Can one construct a harmonic function $f$ defined in the unit disk with the condition $f(0)≥1$ such that area of $\{z∈D:f(z)>0\}$ is small enough, i. e. for every $\epsilon>0$ does there exist a function harmonic in the unit disk with the condition $f(0) \ge 1$ s.t. the Lebesgue planar measure of $\{z∈D:f(z)>0\}$ is less than $\epsilon$ ? That was asked [here](https://math.stackexchange.com/questions/982027/area-growth-of-harmonic-functions), but not answered.



Does there exist harmonic function with that property?
Yes you can. First of all the condition $|f(0)|\geq 1$ is irrelevant: you can always multiply your function on a positive constant. 


Take an entire function $F$ for which the set $\{ z:|F(z)|>1\}$ is contained in the strip
$\{ x+iy:|y|<\pi\}$. Such a function is constructed in Hayman's book Meromorphic functions, section 4.1.1, and it is called $E\_0$ there. (Other people call it Mittag-Leffler function). Make sure that $F(0)=2$ by an appropriate shift.
Then consider $F(kz)$ where $k$ is very large.
Then the set in the unit disk where $|F(kz)|>1$ will be as small as you wish, but will contain $0$. Finally take real part of $F(kz)$. 


