Vtyjkyuk

Let $f:BbbRtoBbbR$ be a $C^1$ function. Let $$g:BbbR^2toBbbR^2$$
$$(x,y)mapsto g(x,y)=(x+f(y),y+f(x)).$$
Suppose $suplimits_xinBbbR|f'(x)|<1.$

$i.)$ Prove that $forall;(x_0,y_0)in BbbR^2,;;g'(x_0,y_0)in ISO(BbbR^2)$

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

$iii.)$Prove that $g(BbbR^2)$ is clopen in $BbbR^2.$

$iv.)$Conclude

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

MY WORK:

It suffices to show that $overlineg(BbbR^2)subset g(BbbR^2).$

Let $(t,z)in overlineg(BbbR^2)$, then $exists;(x_j)_j,(y_j)_jsubset BbbR^2$ such that beginaligng(x_j,y_j)to(t,z)endalign
Then $(g(x_j,y_j))_j$ is a Cauchy sequence in $BbbR^2$. So, $forall;epsilon>0,;exists;j_0inBbbN$ such that $forall;j,kgeq j_0$
beginalign|g(x_j,y_j)-g(x_k,y_k)|<epsilon.endalign
beginalign|(x_j+f(y_j),y_j+f(x_j))-(x_k+f(y_k),y_k+f(x_k))|<epsilon,;;forall;j,kgeq j_0;endalign
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!

This is rather an extended comment.

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

For $(x,y), (xi, eta) in mathbbR^2$ 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
$$
lVert bigl(f(eta) - f(y), f(xi) - f(x)bigr) rVert = sqrt(f(eta) - f(y))^2 + (f(xi) - f(x))^2 \
le sqrtmu^2 bigl((eta - y)^2 + (xi - x)^2bigr) = mu lVert (x,y) - (xi,eta) rVert.
$$
Consequently
$$
lVert (x,y) - (xi,eta) rVert le frac11 - mu lVert g(x,y) - g(xi,eta) rVert.
$$
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 $(tildex, tildey)$, and, by continuity, $g(tildex, tildey) = (t, z)$.

And a conclusion is that $g$ is a homeomorphism (even more, a $C^1$ diffeomorphism) of $mathbbR^2$ onto itself.