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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


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.

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I'd just like to express my appreciation for your providing an explanation of what you've thought about so far. It is far more satisfying for people to help someone who has first tried to do it on their own. Please keep it up! – Zev Chonoles May 25 '12 at 23:59
Two suggestions: (a) don't designate $v$ as parameter. Keep both variables on equal footing. (b) The region $M$ is a square-shaped neighborhood of $(1,-1)$. This suggests that you should compare $f$ to its linearization at $(1,-1)$, namely the map $g(x,y)=f(1,-1)+f_x(1,-1)(x-1)+f_y(1,-1)(y+1)$. The Lipschitz constant of $f-g$ should be small in $M$ precisely because $|x-1|$ and $|y+1|$ are small. – user31373 May 26 '12 at 1:30
up vote 2 down vote accepted

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.)

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Just to make sure I understand; is your intent here is to show that $||\tilde{f}(z)-\tilde{f}(z')|| \geq \alpha ||z-z'||$ for some $\alpha >0$ (on $M$)? – copper.hat May 30 '12 at 7:19
@copper.hat Yes, I did not write that part since it's present in the statement of the problem. The sum of the identity map with a k-Lipschitz map (k<1) is injective, since it satisfies the reverse Lipschitz inequality that you mentioned. – user31373 May 30 '12 at 16:38

$\newcommand{\ve}{\varepsilon}\newcommand{\abs}[1]{|{#1}|}$ I've tried a different approach, a few times I made a mistake before I got this version, so I don't guarantee that there's not still one more error which I did not notice.

We have function $f(x,y)=(f_1(x,y),f_2(x,y))$ where $$ \begin{align} f_1(x,y)&=\frac{1}{2}x^2+y^2+2y;\\ f_2(x,y)&=x^2-2x+y^3. \end{align} $$ We want to find out whether this function is injective on a small region around the point $x=1$, $y=-1$.

Let us try to use directly the definition of injective function. Let's have a look what can be said if $f(x_1,y_1)=f(x_2,y_2)$.

Notice that $$\begin{align} f_2(x,y)-2f_1(x,y)&=y^3-2y^2-4y-2x\\ 2f_1(x,y)+2&=x^2+2(y+1)^2 \end{align}$$

So we have $$ \begin{align} y_1^3-2y_1^2-4y_1-2x_1&=y_2^3-2y_2^2-4y_2-2x_2\\ x_1^2+2(y_1+1)^2&=x_2^2+2(y_2+1)^2 \end{align} $$ which gives $$ \begin{align} 2(x_1-x_2)&=(y_1^3-2y_1^2-4y_1)-(y_2^3-2y_2^2-4y_2) \\ (x_1-x_2)(x_1+x_2)&=2[(y_2+1)^2-(y_1+1)^2] \end{align} $$

Now we will try to use that $\abs{y+1}<\ve$ and $\abs{x-1}<\ve$.

For the function $g(y)=y^3-2y^2-4y$ we have $g'(y)=3y^2-4y-4=3(y+1)^2-10(y+1)+3$. Hence $\abs{g'(y)-3}\le 10\ve+3\ve^2$ and $\abs{g'(y)}\le 3+11\ve$ if $\ve$ is small enough. By mean value theorem we get $$2\abs{x_1-x_2} \le \abs{y_1-y_2}(3+11\ve)$$

We also have $\abs{x_1+x_2}\le \abs{x_1}+\abs{x_2}\le 2(1+\ve)$, which gives $$\abs{(x_1-x_2)(x_1+x_2)}\le (3+11\ve)(1+\ve)\abs{y_1-y_2}.$$ Again, for small enough $\ve$, we get $$\abs{(x_1-x_2)(x_1+x_2)}\le (3+16\ve)\abs{y_1-y_2}.$$

On the other hand, using mean value theorem for the function $y\mapsto (y+1)^2$ we can find out that $$\abs{(y_2+1)^2-(y_1+1)^2} \ge 2(2-\ve)\abs{y_2-y_1}$$

If both these estimates are true for $\abs{y_2-y_1}\ne0$ than $$3+16\ve\ge 8-4\ve$$ which gives $\ve\ge\frac5{20}=\frac14$.

So for $\ve=\frac16$ this cannot be true. (I believe $\ve\le\frac16$ was sufficient in all estimates above, but you should double check this.)

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