# Existence of a global involution extension.

I'm studying the paper "Local and simultaneous structural stability of certain diffeomorphisms. - Marco Antonio Teixeira". At the beginning of the paper, the author gives the following definition

Besides that, the author states the following lemma:

However, he does not demonstrate such a result, just saying "the lemma is easy to proof".

I know how to find an extension $$\omega:\mathbb{R}^2\to \mathbb{R}^2$$ of $$\left. \varphi \right|_{U}$$ such that $$\varphi'(0) + \beta$$ where $$\beta \in \mathcal C_b^0 (\mathbb{R}^2)$$ is Lipschitz with bounded constant by $$\varepsilon$$. However, I don't have the faintest idea how to guarantee $$\omega\circ \omega= \mathrm{Id}.$$

Can anyone help me?

## How do I constructed the function $$\omega$$

Define $$F= \varphi$$ and $$L = \mathrm{d} \varphi(0)$$.

First, choose a $$\mathcal{C}^\infty$$ real function $$\beta:\mathbb{R} \to \mathbb{R}$$ satisfying:

• $$\beta(x) = 0$$, $$\forall$$ $$x$$ $$\in$$ $$\mathbb{R} \setminus [-1,1]$$.

• $$\beta(x) =1$$, $$\forall$$ $$x$$ $$\in$$ $$\left[-\frac{1}{2},\frac{1}{2}\right].$$

• $$\beta (x)$$ is a increasing function in $$\left[-1,-\frac{1}{2}\right]$$ and $$\beta(x)$$ is a decreasing function in $$\left[\frac{1}{2}, 1\right]$$.

Note that, $$\beta$$ is compact support $$\mathcal{C}^\infty$$, so, there exists $$M$$ $$\in$$ $$\mathbb{R}$$, such that, $$M = \sup \{|D\beta(x)|; \hspace{0.1cm} x \in \mathbb{R}\}$$.

We define $$\phi: U \rightarrow \mathbb{R}^2$$ as $$\phi(x) = F(x) - Lx$$. Observe that the function $$\phi$$ is a $$\mathcal{C}^\infty$$, moreover, $$D\phi(0) = 0$$. Therefore, for the real number $$\widetilde{\varepsilon} = \min\left\{ \frac{\varepsilon}{2M} , \frac{\varepsilon}{2}\right\} > 0$$ there exists $$1 \geq r >0$$ such that $$x \in B_r (0) \subset U \Rightarrow \Vert D\phi(x) \Vert < \widetilde{\varepsilon}$$ implying, by the mean value theoream that, if $$x \in B_r (0) \Rightarrow \Vert \phi(x) \Vert < \widetilde{\varepsilon}\Vert x\Vert$$.

Now, consider $$\omega: \mathbb{R}^2 \rightarrow \mathbb{R}^2$$ defined as: $$\omega(x) = \left\{\begin{array}{l} Lx + \beta\left(\frac{\Vert x \Vert}{r}\right)\phi(x), \hspace{0.1cm} \text{if} \hspace{0.1cm} x \in U,\\ Lx, \hspace{0.1cm} \text{if} \hspace{0.1cm} x \in \mathbb{R}^n \setminus U. \end{array} \right.$$

Thus, defining $$r_1 = r$$ and $$r_2 = \frac{r}{2}$$. It is easy to verify that

• $$\omega = F$$ in $$B_{r_2} (0)$$.
• $$\omega = L$$ outside $$B_{r_1}(0)$$.
• The function $$\alpha = G - L$$ is bounded by $$\varepsilon$$.
• $$\alpha$$ is $$\varepsilon$$-Lipschitz.

Once $$\omega = L +\alpha$$, we constructed such an extension. However, $$\omega \circ \omega$$ is not necessarily equal to $$\mathrm{Id}$$.

I made some edits in the image below since the content that answers my question is just a part of the proof of Lemma $$4.2$$. I have erased all the information that is not related to the problem that I wanted to solve.