Lemma 15.19.24240B8EBF50C3

Proof. Replace $g$ by $g/M$. Its area is $a=A/M\leq 1$, and it is bounded above by $h$. We prove the estimate for this normalized metric.

Choose a finite cover by holomorphic coordinate disks $U_ i$ represented by $\{ |z|<1\} $, with each coordinate defined on the larger disk $\{ |z|<2\} $. Write $g=\rho _ i|\mathrm{d}z|^2$ there. There is a common $L>0$ such that $0<\rho _ i\leq L$ on all the closed unit disks, and

where $\lambda _2$ is Euclidean area. For $p,z\in U_ i$, let $\ell _ p(z)$ be the $g$-length of their straight coordinate segment, which stays in $U_ i$. Polar coordinates about $p$ and Fubini give

Indeed each ray meets the convex unit disk in an interval of length $R\leq 2$. Interchanging the radial integrals multiplies the length integrand at radius $s$ by $(R^2-s^2)/2\leq 2$.

For $0<a\leq 1/4$, put $t=\sqrt a$. The integral on the right, without its factor two, is bounded by

Thus for any $p,q\in U_ i$, averaging $\ell _ p(z)+\ell _ q(z)$ over $z\in U_ i$ gives a two-segment path of length at most $K\sqrt{a(1+\log (1/a))}$. The constant is common to the finite cover.

The overlap graph of this cover is connected because $C$ is connected. Choose points in consecutive overlaps along a simple path in this finite graph. The local estimates join any two points of $C$ through at most the number of covering disks, proving the global bound for $a\leq 1/4$. For $1/4\leq a\leq 1$, the comparison $g\leq h$ gives $\operatorname {diam}(C,g)\leq \operatorname {diam}(C,h)$, and increasing $K_ h$ gives the same estimate. Restoring the factor $M$ scales lengths by $\sqrt M$ and proves the assertion. $\square$