B_{3\epsilon})}<C\epsilon^2.$$

Choose $\vartheta$ and $\xi$ positive $\CC^{\infty}$ functions so that

$$\label{e11}
\begin{cases}
\vartheta=\vartheta(s),~~~\xi=\xi(s),\\
\vartheta=0~{\rm on}~B_{\epsilon}~{\rm and}~\vartheta=1
~{\rm on}~\R^m \setminus B_{{{4}\over {3}} \epsilon},\\
\xi=1 ~{\rm on}~B_{{{4}\over {3}} \epsilon},
\xi=0 ~{\rm on}~~\R^m \setminus B_{{{5}\over {3}} \epsilon}.
\end{cases}$$

Define the functions $b_j$ as follows, $$b_1=d \vartheta+(1-\vartheta)a_1$$ and $b_j$ for $1<j\leq m$ is given by $$b_j=a_j+\xi\vartheta (d-a_1)+(1-\xi)(d-a_j),$$ where $d$ is given by \[e7\]. We see that $b_j=a_j$ if $s\in B_{\epsilon}$ and $b_j=d$ if $s \in B_{3\epsilon}
\setminus B_{2\epsilon}$.

We need to show now that a point $(z,s+ib(z,s))$ is in the wedge $\W$. Write $ (z,s+ib(z,s))=(z,s+ia)+(0,iv)$ where $v$ is given by $$v=\vartheta(d-a_1) \left(1,...,1\right )+
(1-\xi)\left (0,a_1-a_2,...,a_1-a_j,...,a_1-a_m\right )$$

By \[e10\] $a_j=O(\epsilon^2)$, so if $\epsilon$ is small enough we see that $ (z,s+ib(z,s))\in \W$ To conclude the proof of lemma \[l1\], we need to make sure that $\epsilon$ is also small enough so that $(z,s+id) \in B_{\eta}(0)
$ when $(z,s)\in \B_{3\epsilon}
\times B_{3\epsilon}$, which we can do by \[e7\]. $\blacksquare$

The proof of proposition \[p1\] is now immediate, the CR extension of $f$ to $\ti \n$ is given by the restriction of the holomorphic extension of $f$ to $\ti
\n$, which is possible by the preceding lemma. $\blacksquare$

[**(Abuse of) Notation.**]{} From henceforth we shall be working 