PomeLemma 17.10.45.488B7E9EA42C The function $d_1$ is a metric on $\mathcal{E}^1(X,\theta ;\phi )$.
Lemma 17.10.45488B7E9EA42C
Proof. There are two facts to prove, as in the two steps below.
Step 1. Let $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$. Assume that $d_1(\varphi ,\psi )=0$. Then we will show that $\varphi =\psi $.
We first treat the case $\psi \leq \varphi $. Then $d_1(\varphi ,\psi )=E_\theta ^\phi (\varphi )-E_\theta ^\phi (\psi )=0$, and proposition 16.5.30 gives
\[ \int _ X (\varphi -\psi )\, \theta _{\psi }^ n=0. \]
We conclude that $\psi =\varphi $ using the domination principle theorem 13.13.43.
In general, put $\gamma =\varphi \land \psi $. Since
\[ d_1(\varphi ,\psi )=d_1(\varphi ,\gamma )+d_1(\psi ,\gamma ), \]
both terms on the right vanish. Applying the comparable case to the pairs $(\varphi ,\gamma )$ and $(\psi ,\gamma )$ gives $\varphi =\gamma =\psi $.
Step 2. Let $\varphi ,\psi ,\gamma \in \mathcal{E}^1(X,\theta ;\phi )$. We prove the triangle inequality:
\[ d_1(\varphi ,\psi )\leq d_1(\varphi ,\gamma )+d_1(\psi ,\gamma ). \]
This can be translated to
\[ E_{\theta }^{\phi }(\varphi \land \gamma )-E_{\theta }^{\phi }(\varphi \land \psi )\leq E_{\theta }^{\phi }(\gamma )-E_{\theta }^{\phi }(\psi \land \gamma ). \]
We just have to compute using proposition 17.10.43:
\[ \begin{aligned} E_{\theta }^{\phi }(\gamma )-E_{\theta }^{\phi }(\psi \land \gamma )\geq & E_{\theta }^{\phi }(\varphi \land \gamma )-E_{\theta }^{\phi }(\varphi \land \psi \land \gamma )\\ \geq & E_{\theta }^{\phi }(\varphi \land \gamma )-E_{\theta }^{\phi }(\varphi \land \psi ), \end{aligned} \]
where the second line follows from 16.5.14. $\square$
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