Lemma 19.6.921BC38898B4A

Proof. First suppose that $\Delta _{\bullet }\in \mathrm{TC}(\Delta )$. We want to verify that $G[\Delta _{\bullet }]$ satisfies the conditions in definition 19.6.5.

We first verify the concavity. Take $a,b\in \Delta $. We want to prove that for any $t\in (0,1)$,

There is nothing to prove if $G[\Delta _{\bullet }](a)$ or $G[\Delta _{\bullet }](b)$ is $-\infty $. So we assume that both are finite. In this case, by (19.6.8),

Thus,

We deduce that

Therefore, (19.6.10) follows.

It is clear that $G[\Delta _{\bullet }]$ is finite on the interior of $\Delta $. It remains to argue that $G[\Delta _{\bullet }]$ is upper semicontinuous.

Let $(a_ i)_{i\geq 1}$ be a sequence in $\Delta $ with limit $a\in \Delta $. Define $\tau _ i=G[\Delta _{\bullet }](a_ i)$. Let $\tau =\varlimsup _ i \tau _ i$. We need to show that

There is nothing to prove if $\tau =-\infty $. We assume that this is not the case. Passing to a subsequence, we may assume that $\tau _ i\to \tau $. In particular, we can assume that $\tau _ i\neq -\infty $ for all $i\geq 1$. It follows from (19.6.8) that $a_ i\in \Delta _{\tau _ i}$ for all $i\geq 1$. If $\tau <\Delta _{\max }$, then $\Delta _{\tau _ i}\xrightarrow {d_{\mathrm{Haus}}}\Delta _{\tau }$, so theorem 19.7.6 implies that $a\in \Delta _{\tau }$. If $\tau =\Delta _{\max }$, then the same conclusion follows from the definition $\Delta _{\Delta _{\max }}=\bigcap _{\tau '<\Delta _{\max }}\Delta _{\tau '}$. Thus, (19.6.11) follows.

Conversely, suppose that $G\colon \Delta \rightarrow [-\infty ,\infty )$ is a test function. We argue that $\Delta [G]_{\bullet }$ is an Okounkov test curve. We verify the conditions in definition 19.6.1.

Firstly, for each $\tau < \sup _{\Delta }G$, the set $\Delta [G]_{\tau }$ is a convex body as $G$ is concave and usc. Moreover, $\Delta [G]_{\tau }$ is clearly decreasing in $\tau $.

Secondly, for each $a\in \Delta $, we can write $a=\lim _{i}a_ i$ with $a_ i\in \operatorname{Int}\Delta $. By assumption, $G$ is finite at $a_ i$. Thus,

By theorem 19.7.6, $\Delta [G]_{\tau }\xrightarrow {d_{\mathrm{Haus}}}\Delta $ as $\tau \to -\infty $.

Thirdly, $\Delta [G]$ is concave. To see this, take $\tau ,\tau '< \Delta _{\max }$. We need to prove that for any $t\in (0,1)$,

Let $a\in \Delta [G]_{\tau }$ and $b\in \Delta [G]_{\tau '}$. We have $G(a)\geq \tau $ and $G(b)\geq \tau '$. As $G$ is concave, we have $G(ta+(1-t)b)\geq t\tau +(1-t)\tau '$. Thus,

and (19.6.12) follows. $\square$