Proposition 13.22.5412DBAD823B9

Proof. Put $s=d_\theta (v,w)$. If $s\geq \delta /2$, the inequality follows from $d_\theta (u,w)\leq 2M$. If $0<s<\delta /2$, set $t=(2s/\delta )^{1/n}$. Lemma 13.18.1, applied to $v\preceq \max (v,w)$, gives $h\in \mathrm{PSH}(X,\theta )$ with

This inequality also gives $h\leq v$. Set $r=\max (u,v)$ and $z=\max (u,v,w)$. Where $z=\max (v,w)$, the inequality above bounds $(1-t)z+th$ by $v\leq r$; where $z=u$, both $u$ and $h$ lie below $r$. Thus

Addition, homogeneity and volume monotonicity give

Since $\max (u,w)\leq z$, it follows that

Here the last step uses (13.22.2). For $s=0$, the same proof works with every $t\in (0,1)$; let $t\downarrow 0$. This proves (13.22.6). Applying it with the limit as the middle point shows that a convergent sequence is Cauchy. If a sequence converges to two limits, applying it with a member of the sequence as the middle point shows that the two limits have volume comparison zero; by proposition 13.22.4, they are equal. $\square$