\frac
{\partial \beta^s}{\partial m}  \\ \end{array} \right)+O(\delta)+
O(\rho)\leq \mu\cdot  \mLip \beta^s + O(\delta),$$

$$$$ since $\rho=O(\delta)$. Combining this estimate with Proposition \[prop:composition\] of the Appendix, we see that: $$\mLip {\overline\beta^s} \leq (\mu \cdot \mLip \beta^s +
O(\delta))\cdot (L + O(\delta))\cdot (1+\textrm{Lip }\overline\b^s).$$ Since $\mLip \beta^s \leq D\delta/2$, the above gives us: $$\mLip {\overline\beta^s} \leq \frac {\mu L \cdot D\delta/2+L\cdot
O(\delta)+O(\delta^2)}{1-\mu L \cdot D\delta/2-L\cdot
O(\delta)-O(\delta^2)}.$$ By assumption , we have $\mu L<1$. Therefore, if we pick the constant $D$ large enough (but still requiring that $D\delta<<1$), the right hand side of the above will be $\leq D\delta/2$. This proves .\
Now that we have proved $\g$ and $\g_b$ to be well-defined, let us pass to proving and . As we said before, the second inequality implies the first, so we will only prove the second one. As above, write $\overline\b^s_0=\g_b(\beta^s_0)$ and $\overline\b^s_1=\g_b(\beta^s_1)$. From , we see that:

$$\label{eqn:estimates0}
||\overline\b^s_0-\overline\b^s_1|| \leq T_1+T_2,$$

where:

$$T_1 =||\pi_u \circ \G_{b}^{-1} \circ \ga({\b^s_0})\circ G_{\overline\b_0,b} -\pi_u \circ \G_{b}^{-1}
\circ \ga({\b^s_1})\circ G_{\overline\b_0,b}||,$$

$$T_2=||\pi_u \circ \G_{b}^{-1} \circ \ga({\b^s_1}) \circ G_{\overline\b_0,b} - \pi_u \circ \G_{b}^{-1}
\circ \ga({\b^s_1}) \circ G_{\overline\b_1,b}||.$$ As it will soon be clear, $T_1$ is the dominant term:

$$T_1 \leq \textrm{Lip }(\pi_u \circ \G_{b}^{-1}) \cdot
||\ga({\b^s_0}) -\ga({\b^s_1})||.$$ The 