Proposition 17.15.11.6DDF81D7BEFF Let $\Gamma \in \mathrm{TC}(X,\theta )_{>0}$, $\Gamma '\in \mathrm{TC}(X,\theta ')_{>0}$ and $C,C'\in \mathbb {R}$. Then
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$(\Gamma +\Gamma ')+C=\Gamma +(\Gamma '+C)=(\Gamma +C)+\Gamma '$;
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$\Gamma +(C+C')=(\Gamma +C)+C'$.
Proof.
1 We first observe that
\[ \left((\Gamma +\Gamma ')+C\right)_{\max }=\left( \Gamma +(\Gamma '+C) \right)_{\max }=\left( (\Gamma +C)+\Gamma ' \right)_{\max }=\Gamma _{\max }+\Gamma '_{\max }+C. \]
Take any $\tau \in \mathbb {R}$ less than this common value. We compute
\[ \begin{aligned} \left((\Gamma +\Gamma ')+C\right)_{\tau }=& (\Gamma +\Gamma ')_{\tau -C} = P_{\theta +\theta '}\left[ \sup _{\delta \in \mathbb {R}}\left( \Gamma _\delta +\Gamma '_{\tau -C-\delta } \right) \right],\\ \left( \Gamma +(\Gamma '+C) \right)_{\tau }=& P_{\theta +\theta '}\left[ \sup _{\delta \in \mathbb {R}}\left( \Gamma _\delta +(\Gamma '+C)_{\tau -\delta } \right) \right]\\ =& P_{\theta +\theta '}\left[ \sup _{\delta \in \mathbb {R}}\left( \Gamma _\delta +\Gamma '_{\tau -C-\delta } \right) \right],\\ \left((\Gamma +C)+\Gamma '\right)_{\tau }=& P_{\theta +\theta '}\left[ \sup _{\delta \in \mathbb {R}}\left( (\Gamma +C)_{C+\delta }+\Gamma '_{\tau -C-\delta } \right) \right]\\ =& P_{\theta +\theta '}\left[ \sup _{\delta \in \mathbb {R}}\left( \Gamma _\delta +\Gamma '_{\tau -C-\delta } \right) \right]. \end{aligned} \]
2 Observe that
\[ \left( \Gamma +(C+C') \right)_{\max }=\left( (\Gamma +C)+C' \right)_{\max }=\Gamma _{\max }+C+C'. \]
For any $\tau \in \mathbb {R}$ less than this value, we have
\[ \left( \Gamma +(C+C') \right)_{\tau }=\Gamma _{\tau -C-C'}=\left( (\Gamma +C)+C' \right)_{\tau }. \]
This completes the proof.
$\square$
Pome
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