Say we have $f:\Bbb{R}\rightarrow \Bbb{R}$ s.t $f\in C^\infty$ and for every $x\in \Bbb{R}$, $|f'(x)|<1$. I have to prove that the function $F(x,y)=(x+f(y), y+f(x))$ is locally invertible for every $x_0\in\Bbb{R}$. Moreover, if also $|f'(x)|<1-\varepsilon$, then $F$ is globally invertible.
First part was quite easy, the Jacobian matrix at $(x_0,y_0)$ is:\begin{pmatrix} 1 & f'(y_0) \\ f'(x_0) & 1 \end{pmatrix}
Which is clearly invertible because $det(J)=1-f'(y_0)f'(x_0)>0$. However, I don't see how to do the second part.
Any help would be apperciated.
For the second part: Let $L := 1 - \varepsilon \in (0, 1)$. Note that for $u, v \in \mathbb{R}$ we have: $$ \lvert f(u) - f(v) \rvert = \left \lvert \int^u_v f'(s)~\mathrm{d}s \right \rvert \leq \left \lvert \int^u_v \lvert f'(s) \rvert~\mathrm{d}s \right \rvert \leq L \left \lvert \int^u_v 1~\mathrm{d}s \right \rvert = L \lvert u - v\rvert $$ Now let $G(x, y) := (f(y), g(x))$. Then we have for $x, y, u, v \in \mathbb{R}$ by the above: $$ \lVert G(x, y) - G(u, v) \rVert _2 \leq \sqrt{L^2 \lvert y - v\rvert^2 + L^2 \lvert x - u\rvert^2} = L \lVert (x, y) - (u, v)\rVert_2 $$
Now we want to prove injectivity of $F$: Let $x, y, u, v \in \mathbb{R}$ such that $F(x, y) = F(u, v)$. Then we have: $$ \lVert (x, y) - (u, v) \rVert_2 = \lVert F(x, y) - G(x, y) - F(u, v) + G(u, v) \rVert_2 = \lVert G(u, v) - G(x, y) \rVert_2 \leq L \lVert (x, y) - (u, v) \rVert_2 $$ Since $L < 1$, this inequality is only fulfilled, if and only if $$ \lVert (x, y) - (u, v) \rVert_2 = 0, $$ i.e. $(x, y) = (u, v)$. This means that $F$ is injective.
Now we need to prove surjectivity: Let $\bar x, \bar y \in \mathbb{R}$. We want to find $(c, d) \in \mathbb{R}$ such that $F(c, d) = (\bar x , \bar y)$. We define $\bar G(x, y) := G(\bar x-x, \bar y-y)$. Then, we know that for all $x, y, u, v \in \mathbb{R}$ we have: $$ \lVert \bar G(x, y) - \bar G(u, v) \rVert_2 \leq L (\bar x - x, \bar y - y) - (\bar x - u, \bar y - v) \rVert_2 = L\lVert (x, y) - (u, v) \rVert_2 $$ So $\bar G$ is a contraction. According to Banach's fixed point theorem there exist $a, b \in \mathbb{R}$ such that $$ G(\bar x - a, \bar y - b) = (a, b). $$ Let $(c, d) := (\bar x - a, \bar y - b)$. Then: $$ G(c, d) = (\bar x - c, \bar y - d) ~ \iff ~ G(c, d) + (c, d) = (\bar x, \bar y) ~ \iff ~ F(c, d) = (\bar x, \bar y) $$ This means that $F$ is surjective. But then $F$ is bijective, hence invertible.