Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So, I saw the following question concerning complex analysis in one variable:

Can you find a continuos, surjective function $f:D\rightarrow D$, where $D$ is the closed unit disk $\overline{B_1(0)}\subseteq \mathbb{C}$, such that the following holds?

$f$ is holomorphic on the open unit disk $B_1(0)$, but there is no point $z$ on the boundary $\partial B_1(0)$ with the property that there is an open neighborhood $U$ and a holomorphic function $g:U\rightarrow\mathbb{C}$, such that $g$ and $f$ agree on $U\cap B_1(0)$.

I have already found a power series which has the latter property, but is sadly only defined on the open unit disk. However, I have no clue whether such a function can be found at all.

How can I tackle that exercise? Might Brouwer's fixed-point theorem be somehow useful here?

share|improve this question
"Sphere" should be "ball" or "disc" throughout. Why do you think Brouwer's theorem might be useful? –  Chris Eagle Apr 20 '12 at 20:14
Look up Hadamards theorem on lacunary series. See also this answer. The function $f(z) = 1+ 2z + \sum_{n=1}^{\infty} 2^{-n^2} z^{2^n}$ is analytic on the open unit disk and $C^\infty$ on the closed disk but cannot be analytically continued at any point of the unit circle. –  t.b. Apr 20 '12 at 20:15
@t.b. But clearly $f(1) = \frac{5}{2} + \frac{1}{2} \theta_3\left(\frac{1}{2}\right)$. Did you mean to give a different expression? –  Sasha Apr 20 '12 at 22:03
@Sasha: I think the expression is the one I intended. It is supposed to converge on the unit circle. But the function it cannot be extended analytically by Hadamard's theorem, how does your identification of $f(1)$ contradict that? –  t.b. Apr 20 '12 at 22:16
@Leonid: Sorry about that, but I don't think my suggestion and my example answer the question, since the example I give doesn't map the closed unit disk onto the closed unit disk. It's only an example of a smooth function defined on the unit disk that cannot be extended analytically over the boundary. I suspect an application of Schwarz's lemma should give a negative answer to the question but I haven't really thought about it and I don't have the time to do so these days. –  t.b. Jun 21 '12 at 16:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.