calculus - Let $g:Bbb{R}^2toBbb{R}^2,;(x,y)mapsto g(x,y)=(x+f(y),y+f(x)).$ Prove that $g(Bbb{R}^2)$ is clopen in $Bbb{R}^2.$

Let $f:\Bbb{R}\to\Bbb{R}$ be a $C^1$ function. Let $$g:\Bbb{R}^2\to\Bbb{R}^2$$
$$(x,y)\mapsto g(x,y)=(x+f(y),y+f(x)).$$
Suppose $\sup\limits_{x\in\Bbb{R}}|f'(x)|<1.$

$i.)$ Prove that $\forall\;(x_0,y_0)\in \Bbb{R}^2,\;\;g'(x_0,y_0)\in ISO(\Bbb{R}^2)$

$ii.)$Prove that $g$ is one-one

$iii.)$Prove that $g(\Bbb{R}^2)$ is clopen in $\Bbb{R}^2.$

$iv.)$Conclude

Part $i.)$ and $ii.)$ are not problems at all. I have been able to show that $g(\Bbb{R}^2)$ is open by Local Inverse Mapping Theorem. So, I want to show that it is closed.

MY WORK:

It suffices to show that $\overline{g(\Bbb{R}^2)}\subset g(\Bbb{R}^2).$

Let $(t,z)\in \overline{g(\Bbb{R}^2)}$, then $\exists\;(x_j)_j,(y_j)_j\subset \Bbb{R}^2$ such that $$\begin{align}g(x_j,y_j)\to(t,z)\end{align}$$

Then $(g(x_j,y_j))_j$ is a Cauchy sequence in $\Bbb{R}^2$. So, $\forall\;\epsilon>0,\;\exists\;j_0\in\Bbb{N}$ such that $\forall\;j,k\geq j_0$
$$\begin{align}|g(x_j,y_j)-g(x_k,y_k)|<\epsilon.\end{align}$$
$$\begin{align}|(x_j+f(y_j),y_j+f(x_j))-(x_k+f(y_k),y_k+f(x_k))|<\epsilon,\;\;\forall\;j,k\geq j_0;\end{align}$$
I think we should be able to use the fact that $f$ is $C^1$ implies that it is Lipschitz, to move forward but I don't know how-to. Please, can someone help me? I also would need a conclusion, drawn from all these! Thanks for your time and efforts!

Answer

This is rather an extended comment.

Assume that $\lVert \cdot \rVert$ is Euclidean norm on $\mathbb{R}^2$. Put $\mu := \sup\lvert f'\rvert < 1$. By the mean value theorem, for any $x, \xi \in \mathbb{R}$ one has $\lvert f(x) - f(\xi) \rvert \le \mu \lvert x - \xi \rvert$.

For $(x,y), (\xi, \eta) \in \mathbb{R}^2$ we write

$$\lVert (x,y) - (\xi,\eta) \rVert = \lVert (x - \xi, y - \eta) \rVert \\ \le \lVert \bigl((x + f(y)) - (\xi + f(\eta)), (y+ f(x)) - (\eta + f(\xi)\bigr) \rVert + \lVert \bigl(f(\eta) - f(y), f(\xi) - f(x)\bigr) \rVert.$$
The first term on the right-hand side is just $\lVert g(x,y) - g(\xi,\eta) \rVert$. Regarding the second term, we estimate
$$\lVert \bigl(f(\eta) - f(y), f(\xi) - f(x)\bigr) \rVert = \sqrt{(f(\eta) - f(y))^2 + (f(\xi) - f(x))^2} \\ \le \sqrt{\mu^2 \bigl((\eta - y)^2 + (\xi - x)^2\bigr)} = \mu \lVert (x,y) - (\xi,\eta) \rVert.$$
Consequently

$$\lVert (x,y) - (\xi,\eta) \rVert \le \frac{1}{1 - \mu} \lVert g(x,y) - g(\xi,\eta) \rVert.$$
Therefore, for a Cauchy sequence $(g(x_j,y_j))_j$, $((x_j,y_j))_j$ is a Cauchy sequence, too. By completeness, it converges to some $(\tilde{x}, \tilde{y})$, and, by continuity, $g(\tilde{x}, \tilde{y}) = (t, z)$.

And a conclusion is that $g$ is a homeomorphism (even more, a $C^1$ diffeomorphism) of $\mathbb{R}^2$ onto itself.