## Appendix A Proofs of the Two Main Theorems
### Proof of the Tempering Theorem
#### a.2.1 Taylor Development
The next term to control is \(\frac{1}{(1+b)^{2}}\). Since \(b\in\Big{[}0,\Big{(}\frac{1}{T}-1\Big{)}\int_{z}p\,\ln p\,e^{a}\Big{]}\), then \(|b|\leq\Big{(}\frac{1}{T}-1\Big{)}\!\!\!\sup_{\theta\in K}\,\,\int_{z}p\,\ln p \,e^{a}\). We already established that for all \(n\geq N\), \(\sup_{\theta\in K}\,\big{|}\int_{z}p\,\ln p\,e^{a}\big{|}\leq C^{\prime}<\infty\), hence \(\sup_{\theta\in K}\,|b(\theta,T_{n})|\underset{T_{n}\longrightarrow 1}{ \longrightarrow}0\). In particular, \(\exists N^{\prime}\in\mathbb{N},\forall n\geq N^{\prime},\forall\theta\in K\) we have \(|b(\theta,T_{n})|\leq\frac{1}{2}\). In that case:  
\[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]5 \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]6  
In the end, when \(n\geq max(N,N^{\prime})\), for any \(\theta\in K\):  
\[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]7 \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]8 \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]9 \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]0 \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]1  
This upper bound does not depend en \(\theta\) anymore and the part in \(T_{n}\) simply converges towards \(0\) when \(T_{n}\longrightarrow 1\). Treating the term \(B(\theta,T_{n})=\bigg{(}\frac{\int_{z}p\,\ln p\,e^{a}}{(1+b)^{2}}\bigg{)}^{2} \leq 16\Bigg{(}\sup_{\theta\in K}\,\,\int_{z}p\,\ln p\,e^{a}\Bigg{)}^{2}\leq 1 6C^{\prime 2}=:C_{2}\) is immediate after having dealt with \(A(\theta,T_{n})\). We now treat the term \(\int_{z}p\,e^{a}\ln^{2}\,p\) in the exact same fashion as we did \(A(\theta,T_{n})\):  
\[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]2 \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]3 \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]4  
We control those two terms as previously. First we apply Lemma A (its first result this time) with \(\alpha\in\overline{B}(1,\epsilon)\). \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]5  
Thanks to the hypotheses (ii), (iii), (iv) and (v), we can once again take the supremum of the bound over \(\theta\in K\), then over \(\alpha\in\overline{B}(1,\epsilon)\) and conserve finite quantities:\[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]6 \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]7 \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]8 \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]9  
The previous result is true for \(\alpha=1\), and since once again \(\exists N^{\prime\prime},\,\forall n\geq N^{\prime\prime},\,\frac{2}{T_{n}}-1 \in\overline{B}(1,\epsilon)\cap\mathbb{R}_{+}^{*}\), it is also true for \(\alpha=\frac{2}{T_{n}}-1\) when \(n\) is large enough. \(C_{3}\) is independent of \(z,\theta\) and \(T_{n}\). In the end \(\forall n\geq N^{\prime\prime},\,\int_{z}p\,e^{2\alpha}\ln^{2}\,p\leq C_{3}<\infty\). We replace the three terms \(A(\theta,T_{n}),\,B(\theta,T_{n})\) and \(\int_{z}p\,e^{2\alpha}\ln^{2}\,p\) by their upper bounds in the inequality (A11). When \(n\geq max(N,N^{\prime},N^{\prime\prime})\):  
\[|p\,\ln p\,e^{a}|\leq max\left(|p\,\ln p|,\left|p^{\frac{1}{T}}\ln p\right| \right).\]0  
Which converges towards \(0\) when \(T_{n}\longrightarrow 1\), i.e., when \(n\longrightarrow\infty\). This concludes the proof of the lemma.