Holomorphic function with a.e. vanishing radial boundary limits - MathOverflow

Holomorphic function with a.e. vanishing radial boundary limits

2011-01-17T07:49:31Z

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

Answer by Andrey Rekalo for Holomorphic function with a.e. vanishing radial boundary limits

2011-01-17T09:57:59Z

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).

Answer by GH for Holomorphic function with a.e. vanishing radial boundary limits

2011-01-17T10:38:36Z

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.

Answer by David Hansen for Holomorphic function with a.e. vanishing radial boundary limits

2011-01-17T22:38:51Z

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

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