$f: \mathbb{R}^2\to \mathbb{R}^2$ is differentiable, and satisfies an inequality that involves its partials - show that f is a bijection.

Question

Suppose that $f: \mathbb{R}^2\to \mathbb{R}^2$ is differentiable, and the partial derivatives of the components $f_1$, $f_2$ satisfy $$max(|\frac{\ df_1}{dx} -1|, |\frac{df_1}{d_y}|, |\frac{df_2}{d_x}|, |\frac{df_2}{d_y}-1|) <10^{-10}.$$

Prove that f is a bijection. Note: f is not assumed to be continuously differentiable.

Any ideas on how to tackle this problem?

We don't have an explicit function given for f, unfortunately. And I can't assume that f $\in$ $C^1$, let alone assume that f is as smooth as we want it to be, so that rules out the usage of the Inverse and Implicit Function Theorems.

Thanks in advance,

edit: I can view the objects that we are taking the maximum of...as a row vector. Then express this row vector as a 2x2 matrix of the form:

$$ \begin{bmatrix} \frac{\ df_1}{dx} -1 & \frac{df_1}{d_y} \\ \frac{df_2}{d_x}& \frac{df_2}{d_y}-1 \\ \end{bmatrix} $$

I observe that this is just the total derivative matrix of the function f, minus the identity matrix, i.e. we have g:= Df - I.

Using the inequality that we are given, we have that the sup norm of the matrix < $10^{-10}$.

Where can I go from here?

Answer 1

This is really a modification of the proof of inverse function theorem, using contraction mapping theorem (fixed point theorem). Note that the matrix $Df$ is uniformly closed to the identity matrix. (I am just imitating the proof of IFT in Rudin's Principles of Mathematical Analysis).

For each $y \in \mathbb R^2$, let $\phi :\mathbb R^2 \to \mathbb R^2$ be given by

$$\phi (x) = x + y - f(x)$$

Then $D\phi (x) = I - Df(x)$ and so $||D\phi (x)|| < C<1$ for some constant $C$ (Note that we need that $Df $ is uniformly close to $I$ so that this $C$ can be chosen independent of $x$). Now consider the function $g: [0,1] \to \mathbb R^2$, where

$$g(t) = \phi (t x_1 + (1-t) x_2)$$

Then by MVT there is $t_0\in [0,1]$ so that

$$ \frac{g(1) - g(0)}{1-0} = g'(t_0) = D\phi (c(t_0)) \cdot c'(t_0)$$

By Chain rule, where $c(t) = tx_1+ (1-t)x_2$ (and $g(t) = \phi(c(t))$). This is the same as

$$\phi(x_1) - \phi(x_2) = D\phi(c(t_0)) (x_1 - x_2)$$

$$\Rightarrow (*)\ \ \ ||\phi(x_1) - \phi(x_2) || \leq ||D\phi (c(t_0))||\ ||x_1 - x_2|| \leq C||x_1 - x_2||$$

for all $x_1, x_2\in \mathbb R^2$. Thus $\phi$ is a contraction so that there is an unique fixed point $x$ so that

$$\phi(x) = x \Leftrightarrow y = f(x)$$

Thus $f$ is bijective.