## Appendix A Proofs of the Two Main Theorems
### Proof of the Tempering Theorem
#### a.2.1 Taylor Development
\(\sup_{\theta\in K}\ \psi(\theta)<\infty\)__,
3. \(\sup_{\theta\in K}\ \left\|\phi(\theta)\right\|<\infty\)__,
4. \(\forall\alpha\in\overline{B}(1,\epsilon),\ \ \sup_{\theta\in K}\ \int_{z}p_{\theta}^{\alpha}(z)dz<\infty\)__,
5. \(\forall\alpha\in\overline{B}(1,\epsilon),\ \forall u,\ \ \sup_{\theta\in K}\ \int_{z}S_{u}^{2}(z)p_{\theta}^{\alpha}(z)dz<\infty\)__. _then_  
\[f(x)=\frac{1}{1+x},\quad f^{\prime}(x)=-\frac{1}{(1+x)^{2}},\quad f(x)=1- \frac{x}{(1+a)^{2}},\ a\in[0,x]\,.\]2  
Proof. : Provided that the following integrals exist, we have, thanks to the [PERSON] development:  
\[ \begin{split}\int_{z}\frac{1}{p}\left(\frac{p^{\frac{1}{\tau}}}{ \int_{z}p^{\frac{1}{\tau}}}-p\right)^{2}&\leq 2\int_{z} \left(\frac{1}{T}-1\right)^{2}p\Big{(}(\ln p\,e^{\alpha})^{2}A+B\Big{)}\\ &=2\left(\frac{1}{T}-1\right)^{2}A\int_{z}pe^{2\alpha}\ln^{2}\ p \,+2\left(\frac{1}{T}-1\right)^{2}B\,.\end{split} \] (A11)  
In this proof, we find finite upper bounds independent of \(\theta\) and \(T_{n}\) for \(A(\theta,T_{n})\), \(B(\theta,T_{n})\) and \(\int_{z}pe^{2\alpha}\ln^{2}\ p\), then--since \(\left(\frac{1}{T_{n}}-1\right)\longrightarrow 0\)--we have the desired result. We start by studying \(A(\theta,T_{n})=\left(1-\left(\frac{1}{1}-1\right)\frac{f_{\perp}\,p\,\ln p\,e^{a} }{(1+b)^{2}}\right)^{2}\). The first term of interest here is \(\int_{z}p\,\ln p\,e^{a}\). We have:  
\[a \in\left[0,\left(\frac{1}{T}-1\right)\ln p\right],\] \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\] \[p\,\ln p\,e^{a} \in\left[p\,\ln p,p^{\frac{1}{T}}\ln p\right].\]  
where we recall that the interval is to be taken in a flexible sense, since we do not now a priory which bound is the largest and which is the smallest. What we have without doubt though is:  
\[|p\,\ln p\,e^{a}|\leq max\left(|p\,\ln p|,\left|p^{\frac{1}{T}}\ln p\right| \right).\]  
We find an upper bound on both those term. Let \(\alpha\in\overline{B}(1,e)\), the second result of Lemma A1 provides us:  
\[\int_{z}p_{\theta}^{a}(z)\,|\ln p_{\theta}(z)|dz\leq|\psi(\theta)|\int_{z}p_{ \theta}^{a}(z)dz+\|\phi(\theta)\|.\left(\sum_{u}\int_{z}S_{u}^{2}(z)p_{\theta }^{a}(z)\,\int_{z}p_{\theta}^{a}(z)\right)^{\frac{1}{2}}.\]  
Thanks to the hypotheses (ii), (iii), (iv) and (v), we have:  
\[\int_{z}p_{\theta}^{a}(z)\,|\ln p_{\theta}(z)|dz \leq\sup_{\theta\in K}\,|\psi(\theta)|.\sup_{\theta\in K}\,\,\int _{z}p_{\theta}^{a}(z)dz\] \[\quad+\sup_{\theta\in K}\,\|\phi(\theta)\|.\sum_{u}\left(\sup_{ \theta\in K}\,\,\int_{z}S_{u}^{2}(z)p_{\theta}^{a}(z)\right)^{\frac{1}{2}}. \left(\sup_{\theta\in K}\,\,\int_{z}p_{\theta}^{a}(z)\right)^{\frac{1}{2}}\] \[=:C(\alpha)\] \[<\infty\,.\]  
The upper bound \(C(\alpha)\) in the previous inequality is independent of \(\theta\) and \(z\) but still dependant of the exponent \(\alpha\). However, since \(\overline{B}(1,e)\) is closed ball, hypotheses (iv) and (v) can be rephrased as:  
\[(iv)\,\sup_{\alpha\in\overline{B}(1,e)}\,\sup_{\theta\in K}\,\, \int_{z}p_{\theta}^{a}(z)dz<\infty\,,\] \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]0  
Hence we can actually take the supremum over \(\alpha\) in the right term of the inequation as well. We have: \(\int_{z}p_{\theta}^{a}(z)\,|\ln p_{\theta}(z)|dz\)  
\[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]1 \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]2 \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]3 \[e^{a} \in\left[1,p^{\frac{1}{1}-1}\right],\]4  
This new upper bound \(C^{\prime}\) is independent of \(\alpha\). Since \(T_{n}\mapsto 1\), then \(\exists N\in\mathbb{N}\), \(\forall n\geq N\), \(\frac{1}{T_{n}}\in\overline{B}(1,\epsilon)\). Hence for \(n\geq N\), we can apply the previous inequation to either \(\alpha=1\) or \(\alpha=\frac{1}{T_{n}}\), which provides us that \(\int_{z}p_{\theta}(z)\,|\ln p_{\theta}(z)|\), \(\int_{z}p_{\theta}^{\frac{1}{T_{n}}}(z)\,|\ln p_{\theta}(z)|\) and their supremum in \(\theta\) are all finite, all of them upper bounded by \(C^{\prime}\). In the end, when \(n\geq N\), we have the control \(\sup_{\theta\in K}\,\big{|}\int_{z}p\,\ln p\,e^{a}\big{|}<C^{\prime}\).