then the derivative does not even converge on the boundary. This makes things simpler. – Anweshi Jan 8 '10 at 21:13

Your Answer


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.