PomeProposition 17.10.79.5408F4A1486E Let $(\varphi _ t)_{t\in [0,1]}$, $(\psi _ t)_{t\in [0,1]}$ be geodesics in $\mathcal{E}^1(X,\theta ;\phi )$. Then the distance $d_1(\varphi _ t,\psi _ t)$ is a convex function of $t\in [0,1]$.
Proposition 17.10.795408F4A1486E
Proof. By definition of $d_1$ and proposition 17.10.65, it suffices to show the concavity of
\[ [0,1]\ni t\mapsto E_{\theta }^{\phi }(\varphi _ t\land \psi _ t). \]
Let $(\eta _ t)_{t\in [0,1]}$ be the geodesic from $\varphi _0\land \psi _0$ to $\varphi _1\land \psi _1$. Then for each $t\in [0,1]$, we have $\eta _ t\leq \varphi _ t\land \psi _ t$. Hence, by the monotonicity of $E_\theta ^\phi $ and proposition 17.10.65,
\[ E_{\theta }^{\phi }( \varphi _ t\land \psi _ t)\geq E_{\theta }^{\phi }(\eta _ t)=(1-t)E_{\theta }^{\phi }(\varphi _0\land \psi _0)+tE_{\theta }^{\phi }(\varphi _1\land \psi _1) \]
for each $t\in [0,1]$. Applying the same argument to the restrictions of $(\varphi _ t)_ t$ and $(\psi _ t)_ t$ to any subinterval of $[0,1]$, which are again geodesics by proposition 17.10.24, gives a similar inequality on every subinterval. Hence $t\mapsto E_\theta ^\phi (\varphi _ t\land \psi _ t)$ is concave, and our assertion follows. $\square$
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