PomeCorollary 16.5.25.B5D23DBE0536 The sets $\mathrm{PSH}(X,\theta ;\phi )$, $\mathcal{E}(X,\theta ;\phi )$, $\mathcal{E}^1(X,\theta ;\phi )$, $\mathcal{E}^{\infty }(X,\theta ;\phi )$ are all convex.
Corollary 16.5.25B5D23DBE0536
Proof. The sets $\mathrm{PSH}(X,\theta ;\phi )$ and $\mathcal{E}^{\infty }(X,\theta ;\phi )$ are trivially convex.
We next handle $\mathcal{E}(X,\theta ;\phi )$. In this case, suppose that $\varphi ,\psi \in \mathcal{E}(X,\theta ;\phi )$ and $t\in (0,1)$. Then we compute
\[ \begin{split} \int _ X \theta _{t\varphi +(1-t)\psi }^ n=\sum _{j=0}^ n \binom {n}{j}t^ j (1-t)^{n-j} \int _ X \theta _{\varphi }^ j\wedge \theta _{\psi }^{n-j}\\ =\sum _{j=0}^ n \binom {n}{j}t^ j (1-t)^{n-j} \int _ X \theta _{\phi }^ n=\int _ X \theta _{\phi }^ n, \end{split} \]
where we applied proposition 16.3.10 in the second equality. Therefore, $t\varphi +(1-t)\psi \in \mathcal{E}(X,\theta ;\phi )$.
Finally, let us consider $\mathcal{E}^1(X,\theta ;\phi )$. In this case, suppose that $\varphi ,\psi \in \mathcal{E}^1(X,\theta ;\phi )$ and $t\in (0,1)$. We have already shown that $t\varphi +(1-t)\psi \in \mathcal{E}(X,\theta ;\phi )$. Furthermore, the energy of this potential is finite due to corollary 16.5.23. Therefore, $t\varphi +(1-t)\psi \in \mathcal{E}^1(X,\theta ;\phi )$ in view of proposition 16.5.22. $\square$
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