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!



Growth of a harmonic function on the disc
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.


