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Let $f(z)$ be a holomorphic function on the unit disc, with logarithmic growth at the boundary:

$$ |f(z)| = \mathcal O\bigg(\log\Big(\frac{1}{1-|z|}\Big)\bigg). $$

Does it follow that the (distributional) boundary value of $f$ is in $L^2$?

An equivalent, more elementary, way of asking my question is:
Do the functions $g_\varepsilon:S^1\to \mathbb C$ given by $g_\varepsilon(z) := f((1-\varepsilon)z)$ form a Cauchy sequence in $L^2(S^1)$?

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    $\begingroup$ Certainly not. You can easily write a lacunary series $\sum_{k\ge 1}z^{n_k} $ that grows as slowly as you want. $\endgroup$
    – fedja
    Commented Nov 25, 2021 at 15:14
  • $\begingroup$ @fedja. Yes of course! Stupid me! Hmm... I have certain functions which came up in my work. They satisfy $|f(z)| = \mathcal O(\log(\tfrac{1}{1-|z|}))$, and I would really like them to have $L^2$ boundary values. Are there any other conditions which I could try to check and that would imply that my functions have have $L^2$ boundary values? PS: you may post you comment as an answer and I will accept it. $\endgroup$
    – André Henriques
    Commented Nov 25, 2021 at 20:39
  • $\begingroup$ I have certain functions which came up in my work. Then why don't you just tell us what they are? $\endgroup$
    – fedja
    Commented Nov 25, 2021 at 20:44
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    $\begingroup$ @fedja Sure. Here you go. Btw, my function is not scalar-valued. It's Hilbert space valued. Let $V$ be a unitary VOA, and let $H$ be the Hilbert space completion of $V$. Let $v\in V$ be in degree $n$, and let $\xi \in H$ be any vector. My function is the $n$th antiderivative of $z\mapsto Y(v,z)z^{2L_0}\xi$ (this function is not defined on the unit disc but only on the universal cover of the punctured unit disc). [Now you see why I didn't just tell us what these functions are ;-) ] $\endgroup$
    – André Henriques
    Commented Nov 25, 2021 at 21:50
  • $\begingroup$ PS: Really, what I know is that $f(z):=Y(v,z)z^{2L_0}\xi$ satisfies $|f(z)|\le \frac{1}{(1-|z|)^n}$. And I what I want to be able to deduce is that its boundary values lies in the ($H$-valued) negative Sobolev space $W^{-n,2}(S^1)$. $\endgroup$
    – André Henriques
    Commented Nov 25, 2021 at 21:58

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