Proposition 17.10.29

Proposition 17.10.29.DE479E2BD255 Let $\varphi _1,\varphi _0\in \mathcal{E}(X,\theta ;\phi )$ with $\varphi _1\preceq \varphi _0$. Let $(\varphi _ t)_{t\in (0,1)}$ be the geodesic from $\varphi _0$ to $\varphi _1$. Then

8AE427F2EF68 17.10.30

\begin{equation} s\sup _{\{ \varphi _0\neq -\infty \} }(\varphi _1-\varphi _0)=\sup _{\{ \varphi _0\neq -\infty \} }(\varphi _ s-\varphi _0) \end{equation}

for all $s\in [0,1]$.

Proof. The notations in the proof are indicated in 17.10.31.³

We may assume that $s\in (0,1)$ since there is nothing to prove when $s=0$ or $s=1$.

After replacing $\varphi _ t$ by $\varphi _ t-C't$ for some large enough $C'>0$, we may assume that $\varphi _1\leq \varphi _0$. This procedure preserves the geodesic property by example 17.10.23. Then $\varphi _ t$ is decreasing in $t$ by 17.10.28.

Let

FEE6016E13FF 17.10.32

\begin{equation} C=\sup _{\{ \varphi _1\neq -\infty \} } \left(\varphi _1-\varphi _0\right)=\sup _{\{ \varphi _0\neq -\infty \} } \left(\varphi _1-\varphi _0\right)\leq 0, \end{equation}

where the equality is due to corollary 5.15.21. Then by proposition 5.14.6, we have

\[ \varphi _1\leq \varphi _0+C. \]

So $(\varphi _1-C(1-t))_{t\in (0,1)}$ is a candidate in (17.10.21) and hence

18547F185BC3 17.10.33

\begin{equation} \varphi _1-C(1-t)\leq \varphi _ t,\quad t\in (0,1). \end{equation}

By proposition 17.10.24, we have $\varphi _ t\xrightarrow {L^1}\varphi _1$ as $t\to 1-$. Since $\varphi _ t$ is decreasing in $t\in (0,1)$, it follows that $\varphi _1=\inf _{t\in (0,1)}\varphi _ t$. Therefore, we can find a pluripolar set $Z\subseteq X$ such that $\varphi _ t(x)\rightarrow \varphi _1(x)>-\infty $ as $t\to 1-$ for all $x\in X\setminus Z$.

Similarly, since $\varphi _0=\operatorname *{sup}_{t\in (0,1)}{}\! {}^{*} \varphi _ t$, after enlarging $Z$, we may also guarantee that $\varphi _ t(x)\rightarrow \varphi _0(x)>-\infty $ as $t\to 0+$ for all $x\in X\setminus Z$ by proposition 5.14.5.

For any such $x\in X\setminus Z$, the function $t\mapsto \varphi _ t(x)$ is a real-valued continuous convex function on $[0,1]$. In particular, $t\mapsto \varphi _ t(x)$ is absolutely continuous on $[0,1]$. Hence, for any $s\in [0,1)$, we have

16DA31D36BD2 17.10.34

\begin{equation} \varphi _1(x)-\varphi _ s(x)=\int _ s^1 \frac{\mathrm{d}}{\mathrm{d}t}\varphi _ t(x)\, \mathrm{d}t\leq (1-s)\lim _{t\to 1-}\frac{\varphi _1(x)-\varphi _ t(x)}{1-t}\leq (1-s)C, \end{equation}

where the second inequality follows from (17.10.33).

Taking supremum in (17.10.34), we find that

1B22CF257482 17.10.35

\begin{equation} \sup _{X\setminus Z} (\varphi _1-\varphi _ s)\leq (1-s)\adjustlimits \sup _{x\in X\setminus Z}\lim _{t\to 1-}\frac{\varphi _1(x)-\varphi _ t(x)}{1-t}\leq (1-s)C. \end{equation}

When $s=0$, we deduce from corollary 5.15.21 and (17.10.32) that

\[ C=\sup _{\{ \varphi _1\neq -\infty \} } \left(\varphi _1-\varphi _0\right)=\adjustlimits \sup _{x\in X\setminus Z}\lim _{t\to 1-}\frac{\varphi _1(x)-\varphi _ t(x)}{1-t}. \]

But this equality works equally well for the geodesic $(\varphi _{(1-s)t+s})_{t\in [0,1]}$ with the same argument. It follows that

\[ \sup _{\{ \varphi _1\neq -\infty \} } \left(\varphi _1-\varphi _ s\right)=(1-s)\adjustlimits \sup _{x\in X\setminus Z}\lim _{t\to 1-}\frac{\varphi _1(x)-\varphi _ t(x)}{1-t}=(1-s)C. \]

Therefore, invoking corollary 5.15.21 again, we deduce that all inequalities in (17.10.35) are in fact equalities. In other words,

5825ABE1D62B 17.10.36

\begin{equation} \sup _{\{ \varphi _1\neq -\infty \} }(\varphi _1-\varphi _0)= \adjustlimits \sup _{x\in X\setminus Z}\lim _{t\to 1-}\frac{\varphi _1(x)-\varphi _ t(x)}{1-t} =\sup _{\{ \varphi _1\neq -\infty \} }\frac{\varphi _1-\varphi _ s}{1-s}. \end{equation}

On the other hand, we have the trivial inequality when $s\in (0,1)$:

\[ \sup _{\{ \varphi _1\neq -\infty \} }(\varphi _1-\varphi _0)\leq s \sup _{\{ \varphi _1\neq -\infty \} }\frac{\varphi _ s-\varphi _0}{s}+ (1-s) \sup _{\{ \varphi _1\neq -\infty \} }\frac{\varphi _1-\varphi _ s}{1-s}. \]

Together with (17.10.36), we find that for $s\in (0,1)$,

\[ \sup _{\{ \varphi _1\neq -\infty \} }(\varphi _1-\varphi _0)\leq \sup _{\{ \varphi _1\neq -\infty \} }\frac{\varphi _ s-\varphi _0}{s}. \]

The reverse inequality follows from the convexity, and hence for $s\in (0,1)$, we have

\[ \sup _{\{ \varphi _1\neq -\infty \} }\frac{\varphi _ s-\varphi _0}{s}=\sup _{\{ \varphi _1\neq -\infty \} }(\varphi _1-\varphi _0). \]

Using corollary 5.15.21, we conclude (17.10.30). $\square$

Proposition 17.10.29`DE479E2BD255`

Comments

Proposition 17.10.29DE479E2BD255

Comments

Proposition 17.10.29`DE479E2BD255`