Harmonic function and Poincaré metric

Question

Let $u$ be a harmonic function on the unit disk $\Delta$, taking values in $[0,1]$.

1. Is it true that this implies that $u$ is Lipschitz for the Poincaré metric ?

2. If not, what can be said about a harmonic function satisfying this property ?

I am thinking the answer to 1. might be true, but I'm not sure. I was thinking that since the unit disk is simply connected, we can find globally a holomorphic function of which $u$ is the real part and maybe somehow manage to apply Schwarz Pick lemma. But I'm not convinced it always possible to do this, and I would be interested in a characterization (ideally) or necessary conditions on $u$ to verify this.

Answer 1

As mfl pointed out, this is true. In fact, the result is not specific to complex analysis: it holds in higher dimensions too. This is how it is presented in the book Harmonic Function Theory (free download) by Axler, Bourdon, and Ramey:

Theorem 6.26: Suppose $u$ is a real-valued harmonic function on $B_n$ (unit ball in $\mathbb R^n$) and $|u|<1$ on $B_n$. Then $$|\nabla u(0)|\le \frac{2V(B_{n-1})}{V(B_n)} \tag1$$

The constant on the right in (1) is not so important: we only need to know that it depends only on $n$. If we have harmonic $u:B_n(r)\to (-1,1)$, where $r$ is the radius of the ball, then applying the above to $v(x)=u(rx)$ yields $$r |\nabla u(0)|\le \frac{2V(B_{n-1})}{ V(B_n)} \tag2$$

Returning to the unit ball and an arbitrary point $x\in B_n$, apply (2) in the ball of radius $1-|x|$ centered at $x$. The result is $$(1-|x|) |\nabla u(x)|\le \frac{2V(B_{n-1})}{ V(B_n)} \tag3$$ The quantity on the left is almost the gradient of $u$ in the Poincaré metric; just multiply by $(1+|x|)$, which is at most $2$.