Why is the Harnack distance continuous? Let $D\subseteq \mathbb{C}$ be a domain. Given $z,w\in D$ we define the Harnack distance $\tau_D(z,w)$ as the smallest number such that $\tau_D(z,w)^{-1}h(w)\leq h(z)\leq \tau_D(z,w)h(w)$ for all positive harmonic functions $h$ on $D$.
Question: Why is the Harnick distance continuous? Is it trivial?
 A: Let $z,w$ be a point in $D$. First of all, try and check the definition of $\tau_D$ to notice that


*

*$\tau_D(z,w) \geq 1$ 

*$\tau_D(z,z) = 1$.

*$\tau_D(z,w) \leq \tau_D(z, \zeta)\tau_D(\zeta, w)$
Now let $\delta$ be a positive real number such that the ball
$$B_\delta(z) = \{w \in \mathbb{C} \mid |z - w| < \delta\}$$
is contained in $D$. 
Every positive harmonic function $h:D\to\mathbb{R}$ restricts itself to a positive harmonic function on $B_\delta(z)$. In particular can apply Harnack's Inequality on on the disk which tells us that if $|w-z| = r$ then
$$\frac{\delta - r}{\delta + r}h(z) \leq h(w) \leq \frac{\delta + r}{\delta - r}h(z).$$
This holds for every $h$ so, by definition of $\tau_D$ we have
$$1 \leq \tau_D(z,w) \leq \frac{\delta + r}{\delta - r}.$$
Now making $w \to z$ we can see that the right side goes to 1 and therefore 
$$\lim_{w \to z}\tau_D(z,w) = 1.$$
Now use that 
$$|\log\tau_D(z,w) - \log\tau_D(z',w')| \leq |\log\tau_D(z,w) - \log\tau_D(z',w)| + |\log\tau_D(z',w) - \log\tau_D(z',w')|$$
By the third property we derived, and show that $\log\tau_D$ is continuous since the RHS goes to zero as $(z,w) \to (z',w')$. Conclude that $\tau$ is also continuous. 
I wouldn't call this trivial, specially if you're new to the subject.
