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Here is my question : I have a harmonic function $h$ on the open unit disc in $D \subset \mathbb{C}$, such that $\iint_D e^{2h} d\lambda(z) \leq A < \infty$ ($d\lambda$ is the Lebesgue measure on $\mathbb{C}$).

Can one have an upper bound of $e^h$ near the boundary of the disc ?

With elementary tools (mean value property for harmonic functions + Jensen's inequality with the exponential map) I can only obtain $e^{h(z)} \leq \frac{C^{te}}{d(z,S^1)}$ ($d(z,S^1)$ is the distance between $z$ and the boundary of $D$), but I would like something better (for example, $e^{h(z)} \leq \frac{C^{te}}{d(z,S^1)^{1/2}}$ or something like that would be nice).

If not, can one have an upper bound on the integral of $e^h$ over a ray passing through 0 ($\int_0^1 e^{h(r)} dr$ for example) ?

Thank you!

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Function $f(z)=(1-z)^\alpha$ with $\alpha=-1+\epsilon$ is zero-free in the unit disk, and its coefficients satisfy $a_n\sim cn^{-\alpha-1}=cn^{-\epsilon}$. Setting $u=\log|f|$, we obtain, using Parseval, $$\int_{|z|<1}e^{2u}d\lambda=\int_{|z|<1}|f|^2d\lambda=\sum_n\frac{1}{2n+1}|a_n|^2<\infty.$$ So your trivial estimate is close to the best possible, at least the order cannot be improved.

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A theorem of F.W. Gehring, On the radial order of subharmonic functions, J. Math. Soc. Japan 9 (1957), 77-79 asserts that a nonnegative subharmonic function $u$ on the unit disk such that $\int_{|z|<1}u^p(z)\,d\lambda(z)<\infty$ for some $p>1$ satisfies, for a.e. $\theta\in\mathbb R$, $u(z) = o\left((1-|z|)^{-1/p}\right)$ as $z\to e ^{i\theta}$ uniformly in nontangential approach regions. Taking $u(z)= e^{h(z)}$, one obtains $e^{h(z)} = o\left((1-|z|)^{-1/2}\right)$ as $z\to e ^{i\theta}$ in the conditions just stated.

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  • $\begingroup$ This is for almost all $\theta$ and the question was for all $\theta$ if I understood it correctly. $\endgroup$ – Alexandre Eremenko Mar 22 '15 at 1:22
  • $\begingroup$ Thank you for this reference ; the paper is interesting. Even though I was trying to find a uniform constant in the majoration (I mean a uniform $o$ in the formula). $\endgroup$ – Clem. Mar 23 '15 at 10:42

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