# Positive harmonic function tending to infinity on the integers tends to infinity everywhere

Let $u(x+iy)$ be a positive harmonic function defined on $-1<y<1$ such that $u(n) \to \infty$. Prove that $u(x+iy) \to \infty$ as $x \to \infty$ for every $y$.

I believe that I almost got the solution: Montel's theorem for positive harmonic functions says that the family of functions $u_n(z)=u(z+n)$ converges (locally uniformly) to a harmonic function everywhere. Since this is a limit of positive functions, it is a nonnegative function. A nonnegative harmonic function must be zero by a connectedness argument, so $u$ is everywhere zero. In other words, for every $x,y$, $u(x+n+iy) \to \infty$. However, I'm not sure how to refine this argument to conclude that for every $y$, $f(x):=u(x+iy)$ tends to $\infty$ in $x$.

One approach was proved here not to work.

An entirely different approach I tried was to map the domain to the unit disk using the map $2 \operatorname{arctanh} (\frac\pi 4 z)$ and use Harnack's principle, which is the first thing that comes to mind for positive harmonic functions, but this map sends the integer points very close to the boundary of the unit disk, so Harnack is not effective in its inequality there.

Harnack's inequality is quite effective here. No need for a conformal map. Given $x+iy$, apply Harnack's inequality in the disk $D$ of radius $1$ centered at $x$. It tells you that $$u(x+iy) \ge \frac{1-|y|}{1+|y|}u(x) \tag1$$ Let $n$ be the integer nearest to $x$, and note that $|n-x|\le 1/2$. By Harnack again, $$u(x) \ge \frac{1-1/2}{1+1/2}u(n) \tag2$$ Combine (1) and (2), and you are done.
A key point is that $y$ is being held constant while $x$ changes; otherwise (1) would not help us much.
Similarly, $u(n)\to 0$ implies $u(x+iy)\to 0$ for a fixed $y$. Just use the other sides of Harnack's inequality: $$u(x+iy) \le \frac{1+|y|}{1-|y|}u(x)$$ and $$u(x) \le \frac{1+1/2}{1-1/2}u(n)$$