enough, the graph transform map $\g$ has a unique fixed point in $\Ga^s$.

The statements about the global graph transform immediately follow from the corresponding statements in the pointwise case. So let us start by proving that $\g_b$ maps $\B^s$ to itself. Take $b\in \La$, $\b^s \in \B^s$ and let $\overline{\beta}^s = \g_b(\b^s)$. We need to prove that: $$\label{eqn:cnorm} \| \overline{\b}^s \| \le \frac {\delta}2,$$ $$\label{eqn:lip} \mLip \overline{\b}^s \le \frac
{D\delta}2.$$ Recall that $\gamma_\b $ is the map of $Q^s \times M$ onto the graph of $\b^s$, see . In the Appendix we prove that for any $\b = \b^s$ that satisfies , there exists a Lipshitz homeomorphism $G_{\bar \b ,
b}: Q^s \times M \to Q^s \times M$, see , such that $$\overline{\b}^s = \pi_u \circ \G_b^{-1} \circ \gamma (\b^s) \circ
G_{\overline\b, b}.$$ Here $\G_b ={(\ph_{h(b)}\times
\textrm{Id})}^{-1} \circ \G \circ (\ph_b \times \textrm{Id})$. Note that $${||\overline{\b}^s||} \le ||\pi_u \circ \G_b^{-1} \circ
\gamma(\b^s)||,$$ because the shift in the argument of the right hand side does not change the $C^0$ norm. Therefore, by , we have: $$||{\overline{\b}}^s|| \le
||\pi_u \circ \F_b^{-1} \circ \gamma(\b^s)|| + O(\rho)  = ||\pi_u
\circ (h^{-1})_b \circ \gamma(\b^s)|| + O(\rho).$$ By , we can further estimate the above: 