Let $f:\mathbb{R}\to\mathbb{R}$ be a $C^1$ function such that $|f'(t)|\leq k<1$ for all $t\in \mathbb{R}$.

Let $\varphi:\mathbb{R}^2\to\mathbb{R}^2$ be the function given by $\varphi(x,y)=(x+f(y),f(x)+y)$.

The problem is to show that $\varphi$ is a diffeomorphism. Notice that

$$\det J_\varphi (x,y)=0\quad\Rightarrow \quad f'(x)f'(y)=1\quad \Rightarrow\quad 1=|f'(x)||f'(y)|\leq k^2<1.$$

So, $\det J_\varphi (x,y)\neq 0$ for all $(x,y)\in\mathbb{R}^2$ and thus $\varphi$ is a local diffeomorphism. Therefore, to show that $\varphi$ is a diffeomorphism it's enough to show that $\varphi$ is injective and $\varphi(\mathbb{R}^2)=\mathbb{R}^2$.

  • $\varphi$ is injective.

$$\varphi(x_1,y_1)=\varphi(x_2,y_2)\quad\Rightarrow\quad x_1+f(y_1)=x_2+f(y_2)\text{ and }f(x_1)+y_1=f(x_2)+y_2$$

How to conclude that $x_1=x_2$ and $y_1=y_2$?

  • $\varphi$ is surjective.

Let $(a,b)\in\mathbb{R}^2$. We need to show that there exists $(x,y)\in \mathbb{R}^2$ such that $$\left\{\begin{align*} x+f(y)=a\\ f(x)+y=b \end{align*}\right.$$ How can we do it?



First, here is a proof of injectivity.

Let $$x_1+f(y_1)=x_2+f(y_2)$$


then $$x_1-x_2=f(y_2)-f(y_1)=f^{\prime}(y)(y_2-y_1)$$



Now since $|f^{\prime}(x)f^{\prime}(y)|< 1$ we have $y_1=y_2$ and similar for $x_1=x_2$.

For surjectivity, define a sequence $x_n$, $y_n$ via



Now $$|x_{n+1}-x_n|=|f(y_{n})-f(y_{n-1})|=|f^{\prime}(y)||y_{n}-y_{n-1}|\leq k|y_{n}-y_{n-1}|$$ and similarily

$$|y_{n+1}-y_n|\leq k|x_{n}-x_{n-1}|$$ This shows that $\{x_n\}$ and $\{y_n\}$ are Cauchy and their limits satisfy the requirements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.