Proposition 13.27.244E1935166EC

Proof. Put $s=d_\theta (p,q)$ and $m=\max (p,q)$. If $s\geq \delta /2$, the bound follows from $|L(p)-L(q)|\leq M$, and if $s=0$, then $p=q$ by proposition 13.22.4. Suppose $0<s<\delta /2$ and take $t=(2s/\delta )^{1/n}$. The upper-volume deficit from either $p$ or $q$ to $m$ is at most $s$, while $L(m)\geq \delta $. Thus lemma 13.18.1 gives a potential $h\in \mathrm{PSH}(X,\theta )$ with

Volume addition and monotonicity imply $L(p)\geq (1-t)^ nL(m)\geq (1-t)^ nL(q)$, so $L(q)-L(p)\leq (1-(1-t)^ n)M\leq nMt$. Interchanging $p$ and $q$ proves the bound.

For the last assertion, put $s_ j=d_\theta (p_ j,p)$. Whenever $0<s_ j<L(p)/2$, the same stretching argument applied to $\max (p_ j,p)$ gives

If $s_ j=0$, then $p_ j=p$. Hence $L(p_ j)\geq L(p)/2$ for all sufficiently large $j$, and (13.27.3) with $\delta =L(p)/2$ gives $L(p_ j)\to L(p)$. $\square$