real analysis - Proving Injectivity

The problem is to show the function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by

$$f(x,y)=(\tfrac{1}{2}x^2+y^2+2y,\,x^2-2x+y^3)$$ is injective on the set

$$M=\{(x,y)\in\mathbb{R}^2:|x-1|+|y+1|<\tfrac{1}{6}\}.$$

My idea is to consider the following map (here, $u,v\in\mathbb{R}^2$):

$$\phi_v(u)=u-f(u)+v,\quad v\in M$$

If I manage to show that

1. $\phi_v:D\rightarrow D$ is well-defined for some closed sets




2. $\phi_v$ is a contraction (Lipschitz constant $<1$) on $D$

then by the Contraction Mapping Theorem, $\phi_v$ has a unique fixed point. Hence, $v$ has a unique preimage $u$ for each $v$. i.e. $f$ is injective as desired.

But I ran into troubles when I attempted to find a suitable closed set $D$. Obviously it depends on the domain $M$. $M$ given here is really weird so I am not too sure how to proceed.

Answer

I'll present an approach along the lines of my comment. First, I'll normalize the derivatives at $(1,-1)$ by dividing the second component by $3$:
$$\tilde f(x,y) = (x^2/2+y^2+2y, (x^2-2x+y^3)/3)$$
This is done so that the Jacobian matrix of $\tilde f$ at $(1,-1)$ is the identity. Now split $\tilde f=L+g$ where $L(x,y)=(x-3/2,y+1/3)$ is the linear part and $g$ is the rest. If we can show that $g$ is Lipschitz with a constant less than 1, we are done.

The first component of $g$ is $g_1(x,y)=x^2/2+y^2+2y-x+3/2$, with the gradient $\nabla g_1=\langle (x-1),2(y+1)\rangle$. We estimate the gradient by $|\nabla g_1|< \sqrt{5}/6<1/2$.

The second component of $g$ is $g_2(x,y)= (x^2-2x+y^3-3y-1)/3$, with the gradient $\nabla g_2=\langle 2(x-1)/3,y^2-1\rangle$. Since $|y^2-1|\le (|y+1|+2)|y+1|<13/36$, we obtain $|\nabla g_2|\le \sqrt{1/81+(13/36)^2}<1/2$.

Since both components have Lipschitz constant $<1/2$, the map $g$ has Lipschitz constant $<1$. (Of course a more precise bound can be given, but this suffices.)