MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hello everybody.

I'm looking for an "easy" example of a (non-zero) holomorphic function $f$ with almost everywhere vanishing radial boundary limits: $\lim\limits_{r \rightarrow 1-} f(re^{i\phi})=0$.

Does anyone know such an example.

Best CJ

share|cite|improve this question
The function is not going to be in any Hardy space... So "easy"? :) – Mariano Suárez-Alvarez Jan 17 '11 at 7:59
Nitpick/bad joke: the words "not identically zero" are missing – Yemon Choi Jan 17 '11 at 8:40
Yemon, I was about to write that... Otherwise $f\equiv 0$ is a VERY easy example... – diverietti Jan 17 '11 at 9:40
Yes, thank you. I ment a non-zero example. – CPJ Jan 17 '11 at 16:24
up vote 7 down vote accepted

I am not sure if this would qualify as 'easy' but the first example of such a function was constructed by Lusin. It can be found in N. Lusin, J. Priwaloff, Sur l'unicité et la multiplicité des fonctions analytiques, Ann. Sci. École Norm. Sup. (3), 1925, p. 143-191 (see p. 185).

share|cite|improve this answer

According to a footnote in the famous Hardy-Ramanujan paper "Asymptotic formulae in combinatory analysis", the function $f(q)=\prod_{n=1}^{\infty}\frac{1}{1-q^n}$ vanishes like $f(re^{i\theta})=o((1-r)^{1/4-\varepsilon})$ for almost all $\theta$. No proof is given, though I can't imagine Hardy would have made a statement like this without a proof in his pocket.

Edit: This isn't actually hard to guess at. By Euler's pengatonal number theorem, we have $f(q)^{-1}=\sum_{n\in \mathbf{Z}}(-1)^{n}q^{n(3n-1)/2}$, so Plancherel gives

$\int_{0}^{2\pi}|f(re^{i\theta})|^{-2}d\theta=2\pi\sum_{n\in \mathbf{Z}}r^{n(3n-1)} \sim 2 \pi^{3/2}3^{-1/2}(1-r)^{-1/2}.$

share|cite|improve this answer
That's impressive. I was thinking of such an example myself, but then the results on the possible sets of good $\theta$'s scared me away (e.g. full measure implies first Baire category). – GH from MO Jan 17 '11 at 22:47
@David: Could you possibly indicate the page containing the footnote? I did not find it. – Andrey Rekalo Jan 18 '11 at 0:13
@Andrey: It's the asterisked footnote on the first page of Section 1.5 of the paper, which is page 281 in the AMS-Chelsea edition of Ramanujan's collected papers. – David Hansen Jan 18 '11 at 0:18
@David: Thank you. – Andrey Rekalo Jan 18 '11 at 0:35
@David: Thats a very nice example. Thank you! – CPJ Jan 18 '11 at 8:14

To complement Andrey Rekalo's response, Lusin's construction was generalized by Bagemihl and Seidel (Math. Zeitschrift 61 (1954), online here). See their Corollary 4 whose proof takes about 2 pages, much less than the original one by Lusin-Priwaloff. Of course the proof relies on Mergelyan's famous approximation theorem for which see Section 20.5 in Rudin: Real and Complex Analysis.

EDIT: Lvriemsurf asked in a comment if we can replace "almost every angle" by "every angle" in the construction. The answer is "no", as follows from the Lusin-Priwaloff theorems.

share|cite|improve this answer
Unfortunately I can give the credit to just one comment. Thanks to both of you. Especially for the links. – CPJ Jan 17 '11 at 16:36
The question and the answers / links are great ! My question is : what if I replace almost all/every $\theta$ by all/every $ theta $ ? Then can we still have such a holomorphic function on $D$ whose all radial limits are zero ? – Analysis Now Jul 30 '11 at 3:39
@Lvriemsurf: I edited my response and included an answer to your question. – GH from MO Jul 31 '11 at 19:15
@ GH : thank you very much. – Analysis Now Aug 7 '11 at 21:17

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.