## 2 Problem Formulation and Lipschitz Constant Estimates
For any \(\lambda>0\), we have the following estimate:  
\[\|x_{2}-x_{1}\| =\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{1}- \lambda F(t_{1},x_{1}))\|\] \[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\] \[\qquad+\|p_{K(t_{1})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}( x_{1}-\lambda F(t_{1},x_{1}))\|\] \[\leq M|t_{1}-t_{2}|+\|(x_{2}-\lambda F(t_{2},x_{2}))-(x_{1}- \lambda F(t_{1},x_{1}))\|. \tag{10}\]  
Moreover, we have \[\|(x_{2} -\lambda F(t_{2},x_{2}))-(x_{1}-\lambda F(t_{1},x_{1}))\|^{2}\] \[=\|x_{2}-x_{1}\|^{2}+\lambda^{2}\|F(t_{2},x_{2})-F(t_{1},x_{1})\|^{ 2}-2\lambda\langle x_{2}-x_{1},F(t_{2},x_{2})-F(t_{1},x_{1})\rangle\] \[=\|x_{2}-x_{1}\|^{2}+\lambda^{2}\|G(x_{2})-G(x_{1})+H(t_{2})-H(t_{ 1})\|^{2}-2\lambda\langle x_{2}-x_{1},G(x_{2})-G(x_{1})\] \[\qquad-2\lambda\langle x_{2}-x_{1},H(t_{2})-H(t_{1})\rangle\] \[\leq\|x_{2}-x_{1}\|^{2}+\lambda^{2}(L_{x}\|x_{2}-x_{1}\|+L_{t}|t_ {2}-t_{1}|)^{2}-2\lambda\langle x_{2}-x_{1},G(x_{2})-G(x_{1})\] \[\qquad-2\lambda\langle x_{2}-x_{1},H(t_{2})-H(t_{1})\rangle\] \[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]0 \[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]1  
Since \(\alpha\leq L_{x}\), we have  
\[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]2  
hence \(\hat{s}:=\sqrt{1+\lambda^{2}L_{x}^{2}-2\lambda\alpha}\) is well defined. Notice that if \(\alpha<L_{x}\), then \(s>0\) for any \(\lambda>0\), while if \(\alpha=L_{x}\), then \(s>0\) for any \(\lambda\in(0,1/L_{x})\cup(1/L_{x},+\infty)\). Moreover, we have  
\[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]3  
thus, if \(\hat{s}>0\), we have  
\[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]4  
It follows from (11) and (12) that  
\[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]5 \[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]6 \[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]7  
hence  
\[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]8  
It follows from (10) and (13) that  
\[\leq\|p_{K(t_{2})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}(x_{ 2}-\lambda F(t_{2},x_{2}))\|\]9  
Notice that \(1-\hat{s}>0\) if and only if \(\lambda\in(0,2\alpha/L_{x}^{2})\). Therefore, in the case \(\alpha<L_{x}\), we obtain  
\[\qquad+\|p_{K(t_{1})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}( x_{1}-\lambda F(t_{1},x_{1}))\|\]0while in the case \(\alpha=L_{x}\), we obtain  
\[\qquad+\|p_{K(t_{1})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}( x_{1}-\lambda F(t_{1},x_{1}))\|\]1  
If we set \(z=\lambda L_{x}\), \(\hat{\alpha}=\alpha/L_{x}\) and \(\hat{L}=L_{t}/L_{x}\), then we can write \(\hat{s}=\sqrt{z^{2}-2\hat{\alpha}z+1}\), and the estimate (14), in the case \(\hat{\alpha}<1\), reads as  
\[\qquad+\|p_{K(t_{1})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}( x_{1}-\lambda F(t_{1},x_{1}))\|\]2  
In the case \(\hat{\alpha}=1\), we have \(\hat{s}=|z-1|\) and the estimate (15) reads as  
\[\qquad+\|p_{K(t_{1})}(x_{2}-\lambda F(t_{2},x_{2}))-p_{K(t_{1})}( x_{1}-\lambda F(t_{1},x_{1}))\|\]3  
where  
\[f(z)=\frac{M}{1-|z-1|}+\frac{\hat{L}z(1+z)}{|z-1|(1-|z-1|)}= \begin{cases} \frac{M}{z}+\frac{\hat{L}(1+z)}{1-z}&\text{if }z\in(0,1),\\ \frac{M}{2-z}+\frac{\hat{L}z(1+z)}{(z-1)(2-z)}&\text{if }z\in(1,2).\end{cases}