Proof.
Step 1. We first establish (17.13.22).
It follows from 17.13.11 that the forward map is well-defined. The inverse map is given by (17.13.14). We show that the inverse map is also well-defined. Given $\ell \in \mathcal{R}(X,\theta ;\phi )$, we know from 17.13.15 that $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$. We need to show that $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$.
The equality $\ell ^*_{-\infty }=\phi $ was proved in 17.13.15. Therefore, $\ell ^*\in \mathrm{TC}(X,\theta ;\phi )$ as expected.
The two operations are inverse to each other thanks to corollary 21.11.26. Hence, (17.13.22) is established.
Step 2. Next we consider the bounded situation. Namely, we want to establish the second half of (17.13.23).
Suppose that $\Gamma \in \mathrm{TC}^{\infty }(X,\theta ;\phi )$. Take $\tau _0\in \mathbb {R}$ so that $\Gamma _{\tau }=\phi $ for all $\tau \leq \tau _0$. It follows from (17.13.2) that
\[ \Gamma _ t^*\geq \phi +t\tau _0 \]
for all $t>0$. Therefore, $\Gamma ^*_ t\sim \phi $ for all $t>0$ and hence $\Gamma ^*\in \mathcal{R}^{\infty }(X,\theta ;\phi )$.
Conversely, suppose that $\ell \in \mathcal{R}^{\infty }(X,\theta ;\phi )$. Thanks to proposition 17.10.37, there is a constant $C>0$ such that
\[ \ell _ t\geq \phi -Ct. \]
Therefore, according to (17.13.14), we have
\[ \ell ^*_{\tau }\geq \inf _{t>0} \Bigl ( \phi -(C+\tau )t \Bigr )=\phi \]
if $\tau \leq -C$. Therefore, $\ell ^*_{\tau }=\phi $ for all $\tau \leq -C$.
Step 3. We establish (17.13.24) and the first half of (17.13.23).
Step 3.1. We reduce to the case where $\Gamma _{\max }=0$.
Suppose that we define
\[ \Gamma '_{\tau }=\Gamma _{\tau +\Gamma _{\max }},\quad \forall \tau <0. \]
Then $\Gamma '\in \mathrm{TC}(X,\theta ;\phi )$ as well and for all $t>0$,
\[ \Gamma '^*_ t=\sup _{\tau <0} \left( t\tau +\Gamma '_{\tau } \right)=\sup _{\tau <\Gamma _{\max }} \left( t\tau +\Gamma _{\tau } \right)-t\Gamma _{\max }=\Gamma ^*_ t-t\Gamma _{\max }. \]
Therefore,
\[ \mathbf{E}^{\phi }\left(\Gamma '^*\right)=\mathbf{E}^{\phi }\left(\Gamma ^*\right)-\Gamma _{\max }\int _ X \theta _{\phi }^ n. \]
by (16.5.13). Using (17.12.12), we also have
\[ \begin{aligned} \mathbf{E}^{\phi }(\Gamma ')=& \int _{-\infty }^0 \left( \int _ X \theta _{\Gamma '_{\tau }}^ n-\int _ X \theta _{\phi }^ n \right)\, \mathrm{d}\tau \\ =& \int _{-\infty }^{\Gamma _{\max }}\left( \int _ X \theta _{\Gamma _{\tau }}^ n-\int _ X \theta _{\phi }^ n \right)\, \mathrm{d}\tau \\ =& \mathbf{E}^{\phi }(\Gamma )-\Gamma _{\max }\int _ X \theta _{\phi }^ n. \end{aligned} \]
Therefore, it suffices to establish (17.13.24) for $\Gamma '$ in place of $\Gamma $.
Step 3.2. We assume that $\Gamma _{\max }=0$ and $\Gamma \in \mathrm{TC}^{\infty }(X,\theta ;\phi )$. We prove (17.13.24).
For $N\in \mathbb {Z}_{>0}$, $M \in \mathbb {Z}$, we introduce the following:
\[ \Gamma ^{*,N,M}_ t \coloneqq \max _{\substack {k \in \mathbb {Z} \\ k \leq M}} \left(\Gamma _{k/2^ N} + tk/2^ N\right)\in \mathcal{E}^{\infty }(X,\theta ;\phi ),\quad t>0. \]
We first claim that for all $t>0$, $N\in \mathbb {Z}_{>0}$ and $M\in \mathbb {Z}_{<0}$,
1C55A624E61E
17.13.25
\begin{equation} \frac{t}{2^ N} \int _ X \theta ^ n_{\Gamma _{(M+1)/2^ N}}\leq E_{\theta }^{\phi }\left(\Gamma _ t^{*,N,M+1}\right)-E_{\theta }^{\phi }\left(\Gamma _ t^{*,N,M}\right)\leq \frac{t}{2^ N} \int _ X \theta ^ n_{\Gamma _{M/2^ N}}. \end{equation}
Assuming this, let us prove (17.13.24).
Fixing $N$, let $M=\lfloor 2^{N}\Gamma _{\min }\rfloor $. Recall that $\Gamma _{\min }$ is defined in (17.12.11). Then repeated applications of (17.13.25) yield
\[ \sum _{j=M+1}^0 \frac{t}{2^ N} \int _ X \theta ^ n_{\Gamma _{j/2^ N}}\leq E^{\phi }_{\theta }\left(\Gamma _ t^{*,N,0}\right)-E^{\phi }_{\theta }\left(\Gamma _ t^{*,N,M}\right)\leq \sum _{j=M}^{-1} \frac{t}{2^ N} \int _ X \theta ^ n_{\Gamma _{j/2^ N}}. \]
Since $M \leq 2^{N}\Gamma _{\min }$, we have
\[ \Gamma ^{*,N,M}_ t =\phi +t M/2^ N. \]
Using (16.5.13), we can continue to write
\[ \sum _{j=M+1}^0 \frac{t}{2^ N} \left(\int _ X \theta ^ n_{\Gamma _{j/2^ N}} - \int _ X \theta ^ n_{\phi }\right)\leq E_{\theta }^{\phi }\left(\Gamma _ t^{*,N,0}\right)\leq \sum _{j=M}^{-1} \frac{t}{2^ N} \left(\int _ X \theta ^ n_{\Gamma _{j/2^ N}} - \int _ X \theta ^ n_{\phi }\right). \]
We now notice that we have Riemann sums on both the left and right of the above inequality. Moreover, as $N\to \infty $, the potentials $\Gamma _ t^{*,N,0}$ increase to $\Gamma _ t^*$, hence their energies converge to $E^\phi _\theta (\Gamma _ t^*)$ by proposition 16.5.37. Using proposition 17.12.9, it is possible to let $N \to \infty $ and obtain
85E0D04C585D
17.13.26
\begin{equation} E_{\theta }^{\phi }(\Gamma _ t^{*})=t\mathbf{E}^{\phi }(\Gamma ) \end{equation}
So (17.13.24) follows as desired.
It remains to argue (17.13.25). Fix $t>0$, $N\in \mathbb {Z}_{>0}$ and $M\in \mathbb {Z}$. By proposition 16.5.36,
877135CB19C2
17.13.27
\begin{equation} \begin{aligned} \int _ X \left(\Gamma _ t^{*,N,M+1}-\Gamma _ t^{*,N,M}\right)\, \theta ^ n_{\Gamma _ t^{*,N,M+1}}\leq & E_{\theta }^{\phi }\left(\Gamma _ t^{*,N,M+1}\right)-E_{\theta }^{\phi }\left(\Gamma _ t^{*,N,M}\right)\\ \leq & \int _ X \left(\Gamma _ t^{*,N,M+1}-\Gamma _ t^{*,N,M}\right)\, \theta ^ n_{\Gamma _ t^{*,N,M}}. \end{aligned} \end{equation}
Clearly $\Gamma _ t^{*,N,M+1} \geq \Gamma _ t^{*,N,M}$. Moreover, since $\mathbb {R}\ni \tau \mapsto \Gamma _{\tau }+t\tau $ is concave, we notice that
\[ U_ t \coloneqq \left\{ \Gamma ^{*,N,M+1}_ t>\Gamma ^{*,N,M}_ t\right\} = \left\{ \Gamma _{(M+1)/2^ N} + 2^{-N}t >\Gamma _{M/2^ N}\right\} , \]
and on $U_ t$ we have
23E6778F003A
17.13.28
\begin{equation} \Gamma _ t^{*,N,M+1} = \Gamma _{(M+1)/2^ N} + t (M+1)/2^{N},\quad \Gamma _ t^{*,N,M} = \Gamma _{M/2^ N} + t M /2^{N}. \end{equation}
We also note that $U_ t$ is $\mathcal{F}$-open by corollary 5.15.19. So from the lower bound in (17.13.27), we have
\[ \begin{aligned} E_{\theta }^{\phi }\left(\Gamma _ t^{*,N,M+1}\right)-E_{\theta }^{\phi }\left(\Gamma _ t^{*,N,M}\right)\geq & \int _{U_ t} \left(\Gamma _ t^{*,N,M+1}-\Gamma _ t^{*,N,M}\right)\, \theta ^ n_{\Gamma _ t^{*,N,M+1}}\\ = & \int _{U_ t} \left(\Gamma _{(M+1)/2^ N}-\Gamma _{M/2^ N}+t2^{-N}\right)\, \theta ^ n_{\Gamma _{(M+1)/2^ N}}\\ \geq & \int _{\left\{ \Gamma _{(M+1)/2^ N}=0\right\} }t2^{-N}\, \theta ^ n_{\Gamma _{(M+1)/2^ N}}, \end{aligned} \]
where on the second line, we applied (17.13.28) and proposition 13.11.4, on the third line, we applied the fact that $\theta _{\Gamma _{(M+1)/2^ N}}^ n$ is supported on the set (cf. theorem 16.3.12)
\[ \left\{ \Gamma _{(M+1)/2^ N}=0\right\} \subseteq U_ t\cap \left\{ \Gamma _{M/2^ N}=0\right\} . \]
When $(M+1)/2^ N=0$, we use proposition 16.4.3 to see that $P_\theta [\Gamma _0]=\Gamma _0$, since $\Gamma _0$ is the decreasing limit of the model potentials $\Gamma _\tau $ as $\tau \nearrow 0$. Thus theorem 16.3.12 applies to this endpoint as well.
We have deduced the first inequality in (17.13.25). Next, we apply the upper bound part in (17.13.27) and compute similarly
\[ \begin{aligned} E_{\theta }^{\phi }(\Gamma _ t^{*,N,M+1})-E_{\theta }^{\phi }(\Gamma _ t^{*,N,M})\leq & \int _ X \left(\Gamma _ t^{*,N,M+1}-\Gamma _ t^{*,N,M}\right)\, \theta ^ n_{\Gamma _ t^{*,N,M}}\\ =& \int _{U_ t} \left(\Gamma _{(M+1)/2^ N}-\Gamma _{M/2^ N}+t2^{-N}\right)\, \theta ^ n_{\Gamma _{M/2^ N}}\\ \leq & \int _{\{ \Gamma _{M/2^ N}=0\} \cap U_ t}\left(\Gamma _{(M+1)/2^ N}+t2^{-N}\right)\, \theta ^ n_{\Gamma _{M/2^ N}} \\ \leq & \int _{\{ \Gamma _{M/2^ N}=0\} \cap U_ t} t2^{-N}\, \theta ^ n_{\Gamma _{M/2^ N}}. \end{aligned} \]
We conclude the latter half of (17.13.25).
Step 3.3. We assume that $\Gamma _{\max }=0$. Now $\Gamma \in \mathrm{TC}(X,\theta ;\phi )$ only.
For each $\epsilon >0$, we introduce $\Gamma ^{\epsilon }\in \mathrm{TC}^{\infty }(X,\theta ;\phi )$ as follows:
1E08B0569766
Let $\Gamma ^{\epsilon }_{\max }=0$, and
70863808F7A7
we set
\[ \Gamma ^{\epsilon }_{\tau }= \begin{cases} \phi ,& \textup{if }\tau \leq -\epsilon ^{-1};\\ P_{\theta }\left[ (1+\epsilon \tau )\Gamma _{\tau }-\epsilon \tau \phi \right],& \textup{if }\tau \in \left(-\epsilon ^{-1},0\right). \end{cases} \]
The concavity and monotonicity in $\tau $ follow from proposition 16.4.2. The value at $-\infty $ is $\phi $. Moreover, $\Gamma ^\epsilon _\tau =\phi $ for $\tau \leq -\epsilon ^{-1}$, so $\Gamma ^\epsilon \in \mathrm{TC}^\infty (X,\theta ;\phi )$.
It follows from corollary 16.9.38 and corollary 16.9.14 that for each $\tau <0$, the potentials $\Gamma ^{\epsilon }_{\tau }$ decrease to $\Gamma _{\tau }$ as $\epsilon \searrow 0$. Applying proposition 16.4.4, we have
\[ \lim _{\epsilon \to 0+} \int _ X \left(\theta +\mathrm{dd}^{\mathrm{c}}\Gamma ^{\epsilon }_{\tau }\right)^ n=\int _ X \left(\theta +\mathrm{dd}^{\mathrm{c}}\Gamma _{\tau }\right)^ n \]
for all $\tau <0$. Hence, by the monotone convergence theorem and Step 3.2, we find
DBCF50A40C12
17.13.29
\begin{equation} \mathbf{E}^{\phi }(\Gamma )=\lim _{\epsilon \to 0+}\mathbf{E}^{\phi }(\Gamma ^{\epsilon })=\lim _{\epsilon \to 0+}\mathbf{E}^{\phi }(\Gamma ^{\epsilon *})=\lim _{\epsilon \to 0+} E_{\theta }^{\phi }(\Gamma ^{\epsilon *}_1), \end{equation}
where the last equality follows from (17.13.26). Furthermore, according to proposition 21.11.36, we have
\[ \Gamma ^*_ t=\inf _{\epsilon >0} \Gamma ^{\epsilon *}_ t \]
for all $t>0$. Note that we do not have to take the closure of the right-hand side since it is automatically upper semicontinuous in $t$.
Now suppose that $\Gamma \in \mathrm{TC}^1(X,\theta ;\phi )$. Then by (17.13.29), as $\epsilon \to 0+$, $(\Gamma _ t^{\epsilon *})_{\epsilon }$ is a decreasing Cauchy net in $\mathcal{E}^1(X,\theta ;\phi )$ and hence by theorem 17.10.63 for each $t>0$,
\[ E_{\theta }^{\phi }\left(\Gamma ^*_ t\right)=\lim _{\epsilon \to 0+}E_{\theta }^{\phi }\left(\Gamma ^{\epsilon ,*}_ t\right)=t\mathbf{E}^{\phi }(\Gamma )>-\infty , \]
where we have applied (17.13.26) and (17.13.29). Hence, $\Gamma ^*\in \mathcal{R}^1(X,\theta ;\phi )$. Moreover, (17.13.24) follows.
Conversely, suppose that $\Gamma ^*\in \mathcal{R}^1(X,\theta ;\phi )$. Then (17.13.29) implies that
\[ \mathbf{E}^{\phi }(\Gamma )=\lim _{\epsilon \to 0+} E_{\theta }^{\phi }\left(\Gamma ^{\epsilon *}_1\right) \geq E_{\theta }^{\phi }\left(\Gamma ^*_1\right)>-\infty . \]
Hence, $\Gamma \in \mathrm{TC}^1(X,\theta ;\phi )$.
$\square$
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