I am trying to figure out the following problem:
Let $X$ be a Riemann surface and $\mathcal{F}$ a family of harmonic functions that are bounded below by $-1$. Assume that there is a compact set $A\subset X$ such that $$\min \{u(z):z\in A\}\leq 1$$ for all $u\in \mathcal{F}$. Show that $\mathcal{F}$ is locally bounded.
My try:
It is obvious that we only need to show
For every $x\in X$ we can found a neighborhood $U\subset X$ and a constant $C$ such that $u(z)\leq C$ for every $z\in U$ and $u\in \mathcal{F}$.
However, all the given conditions are about the minimum. So I have no idea how can I use this to prove there is an upper bound.
The only relevant property of harmonic functions is
If $h$ is a non-negative harmonic function in $D_R(0)$, then for $z\in D_r(0)$, we have
$$\frac{R-r}{R+r}h(0)\leq h(z)\leq \frac{R+r}{R-r}h(0).$$
Since these functions are bounded below by $-1$, so we can add $2$ to every function to make them non-negative. Then we can use this property. But we cannot get a uniform conclusion even we can get an upper bound for every function. Then I got stuck.
Any hints and answers are welcomed! Thanks!
Yes, Harnack's inequality is the way to go. For each point $x\in X$ there is a chain of disks $B_1, \dots, B_n$ such that the $B_1$ contains a neighborhood of $x$, the set $A$ is covered by the union of $B_k$, and $B_k \cap B_{k+1}$ is nonempty. The existence of such "Harnack chain" is a consequence of $X$ being connected.
By doubling the size of each $B_k$, we can ensure that $B_k$ and $B_{k+1}$ overlap substantially: each contains the center of the other one.
After adding $1$ to each harmonic function, we have nonnegative harmonic functions. By Harnack's inequality, for any such function $h$, its minimum on $B_k$ and its minimum on $B_{k+1}$ are comparable to each other, with a constant independent of $h$. (Indeed, the minimum on one disk controls the maximum on 90% of that disk, which includes some points of the other disk.) Since $A$ is covered by these disks, there exists $k$ such that $\min_{B_k}h\le 2$. Following the chain to $B_1$ yields $h\le C$ in a neighborhood of $x$.