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
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1$\begingroup$ The function is not going to be in any Hardy space... So "easy"? :) $\endgroup$– Mariano Suárez-ÁlvarezCommented Jan 17, 2011 at 7:59
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1$\begingroup$ Nitpick/bad joke: the words "not identically zero" are missing $\endgroup$– Yemon ChoiCommented Jan 17, 2011 at 8:40
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$\begingroup$ Yemon, I was about to write that... Otherwise $f\equiv 0$ is a VERY easy example... $\endgroup$– diveriettiCommented Jan 17, 2011 at 9:40
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$\begingroup$ Yes, thank you. I ment a non-zero example. $\endgroup$– CPJCommented Jan 17, 2011 at 16:24
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3 Answers
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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).
answered Jan 17, 2011 at 9:57
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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
$\int_{0}^{2\pi}|f(re^{i\theta})|^{-2}d\theta=2\pi\sum_{n\in \mathbf{Z}}r^{n(3n-1)} \sim 2 \pi^{3/2}3^{-1/2}(1-r)^{-1/2}.$
answered Jan 17, 2011 at 22:38
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$\begingroup$ 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). $\endgroup$ Commented Jan 17, 2011 at 22:47
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$\begingroup$ @David: Could you possibly indicate the page containing the footnote? I did not find it. $\endgroup$ Commented Jan 18, 2011 at 0:13
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$\begingroup$ @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. $\endgroup$ Commented Jan 18, 2011 at 0:18
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$\begingroup$ @David: Thats a very nice example. Thank you! $\endgroup$– CPJCommented Jan 18, 2011 at 8:14
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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.
answered Jan 17, 2011 at 10:38
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1$\begingroup$ Unfortunately I can give the credit to just one comment. Thanks to both of you. Especially for the links. $\endgroup$– CPJCommented Jan 17, 2011 at 16:36
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1$\begingroup$ 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 ? $\endgroup$ Commented Jul 30, 2011 at 3:39
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$\begingroup$ @Lvriemsurf: I edited my response and included an answer to your question. $\endgroup$ Commented Jul 31, 2011 at 19:15
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